3. LINEAR EQUATIONS IN TWO VARIABLES Exercise : 3A

Chapter:3.LINEAREQUATIONSINTWOVARIABLES

Exercise:3AQuestion:1

Solveeachofthe

Solution:

Forequation,2x+3y=2

First,takex=0andfindthevalueofy.

Then,takey=0andfindthevalueofx.

Nowsimilarlysolveforequation,x-2y=8

Plotthevaluesinagraphandfindtheintersectingpointforthesolution.

Hence,thesolutionsoobtainedfromthegraphis(4,-2),whichistheintersectingpointofthetwolines.

Question:2

Solveeachofthe

Solution:

Forequation,3x+2y=4


Nowsimilarlysolveforequation,2x-3y=7



Question:3

Solveeachofthe

Solution:

Wecanrewritetheequationsas:

2x+3y=8

x-2y=-3

Forequation,2x+3y=8



Nowsimilarlysolveforequation,x-2y=-3


Hence,thesolutionsoobtainedfromthegraphis(1,2),whichistheintersectingpointofthetwolines.

Question:4

Solveeachofthe

Solution:


2x–5y=-4

&2x+y=8

Forequation,2x–5y=-4



Nowsimilarlysolveforequation,2x+y=8



Question:5

Solveeachofthe

Solution:

Forequation,3x+2y=12



Nowsimilarlysolveforequation,5x–2y=4



Question:6

Solveeachofthe

Solution:


3x+y=-1

&2x–3y=-8

Forequation,3x+y=-1



Nowsimilarlysolveforequation,2x–3y=-8


Hence,thesolutionsoobtainedfromthegraphis(-1,2),whichistheintersectingpointofthetwolines.

Question:7

Solveeachofthe

Solution:


2x+3y=-5

&3x+2y=12

Forequation,2x+3y=-5



Nowsimilarlysolveforequation,3x+2y=12



Question:8

Solveeachofthe

Solution:


2x–3y=-13

&3x–2y=-12




Nowsimilarlysolveforequation,3x–2y=-12



Question:9

Solveeachofthe

Solution:


2x+3y=4

&3x–y=-5

Forequation,2x+3y=4



Nowsimilarlysolveforequation,3x–y=-5



Question:10

Solveeachofthe

Solution:


x+2y=-2

&3x+2y=2

Forequation,x+2y=-2






Question:11

Solveeachofthe

Solution:


x–y=-3

&2x+3y=4

Forequation,x–y=-3






TheverticesoftheformedtriangleABCbytheselinesandthex-axisinthegraphareA(-1,2),B(-3,0)andC(2,0).

Clearly,fromthegraphwecanidentifybaseandheightofthetriangle.

Now,weknow

AreaofTriangle=1/2×base×height

Thus,Area(∆ABC)=

[∵Base=BO+OC=3+2=5units&height=2units]

Area(∆ABC)=5sq.units

Question:12

Solveeachofthe

Solution:


2x–3y=-4

&x+2y=5




Nowsimilarlysolveforequation,x+2y=5



TheverticesoftheformedtriangleABCbytheselinesandthex-axisinthegraphareA(1,2),B(-2,0)andC(5,0).


Now,weknow


Thus,Area(∆ABC)=



Question:13

Solveeachofthe

Solution:


4x–3y=-4

&4x+3y=20







TheverticesoftheformedtriangleABCbytheselinesandthex-axisinthegraphareA(2,4),B(-1,0)andC(5,0).


Now,weknow


Thus,Area(∆ABC)=



Question:14

Solveeachofthe

Solution:


x–y=-1

&3x+2y=12

Forequation,x–y=-1






Theverticesoftheformedtrianglebytheselinesandthex-axisinthegraphareA(2,3),B(-1,0)andC(4,0).


Now,weknow


Thus,Area(∆ABC)=


Area(∆ABC)=7.5sq.units

Question:15

Solveeachofthe

Solution:


x–2y=-2

&2x+y=6

Forequation,x–2y=-2






Theverticesoftheformedtrianglebytheselinesandthex-axisinthegraphareA(2,2),B(-2,0)andC(3,0).


Now,weknow


Thus,Area(∆ABC)=1/2×5×2



Question:16

Solveeachofthe

Solution:


2x–3y=-6

&2x+3y=18







Theverticesoftheformedtrianglebytheselinesandthey-axisinthegraphareA(3,4),B(0,6)andC(0,2).


Now,weknow


Thus,Area(∆ABC)=

[∵Base=OB–OC=6–2=4units&height=3units]


Question:17

Solveeachofthe

Solution:


4x–y=4

&3x+2y=14

Forequation,4x–y=-2






Theverticesoftheformedtrianglebytheselinesandthey-axisinthegraphareA(2,4),B(7,0)andC(0,-4).


Now,weknow


Thus,Area(∆ABC)=

[∵Base=OB+OC=7+4=11units&height=4units]


Question:18

Solveeachofthe

Solution:


x–y=5

&3x+5y=15

Forequation,x–y=5








Now,weknow





Question:19

Solveeachofthe

Solution:


2x–5y=-4

&2x+y=8







Theverticesoftheformedtrianglebytheselinesandthey-axisinthegraphareA(3,2),B(0,8)andC(0,0.8).


Now,weknow


Thus,Area(∆ABC)=1/2×7.2×3

[∵Base=OB–OC=8-0.8=7.2units&height=3units]

Area(∆ABC)=10.8sq.units

Question:20

Solveeachofthe

Solution:


5x–y=7

&x–y=-1

Forequation,5x–y=7



Nowsimilarlysolveforequation,x–y=-1





Now,weknow



[∵Base=OB+OC=1+7=8units&height=2unitsfromthey-axistothepointA]


Question:21

Solveeachofthe

Solution:


2x–3y=12

&x+3y=6

Forequation,2x–3y=12



Nowsimilarlysolveforequation,x+3y=6





Now,weknow





Question:22

Showgraphically

Solution:

Forequation,2x+3y=6





Thelinescoincideoneachother,thisindicatesthattherearenumberofintersectionpointsonthelinesincealineconsistsofinfinitepoints.

Hence,thegraphshowsthatthesystemofequationshaveinfinitenumberofsolutions.

Question:23

Showgraphically

Solution:

Forequation,3x–y=5



Nowsimilarlysolveforequation,6x-2y=10




Question:24

Showgraphically

Solution:

Forequation,2x+y=6







Question:25

Showgraphically

Solution:

Forequation,x–2y=5







Question:26

Showgraphically

Solution:

Forequation,x–2y=6





Theequationlinex-2y=6willpassthroughpoints(0,-3)and(6,0).

Buttheequationline3x-6y=0willpassthroughx-axisandy-axis,whichdoesnotactuallyintersecttheline,x-2y=6.Hence,thegraphshowsthatthesystemofequationshavenosolutions.

Question:27

Showgraphically

Solution:

Forequation,2x+3y=4





Thesetofequationsareparalleltoeachotherinthegraph.

Parallellinesnevermeeteachothereveniftheyareextended.

Hence,thegraphshowsthatthesystemofequationshavenosolutions.

Question:28

Showgraphically

Solution:

Forequation,2x+y=6





Thesetofequationsareparalleltoeachotherinthegraph.

Parallellinesnevermeeteachothereveniftheyareextended.

Hence,thegraphshowsthatthesystemofequationshavenosolutions.

Question:29

Drawthegraphso

Solution:

Forequation,2x+y=2





Since,theline2x+y=6cutstheliney-axisatA(0,6)andx-axisatB(3,0)

&theline2x+y=2cutsthex-axisatC(1,0)andy-axisatD(0,2).

Thus,itisclearfromthegraphthatABCDformsatrapezium.

Andthecoordinatesjoiningthistrapeziumare(0,6),(3,0),(1,0)and(0,2).

WecanfindtheareaoftrapeziumABCD.

TheformulatocalculateareaofatrapeziumABCDis:

Area(trap.ABCD)=Area(∆OAB)–Area(∆OCD)

=(1/2×3×6)–(1/2×1×2)

[∵base(∆OAB)=3units&height(∆OAB)=6units

=9-1base(∆OCD)=1units&height(∆OCD)=2units]

=8sq.units

Exercise:3BQuestion:1

Solveforxandy

Solution:

Wehave,

x+y=3…eq.1

4x–3y=26…eq.2

Tosolvetheseequations,weneedtomakeoneofthevariablesineachequationhavesamecoefficient.

Letsmultiplyeq.1by4,sothatvariablexinboththeequationshavesamecoefficient.

Recallingequations1&2,

x+y=3[×4]

4x–3y=26

⇒4x+4y=12

4x–3y=26

Onsolvingthetwoequationsweget,

7y=-14

⇒7y=-14

⇒y=-2

Substitutey=-2ineq.1/eq.2,asperconvenienceofsolving.

Thus,substitutingineq.1,weget

x+(-2)=3

⇒x=3+2

⇒x=5

Hence,wehavex=5andy=-2.

Question:2

Solveforxandy

Solution:

Wehave,

x–y=3…eq.1

…eq.2

Letusfirstsimplifyeq.2,bytakingLCMofdenominator,

⇒

⇒2x+3y=36…eq.3

Tosolvetheseequations,weneedtomakeoneofthevariables(inboththeequations)havesamecoefficient.

Lets,multiplyeq.1by2,sothatvariablexinboththeequationshavesamecoefficient.


x–y=3[×2

2x+3y=36

⇒2x–2y=6

2x+3y=36

Onsolvingweget,

⇒-5y=-30

⇒y=6

Substitutey=6ineq.1/eq.3,asperconvenienceofsolving.


x–(6)=3

⇒x=3+6

⇒x=9

Hence,wehavex=9andy=6.

Question:3

Solveforxandy

Solution:

Wehave,

2x+3y=0…eq.1

3x+4y=5…eq.2


Lets,multiplyeq.1by3andeq.2by2,sothatvariablexinboththeequationshavesamecoefficient.


2x+3y=0[×3

3x+4y=5[×2

⇒6x+9y=0

6x+8y=10


y=-10

Substitutey=-10ineq.1/eq.2,asperconvenienceofsolving.


2x+3(-10)=0

⇒2x=30

⇒x=15


Question:4

Solveforxandy

Solution:

Wehave,

2x–3y=13…eq.1

7x–2y=20…eq.2


Letsmultiplyeq.1by2andeq.2by3,sothatvariableyinboththeequationshavesamecoefficient.


2x–3y=13[×2]

7x–2y=20[×3]

⇒4x–6y=26

21x–6y=60


-17x=-34

⇒x=2

Substitutex=2ineq.1/eq.2,asperconvenienceofsolving.


2(2)–3y=13

⇒-3y=13–4

⇒-3y=9

⇒y=-3


Question:5

Solveforxandy

Solution:

Rearrangingtheequations,wehave

3x–5y=19(1)

-7x+3y=-1(2)


Letsmultiply(1)by7and(2)by3,sothatvariablexinboththeequationshavesamecoefficient.

(3x–5y=19)×7

(-7x+3y=-1)×3

21x-35y=133(3)-21x+9y=-3(4)

adding(3)and(4),weget

⇒-26y=130

⇒y=-5

Substitutey=-5in(1)3x-5(-5)=19⇒3x+25=19⇒3x=-6⇒x=-2Hence,x=-2andy=-5isthesolutionofgivenpairofequations.

Question:6

Solveforxandy

Solution:

Rearrangingtheequations,wehave

2x–y=-3…eq.1

3x–7y=-10…eq.2


Letsmultiplyeq.1by7,sothatvariableyinboththeequationshavesamecoefficient.


2x–y=-3[×7

3x–7y=-10

⇒14x–7y=-21

3x–7y=-10

Onsolvingtheabovetwoequationsweget,

⇒11x=-11

⇒x=-1

Substitutex=-1ineq.1/eq.2,asperconvenienceofsolving.


2(-1)–y=-3

⇒-2–y=-3

⇒y=-2+3

⇒y=1

Hence,wehavex=-1andy=1.

Question:7

Solveforxandy

Solution:

Wehave,

…eq.1

…eq.2

Letusfirstsimplifyeq.1&eq.2,bytakingLCMofdenominators,

Eq.1⇒

⇒

⇒9x–2y=108…eq.3

Eq.2⇒

⇒

⇒3x+7y=105…eq.4




9x–2y=108[×7

3x+7y=105[×2

⇒63x–14y=756

6x+14y=210

Onaddingtheabovethetwoequationsweget,

69x+0=966

⇒69x=966

⇒x=14



3(14)+7y=105

⇒7y=105-42

⇒7y=63

⇒y=9


Question:8

Solveforxandy

Solution:

Wehave,

…eq.1

…eq.2

Letusfirstsimplifyeq.1&eq.2,bytakingLCMofdenominators,

Eq.1⇒

⇒

⇒4x+3y=132…eq.3

Eq.2⇒

⇒

⇒5x–2y=-42…eq.4




4x+3y=132[×2

5x–2y=-42[×3

⇒8x+6y=264

15x–6y=-126

23x+0=138

⇒23x=138

⇒x=6



5(6)–2y=-42

⇒30–2y=-42

⇒2y=30+42

⇒2y=72

⇒y=36


Question:9

Solveforxandy

Solution:

Wehave,

4x–3y=8…eq.1

…eq.2

Letusfirstsimplifyeq.2bytakingLCMofdenominator,

Eq.2⇒

⇒18x–3y=29…eq.3


Anditissothattheequations1&3havevariableyhavingsamecoefficientalready,soweneednotmultiplyordivideitwithanynumber.


4x–3y=8

18x–3y=29

⇒4x–3y=8

18x–3y=29

Onsolvingtheaboveequationsweget,

⇒-14x=-21

⇒

⇒

Substitute ineq.1/eq.3,asperconvenienceofsolving.


4 –3y=8

⇒6–3y=8

⇒3y=6–8

⇒3y=-2

⇒

Hence,wehave and

Question:10

Solveforxandy

Solution:

Wehave,

…eq.1

5x=2y+7or5x–2y=7…eq.2


Eq.2⇒

⇒8x–3y=12…eq.3




5x–2y=7[×3]

8x–3y=12[×2]

⇒15x–6y=21

16x–6y=24

Onsolvingtheaboveequationsweget,

-x–0=-3

⇒-x=-3

⇒x=3



5(3)–2y=7

⇒15–2y=7

⇒2y=15–7

⇒2y=8

⇒y=4

Hence,wehavex=3andy=4

Question:11

Solveforxandy

Solution:

Wehave,

…eq.1

…eq.2

Letusfirstsimplifyeq.1&eq.2bytakingLCMofdenominators,

Eq.1

⇒6x+15y=8…eq.3

Eq.2

⇒18x–12y=5…eq4


Letsmultiplyeq.3by18andeq.4by6,sothatvariablexinboththeequationshavesamecoefficient.


6x+15y=8[×18]

18x–12y=5[×6]

⇒108x+270y=144

108x–72y=30

Onsolvingthesetwoequationsweget,

⇒342y=114

⇒

⇒

Substitute ineq.3/eq.4,asperconvenienceofsolving.


6x+ =8

⇒6x+5=8

⇒6x=8–5

⇒6x=3

⇒

Hence,wehave and

Question:12

Solveforxandy

Solution:

Afterrearrangement,wehave

2x+3y=-1…eq.1

…eq.2


Eq.1⇒

⇒7–4x=3y

⇒4x+3y=7…eq.3


Anditissothattheequations1&3havevariableyhavingsamecoefficientalready,soweneednotmultiplyordivideitwithanynumber.


2x+3y=-1

4x+3y=7

Onsolvingthesetwoequationsweget,

⇒x=4



4(4)+3y=7

⇒16+3y=7

⇒3y=7–16

⇒3y=-9

⇒y=-3

Hence,wehavex=4andy=-3

Question:13

Solveforxandy

Solution:

Wehave

0.4x+0.3y=1.7

0.7x–0.2y=0.8

Letssimplifytheseequations.Wecanrewritethemas,

⇒4x+3y=17…eq.1

⇒7x–2y=8…eq.2


Letsmultiplyeq.1by2&eq.2by3,sothatvariableyinboththeequationshavesamecoefficient.


4x+3y=17,onmultiplyingequationwith2

7x–2y=8,onmultiplyingequationwith3

Weget,

8x+6y=34

21x–6y=24

Onsolvingtheequation,weget,

x=2



7(2)–2y=8

⇒14–2y=8

⇒2y=14–8

⇒2y=6

⇒y=3


Question:14

Solveforxandy

Solution:

Wehave

0.3x+0.5y=0.5

0.5x+0.7y=0.74

Letssimplifytheseequations.Wecanrewritethemas,

⇒3x+5y=5…(i)

⇒50x+70y=74…(ii)


Letsmultiplyequation(i)by14,sothatvariableyinboththeequationshavesamecoefficient.

Recallingequations(i)&(ii),

3x+5y=5[×14

50x+70y=74

⇒-8x=-4

⇒

⇒

⇒x=0.5

Substitute ineq.(i)/eq.(ii),asperconvenienceofsolving.

Thus,substitutinginequation(i),weget

⇒

⇒3+10y=10

⇒10y=10–3

⇒10y=7

⇒

⇒y=0.7

Hence,wehavex=0.5andy=0.7

Question:15

Solveforxandy

Solution:

Wehave

7(y+3)–2(x+2)=14

4(y–2)+3(x–3)=2

Letssimplifytheseequations.Wecanrewritethem,

7(y+3)–2(x+2)=14

⇒7y+21–2x–4=14

⇒7y–2x+17=14

⇒2x–7y=3…(i)

4(y–2)+3(x–3)=2

⇒4y–8+3x–9=2

⇒3x+4y–17=2

⇒3x+4y=19…(ii)


Letsmultiplyeq.(i)by3andeq.(ii)by2,sothatvariablexinboththeequationshavesamecoefficient.


2x–7y=3[×3

3x+4y=19[×2

⇒-29y=-29

⇒y=1

Substitutey=1ineq.(i)oreq.(ii),asperconvenienceofsolving.


2x–7(1)=3

⇒2x–7=3

⇒2x=7+3

⇒2x=10

⇒x=5


Question:16

Solveforxandy

Solution:

Since,ifa=b=c⇒a=b&b=c

Thus,wehave

6x+5y=7x+3y+1

2(x+6y–1)=7x+3y+1


6x+5y=7x+3y+1

⇒7x–6x+3y–5y=-1

⇒x–2y=-1…(i)

2(x+6y–1)=7x+3y+1

⇒2x+12y–2=7x+3y+1

⇒7x–2x+3y–12y=-2–1

⇒5x–9y=-3…(ii)


Letsmultiplyeq.(i)by5,sothatvariablexinboththeequationshavesamecoefficient.


x–2y=-1[×5

5x–9y=-3

⇒-y=-2

⇒y=2

Substitutey=2ineq.(i)/eq.(ii),asperconvenienceofsolving.

Thus,substitutingineq.(i),weget

x–2(2)=-1

⇒x–4=-1

⇒x=-1+4

⇒x=3


Question:17

Solveforxandy

Solution:

Since,ifa=b=c⇒a=b&b=c

Thus,wehave

and


⇒3(x+y–8)=2(x+2y–14)

⇒3x+3y–24=2x+4y–28

⇒3x–2x+3y–4y=-28+24

⇒x–y=-4…(i)

⇒3(3x+y–12)=11(x+2y–14)

⇒9x+3y–36=11x+22y–154

⇒11x–9x+22y–3y=154–36

⇒2x+19y=118…(ii)


Letsmultiplyeq.(i)by19,sothatvariableyinboththeequationshavesamecoefficient.


x–y=-4[×19

2x+19y=118

⇒21x=42

⇒x=2

Substitutex=2ineq.(i)/eq.(ii),asperconvenienceofsolving.


2–y=-4

⇒y=2+4

⇒y=6


Question:18

Solveforxandy

Solution:

Wehave

and

Letssimplifytheseequations.Assuming1/x=z,wecanrewritethem,

⇒5z+6y=13…(i)

⇒3z+4y=7…(ii)


Letsmultiplyeq.(i)by3andeq.(ii)by5,sothatvariablezinboththeequationshavesamecoefficient.


5z+6y=13[×3

3z+4y=7[×5

⇒-2y=4

⇒y=-2

Substitutey=-2ineq.(i)/eq.(ii),asperconvenienceofsolving.

Thus,substitutingineq.(ii),weget

3z+4(-2)=7

⇒3z–8=7

⇒3z=7+8

⇒3z=15

⇒z=5

Thus,z=5andy=-2

Asz=1/x,

⇒5=1/x

⇒x=1/5

Hence,wehavex=1/5andy=-2

Question:19

Solveforxandy

Solution:

Wehave

and

Letssimplifytheseequations.Assuming1/y=z,wecanrewritethem,

⇒x+6z=6…(i)

⇒3x–8z=5…(ii)


Letsmultiplyeq.(i)by3,sothatvariable“x”inboththeequationshavesamecoefficient.


x+6z=6[×3

3x–8z=5

⇒26z=13

⇒z=13/26

⇒z=1/2

Substitutez=1/2ineq.(i)/eq.(ii),asperconvenienceofsolving.


x+6(1/2)=6

⇒x+3=6

⇒x=3

Thus,z=1/2andx=3

Asz=1/y,

⇒

⇒y=2


Question:20

Solveforxandy

Solution:

Wehave

and

wherey≠0

Letssimplifytheseequations.Assuming1/y=z,wecanrewritethem,

⇒2x–3z=9…(i)

⇒3x+7z=2…(ii)


Letsmultiplyeq.(i)by3andeq.(ii)by2,sothatvariablexinboththeequationshavesamecoefficient.


2x–3z=9[×3

3x+7z=2[×2

⇒-23z=23

⇒z=-1

Substitutez=-1ineq.(i)/eq.(ii),asperconvenienceofsolving.


2x–3(-1)=9

⇒2x+3=9

⇒2x=6

⇒x=3

Thus,z=-1andx=3

Asz=1/y,

⇒-1=1/y

⇒y=-1

Hence,wehavex=3andy=-1

Question:21

Solveforxandy

Solution:

Wehave

and

wherex≠0andy≠0

Letssimplifytheseequations.Assuming1/x=pand1/y=q,wecanrewritethem,

⇒3p–q=-9…(i)

⇒2p+3q=5…(ii)


Letsmultiplyeq.(i)by3,sothatvariableqinboththeequationshavesamecoefficient.


3p–q=-9[×3

2p+3q=5

⇒11p=-22

⇒p=-2

Substitutep=-2ineq.(i)/eq.(ii),asperconvenienceofsolving.


3(-2)–q=-9

⇒-6–q=-9

⇒q=9–6

⇒q=3

Thus,p=-2andq=3

Asp=1/x,

⇒-2=1/x

⇒x=-1/2

Andq=1/y

⇒3=1/y

⇒y=1/3

Hence,wehavex=-1/2andy=1/3

Question:22

Solveforxandy

Solution:

Wehave

and

wherex≠0andy≠0


⇒9p–4q=8…(i)

⇒13p+7q=101…(ii)


Letsmultiplyeq.(i)by7andeq.(ii)by4,sothatvariableqinboththeequationshavesamecoefficient.


9p–4q=8[×7

13p+7q=101[×4

⇒115p=460

⇒p=4

Substitutep=4ineq.(i)/eq.(ii),asperconvenienceofsolving.


9(4)–4q=8

⇒36–4q=8

⇒4q=36–8=28

⇒q=7

Thus,p=4andq=7

Asp=1/x,

⇒4=1/x

⇒x=1/4

Andq=1/y

⇒7=1/y

⇒y=1/7

Hence,wehavex=1/4andy=1/7

Question:23

Solveforxandy

Solution:

Wehave

and

wherex≠0andy≠0


⇒5p–3q=1…(i)

⇒

⇒9p+4q=30…(ii)

Tosolvetheseequations,weneedtomakeoneofthevariables(inboththeequations)havesame

coefficient.



5p–3q=1[×4

9p+4q=30[×3

⇒47p=94

⇒p=2



5(2)–3q=1

⇒10–3q=1

⇒3q=10–1=9

⇒q=3

Thus,p=2andq=3

Asp=1/x,

⇒2=1/x

⇒x=1/2

Andq=1/y

⇒3=1/y

⇒y=1/3


Question:24

Solveforxandy

Solution:

Wehave

and

wherex≠0andy≠0


⇒3p+2q=12…(i)

⇒2p+3q=13…(ii)


Letsmultiplyeq.(i)by2andeq.(ii)by3,sothatvariablepinboththeequationshavesamecoefficient.


3p+2q=12[×2

2p+3q=13[×3

⇒-5q=-15

⇒q=3

Substituteq=3ineq.(i)/eq.(ii),asperconvenienceofsolving.


3p+2(3)=12

⇒3p+6=12

⇒3p=12–6=6

⇒p=2

Thus,p=2andq=3

Asp=1/x,

⇒2=1/x

⇒x=1/2

Andq=1/y

⇒3=1/y

⇒y=1/3


Question:25

Solveforxandy

Solution:

Wehave

4x+6y=3xy

and8x+9y=5xy

wherex≠0andy≠0

Letssimplifytheseequations.

4x+6y=3xy

Dividingtheequationbyxythroughout,

⇒

Assumingp=1/yandq=1/x,weget

4p+6q=3…(i)

Also,8x+9y=5xy


⇒


⇒8p+9q=5…(ii)


Letsmultiplyeq.(i)by2,sothatvariablepinboththeequationshavesamecoefficient.


4p+6q=3[×2

8p+9q=5

⇒3q=1

⇒q=1/3

Substituteq=1/3ineq.(i)/eq.(ii),asperconvenienceofsolving.


4p+6(1/3)=3

⇒4p+2=3

⇒4p=3–2=1

⇒p=1/4

Thus,p=1/4andq=1/3

Asq=1/x,

⇒1/3=1/x

⇒x=3

Andp=1/y

⇒1/4=1/y

⇒y=4


Question:26

Solveforxandy

Solution:

Wehave

x+y=5xy

and3x+2y=13xy

wherex≠0andy≠0


x+y=5xy


⇒


p+q=5…(i)

Also,3x+2y=13xy


⇒


⇒3p+2q=13…(ii)


Letsmultiplyeq.(i)by2,sothatvariableqinboththeequationshavesamecoefficient.


p+q=5[×2]

3p+2q=13

⇒-p=-3

⇒p=3



3+q=5

⇒q=5–3

⇒q=2

Thus,p=3andq=2

Asq=1/x,

⇒2=1/x

⇒x=1/2

Andp=1/y

⇒3=1/y

⇒y=1/3


Question:27

Solveforxandy

Solution:

Wehave

and

Letssimplifytheseequations.Assumingp=1/(x+y)andq=1/(x–y),

5p–2q=-1…(i)

Also,

⇒15p+7q=10…(ii)




5p–2q=-1[×3

15p+7q=10

⇒-13q=-13

⇒q=1



5p–2(1)=-1

⇒5p–2=-1

⇒5p=2–1=1

⇒p=1/5

Thus,p=1/5andq=1

Asp=1/(x+y),

⇒

⇒x+y=5…(iii)

Andq=1/(x–y)

⇒

⇒x–y=1…(iv)

Addingequations(iii)and(iv)toobtainxandy,

(x+y)+(x–y)=5+1

⇒2x=6

⇒x=3

Puttingthevalueofxinequation(iii),weget

3+y=5

⇒y=2


Question:28

Solveforxandy

Solution:

Wehave

and


3p+2q=2…(i)

Also,

⇒9p–4q=1…(ii)




3p+2q=2[×3

9p–4q=1

⇒10q=5

⇒q=1/2



3p+2(1/2)=2

⇒3p+1=2

⇒3p=2–1=1

⇒p=1/3

Thus,p=1/3andq=1/2

Asp=1/(x+y),

⇒

⇒x+y=3…(iii)

Andq=1/(x–y)

⇒

⇒x–y=2…(iv)


(x+y)+(x–y)=3+2

⇒2x=5

⇒x=5/2


5/2+y=3

⇒y=3–5/2

⇒y=1/2


Question:29

Solveforxandy

Solution:

Wehave

and

wherex≠-1andy≠1

Letssimplifytheseequations.Assumingp=1/(x+1)andq=1/(y–1),

5p–2q=1/2

10p–4q=1…(i)

Also,

⇒10p+2q=5/2

⇒20p+4q=5…(ii)


Thevariableqinboththeequationshavesamecoefficient.

⇒30p=6

⇒p=1/5

Substitutep=1/5ineq.(i)/eq.(ii),asperconvenienceofsolving.


10(1/5)–4q=1

⇒2–4q=1

⇒4q=2–1=1

⇒q=1/4

Thus,p=1/5andq=1/4

Asp=1/(x+1),

⇒

⇒x+1=5

⇒x=4

Andq=1/(y–1)

⇒

⇒y–1=4

⇒y=5


Question:30

Solveforxandy

Solution:

Wehave

and


44p+30q=10…(i)

Also,

⇒55p+40q=13…(ii)




44p+30q=10[×4

55p+40q=13[×3

⇒11p=1

⇒p=1/11



44(1/11)+30q=10

⇒4+30q=10

⇒30q=10–4=6

⇒q=1/5

Thus,p=1/11andq=1/5

Asp=1/(x+y),

⇒

⇒x+y=11…(iii)

Andq=1/(x–y)

⇒

⇒x–y=5…(iv)


(x+y)+(x–y)=11+5

⇒2x=16

⇒x=8


8+y=11

⇒y=11–8

⇒y=3


Question:31

Solveforxandy

Solution:

Wehave

and


10p+2q=4…(i)

Also,

⇒15p–9q=-2…(ii)




10p+2q=4[×9]

15p–9q=-2[×2]

⇒120p=32

⇒p=4/15



15(4/15)–9q=-2

⇒4–9q=-2

⇒9q=4+2=6

⇒q=2/3

Thus,p=4/15andq=2/3

Asp=1/(x+y),

⇒

⇒4x+4y=15…(iii)

Andq=1/(x–y)

⇒

⇒2x–2y=3…(iv)

Multiplyingeq.(iv)by2,weget

4x–4y=6…(v)

andthenaddingequations(iii)and(v)toobtainxandy,

(4x+4y)+(4x–4y)=6+15

⇒8x=21

⇒x=21/8

Puttingthevalueofxinequation(iv),weget

2(21/8)–2y=3

⇒21/4–2y=3

⇒2y=21/4–3=9/4

⇒y=9/8


Question:32

Solveforxandy

Solution:

Wehave,

71x+37y=253…(i)

37x+71y=287…(ii)

Tosolvetheseequations,weneedtosimplifythem.

So,byaddingequations(i)and(ii),weget

(71x+37y)+(37x+71y)=253+287

⇒(71x+37x)+(37y+71y)=540

⇒108x+108y=540

Nowdividingitby108,weget

x+y=5…(iii)

Similarly,subtractingequations(i)and(ii),

(71x+37y)–(37x+71y)=253–287

⇒(71x–37x)+(37y–71y)=-34

⇒34x–34y=-34

Dividingtheequationby34,weget

x–y=-1…(iv)

Tosolveequations(iii)and(iv),weneedtomakeoneofthevariables(inboththeequations)havesamecoefficient.

Herethevariablesx&yinboththeequationshavesamecoefficients.

⇒2x=4

⇒x=2

Substitutex=2ineq.(iii)/eq.(iv),asperconvenienceofsolving.

Thus,substitutingineq.(iii),weget

2+y=5

⇒y=3


Question:33

Solveforxandy

Solution:

Wehave,

217x+131y=913…(i)

131x+217y=827…(ii)



(217x+131y)+(131x+217y)=913+827

⇒(217x+131x)+(131y+217y)=1740

⇒348x+348y=1740


x+y=5…(iii)


(217x+131y)–(131x+217y)=913–827

⇒(217x–131x)+(131y–217y)=86

⇒86x–86y=86

Dividingtheequationby86,weget

x–y=1…(iv)



⇒2x=6

⇒x=3



3+y=5

⇒y=2


Question:34

Solveforxandy

Solution:

Wehave,

23x–29y=98…(i)

29x–23y=110…(ii)



(23x–29y)+(29x–23y)=98+110

⇒(23x+29x)–(29y+23y)=208

⇒52x–52y=208


x–y=4…(iii)


(23x–29y)–(29x–23y)=98–110

⇒(23x–29x)–(29y–23y)=-12

⇒-6x–6y=-12

Dividingtheequationby-6,weget

x+y=2…(iv)



⇒2x+0=6

⇒2x=6

⇒x=3


Thus,substitutingineq.(iv),weget

3+y=2

⇒y=-1


Question:35

Solveforxandy

Solution:

Wehave

and

Letssimplifytheseequations.Assumingp=1/xandq=1/y,

⇒

2q+5p=6…(i)

Also,

⇒

⇒4q–5p=-3…(ii)


Coefficientsofpinbothequationsarealreadysame.

⇒6q=3

⇒q=1/2



4(1/2)–5p=-3

⇒2–5p=-3

⇒5p=5

⇒p=1

Thus,p=1andq=1/2

Asp=1/x,

⇒1=1/x

⇒x=1

Andq=1/y,

⇒

⇒y=2


Question:36

Solveforxandy

Solution:

Wehave

and

Letssimplifytheseequations.Assumingp=1/(3x+y)andq=1/(3x–y),

p+q=3/4

⇒4p+4q=3…(i)

Also,

⇒p/2–q/2=-1/8

⇒4p–4q=-1…(ii)


Thevariablepandqinboththeequationshavesamecoefficient.

⇒8p=2

⇒p=1/4



4(1/4)–4q=-1

⇒1–4q=-1

⇒4q=2

⇒q=1/2

Thus,p=1/4andq=1/2

Asp=1/(3x+y),

⇒

⇒3x+y=4…(iii)

Andq=1/(3x–y)

⇒

⇒3x–y=2…(iv)


(3x+y)+(3x–y)=4+2

⇒6x=6

⇒x=1


3(1)–y=2

⇒3–y=2

⇒y=1


Question:37

Solveforxandy

Solution:

Wehave

and

Wherex+2y≠0and3x-2y≠0

Letssimplifytheseequations.Assuming and

⇒

Multiplyitwith6,weget

3p+10q=-9…(i)

Also,

⇒

Multiplyitwith20,weget

25p–12q=61/3

⇒75–36q=61…(ii)


Multiplyequation(i)by36andequation(ii)by10,sothatthevariablespandqinboththeequationshavesamecoefficients.

Recallingequations(i)and(ii),

3p+10q=-9[×36

75p–36q=61[×10

⇒858p=286

⇒p=286/858=1/3



3(1/3)+10q=-9

⇒1+10q=-9

⇒10q=-9-1=-10

⇒q=-1

Thus,p=1/3andq=-1

Asp=1/(x+2y),

⇒

⇒x+2y=3…(iii)

Andq=1/(3x–2y)

⇒

⇒2y–3x=1…(iv)

Subtractingequations(iii)and(iv)toobtainxandy,

(x+2y)–(2y–3x)=3–1

⇒x+2y–2y+3x=2

⇒4x=2

⇒x=1/2


2y–3(1/2)=1

⇒4y–3=2

⇒4y=2+3=5

⇒y=5/4


Question:38

Solveforxandy

Solution:

Wehave

and

Letssimplifytheseequations.Assumingp=1/(3x+2y)andq=1/(3x–2y),

2p+3q=17/5

⇒10p+15q=17…(i)

Also,

⇒5p+q=2…(ii)


Multiplyequation(ii)by2,sothatthevariablepinboththeequationshavesamecoefficient.


10p+15q=17

5p+q=2[×2

⇒13q=13

⇒q=1



5p+1=2

⇒5p=1

⇒p=1/5

Thus,p=1/5andq=1

Asp=1/(3x+2y),

⇒

⇒3x+2y=5…(iii)

Andq=1/(3x–2y)

⇒

⇒3x–2y=1…(iv)


(3x+2y)+(3x–2y)=5+1

⇒6x=6

⇒x=1


3(1)–2y=1

⇒3–2y=1

⇒2y=2

⇒y=1


Question:39

Solveforxandy

Solution:

Wehave

3(2x+y)=7xy

And3(x+3y)=11xy


3(2x+y)=7xy

Dividingthroughoutbyxy,andassumingp=1/xandq=1/y,

⇒

⇒

6q+3p=7…(i)

Also,3(x+3y)=11xy

Dividingthroughoutbyxy,andassumingp=1/xandq=1/y,

⇒

⇒

⇒3q+9p=11…(ii)


Multiplyequation(ii)by2,sothatthevariableqinboththeequationshavesamecoefficient.


6q+3p=7

3q+9p=11[×2

⇒-15p=-15

⇒p=1



6q+3(1)=7

⇒6q=7–3

⇒6q=4

⇒q=2/3

Thus,p=1andq=2/3

Asp=1

⇒1=1/x

⇒x=1

Andq=1/y,

⇒

⇒y=3/2

Hence,wehavex=1andy=3/2

Question:40

Solveforxandy

Solution:

Wehave,

x+y=a+b…(i)

ax–by=a2–b2…(ii)


Letsmultiplyequation(i)byb,sothatvariableyinboththeequationshavesamecoefficient.


x+y=a+b[×b

ax–by=a2–b2

⇒bx+ax=ab+a2

⇒(b+a)x=a(a+b)

⇒x=a

Substitutex=ainequations(i)/(ii),asperconvenienceofsolving.


a+y=a+b

⇒y=b

Hence,wehavex=aandy=b.

Question:41

Solveforxandy

Solution:

Wehave,

⇒bx+ay=2ab…(i)



Letsmultiplyequation(i)byband(ii)bya,sothatvariableyinboththeequationshavesamecoefficient.


bx+ay=2ab[×b

ax–by=a2–b2[×a

⇒b2x+a2x=2ab2+a3–ab2

⇒(b2+a2)x=a(2b2+a2–b2)

⇒(b2+a2)x=a(b2+a2)

⇒x=a



ab+ay=2ab

⇒ay=2ab–ab=ab

⇒y=b


Question:42

Solveforxandy

Solution:

Wehave,

px+qy=p–q…(i)

qx–py=p+q…(ii)

Tosolvetheseequations,weneedtomakeoneofthevariables(inboththeequations)havesame

coefficient.

Letsmultiplyequation(i)bypand(ii)byq,sothatvariableyinboththeequationshavesamecoefficient.


px+qy=p–q[×p]

qx–py=p+q[×q]

⇒p2x+q2x=p2+q2

⇒(p2+q2)x=p2+q2

⇒x=1

Substitutex=1inequations(i)/(ii),asperconvenienceofsolving.


p+qy=p–q

⇒qy=-q

⇒y=-1


Question:43

Solveforxandy

Solution:

Wehave,

⇒bx–ay=0…(i)

ax+by=a2+b2…(ii)


Letsmultiplyequation(i)byband(ii)bya,sothatvariableyinboththeequationshavesamecoefficient.


bx–ay=0[×b

ax+by=a2+b2[×a

b2x+a2x=a3+ab2

⇒(b2+a2)x=a(a2+b2)

⇒(b2+a2)x=a(b2+a2)

⇒x=a



ab–ay=0

⇒ay=ab

⇒y=b


Question:44

Solveforxandy

Solution:

Wehave,

6(ax+by)=3a+2b

⇒6ax+6by=3a+2b…(i)

6(bx–ay)=3b–2a

⇒6bx–6ay=3b–2a…(ii)


Letsmultiplyequation(i)byaand(ii)byb,sothatvariableyinboththeequationshavesamecoefficient.


6ax+6by=3a+2b[×a]

6bx–6ay=3b–2a[×b]

⇒6a2x+6b2x+0=3a2+3b2

⇒6(a2+b2)x=3(a2+b2)

⇒2x=1

⇒x=1/2

Substitutex=1/2inequations(i)/(ii),asperconvenienceofsolving.


6a(1/2)+6by=3a+2b

⇒3a+6by=3a+2b

⇒6by=2b

⇒y=1/3

Hence,wehavex=1/2andy=1/3.

Question:45

Solveforxandy

Solution:

Wehave,

ax–by=a2+b2…(i)

x+y=2a…(ii)


Letsmultiplyequation(ii)byb,sothatvariableyinboththeequationshavesamecoefficient.


ax–by=a2+b2

x+y=2a[×b

⇒ax+bx=a2+b2+2ab

⇒(a+b)x=(a+b)2

⇒x=a+b

Substitutex=(a+b)inequations(i)/(ii),asperconvenienceofsolving.

Thus,substitutinginequation(ii),weget

a+b+y=2a

⇒y=2a–a–b

⇒y=a–b

Hence,wehavex=(a+b)andy=(a–b).

Question:46

Solveforxandy

Solution:

Wehave,

⇒b2x–a2y+a2b+ab2=0

⇒a2y–b2x=a2b+ab2…(i)

bx–ay=-2ab

⇒ay–bx=2ab…(ii)


Letsmultiplyequation(ii)byb,sothatvariableyinboththeequationshavesamecoefficient.


a2y–b2x=a2b+ab2

ay–bx=2ab[×b]

⇒a2y–aby=a2b–ab2

⇒(a2–ab)y=a2b–ab2

⇒a(a–b)y=ab(a–b)

⇒y=b

Substitutey=binequations(i)/(ii),asperconvenienceofsolving.


a(b)–bx=2ab

⇒ab–bx=2ab

⇒bx=ab–2ab=-ab

⇒x=-a

Hence,wehavex=-aandy=b.

Question:47

Solveforxandy

Solution:

Wehave,

⇒b2x+a2y=a3b+ab3…(i)

x+y=2ab…(ii)


Letsmultiplyequation(ii)bya2,sothatvariableyinboththeequationshavesamecoefficient.


b2x+a2y=a3b+ab3

x+y=2ab[×a2]

⇒b2x–a2x=-a3b+ab3

⇒(b2–a2)x=ab(b2–a2)

⇒x=ab

Substitutex=abinequations(i)/(ii),asperconvenienceofsolving.


(ab)+y=2ab

⇒y=ab

Hence,wehavex=abandy=ab.

Question:48

Solveforxandy

Solution:

Wehave,

x+y=a+b…(i)



Letsmultiplyequation(i)byb,sothatvariableyinboththeequationshavesamecoefficient.


x+y=a+b[×b

ax–by=a2–b2

⇒bx+ax=ab+a2

⇒(b+a)x=a(b+a)

⇒x=a



a+y=a+b

⇒y=b


Question:49

Solveforxandy

Solution:

Wehave,

a2x+b2y=c2…(i)

b2x+a2y=d2…(ii)



(a2x+b2y)+(b2x+a2y)=c2+d2

⇒(a2x+b2x)+(b2y+a2y)=c2+d2

⇒(a2+b2)x+(a2+b2)y=c2+d2

Nowdividingitby(a2+b2),weget

x+y=(c2+d2)/(a2+b2)…(iii)


(a2x+b2y)–(b2x+a2y)=c2–d2

⇒(a2x–b2x)–(b2y–a2y)=c2–d2

⇒(a2–b2)x–(a2–b2)y=c2–d2

Dividingtheequationby(a2–b2),weget

x–y=(c2–d2)/(a2–b2)…(iv)


Herethevariablesxinboththeequationshavesamecoefficients.

⇒

⇒

⇒

⇒

⇒

Substitute ineq.(iii)/eq.(iv),asperconvenienceofsolving.


⇒

⇒

⇒

⇒

Hence,wehave and .

Question:50

Solveforxandy

Solution:

Wehave,

⇒bx+ay=a2b+ab2…(i)

b2x+a2y=2a2b2…(ii)


Letsmultiplyequation(i)bya,sothatvariableyinboththeequationshavesamecoefficient.


bx+ay=a2b+ab2[×a

b2x+a2y=2a2b2

⇒abx–b2x=a3b–a2b2

⇒b(a–b)x=a2b(a–b)

⇒x=a2

Substitutex=a2inequations(i)/(ii),asperconvenienceofsolving.


b(a2)+ay=a2b+ab2

⇒a2b+ay=a2b+ab2

⇒ay=ab2

⇒y=b2

Hence,wehavex=a2andy=b2.

Exercise:3CQuestion:1

Solveeachofthe

Solution:

Wehave,

x+2y+1=0…(i)

2x–3y–12=0…(ii)

Fromequation(i),wegeta1=1,b1=2andc1=1

Andfromequation(ii),wegeta2=2,b2=-3andc2=-12

Usingcrossmultiplication,

⇒

⇒

⇒

⇒ and

⇒ and

⇒x=3andy=-2

Thus,x=3,y=-2

Question:2

Solveeachofthe

Solution:

Wehave,

3x–2y+3=0…(i)

4x+3y–47=0…(ii)

Fromequation(i),wegeta1=3,b1=-2andc1=3

Andfromequation(ii),wegeta2=4,b2=3andc2=-47


⇒

⇒

⇒

⇒ and

⇒ and

⇒x=5andy=9

Thus,x=5,y=9

Question:3

Solveeachofthe

Solution:

Wehave,

6x–5y–16=0…(i)

7x–13y+10=0…(ii)

Fromequation(i),wegeta1=6,b1=-5andc1=-16

Andfromequation(ii),wegeta2=7,b2=-13andc2=10


⇒

⇒

⇒

⇒ and

⇒ and

⇒x=6andy=4

Thus,x=6,y=4

Question:4

Solveeachofthe

Solution:

Wehave,

3x+2y+25=0…(i)

2x+y+10=0…(ii)

Fromequation(i),wegeta1=3,b1=2andc1=25

Andfromequation(ii),wegeta2=2,b2=1andc2=10


⇒

⇒

⇒

⇒ and

⇒x=5andy=-20

Thus,x=5,y=-20

Question:5

Solveeachofthe

Solution:

Wehave,

2x+5y–1=0…(i)

2x+3y–3=0…(ii)

Fromequation(i),wegeta1=2,b1=5andc1=-1



⇒

⇒

⇒

⇒ and

⇒ and

⇒x=3andy=-1

Thus,x=3,y=-1

Question:6

Solveeachofthe

Solution:

Wehave,

2x+y–35=0…(i)

3x+4y–65=0…(ii)

Fromequation(i),wegeta1=2,b1=1andc1=-35



⇒

⇒

⇒

⇒ and

⇒ and

⇒x=15andy=5

Thus,x=15,y=5

Question:7

Solveeachofthe

Solution:

Wehave,

7x–2y–3=0…(i)

22x–3y–16=0…(ii)

Fromequation(i),wegeta1=7,b1=-2andc1=-3

Andfromequation(ii),wegeta2=22,b2=-3andc2=-16


⇒

⇒

⇒

⇒ and

⇒ and

⇒x=1andy=2

Thus,x=1,y=2

Question:8

Solveeachofthe

Solution:

Wehave,

…(i)

…(ii)

Bysimplifying,weget

Fromequation(i),

⇒5x+2y–120=0…(iii)

Fromequation(ii),

⇒4x–y–57=0…(iv)

Fromequation(iii),wegeta1=5,b1=2andc1=-120

Andfromequation(ii)wegeta2=4,b2=-1andc2=-57


⇒

⇒

⇒

⇒ and

⇒ and

⇒x=18andy=15

Thus,x=18,y=15

Question:9

Solveeachofthe

Solution:

Wehave,

…(i)

…(ii)

Let1/x=pand1/y=q.Now,

Fromequation(i),p+q=7

⇒p+q–7=0…(iii)

Fromequation(ii),2p–3q=17

⇒2p+3q–17=0…(iv)

Fromequation(iii),wegeta1=1,b1=1andc1=-7

Andfromequation(iv),wegeta2=2,b2=3andc2=-17


⇒

⇒

⇒

⇒ and

⇒p=4andq=3

⇒x=1/4andy=1/3[∵p=1/xandq=1/y]

Thus,x=1/4,y=1/3

Question:10

Solveeachofthe

Solution:

Wehave,

…(i)

…(ii)

Let1/(x+y)=pand1/(x-y)=q.Now,

Fromequation(i),5p–2q+1=0…(iii)

Fromequation(ii),15p+7q–10=0…(iv)

Fromequation(iii),wegeta1=5,b1=-2andc1=1

Andfromequation(iv),wegeta2=15,b2=7andc2=-10


⇒

⇒

⇒

⇒ and

⇒ and

⇒p=1/5andq=1

⇒ and [∵p=1/(x+y)andq=1/(x-y)]

Tosolvethese,weneedtotakereciprocaloftheseequations.Bytakingreciprocal,weget

x+y=5andx–y=1

Rearrangingthemagain,

x+y–5=0…(v)

x–y–1=0…(vi)

Fromequation(v),wegeta1=1,b1=1andc1=-5

Andfromequation(vi),wegeta2=1,b2=-1andc2=-1


⇒

⇒

⇒

⇒ and

⇒ and

⇒x=3andy=2

Thus,x=3,y=2

Question:11

Solveeachofthe

Solution:

Wehave,

…(i)

ax–by=2ab…(ii)

Bysimplifying,weget

Fromequation(i),

…(iii)

Fromequation(ii),

ax–by–2ab=0…(iv)

Fromequation(iii),wegeta1=a/b,b1=-b/aandc1=-(a+b)

Andfromequation(iv),wegeta2=a,b2=-bandc2=-2ab


⇒

⇒

⇒

⇒ and

⇒ and

⇒x=bandy=-a

Thus,x=b,y=-a

Question:12

Solveeachofthe

Solution:

Wehave,

2ax+3by–(a+2b)=0…(i)

3ax+2by–(2a+b)=0…(ii)

Fromequation(i),wegeta1=2a,b1=3bandc1=-(a+2b)

Andfromequation(ii),wegeta2=3a,b2=2bandc2=-(2a+b)


⇒

⇒

⇒

⇒ and

⇒ and

⇒ and

Thus, ,

Question:13

Solveeachofthe

Solution:

Wehave,

…(i)

…(ii)

Let1/x=pand1/y=q.Now,

Fromequation(i),ap–bq=0

⇒ap–bq+0=0…(iii)

Fromequation(ii),ab2p–a2bq=(a2+b2)

⇒ab2p–a2bq–(a2+b2)=0…(iv)

Fromequation(iii),wegeta1=a,b1=-bandc1=0

Andfromequation(iv),wegeta2=ab2,b2=-a2bandc2=-(a2+b2)


⇒

⇒

⇒

⇒ and

⇒ and

⇒p=1/aandq=1/b

Thus,x=a,y=b[∵p=1/xandq=1/y]

Exercise:3DQuestion:1

Showthateachof

Solution:

Given:3x+5y=12–eq1

5x+3y=4–eq2

Here,

a1=3,b1=5,c1=-12

a2=5,b2=3,c2=-4

= , =

Here, ≠

∴givensystemofequationshaveuniquesolutions.

Now,

In–eq1

3x=12–5y

X=

Substitutexin–eq2

weget,

5× +3y=4

=4

60–25y+9y=12

60–16y=12

16y=60–12

16y=48

y= =3

∴y=3

Now,substituteyin–eq1

Weget,

3x+5×3=12

3x+15=12

3x=12–15

3x=-3

x= =-1

∴x=-1

∴x=-1andy=3

Question:2

Showthateachof

Solution:

Given:2x–3y=17–eq1

4x+y=13–eq2

Here,

a1=2,b1=-3,c1=17

a2=4,b2=1,c2=13

= , =

Here, ≠

∴Givensystemofequationshaveuniquesolutions.

Now,

In–eq1

2x=17+3y

X=

Substitutexin–eq2

weget,

4× +y=13

=13

68+12y+2y=26

68+14y=26

14y=26–68

14y=-42

y= =-3

∴y=-3


Weget,

2x–3×(-3)=17

2x+9=17

2x=17–9

2x=8

x=8/2=4

∴x=4

∴x=4andy=-3

Question:3

Showthateachof

Solution:

Given: ⇒2x+3y=18–eq1

x-2y=2–eq2

Here,

a1=2,b1=3,c1=18

a2=1,b2=-2,c2=2

= , =

Here, ≠

∴Givensystemofequationshaveuniquesolutions.

Now,

In–eq1

2x=18–3y

X=

Substitutexin–eq2

weget,

–2y=2

=2

18–3y–4y=4

18–7y=4

7y=18–4

7y=14

y= =2

∴y=2


Weget,

x–2×(2)=2

x–4=2

x=2+4

x=6

∴x=6

∴x=6andy=2

Question:4

Findthevalueof

Solution:

Given:2x+3y-5=0–eq1

kx-6y-8=0–eq2

Here,

a1=2,b1=3,c1=-5

a2=k,b2=-6,c2=-8

Givensystemsofequationshasauniquesolution

∴ ≠

2≠

k≠2×2=-4

∴k≠-4

Question:5

Findthevalueof

Solution:

Given:x-ky=2–eq1

3x+2y+5=0–eq2

Here,

a1=1,b1=-k,c1=-2

a2=3,b2=2,c2=5


∴

2≠-3k

-3k≠2

Question:6

Findthevalueof

Solution:

Given:5x-7y-5=0–eq1

2x+ky-1=0–eq2

Here,

a1=5,b1=-7,c1=-5

a2=2,b2=k,c2=-1


∴ ≠

≠

5k≠(-7)×2

5k≠ 14

k≠

∴

Question:4

Findthevalueof

Solution:

Given:4x+ky+8=0–eq1

x+y+1=0–eq2

Here,

a1=4,b1=k,c1=8

a2=1,b2=1,c2=1


∴

≠

4≠k

∴k≠4

Question:8

Findthevalueof

Solution:

Given:4x–5y=k–eq1

2x–3y=12–eq2

Here,

a1=4,b1=-5,c1=-k

a2=2,b2=-3,c2=-12


∴

Here,thesystemofequationshaveuniquesolutions,irrespectiveofthevalueofk.

Thatissolutionofthegivensystemofequationsisindependentofthevalueofk.

∴kisanyrealnumber

Question:9

Findthevalueof

Solution:

Given:kx+3y=(k–3)–eq1

12x+ky=k–eq2

Here,

a1=k,b1=3,c1=k–3

a2=12,b2=k,c2=k


∴ ≠

≠

K2≠36

k≠√36

∴k≠±6

∴k≠6andk≠–6

Thatiskcanbeanyrealnumberotherthan-6and6

∴kisanyrealnumberotherthan6and-6

Question:10

Showthatthesys

Solution:

Given:2x–3y=5–eq1

6x–9y=15–eq2

Here,

a1=2,b1=-3,c1=5

a2=6,b2=-9,c2=15

Here,

= =

= =

= =

Here,

= =

∴Thegivensystemofequationshasinfinitenumberofsolutions.

Question:11

Showthatthesys

Solution:

Given:6x+5y=11–eq1

9x+ y=21⇒18x+15y=42–eq2

Here,

a1=6,b1=5,c1=-11

a2=18,b2=15,c2=-42

Here,

= =

= =

Here,

= ≠

Thatisgivesystemofequationsareparallellines,thatistheydon’thaveanysolutions.

∴Thesystemofequationshasnosolution.

Question:12

Forwhatvalueof

Solution:

(i)Given:kx+2y=5–eq1

3x–4y=10–eq2

Here,

a1=k,b1=2,c1=5

a2=3,b2=-4,c2=10


∴ ≠

≠

-4k 6

k≠

∴k≠

(ii)Given:kx+2y=5–eq1

3x–4y=10–eq2

Here,

a1=k,b1=2,c1=5

a2=3,b2=-4,c2=10

Giventhatsystemsofequationshasnosolution

∴ =

Here,

=

Here,

-4k=6

∴

Question:13

Forwhatvalueof

Solution:

(i)Given:x+2y=5–eq1

3x+ky+15=0–eq2

Here,

a1=1,b1=2,c1=-5

a2=3,b2=k,c2=15


∴ ≠

≠

k≠6

∴k≠6

(ii)Given:x+2y=5–eq1

3x+ky+15=0–eq2

Here,

a1=1,b1=2,c1=-5

a2=3,b2=k,c2=15

Giventhatsystemsofequationshasnosolution

∴ = ≠

Here,

=

Here,

k=6

∴k=6

Question:14

Forwhatvalueof

Solution:

(i)Given:x+2y=3–eq1

5x+ky+7=0–eq2

Here,

a1=1,b1=2,c1=-3

a2=5,b2=k,c2=7


∴ ≠

≠

k≠10

∴k≠10

(ii)Given:x+2y=3–eq1

5x+ky+7=0–eq2

Here,

a1=1,b1=2,c1=-3

a2=5,b2=k,c2=7

Giventhatsystemofequationshasnosolution

∴ = ≠

Here,

=

Here,

k=10

∴k=10

Forthesystemofequationstohaveinfinitelymanysolutions

= =

= = whichiswrong.

Thatis,foranyvalueofkthegivesystemofequationscannothaveinfinitelymanysolutions.

Question:15

Findthevalueof

Solution:

Given:2x+3y=7–eq1

(k-1)x+(k+2)y=3k–eq2

Here,

a1=2,b1=3,c1=7

a2=k-1,b2=k+2,c2=3k

Giventhatsystemofequationshasinfinitelymanysolution

∴ = =

= =

Here,

=

2×(k+2)=3×(k-1)

2k+4=3k–3

3k–2k=4+3

K=7

∴k=7

Question:16

Findthevalueof

Solution:

Given:2x+(k–2)y=k–eq1

6x+(2k–1)y=(2k+5)–eq2

Here,

a1=2,b1=k–2,c1=k

a2=6,b2=2k–1,c2=2k+5


∴ = =

= =

Here,

=

2×(2k–1)=6×(k-2)

4k–2=6k–12

12–2=6k–4k

2k=10

K=5

∴k=5

Question:17

Findthevalueof

Solution:

Given:kx+3y=(2k+1)–eq1

2(k+1)x+9y=(7k+1)–eq2

Here,

a1=k,b1=3,c1=-(2k+1)

a2=2(k+1),b2=9,c2=-(7k+1)


∴ = =

= =

Here,

=

9k=6×(k+1)

9k=6k+6

9K–6k=6

3k=6

K=

K=2

∴k=2

Question:18

Findthevalueof

Solution:

Given:5x+2y=2k–eq1

2(k+1)x+ky=(3k+4)–eq2

Here,

a1=5,b1=2,c1=-2k

a2=2(k+1),b2=k,c2=-(3k+4)


∴ = =

= =

Here,

=

5k=4×(k+1)

5k=4k+4

5K–4k=4

k=4

∴k=4

Question:19

Findthevalueof

Solution:

Given:(k–1)x–y=5–eq1

(k+1)x+(1–k)y=(3k+1)–eq2

Here,

a1=(k-1),b1=-1,c1=-5

a2=(k+1),b2=(1-k),c2=-(3k+1)


∴ = =

= =

Here,

=

(3k+1)=-5×(1-k)

3k+1=-5+5k

5K–3k=1+5

2k=6

k=

k=3

∴k=3

Question:20

Findthevalueof

Solution:

Given:(k–3)x+3y=k–eq1

kx+ky=12–eq2

Here,

a1=(k-3),b1=3,c1=-k

a2=k,b2=k,c2=-12


∴ = =

= =

Here,

=

3×(-12)=-k×(k)

-36=-k2

K2=36

k=√36

k=±6

k=6andk=-6–eq3

Also,

=

K(k-3)=3k

K2-3k=3k

K2-6k=0

K(k-6)=0

K=0andk=6–eq4

From–eq3and–eq4

k=6

∴k=6

Question:21

Findthevalueso

Solution:

Given:(a–1)x+3y=2–eq1

6x+(1–2b)y=6–eq2

Here,

a1=(a-1),b1=3,c1=-2

a2=6,b2=(1-2b),c2=-6


∴ = =

= =

Here,

=

3×(-6)=(1-2b)×(-2)

-18=-2+4b

4b=-18+2

4b=-16

b=

b=-4

Also,

=

-6(a-1)=-2×6

-6a+6=-12

-6a=-12-6

-6a=-18

a=

a=3

∴a=3

∴a=3,b=-4

Question:22

Findthevalueso

Solution:

Given:(2a-1)x+3y=5–eq1

3x+(b-1)y=2–eq2

Here,

a1=(2a-1),b1=3,c1=-5

a2=3,b2=(b-1),c2=-2


∴ = =

= =

Here,

=

-2×(2a-1)=3×(-5)

-4a+2=-15

-4a=-15-2

-4a=-17

b=

∴b=-

Also,

=

3(-2)=-5×(b-1)

-6=-5b+5

5b=5+6

5b=11

b=

∴b=

∴a= ,b=

Question:23

Findthevalueso

Solution:

Given:2x-3y=7–eq1

(a+b)x-(a+b-3)y=4a+b–eq2

Here,

a1=2,b1=-3,c1=-7

a2=(a+b),b2=-(a+b-3),c2=-(4a+b)


∴ = =

= =

Here,

=

-3×(-4a+b)=-7×-(a+b-3)

12a+3b=7a+7b-21

12a-7a=-3b+7b-21

5a=4b-21

5a–4b+21=0 eq3

Also,

=

2×-(4a+b)=-7×(a+b)

-8a–2b=-7a–7b

-8a+7a=2b–7b

-a=-5b

a=5b eq4

substitute–eq4in–eq3

5(5b)–4b+21=0

25b–4b+21=0

21b+21=0

b=

b=-1

substitute‘b’in–eq4

a=5(-1)

a=-5

∴a=-5,b=-1

Question:24

Findthevalueso

Solution:

Given:2x+3y=7–eq1

(a+b+1)x+(a+2b+2)y=4(a+b)+1–eq2

Here,

a1=2,b1=3,c1=-7

a2=(a+b+1),b2=(a+2b+2),c2=-(4(a+b)+1)


∴ = =

= =

Here,

=

3×-(4(a+b)+1)=-7×(a+2b+2)

-12a-12b-3=-7a-14b-14

-12a+7a-3=12b-14b-14

-5a-3=-2b-14

5a-2b-11=0 eq3

Also,

=

2×-(4(a+b)+1)=-7×(a+b+1)

-8a–8b–2=-7a–7b–7

-8a+7a=8b–7b–7+2

-a=b–5

a+b=5

a=5–b eq4


5(5–b)-2b-11=0

25–5b-2b-11=0

-7b+14=0

b=

b=2


a=5-2

a=3

∴a=3,b=2

Question:25

Findthevalueso

Solution:

Given:2x+3y=7–eq1

(a+b)x+(2a-b)y=21–eq2

Here,

a1=2,b1=3,c1=-7

a2=(a+b),b2=(2a–b),c2=-21


∴ = =

= =

Here,

=

3×-21=-7×(2a-b)

-63=-14a+7b

14a-7b-63=0

2a–b–9=0 eq3

Also,

=

2×-21=-7×(a+b)

-42=-7a–7b

7a+7b+42=0

a+b+6=0

a+b=6

a=6–b eq4


2(6–b)–b–9=0

12–2b–b–9=0

-3b+3=0

b=

b=1


a=6-1

a=5

∴a=5,b=1

Question:26

Findthevalueso

Solution:

Given:2x+3y=7–eq1

2ax+(a+b)y=28–eq2

Here,

a1=2,b1=3,c1=-7

a2=2a,b2=(a+b),c2=-28


∴ = =

= =

Here,

=

3×-28=-7×(a+b)

-84=–7a–7b

7a+7b–84=0

a+b–12=0 eq3

Also,

=

2×-28=-7×2a

–56=–14a

14a=56

a=

a=4 eq4


4+b–12=0

a+b–12=0

b–8=0

b=8

∴a=4,b=8

Question:27

Findthevalueof

Solution:

Given:8x+5y=9–eq1

kx+10y=15–eq2

Here,

a1=8,b1=5,c1=-9

a2=k,b2=10,c2=-15

Here,


∴ = ≠

= ≠

Here,

=

8×10=5×k

5k=80

K=

k=16

∴k=16

Question:28

Findthevalueof

Solution:

Given:kx+3y=3–eq1

12x+ky=6–eq2

Here,

a1=k,b1=3,c1=-3

a2=12,b2=k,c2=-6

Here,


∴ = ≠

= ≠

Here,

=

k×k=3×12

k2=√36

K=±6 eq3

Also,

≠

3×-6≠-3×k

-18≠-3k

3k≠18

K≠6 eq4

From eq3and eq4wecanconclude

K= 6

∴k=-6

Question:29

Findthevalueof

Solution:

Given:3x-y-5=0–eq1

6x-2y+k=0–eq2

Here,

a1=3,b1=-1,c1=-5

a2=6,b2=-2,c2=k

Here,


∴ = ≠

= ≠

Here,

≠

-k≠-2×-5

-k≠-10

K≠10

∴k≠-10Therefore,fork=10,systemhasnosolution.

Question:30

Findthevalueof

Solution:

Given:kx+3y=k-3–eq1

12x+ky=k–eq2

Here,

a1=k,b1=3,c1=-(k-3)

a2=12,b2=k,c2=-k

Here,


∴ = ≠

= ≠

Here,

=

k×k=3×12

k2=√36

K=±6 eq3

Also,

≠

3×-k -(k-3)×k

-3k≠-k2+3k

K2-3k-3k≠0

K2 6k≠0

K(k-6)≠0

K≠0andk≠6 eq4

From eq3and eq4wecanconclude

K= 6

∴k=-6

Question:31

Findthevalueof

Solution:

Given:5x-3y=0–eq1

2x+ky=0–eq2

Here,

a1=5,b1=-3,c1=0

a2=2,b2=k,c2=0

Here,

Giventhatsystemofequationshasnonzerosolution.

∴ =

=

Here,

=

5×k=-3×2

5k= 6

K=

∴

Exercise:3EQuestion:1

5chairsand4ta

Solution:

Letthecostofeachchairandeachtablearexandyrespectively.

Accordingtoquestion,

5×(costofeachchair)+4×(costofeachtable)=5600,and

4×(costofeachchair)+3×(costofeachtable)=4340

∴5x+4y=5600.....(1)

4x+3y=4340.....(2)

fromequation(1),weget-

x=(5600-4y)/5.....(3)

substitutingthevalueofxinequation(2),weget-

⇒1/5y=140

∴y=700

substitutingthevalueofyinequation(3),weget-

x=560

ThusthecostofeachchairisRs.560andthatofatableisRs.700.

Question:2

23spoonsand17

Solution:

Letthecostofeachspoonandeachforkarexandyrespectively.

Accordingtoquestion,

23×(costofeachspoon)+17×(costofeachfork)=1770,and

17×(costofeachspoon)+23×(costofeachfork)=1830

∴23x+17y=1770.....(1)

17x+23y=1830.....(2)

fromequation(1),weget-

x=(1770-17y)/23.....(3)


⇒30090+240y=42090

⇒240y=12000

∴y=50


x=40

ThusthecostofeachspoonisRs.40andthatofaforkisRs.50.

Question:3

Aladyhasonly2

Solution:

Lettheno.of25-paisacoinsbex.

thenoof50-paisacoins=50-x

[∵thetotalno.ofcoins=50]

Accordingtothequestion,

totalmoney=Rs.19.50=1950paise

∴25x+50(50-x)=1950

⇒2500-25x=1950

⇒25x=550

∴x=22

Thus,thenoof25-paisacoins=x=22and,

thenoof50-paisacoins=50-x=50-22=28.

Question:4

Thesumoftwonu

Solution:

Letthetwonumbersbexandy.

Accordingtoquestion-

x+y=137.....(1)

x-y=43.....(2)

Addingequations(1)and(2),weget-

2x=180

∴x=90


y=90-43=47

Thus,thenumbersare90and47.

Question:5

Findtwonumbers

Solution:



2x+3y=92.....(1)

4x-7y=2.....(2)

Fromequation(1),weget-

x=(92-3y)/2.....(3)

Substitutingthevalueofxinequation(2),weget-

⇒184-6y-7y=2

⇒13y=182

∴y=14


x=25


Question:6

Findtwonumbers

Solution:



3x+y=142.....(1)

4x-y=138.....(2)


7x=280

∴x=40


y=142-3x=142-120=22


Question:7

If45issubtract

Solution:

Letthegreaternumberbexandthesmallernumberbey.


2x-45=y

⇒2x-y=45.....(1)

and,2y-21=x

⇒x-2y=-21.....(2)


x=(y+45)/2.....(3)


⇒45-3y=-42

⇒3y=87

∴y=29


x=37


Question:8

Ifthreetimesth

Solution:

LettheLargernumberbexandthesmallernumberbey.

Weknowthat-

Dividend=Quotient Divisor+Remainder


3x=4y+8

⇒3x-4y=8.....(1)

and,5y=3x+5

⇒-3x+5y=5.....(2)


x=(4y+8)/3.....(3)


⇒y-8=5

∴y=13


x=20


Question:9

If2isaddedto

Solution:



OnCrossmultiplying,weget-

⇒2x+4=y+2

⇒2x-y=-2.....(1)

and,

⇒11x-44=5y-20

⇒11x-5y=24


x=(y-2)/2


⇒y-22=48

∴y=70


x=34


Question:10

Thedifferencebe

Solution:



x-y=14.....(1)

x2-y2=448.....(2)


x=y+14.....(3)

Substitutethevalueofxinequation(2),weget-

(y+14)2-y2=448

⇒28y+196=448

⇒28y=252

∴y=9

Substitutethevalueofyinequation(3),weget-

x=23


Question:11

Thesumofthedi

Solution:

Letthetwo-digitnumberbexy(i.e.10x+y).

Afterinterchangingthedigitsofthenumberxy,thenewnumberbecomesyx(i.e.10y+x).


sumofthedigitsis12

⇒x+y=12.....(1)

Also,thenumberobtainedbyinterchangingitsdigitsexceedsthegivennumberby18

⇒(10y+x)-(10x+y)=18

⇒-9x+9y=18

⇒-x+y=2.....(2)


x+y-x+y=10+4⇒2y=14⇒y=7


x=5

Thus,therequirednumberis57.

Question:12

Anumberconsisti

Solution:


Afterreversingthedigitsofthenumberxy,thenewnumberbecomesyx(i.e.10y+x).


(10x+y)=7(x+y)

⇒3x=6y

⇒x=2y.....(1)

and,

(10x+y)-27=(10y+x)

⇒9x-9y=27

⇒x-y=3.....(2)

Substitutingequation(1)into(2),weget-

y=3


x=6


Question:13

Thesumofthedi

Solution:


Afterinterchangingthedigitsofthenumberxy,thenewnumberbecomesyx(i.e.10y+x).


x+y=15.....(1)

(10y+x)-(10x+y)=9

⇒-9x+9y=9

⇒-x+y=1.....(2)


y=8


x=7


Question:14

Atwo-digitnum

Solution:




(10x+y)=4(x+y)+3

⇒6x-3y=3

⇒2x-y=1.....(1)

and,

(10x+y)+18=(10y+x)

⇒9x-9y=-18

⇒x-y=-2.....(2)

Subtractingequation(2)from(1),weget-

x=3


y=5


Question:15

Anumberconsists

Solution:



Weknowthat-

Dividend=Quotient×Divisor+Remainder


(10x+y)=6(x+y)

⇒4x=5y

⇒x=(5/4)y.....(1)

and,

(10x+y)-9=(10y+x)

⇒9x-9y=9

⇒x-y=1.....(2)


y=4


x=5

Thus,thenumberis54.

Question:16

Atwo-digitnum

Solution:




xy=35

⇒x=35/y.....(1)

and,

(10x+y)+18=(10y+x)

⇒9x-9y=-18

⇒x-y=-2.....(2)


⇒35-y2=-2y

⇒y2-2y-35=0

⇒y2-7y+5y-35=0

⇒y(y-7)+5(y-7)=0

⇒(y+5)(y-7)=0

∴y=7

[y=-5isinvalidbecausedigitsofanumbercannotbenegative.]

Substitutingthevalueofyinequation(1),weget-

x=5


Question:17

Atwo-digitnum

Solution:




xy=18

⇒x=18/y.....(1)

and,

(10x+y)-63=(10y+x)

⇒9x-9y=63

⇒x-y=7.....(2)


⇒18-y2=7y

⇒y2+7y-18=0

⇒y2+9y-2y-18=0

⇒y(y+9)-2(y+9)=0

⇒(y+9)(y-2)=0

∴y=2

[y=-9isinvalidbecausedigitsofanumbercannotbenegative.]


x=9


Question:18

Thesumofatwo

Solution:




(10x+y)+(10y+x)=121

⇒11x+11y=121

⇒x+y=11.....(1)

and,

x-y=3ory-x=3

[aswedon'tknowwhichdigitisgreateroutofxandy]

⇒x-y=±3.....(2)

AddingEquation(1)and(2),weget-

2x=14or8

⇒x=7or4

Case1.whenx=7

y=4[fromequation(1)]

Case2.whenx=4

y=7[fromequation(1)]

Thus,thepossiblenumbersare47or74.

Question:19

Thesumofthenu

Solution:

Letthefractionbex/y.


x+y=8.....(1)

and,


⇒4x+12=3y+9

⇒4x-3y=-3.....(2)


x=8-y.....(3)


4(8-y)-3y=-3

⇒7y=35

∴y=5


x=3

Thus,therequiredfractionis3/5.

Question:20

If2isaddedto

Solution:




⇒2x+4=y.....(1)

and,


3x=y-1

⇒3x+1=y.....(2)

ComparingL.H.Sofequation(1)andequation(2),weget-

2x+4=3x+1

⇒x=3


y=10


Question:21

Thedenominatoro

Solution:



-x+y=11.....(1)

and,

OnCrossmultiplying,Weget-

⇒4x+32=3y+24

⇒4x-3y=-8.....(2)


x=y-11.....(3)


4(y-11)-3y=-8

⇒y=36


x=25


Question:22

Findafractionw

Solution:




⇒2x-2=y+2

⇒2x-y=4.....(1)

and,


3x-21=y-2

⇒3x-y=19.....(2)

Subtractingequation(1)fromequation(2),weget-

⇒x=15


y=26


Question:23

Thesumofthenu

Solution:



x+y=4+2x

⇒-x+y=4.....(1)

and,


⇒3x+9=2y+6

⇒3x-2y=-3.....(2)


x=y-4.....(3)


3(y-4)-2y=-3

⇒y=9


x=5


Question:24

Thesumoftwonu

Solution:



x+y=16.....(1)

and,

⇒3x+3y=xy.....(2)


x=16-y.....(3)


3(16-y)+3y=(16-y)y

⇒48=16y-y2

⇒y2-16y+48=0

⇒y2-12y-4y+48=0

⇒y(y-12)-4(y-12)=0

⇒(y-4)(y-12)=0

⇒y=4ory=12

Case1.Wheny=4

x=12[fromequation(3)]

Case2.Wheny=12

x=4[fromequation(3)]

Thus,thepossiblevaluesare12and4.

Question:25

Therearetwocla

Solution:

LetinitiallythenumberofstudentsinclassroomAandBbexandyrespectively.


x-10=y+10

⇒x-y=20.....(1)

and,

x+20=2(y-20)

⇒x-2y=-60.....(2)


y=80

substitutethevalueofyinequation(1),weget-

x=100

Thus,No.ofstudentsinclassroomAis100andinBis80.

Question:26

Taxichargesina

Solution:

Letthefixedchargeoftaxibex.

Excludingfixedcharge,amanpaysRs.(1330-x)for80kmandRs.(1490-x)for90kmdistance.

∴Rateperkmisgivenby-


⇒90(1330-x)=80(1490-x)

⇒119700-90x=119200-80x

⇒10x=500

⇒x=50

Hence,thefixedcharge=Rs.50

and,Rateperkm=((1330-50)/80)=(1280/80)=Rs.16

Question:27

Apartofmonthly

Solution:

Lettheperdayfixedchargeofhostelbex.

Excludingfixedcharge,astudentpaysRs.(4500-30x)for25daysandRs.(5200-30x)for30daysmesscharge

Costoffoodperdayisgivenby-


⇒30(4500-30x)=25(5200-30x)

⇒135000-900x=130000-750x

⇒150x=5000

⇒x=(500/15)

Hence,thefixedchargeofhostelpermonth=(500/15)×30=Rs.1000

and,Rateperkm=((4500-1000)/25)=(3500/25)=Rs.140

Question:28

Amaninvestedan

Solution:

Lettheamountinvestedat10%perannumand8%perannumbexandyrespectively.


AnnualInterestonamountx+AnnualInterestofamounty=Rs.1350

⇒10x+8y=135000

⇒5x+4y=67500.....(1)

and,

AnnualInterestonamounty+AnnualInterestofamountx=Rs.1305

∴

⇒8x+10y=130500

⇒4x+5y=65250.....(2)


x=(67500-4y)/5.....(3)


⇒

⇒(9/5)y=11250

∴y=6250


x=8500

Thus,theamountinvestedat10%perannum=Rs.8500and,

theamountinvestedat8%perannum=Rs.6250

Question:29

Themonthlyincom

Solution:

LetthemonthlyincomesofAandBare5xand4xrespectively.Alsotheirmonthlyexpendituresare7yand5yrespectively.


SavingsoffamilyA=5x-7y=9000.....(1)

SavingsoffamilyB=4x-5y=9000.....(2)


x=2y.....(3)


y=3000


x=6000

Thus,themonthlyincomeofA=5x=Rs.30000

and,themonthlyincomeofB=4x=Rs.24000

Question:30

Amansoldachai

Solution:

Letthecostpriceofeachchairandthatofatablebexandyrespectively.


SellingPriceofachair(Profit=25%)+SellingPriceofatable(Profit=10%)=Rs.1520

⇒25x+22y=30400.....(1)

and,

SellingPriceofachair(Profit=10%)+SellingPriceofatable(Profit=25%)=Rs.1535

⇒22x+25y=30700.....(2)


⇒3x-3y=-300

⇒x-y=-100.....(3)


x=y-100


22(y-100)+25y=30700

⇒47y=32900

∴y=700


x=600

Thus,CostpriceofeachchairandthatofatableareRs.600andRs.700respectively.

Question:31

PointsAandBar

Solution:

Letthespeedofthe1stcaratpointAand2ndcaratpointBtravellinginpositivex-axisdirectionbexandyrespectively.

Case1:SameDirection

DistanceTravelledby1stand2ndCarin7hoursare7xand7yrespectively.

BoththecarswillmeetoutsideofthepointsAandBwhichare70kmapart.So,the1stcarwilltravel70kmmoredistancefrom2ndcarmeetingeachotherin7hours.

∴7x-7y=70

⇒x-y=10.....(1)

Case2:OppositeDirection

DistanceTravelledby1stand2ndCarin1hoursarexandyrespectively.

BoththecarswillmeetinbetweenthepointsAandBwhichare70kmapart.So,thesumofdistancetravelledby1stcaranddistancetravelledby2ndcarmeetingeachotherin1hoursisequalto70km.

∴x+y=70.....(2)


2x=80

∴x=40

substitutethevalueofxinequation(2),weget-

y=30

Thus,thespeedof1stcar=40km/h

and,thespeedof2ndcar=30km/h

Question:32

Atraincovereda

Solution:

Letthespeedoftrainbeskmphandthescheduledtimebethours

Also,Letthelengthofjourneybed.

∴s×t=d.....(1)


(s+5)(t-3)=d

⇒st-3s+5t-15=d

⇒3s-5t=-15.....(2)[∵s×t=dfrom(1)]

and,

(s-4)(t+3)=d

⇒st+3s-4t-12=d

⇒3s-4t=12.....(3)[∵s×t=dfrom(1)]


t=27

Substitutingthevalueoftinequation(3),weget-

s=40

∴d=s×t=40×27=1080km

Thus,thelengthofthejourney=1080Km

Question:33

Abdultravelled3

Solution:

Letthespeedofthetrainandthatofthetaxibexkmphandykmphrespectively.


.....(1)

and,

.....(2)


.....(3)

Substituteequation(3)in(2),weget-

⇒y=80


x=100

Thus,thespeedoftrain=100km\handthespeedoftaxi=80km\h.

Question:24

PlacesAandBar

Solution:

Letthespeedofthe1stcaratpointAand2ndcaratpointBtravellinginpositivex-axisdirectionbexandyrespectively.

Case1:SameDirection


BoththecarswillmeetoutsideofthepointsAandBwhichare160kmapart.So,the1stcarwilltravel160kmmoredistancefrom2ndcarmeetingeachotherin8hours.

∴8x-8y=160

⇒x-y=20.....(1)

Case2:OppositeDirection


BoththecarswillmeetinbetweenthepointsAandBwhichare160kmapart.So,thesumofdistancetravelledby1stcaranddistancetravelledby2ndcarmeetingeachotherin2hoursisequalto160km.

∴2x+2y=160

⇒x+y=80.....(2)


2x=100

∴x=50

substitutethevalueofxinequation(2),weget-

y=30

Thus,thespeedof1stcar=50km/h

and,thespeedof2ndcar=30km/h

Question:35

Asailorgoes8k

Solution:

Letthespeedofthesailorinstillwaterbevkmphandthespeedofthecurrentbeukmph.


Speedofthesailorinupstreamdirection=v-u

Speedofthesailorindownstreamdirection=v+u

∴8/(v+u)=2/3

⇒v+u=12.....(1)

and,

⇒8/(v-u)=1

⇒v-u=8.....(2)


v=10

Substitutingthevalueofvin(2),weget-

u=2

Thus,speedofsailorinstillwater=10kmphandspeedofcurrent=2kmph.

Question:36

Aboatgoes12km

Solution:

Letthespeedoftheboatinstillwaterandthespeedofthestreambevkmphandukmphrespectively.

Speedoftheboatinupstreamdirection=v-u

Speedoftheboatindownstreamdirection=v+u


12/(v-u)+40/(v+u)=8

⇒12x+40y=8[Let1/(v-u)=xand1/(v+u)=y]

⇒3x+10y=2.....(1)

and,

16/(v-u)+32/(v+u)=8

⇒16x+32y=8[Let1/(v-u)=xand1/(v+u)=y]

⇒2x+4y=1.....(2)


x=(2-10y)/3.....(3)


⇒4-8y=3

⇒8y=1

∴y=1/8

⇒v+u=8.....(4)


x=1/4

⇒v-u=4.....(5)


v=6

Substitutingthevalueofvinequation(4),weget-

u=2

Thus,Speedoftheboatinstillwater=6kmphandspeedofstream=2kmph.

Question:37

2menand5boys

Solution:

1stMethod

Letthetimetakenbyonemanalonetofinishtheworkandthattakenbyoneboyalonetofinishtheworkbeuandvdaysrespectively.

Timetakenby1mantofinishonepartofthework=1/udays

Timetakenby1boytofinishonepartofthesamework=1/vdays


2menand5boyscanfinishapieceofworkin4days.Therefore,tofinishonepartofworktheywilltake1/4days

.....(1)

Similarly,3menand6boyscanfinishthesameworkin3days.Therefore,tofinishonepartofworktheywilltake1/3days

.....(2)

Multiplyingequation(1)by3andequation(2)by2andusingtheeliminationmethod,wehave

⇒v=36

Substitutingthevalueofvinequation(1),weget-

u=18

Thus,timetakenbyonemantofinishtheworkalone=18days

and,timetakenbyoneboytofinishtheworkalone=36days

Question:38

Thelengthofar

Solution:

Letthelengthandbreadthoftheroombelandbmetersrespectively.


l-b=3.....(1)

and,

lb=(l+3)(b-2)

⇒lb=lb-2l+3b-6

⇒2l-3b=-6.....(2)

Subtractingequation(2)from[3×equation(1)],weget-

l=15

Substitutingthevalueoflinequation(1),weget-

b=12

Thus,Lengthofroom=15metersandbreadthofroom=12meters.

Question:39

Theareaofarec

Solution:

Letthelengthandbreadthofrectanglebelandbmetersrespectively.

areaofrectangle=l×b


(l-5)(b+3)=(l×b)-8

⇒lb+3l-5b-15=lb-8

⇒3l-5b=7.....(1)

and,

(l+3)(b+2)=(l×b)+74

⇒lb+2l+3b+6=lb+74

⇒2l+3b=68.....(2)


l=(5b+7)/3.....(3)

Substitutingthevalueoflinequation(2),weget-

⇒19b+14=204

⇒19b=190

⇒b=10

Substitutingthevalueofbinequation(3),weget-

l=19

Thus,Length=19metersandBreadth=10meters.

Question:40

Theareaofarec

Solution:

Letthelengthandbreadthofrectanglebelandbmetersrespectively.

areaofrectangle=l×b


(l+3)(b-4)=(l×b)-67

⇒lb-4l+3b-12=lb-67

⇒4l-3b=55.....(1)

and,

(l-1)(b+4)=(l×b)+89

⇒lb+4l-b-4=lb+89

⇒4l-b=93.....(2)

SubtractingEquation(1)Fromequation(2),weget-

b=19

Substitutingthevalueofbinequation(2),weget-

l=28

Thus,Length=28metersandBreadth=19meters.

Question:41

Arailwayhalfti

Solution:

LetthebasicfirstclassfullfareandreservationchargebeRs.xandRs.yrespectively.


Fullfare+reservationcharge=Rs.4150

⇒x+y=4150.....(1)

and,

[fullfare+reservationcharge]+[halffare+reservationcharge]=Rs.6255

⇒x+y+(x/2)+y=6255

⇒(3x/2)+2y=6255.....(2)


⇒x/2=2045

⇒x=4090

Substitutingthevalueofxintheequation(1),weget-

y=60

Thus,basicfullfare=Rs.4090

reservationcharge=Rs.60

Question:42

Fiveyearshence,

Solution:

Lettheageofthemanandhissonbexandyyearsrespectively.


Fiveyearshence,aman'sagewillbethreetimestheageofhisson

x+5=3(y+5)

⇒x-3y=10.....(1)

and,

Fiveyearsago,themanwasseventimesasoldashisson

x-5=7(y-5)

⇒x-7y=-30.....(2)

SubtractingEquation(2)from(1),weget-

⇒-3y+7y=10+30⇒4y=40⇒y=10


⇒x-3(10)=10⇒x=30+10⇒x=40

Thus,Man'sage,x=40yearsandson'sage,y=10years

Question:43

Twoyearsago,a

Solution:

Lettheageofthemanandhissonbexandyyearsrespectively.


x-2=5(y-2)

x-5y=-8.....(1)

and,

x+2=3(y+2)+8

⇒x-3y=12.....(2)

SubtractingEquation(1)from(2),weget-

y=10


x=42

Thus,Man'sage=42years

son'sage=10years

Question:44

Iftwicetheson'

Solution:

Lettheageofthefatherandhissonbexandyyearsrespectively.


x+2y=70.....(1)

and,

2x+y=95.....(2)


y=15


x=40

Thus,Ageoffather=40yearsandageofson=15years.

Question:45

Thepresentageo

Solution:

Lettheageofwomanandherdaughterbexandyyearsrespectively.

AccordingtoQuestion-

x=3y+3.....(1)

and,

x+3=10+2(y+3)

⇒x-2y=13.....(2)

Substituteequation(1)intoequation(2),weget-

y=10

Substitutingthevalueofyinequation(2),weget

x=33

Thus,theageofwoman=33years

and,theageofherdaughter=10years.

Question:46

Onsellingatea

Solution:

LettheactualpriceofeachoftheteasetandthelemonsetbeRs.xandRs.yrespectively.


[Sellingpriceofteaset(Loss=5%)+SellingPriceoflemonSet(Profit=15%)]-[costpriceofteaset+costpriceoflemonset]=Rs.7

⇒-5x+15y=700

⇒-x+3y=140.....(1)

and,

[Sellingpriceofteaset(Profit=5%)+SellingPriceoflemonSet(Profit=10%)]-[costpriceofteaset+costpriceoflemonset]=Rs.13

⇒5x+10y=1300

⇒x+2y=260.....(2)


5y=400

∴y=80

Substitutingthevalueof inequation(2),weget-

x=100

Thus,thecostofteaset=Rs.100andthecostoflemontea=Rs.80.

Question:47

Alendinglibrary

Solution:

Letthefixedchargeandthechargeforeachextradaybexandyrespectively.


x+4y=27

and,

x+2y=21


y=3

Substitutingthevalueofyin(2),weget-

x=15

Thus,FixedCharge=15andthechargeforeachextraday=Rs.3perday.

Question:48

Achemisthasone

Solution:

Letthe50%solutionusedbexlitres

Totalvolumeofsolution=10litres(Given)

∴25%solutionused=(10-x)litres

Volumeofacidinmixture=40%of10litres=4litres

but,volumeofacidinmixture=50%ofx+25%of(10-x)

⇒x+10=16

⇒x=6litres

Thus,50%solution=6litresand25%solution=4litres.

Question:49

Ajewellerhasba

Solution:

Lettheweightof18-caratgoldbexg.

∴weightof12-caratgold=(120-x)g

24-caratequals100%gold(Given)

∴%ofgoldin18-caratgold=(100/24)×18=75%

and%ofgoldin12-caratgold=50%

and%ofgoldin16-caratgold=(200/3)%

Now,

75%ofx+50%of(120-x)=(200/3)%of120

⇒x+240=320

⇒x=80

Thus,theweightof18-caratgold=80gandtheweightof12-caratgold=40g.

Question:50

90%and97%pure

Solution:

Letthequantityof90%acidsolutionbexlitres.

∴quantityof97%acidsolution=(21-x)litres

Now,

90%ofx+97%of(21-x)=95%of21

⇒7x=21(97-95)

⇒7x=42

⇒x=6

Thus,90%acidsolution=6litres

97%acidsolution=21-6=15litres

Question:51

Thelargerofthe

Solution:

Letthebiggersupplementaryanglebex°.

andsmallersupplementaryanglebey°.


x°+y°=180°.....(1)

[∵propertiesofsupplementaryangles]

and,

x°-y°=18°.....(2)


x°=99°

Substitutingthevalueofx°inequation(1),weget-

y°=81°

Thus,thetwosupplementaryanglesare81°and99°.

Question:52

InaΔABC,

Solution:

Ina∆ABC,

∠A+∠B+∠C=180°

[∴Inany∆ABC,thesumofalltheanglesis180°]

⇒x°+(3x-2)°+y°=180°

⇒4x°+y°=182°.....(1)

and,

∠C-∠B=9°(Given)

⇒-3x°+y°=7°.....(2)


7x°=175°

⇒x°=25°


y°=82°

Thus,∠A=25°,∠B=73°,∠C=82°

Question:53

Inacyclicquadr

Solution:

Inacyclicquadrilateral,thesumofoppositeanglesis180°andsumofalltheinterioranglesinaquadrilateralis360°.

∠A+∠B+∠C+∠D=360°

⇒(2x+4)°+(y+3)°+(2y+10)°+(4x-5)°=360°

⇒6x°+3y°=348°

⇒2x°+y°=116°.....(1)

and,

∠A+∠C=180°

⇒2x°+2y°=166°

⇒x°+y°=83°.....(2)


⇒x°=33°


y°=50°

Thus,∠A=70°,∠B=53°,∠C=110°,and∠D=127°

Exercise:3FQuestion:1

Writethenumber

Solution:

Therearetwoequationsgiveninthequestion:

x+2y–8=0…(i)

And,2x+4y–16=0…(ii)

Thesegivenequationsareintheforma1x+b1y+c1=0anda2x+b2y+c2=0where,

a1=1,b1=2andc1=–8

Also,a2=2+b2=4andc2=–16

Now,wehave:

And,

∴

Hence,thepairoflinearequationsarecoincidentandthereforehasinfinitelymanysolutions

Question:2

Findthevalueof

Solution:


2x+3y–7=0(i)

And,(k–1)x+(k+2)y–3k=0(ii)

Thesegivenequationsareintheforma1x+b1y+c1=0and

a2x+b2y+c2=0where,

a1=2,b1=3andc1=–7

Also,a2=(k–1),b2=(k+2)andc2=–3k

Now,forthegivenpairoflinearequationshavinginfinitelymanysolutionswemusthave:

, and

2(k+2)=3(k–1),3×3k=7(k+2)and2×3k=7(k–1)

2k+4=3,9k=7k+14and6k=7k–7

∴k=7,k=7andk=7

Hence,thevalueofkis7

Question:3

Forwhatvalueof

Solution:


10x+5y–(k–5)=0…(i)

And,20x+10y–k=0…(ii)


a1=10,b1=5andc1=–(k–5)

Also,a2=20,b2=10andc2=–k

Now,forthegivenpairoflinearequationshavinginfinitemanysolutionswemusthave:

2k–10=k

∴k=10


Question:4

Forwhatvalueof

Solution:


2x+3y–9=0(i)

And,6x+(k–2)y–(3k–2)=0(ii)


a1=2,b1=3andc1=–9

Also,a2=6,b2=(k–2)andc2=–(3k–2)

Now,forthegivenpairoflinearequationshavingnosolutionwemusthave:

,

k=11,

k=11,3(3k–2) 9(k–2)

∴k=11and1 3(True)


Question:5

Writethenumber

Solution:


x+3y–4=0…(i)

And,2x+6y–7=0…(ii)


a1=1,b1=3andc1=–4

Also,a2=2+b2=6andc2=–7

Now,wehave:

And,

∴

Hence,thegivenpairoflinearequationhasnosolution

Question:6

Writethevalueo

Solution:


3x+ky=0…(i)

And,2x–y=0…(ii)


a1=3,b1=kandc1=0

Also,a2=2+b2=–1andc2=0

Now,forthegivenpairhaveauniquesolutionwemusthave:

Hence,

Question:7

Thedifferencebe

Solution:

Letusassumethetwonumbersbexandy,wherex>y

So,accordingtoquestionwehave:

x–y=5…(i)

x2–y2=65…(ii)

Now,bydividing(ii)by(i)weget:

x+y=13…(iii)

Now,adding(i)and(ii)weget:

2x=18

Puttingthevalueofxin(iii),weget:

9+y=13

y=13–9

y=4

∴Thetwonumbersare9and4

Question:8

Thecostof5pen

Solution:

Letusassumethecostof1penisRsxandthatofpencilisRsy

Accordingtothequestion,wehave

5x+8y=120…(i)

8x+5y=153…(ii)

Now,addingboththeequationsweget:

13x+13y=273

13(x+y)=273

x+y=21…(iii)

Now,bysubtracting(i)from(ii)weget:

3x–3y=33

x–y=11…(iv)

Byadding(iii)and(iv),weget:

2x=32

Puttingthevalueofxin(iii),weget

16+y=21

y=21–16

y=5

∴Thecostof1penisRs.16andthatof1pencilisRs.5

Question:9

Thesumoftwonu

Solution:

Letusassumethelargernumberbexandthesmallernumberbey

Accordingtothequestion,wehave:

x+y=80…(i)

x=4y+5

x–4y=5…(ii)

Now,bysubtracting(ii)form(i)weget

5y=75

y=15

Puttingthevalueofyin(i),weget

x+15=80

x=80–15

x=65

Hence,thetwonumbersbe65and15

Question:10

Anumberconsists

Solution:

Letusassumetheonesdigitbexandthetensdigitbey


x+y=10…(i)

(10y+x)–18=10x+y

x–y=–2…(ii)


2x=8

Nowbyputtingthevalueofxin(i),weget

4+y=10

y=10–4

y=6

Hencetherequirednumberis64

Question:11

Amanpurchased4

Solution:

Letusassumethenumberofstampsof20pand25pbexandyrespectively


x+y=47…(i)

0.20x+0.25y=10

Also,4x+5y=200…(ii)

Fromequation(i),wehave

y=47–x

Now,puttingthevalueofyin(ii),weget

4x+5(47–x)=200

4x–5x+235=200

x=235–200

x=35

Nowputtingthevalueofxin(i),weget:

35+y=47

∴y=47–35

y=12

Hence,thenumberof20pstampsare35andthenumberof25pstampsare12

Question:12

Amanhassomehe

Solution:

Letusassumethenumberofhensbexandthatofcowsbey


x+y=48…(i)

2x+4y=140

x+2y=70…(ii)

Now,subtracting(i)from(ii)weget:

y=22

Hence,thenumberofcowsis22

Question:13

If …(i)

…(ii)

Now,multiplying(i)and(ii)byxyweget:

3x+2y=9…(iii)

9x+4y=21…(iv)

Now,multiplying(iii)by2andsubtractingitfrom(iv)weget:

9x–6x=21–18

3x=3

Now,puttingthevalueofxin(iii)weget:

3×1+2y=9

3+2y=9

2y=6

y=3

Hence,thevalueofx=1andy=3

Question:14

If …(i)

…(ii)

Now,multiplying(i)by12and(ii)by4weget:

3x+4y=5…(iii)

2x+4y=4…(iv)

Now,subtracting(iv)from(iii)weget:

x=1

Now,puttingthevalueofxin(iv)weget:

2+4y=4

4y=2

=

=

Hence,thevalueof(x+y)is

Question:15

If12x+17y=53

Solution:

Wehavethegivenpairofequationsare:

12x+17y=53…(i)

17x+12y=63…(ii)


29x+29y=116

29(x+y)=116

(x+y)=4

∴Thevalueof(x+y)is4

Question:16

Findthevalueof

Solution:

Thegiventwoequationsare:

3x+5y=0…(i)

kx+10y=0…(ii)

Thegivenequationisahomogenoussystemoflineardifferentialequationsoitalwayshasazerosolution

Weknowthat,forhavinganon–zerosolutionitmusthaveinfinitelymanysolutions

∴

k=6


Question:17

Findkforwhich

Solution:


kx–y–2=0…(i)

6x–2y–3=0…(ii)

Here,wehave:

a1=k,b1=–1andc1=–2

a2=6,b2=–2andc2=–3

Weknowthat,forthesystemhavingauniquesolutionwemusthave

∴

Question:18

Findkforwhich

Solution:


2x+3y–5=0…(i)

4x+ky–10=0…(ii)

Here,wehave:

a1=2,b1=3andc1=–5

a2=4,b2=kandc2=–10

Weknowthat,forthesystemhavingainfinitenumberofsolutionswemusthave

∴k=6


Question:19

Showthatthesys

Solution:


2x+3y–1=0…(i)

4x+ky–10=0…(ii)

Here,wehave:

a1=2,b1=2andc1=–1

a2=4,b2=6andc2=–4

Now,wehave

∴

Hence,thegivensystemhasnosolution

Question:20

Findkforwhich

Solution:


x+2y–3=0…(i)

5x+ky+7=0…(ii)

Here,wehave:

a1=1,b1=2andc1=–3

a2=5,b2=kandc2=7

Weknowthat,forthesystemtobeconsistentwemusthave

k=10


Question:21

Solve:

Solution:


…(i)

…(ii)

Now,substituting and in(i)and(ii)thegivenequationwillchangedto:

3u+2v=2…(iii)

9u–4v=1…(iv)

Now,bymultiplying(i)by2andaddingitwith(ii)weget:

15u=4+1

Also,bymultiplying(i)by3andsubtractingitfrom(ii)weget:

6u+4v=6–1

∴x+y=3…(v)

And,x–y=2…(vi)

Now,adding(v)and(vi)weget:

2x=5

Nowsubstitutingthevalueofxin(v),weget:

Exercise:MULTIPLECHOICEQUESTIONS(MCQ)Question:1

If2x+3y=12a

Solution:

Wehave:

2x+3y=12…(i)

3x–2y=5…(ii)

Now,bymultiplying(i)by2and(ii)by3andthenaddingthemweget:

4x+9x=24+15

13x=39

Nowputtingthevalueofxin(i),weget

2×3+3y=12

∴

Hence,optionCiscorrect

Question:2

Ifx–y=2and

Solution:

Wehave:

x–y=2…(i)

x+y=10…(ii)


2x=12

x=6

Puttingthevalueofxin(ii),weget

6+y=10

y=10–6

y=4


Question:3

If …(i)

…(ii)

Now,multiplying(i)and(ii)by6weget:

4x–3y=–1…(iii)

3x+4y=18…(iv)

Now,multiplying(iii)by4and(iv)by3andaddingthemweget:

16x+9x=–4+54

Puttingthevalueofxin(iv)weget:

3×2+4y=18

y=3

Hence,optionAiscorrect

Question:4

If …(i)

…(ii)



Hence,optionDiscorrect

Question:5

If and

Bysimplifyingaboveequations,weget:

3(2x+y+2)=5(3x–y+1)

6x+3y+6=15x–5y+5

9x–8y=1…(i)

And,6(3x–y+1)=3(3x+2y+1)

18x–6y+6=9x+6y+3

3x–4y=–1…(ii)

Now,multiplying(ii)by2andthensubtractingitfrom(i)weget:

9x–6x=1+2

∴x=1

Puttingthevalueofxin(ii),weget

3×1–4y=–1

∴


Question:6

If …(i)

…(ii)

Now,substituting and in(i)and(ii)weget:

3u+2v=2…(iii)

9u–4v=1…(iv)

Multiplying(iii)by2andaddingitwith(iv)weget:

6u+9u=4+1

Multiplyingagain(iii)by2andthensubtractingitfrom(iv),weget:

6v+4v=6–1

∴x+y=3…(v)

And,x–y=2…(vi)

Now,byadding(v)and(vi)weget:

2x=3+2

Substitutingthevalueofxin(v),weget

Hence,optionBiscorrect

Question:7

If4x+6y=3xy

Solution:

Wehave,

4x+6y=3xy…(i)

8x+9y=5xy…(ii)

Now,dividing(i)and(ii)byxyweget:

…(iii)

Also, …(iv)

Now,multiplying(iii)by2andthensubtractingitfrom(iv)weget:

∴x=3

Now,substitutingthevalueofxin(iii)weget:

∴y=4


Question:8

If29x+37y=10

Solution:

Wehave,

29x+37y=103…(i)

37x+29y=95…(ii)

Now,addingboththeequationsweget:

66x+66y=198

66(x+y)=198

x+y=3…(iii)

Now,subtracting(i)from(ii)weget:

8x–8y=–8

x–y=–1…(iv)

Nowadding(iii)and(iv),weget

2x=2

∴x=1

Puttingthevalueofxin(iii),weget

1+y=3

∴y=3–1=2


Question:9

If2x+y

Solution:

Wehave,

∴x+y=x–y

Hence,y=o

Thus,optionCiscorrect

Question:10

If …(i)

Also, …(ii)

Now,multiplying(ii)by2andthensubtractingitfrom(ii)weget:

∴y=1

Nowsubstitutingthevalueofyin(ii),weget:


Question:11

Thesystemkx–y

Solution:

Wehave,

kx–y–2=0(i)

6x–2y–3=0(ii)

Here,a1=k,b1=–1andc1=–2

a2=6,b2=–2andc2=–3

Weknowthat,forthesystemhavingauniquesolutionitmusthave:

∴


Question:12

Thesystemx–2y

Solution:

Wehave,

x–2y–3=0

3x+ky–1=0

Thegivenequationisintheform:a1x+b1y+c1=0anda2x+b2y+c2=0

Here,wehave:

a1=1,b1=–2andc1=–3

And,a2=3,b2=kandc2=–1

∴ , and

Thesegraphlineswillintersectatauniquepointwhenwehave:

∴

Hence,khasallrealvaluesotherthan–6

Thus,optionBiscorrect

Question:13

Thesystemx+2y

Solution:

Wehave,

x+2y–3=0

And,5x+ky+7=0

Here,a1=1,b1=2andc1=–3

a2=5,b2=kandc2=7

∴

And,

Weknowthat,forthesystemhavingnosolutionwemusthave:

∴k=10


Question:14

Ifthelinesgive

Solution:

Wehave,

3x+2ky–2=0

And,2x+5y+1=0

Here,a1=3,b1=2kandc1=–2

a2=2,b2=5andc2=1

∴

And,

Weknowthat,forthesystemhavingparallellineswemusthave:


Question:15

Forwhatvalueof

Solution:

Wehave,

kx–2y–3=0

And,3x+y–5=0

Here,a1=k,b1=–2andc1=–3

a2=3,b2=1andc2=–5

∴

And,

Weknowthat,forthesegraphsintersectatauniquepointwemusthave:

Hence,thelinesofthegraphwillintersectatallrealvaluesofkexcept–6

Thus,optionDiscorrect

Question:16

Thepairofequat

Solution:

Wehave,

x+2y+5=0

And,–3x–6y+1=0

Here,a1=1,b1=2andc1=5

a2=–3,b2=–6andc2=1

∴

And,

∴



Question:17

Thepairofequat

Solution:

Wehave,

2x+3y–5=0

And,4x+6y–15=0


a2=4,b2=6andc2=–15

∴

And,

∴



Question:18

Ifapairofline

Solution:

Weknowthat,

Ifapairoflinearequationsisconsistentthentheirgraphlineswilleitherintersectatapointorcoincidence


Question:19

Ifapairofline

Solution:

Weknowthat,

Ifapairoflinearequationsisinconsistentthentheirgraphlinesdonotintersecteachotherandtherewillbenosolutionexists.Hence,thelinesareparallel

Thus,optionAiscorrect

Question:20

InaΔABC,

Solution:

Letusassume,∠A=xoand∠B=yo

∴∠A=3∠B=(3y)o

Weknowthat,sumofallsidesofthetriangleisequalto180o

∴∠A+∠B+∠C=180o

x+y+3y=180o

x+4y=180o(i)

Alsowehave,∠C=2(∠A+∠B)

3y=2(x+y)

2x–y=0(ii)

Now,bymultiplying(ii)by4weget:

8x–4y=0(iii)

Andadding(i)and(iii),weget

9x=180o

x=20

Puttingthevalueofxin(i),weget

20+4y=180

4y=180–20

4y=160

y=40

∴∠B=y=40o


Question:21

Inacyclicquadr

Solution:

Itisgiveninthequestionthat,

IncyclicquadrilateralABCD,wehave:

∠A=(x+y+10)o

∠B=(y+20)o

∠C=(x+y–30)o

∠D=(x+y)o

AsABCDisacyclicquadrilateral

∴∠A+∠C=180oand∠B+∠D=180o

Now,∠A+∠C=180o

(x+y+10)o+(x+y–30)o=180o

2x+2y–20o=180o

x+y=100o(i)

Also,∠B+∠D=180o

(y+20)o+(x+y)o=180o

x+2y+20o=180o

x+2y=160o(ii)

Onsubtracting(i)from(ii),weget

y=(160–100)o

y=60o


x+60o=100o

x=100o–60o

x=40o

∴∠B=(y+20)o

∠B=60o+20o=80o


Question:22

Thesumofthedi

Solution:

Letusassumethetensandtheunitdigitsoftherequirednumberbexandyrespectively

∴Requirednumber=(10x+y)

Accordingtothegivenconditioninthequestion,wehave

x+y=15(i)

Byreversingthedigits,weobtainthenumber=(10y+x)

∴(10y+x)=(10x+y)+9

10y+x–10x–y=9

9y–9x=9

y–x=1(ii)

Now,onadding(i)and(ii)weget:

2y=16

Puttingthevalueofyin(i),weget:

x+8=15

x=15–8

x=7

∴Requirednumber=(10x+y)

=10×7+8

=70+8

=78


Question:23

Inagivenfracti

Solution:

Letthefractionbe

Accordingtothequestion,

2x–2=y+2

y=2x–4…(i)

And,

3x–21=y–2

3x=y+19…(ii)

Using(i)in(ii)

3x=2x–4+19

X=15

Usingvalueofxin(i),weget

y=2(15)–4

y=30–4

y=26

Therefore,requiredfraction=


Question:24

5yearshence,th

Solution:

Letusassumethepresentageofmenbexyears

Also,thepresentageofhissonbeyyears

Accordingtoquestion,after5years:

(x+5)=3(y+5)

x+5=3y+15

x–3y=10…(i)

Also,fiveyearsago:

(x–5)=7(y–5)

x–5=7y–35

x–7y=–30…(ii)

Now,onsubtracting(i)from(ii)weget:

–4y=–40

y=10


x–3×10=10

x–30=10

x=10+30

x=40

∴Thepresentageofmenis40years


Question:25

Thegraphsofthe

Solution:

Wehave,

6x–2y+9=0

And,3x–y+12=0

Here,a1=6,b1=–2andc1=9

a2=3,b2=–1andc2=12

∴ , and

Clearly,

Hence,thegivensystemhasnosolutionandthelinesareparallel

∴OptionBiscorrect

Question:26

Thegraphsofthe

Solution:

Wehave,

2x+3y–2=0

And,x–2y–8=0


And,a2=1,b2=–2andc2=–8

∴ , and

Clearly,

Hence,thegivensystemhasauniquesolutionandthelinesintersectexactlyatonepoint

∴OptionCiscorrect

Question:27

Thegraphsofthe

Solution:

Wehave,

5x–15y–8=0

And,

Here,a1=5,b1=–15andc1=–8

And,a2=3,b2=–9andc2=

∴ , and

Clearly,

Hence,thegivensystemhasauniquesolutionandthelinesarecoincident

∴OptionAiscorrect

Exercise:FORMATIVEASSESSMENT(UNITTEST)Question:1

Thegraphicrepre

Solution:

Given:Twoequations,x+2y=3

⇒x+2y–3=0----(1)

2x+4y+7=0----(2)

Weknowthatthegeneralformforapairoflinearequationsin2variablesxandyisa1x+b1y+c1=0

anda2x+b2y+c2=0.

Comparingwithaboveequations,

wehavea1=1,b1=2,c1=-3;a2=2,b2=4,c2=7

Since

∴Bothlinesareparalleltoeachother.

Question:2

If2x-3y=7an

Solution:

Given:Twoequations,2x–3y=7

⇒2x–3y–7=0

(a+b)x–(a+b–3)y=4a+b

(a+b)x–(a+b–3)y–(4a+b)=0


anda2x+b2y+c2=0.


wehavea1=2,

b1=-3,

c1=-7;

a2=a+b,

b2=-(a+b–3),

c2=-(4a+b)

Since,itisgiventhattheequationshaveinfinitenumberofsolutions,thenlinesarecoincidentand

So,

Letusconsider

Then,bycrossmultiplication,2(a+b–3)=3(a+b)

⇒2a+2b–6=3a+3b

⇒a+b+6=0…(1)

Nowconsider

Then,3(4a+b)=7(a+b–3)

⇒12a+3b=7a+7b–21

⇒5a–4b+21=0…(2)

Solvingequations(1)and(2),

5×(1),(5a+5b+30)–(5a–4b+21)=0

⇒9b+9=0

⇒9b=-9

⇒b=-1

Substitutebvaluein(1),

a-1+6=0

a+5=0

a=-5

∴a=-5;b=-1

Question:3

Thepairofequat

Solution:

Given:2x+y–5=0and3x+2y–8=0


anda2x+b2y+c2=0.


wehavea1=2,b1=1,c1=-5;a2=3,b2=2,c2=-8

Since

Thelinesareintersecting.

∴Thepairofequationshasauniquesolution.

Question:4

Ifx=-yandy

Solution:

Giventhatx=-yandy>0

Letusverifyalltheoptionsbysubstitutingthevalueofx.

OptionA:x2y>0

⇒(-y)2(y)>0

⇒y2(y)>0

⇒y3>0

Sincey>0,y3>0satisfies.

OptionB:x+y=0

⇒(-y)+y=0

0=0

LHS=RHS

Hencesatisfies.

OptionC:xy<0

⇒(-y)(y)<0

⇒-y2<0

Hencesatisfies.

OptionD:

⇒

⇒

Sincey>0,also1/y>0but-2/y<0

Hence,itisnotsatisfied.

Question:5

Showthatthesys

Solution:

Given:-x+2y+2=0and

ToProve:Thesystemofgivenequationshasauniquesolution.

Proof:


anda2x+b2y+c2=0.


wehavea1=-1,

b1=2,

c1=2;

a2=1/2,

b2=-1/2

c2=-1

Since

Thelinesareintersecting.

Thesystemofgivenequationshaveauniquesolution.

Question:6

Forwhatvalueso

Solution:

Given:kx+3y=k-2,

12x+ky=k


anda2x+b2y+c2=0.


wehavea1=k,b1=3,c1=-(k–2);a2=12,b2=k,c2=-k

Forgivenequationstobeinconsistent,

Bycrossmultiplication,k2=36

So,k=±6

Fork=±6,thesystemofequationskx+3y=k-2,12x+ky=kisinconsistent.

Question:7

Showthattheequ

Solution:

Given:9x-10y=21,

ToProve:Thegivenequationshaveinfinitelymanysolutions.

Proof:


anda2x+b2y+c2=0.


wehavea1=9,

b1=-10,

c1=-21;

a2=3/2,

b2=-5/3

c2=-7/2

Since

Thelinesarecoincident.

Thegivenequationshaveinfinitelymanysolutions.

Question:8

Solvethesystem

Solution:

x=4,y=2

Given:x-2y=0…(1)

3x+4y=20…(2)

Byeliminationmethod,

Step1:Multiplyequation(1)by3andequation(2)by1tomakethecoefficientsofxequal.

Then,wegettheequationsas:

3x–6y=0…(3)

3x+4y=20…(4)

Step2:Subtractequation(4)fromequation(3),

(3x–3x)+(4y+6y)=20–0

⇒10y=20

y=2

Step3:Substituteyvaluein(1),

x–2(2)=0

⇒x=4

Thesolutionisx=4,y=2.

Question:9

Showthatthepat

Solution:

Given:x-3y=2and-2x+6y=5

ToProve:Thepathsrepresentedbythegivenequationsareparallel.

Proof:


anda2x+b2y+c2=0.


wehavea1=1,b1=-3,c1=-2;a2=-2,b2=6,c2=-5

Since

Bothlinesareparalleltoeachother.

Question:10

Thedifferencebe

Solution:

Thepairoflinearequationsformedis:

a–b=26…(1)

a=3b…(2)

Wesubstitutevalueofainequation(1),toget

3b–b=26

⇒2b=26

⇒b=13

Substitutingvalueofbinequation(2),

a=3(13)

⇒a=39

Thenumbersare13and39.

Question:11

Solve:23x+29y

Solution:

Thegivenequationsare23x+29y=98,29x+23y=110.


anda2x+b2y+c2=0.


wehavea1=23,b1=29,c1=-98;a2=29,b2=23,c2=-110

Wecansolvebycrossmultiplicationmethodusingtheformula

Substitutingvaluesintheformula,weget

⇒

⇒

⇒ and

⇒x=3andy=1

Thesolutionisx=3andy=1.

Question:12

Solve:6x+3y=

Solution:

Thegivenequationsare6x+3y=7xyand3x+9y=11xy.

Dividingbyxyonbothsidesofthegivenequations,weget

Then,

…(1)

…(2)

Ifwesubstitute and in(1)and(2),weget

3p+6q=7…(3)

9p+3q=11…(4)

Nowbyeliminationmethod,



9p+18q=21…(5)

9p+3q=11…(6)


(9p–9p)+(3q–18q)=11–21

⇒-15q=-10

⇒

Step3:Substituteqvaluein(3),

3p=3

⇒p=1

Weknowthat and .

Substitutingvaluesofpandq,weget

x=1andy=

Thesolutionisx=1andy= .

Question:13

Findthevalueof

Solution:

Thegivensystemofequationsis3x+y=1andkx+2y=5.


anda2x+b2y+c2=0.


wehavea1=3,b1=1,c1=-1;a2=k,b2=2,c2=-5

i)Forthegivensystemofequationstohaveauniquesolution,

⇒

⇒k≠6

Fork≠6,thegivensystemofequationshasauniquesolution.

ii)Forthegivensystemofequationstohavenosolution,

⇒

⇒k=6

Fork=6,thegivensystemofequationshasnosolution.

Question:14

InaABC,∠C

Solution:

Weknowthatthesumofanglesofatriangleis180°

i.e.∠A+∠B+∠C=180°

Thegivenrelationis∠C=3∠B=2(∠A+∠B)…(1)

⇒3∠B=2(∠A+∠B)

⇒3∠B=2∠A+2∠B

⇒2∠A=∠B

⇒∠A=∠B/2

SubstitutingvaluesintermsofBinequation(1),

∠B/2+∠B+3∠B=180°

∠B/2+4∠B=180°

∠B(9/2)=180°

∠B=180×9/2

∠B=40°

SubstitutingBvaluein(1),

∠C=3∠B=3(40)=120°

And∠A=∠B/2=40/2=20°

Themeasuresare∠A=20°,∠B=40°,∠C=120°.

Question:15

5pencilsand7p

Solution:

Letthecostofpencilsbexandcostofpensbey.

Thelinearequationsformedare:

5x+7y=195…(1)

7x+5y=153…(2)


anda2x+b2y+c2=0.


wehavea1=5,b1=7,c1=-195;a2=7,b2=5,c2=-153



⇒

⇒

⇒ and

⇒x=4andy=25

ThecostofeachpencilisRs.4andcostofeachpenisRs.25.

Question:16

Solvethefollowi

Solution:

For2x–3y=1,(Ingraph-redline)

For4x–3y+1=0,(Ingraph–blueline)

Fromtheabovegraph,weobservethatthereisapoint(-1,-1)commontoboththelines.

So,thesolutionofthepairoflinearequationsisx=-1andy=-1.

Thegivenpairofequationsisconsistent.

Question:17

Findtheangleso

Solution:

ItisgiventhatanglesofacyclicquadrilateralABCDaregivenby:

∠A=(4x+20)°,

∠B=(3x-5)°,

∠C=(4y)°

and∠D=(7y+5)°.

Weknowthattheoppositeanglesofacyclicquadrilateralaresupplementary.

∠A+∠C=180°

4x+20+4y=180°

4x+4y–160=0…(1)

And∠B+∠D=180°

3x–5+7y+5=180°

3x+7y-180°=0…(2)

Byeliminationmethod,



12x+12y=480…(3)

12x+16y=540…(4)


(12x–12x)+(16y-12y)=540–480

⇒4y=60

y=15

Step3:Substituteyvaluein(1),

4x–4(15)–160=0

⇒4x–220=0

⇒x=55

Thesolutionisx=55,y=15.

Question:18

Solveforxandy

Solution:

Letusput and .

Onsubstitutingthesevaluesinthegivenequations,weget

35p+14q=19…(1)

14p+35q=37…(2)


anda2x+b2y+c2=0.


wehavea1=35,b1=14,c1=-19;a2=14,b2=35,c2=-37



⇒

⇒

⇒ and

⇒p=1/7andq=1

Since

⇒

⇒x+y=7…(3)andx–y=1…(4)

Addingequations(3)and(4),

(x+x)+(y–y)=7+1

2x=8

x=4

Substitutingxvaluein(4),

4–y=1

y=3

Thesolutionisx=4andy=3.

Question:19

If1isaddedto

Solution:


Giventhat

⇒5x+5=4y+4

⇒5x–4y+1=0…(1)

Alsogiventhat

⇒2x–10=y–5

⇒2x–y–5=0…(2)


anda2x+b2y+c2=0.


wehavea1=5,b1=-4,c1=1;a2=2,b2=-1,c2=-5



⇒

⇒

⇒ and

⇒x=7andy=9

Thefractionis7/9.

Question:20

Solve:

Solution:

Given: …(1)

ax-by=2ab…(2)

Multiplyingbyabto(1)andato(2),weget

a2x–b2y=a2b+ab2…(3)

a2x–aby=2a2b…(4)

Subtractingequation(4)fromequation(3),

(a2x–a2x)+(-aby)–(-b2y)=(2a2b-a2b)–ab2

⇒-aby+b2y=a2b–ab2

⇒by(b–a)=ab(a–b)

⇒y=b(b–a)/ab(a–b)

⇒y=-a

Substituteyvaluein(2),

ax–b(-a)=2ab

⇒ax+ab=2ab

⇒ax=ab

⇒x=b

Thesolutionisx=bandy=-a.

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3. LINEAR EQUATIONS IN TWO VARIABLES Exercise : 3A

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