Linear Equations in Two Variables

Made By:- Anant K

aul

Class : 9

th C

Roll N

o. 6

Linear Equations in Two Variables

may be put in the form

Ax + By = C,

Where A, B, and C are real numbers and A and B are not

both zero.

Solutions to Linear Equations in Two Variables

Consider the equation The equation’s solution set is

infinite because there are an infinite number of x’s and y’s that make it TRUE.

For example, the ordered pair (0, 10) is a solution because

Can you list other ordered pairs that satisfy this equation?

5 2 20x y

5 0 2 10 20 5 2 20x y

Ordered Pairs are listed with the

x-value first and the y-valuesecond.

Input-Output Machines

We can think of equations as input-output machines. The x-values being the “inputs” and the y-values being the “outputs.”

Choosing any value for input and plugging it into the equation, we solve for the output.

y = -2x + 5 y = -2(4) + 5

y = -8 + 5y = -3

x = 4 y = -3

Functions

Function- a relationship between two variables (equation) so that for every INPUT there is EXACTLY one OUTPUT.

To determine (algebraically) if an equation is a function we can examine its x/y table. If it is possible to get two different outputs for a certain input- it is NOT a function. In this case an x-value in the table or ordered pairs would repeat.

This may be determined (graphically) by using the Vertical Line Test. If any vertical line would touch the graph at more than one point- it is NOT a function.

Using Tables to List Solutions

For an equation we can list some solutions in a table.

Or, we may list thesolutions in ordered

pairs .{(0,-4), (6,0), (3,-2), ( 3/2, -3), (-3,-6), (-6,-8), … }

2 3 12x y x y

0 -4

6 0

3 -2

3/2 -3

-3 -6

-6 -8

… …

Graphing a Solution Set

To obtain a more complete picture of a solution set we can graph the ordered pairs from our table onto a rectangular coordinate system.

Let’s familiarize ourselves with the Cartesian coordinate system.

Cartesian Plane

x- axis

y-axis

Quadrant I(+,+)

Quadrant II( - ,+)

Quadrant IV(+, - )

Quadrant III( - , - )

origin

Graphing Ordered Pairs on a Cartesian Plane

x- axis

y-axis

1) Begin at the origin

2) Use the x-coordinate to move right (+) or left (-) on the x-axis

3) From that position move either up(+) or down(-) according to the y-coordinate

4) Place a dot to indicate a point on the plane

Examples: (0,-4)

(6, 0)

(-3,-6)

(6,0)

(0,-4)(-3, -6)

Graphing More Ordered Pairs from our

Table for the equation

x

y

(3,-2)

(3/2,-3)

(-6, -8)

2 3 12x y •Plotting more points

we see a pattern.

•Connecting the pointsa line is formed.

•We indicate that thepattern continues by placing

arrows on the line.

•Every point on this line is asolution of its equation.

Graphing Linear Equationsin Two Variables

The graph of any linear equation in two variables is a straight line.

Finding intercepts can be helpful when graphing.

The x-intercept is the point where the line crosses the x-axis.

The y-intercept is the point where the line crosses the y-axis.

On our previous graph, y = 2x – 3y = 12, find the intercepts.

y

x

Graphing Linear Equationsin Two Variables

On our previous graph, y = 2x – 3y = 12, find the intercepts.

The x-intercept is (6,0).

The y-intercept is (0,-4).

y

x

Finding INTERCEPTS

To find the To find the x-intercept: Plug in x-intercept: Plug in ZERO for y and solve ZERO for y and solve for x.for x.

2x – 3y = 122x – 3y = 12

2x – 3(0) = 122x – 3(0) = 12

2x = 122x = 12

x = 6x = 6Thus, the x-intercept is (6,0).

To find the To find the y-intercept: Plug in y-intercept: Plug in ZERO for x and ZERO for x and solve for y.solve for y. 2(0) – 3y = 122(0) – 3y = 12 2(0) – 3y = 122(0) – 3y = 12

-3y = 12-3y = 12 y = -4y = -4

Thus, the y-intercept is Thus, the y-intercept is (0,-4). (0,-4).

Special Lines y + 5 = 0 x = 3y = -5

y

x

x

y

y = # is a horizontal line x = # is a vertical line

NCERT BOOK

QUESTIONS WITH THEIR SOLUTIONS

Exercise 4.1

Question 1: The cost of a notebook is twice the cost of a pen. Write a linear

equation in two variables to represent this statement.

(Take the cost of a notebook to be Rs x and that of a pen to be Rs y.)

Answer :Let the cost of a notebook and a pen be x and y respectively.

Cost of notebook = 2 × Cost of pen

x = 2y

x − 2y = 0

Exercise 4.1

Question 2: Express the following linear equations in the form ax + by + c = 0

and indicate the values of a, b, c in each case:

(i) 2x + 3y = 9.35

Answer : (i) 2x + 3y = 9.35

2x + 3y – 9.35 = 0

Comparing this equation with ax + by + c = 0,

a = 2, b = 3, c = - 9.35

Exercise 4.2

Question 1: Which one of the following options is true, and why?

y = 3x + 5 has

(i) a unique solution, (ii) only two solutions, (iii) infinitely many solutions

Answer : y = 3x + 5 is a linear equation in two variables and it has infinite

possible solutions. As for every value of x, there will be a value of y satisfying the above equation and vice-versa.

Hence, the correct answer is (iii).

Exercise 4.2

Question 3: Check which of the following are solutions of the equation x − 2y = 4 and

which are not: (i) (0, 2 (ii) (2, 0)

Answer : (i) (0, 2) Putting x = 0 and y = 2 in the L.H.S of the given equation, x − 2y = 0 − 2×2 = − 4 ≠ 4 L.H.S ≠ R.H.S Therefore, (0, 2) is not a solution of this equation. (ii) (2, 0) Putting x = 2 and y = 0 in the L.H.S of the given equation, x − 2y = 2 − 2 × 0 = 2 ≠ 4 L.H.S ≠ R.H.S Therefore, (2, 0) is not a solution of this equation.

Exercise 4.2

Question 4: Find the value of k, if x = 2, y = 1 is a solution of the equation 2x +

3y = k.

Answer : Putting x = 2 and y = 1 in the given equation,

2x + 3y = k

⇒ 2(2) + 3(1) = k

⇒ 4 + 3 = k

 ⇒ k = 7

Therefore, the value of k is 7.

Exercise 4.3

Question 1: Draw the graph of each of the following linear equations in two variable: (i) y = 3x Answer : y = 3x It can be observed that  x = −1, y = −3 and x = 1, y = 3 are solutions of the above equation. Therefore, the solution table is as follows:

The graph of the above equation is constructed as follows.

x -1 1

y -3 3

Exercise 4.3

Question 2: Give the equations of two lines passing through (2, 14). How

many more such lines are there, and why?

Answer : It can be observed that point (2, 14) satisfies the equation

7x − y = 0 and x − y + 12 = 0. Therefore, 7x − y = 0 and x − y + 12 = 0 are two lines

passing through point (2, 14). As it is known that through one point, infinite number of lines

can pass through, therefore, there are infinite lines of such type passing through the given point.

Exercise 4.3

Question 3: If the point (3, 4) lies on the graph of the equation 3y = ax + 7,

find the value of a.

Answer :

Putting x = 3 and y = 4 in the given equation,

3y = ax + 7

3 (4) = a (3) + 7

5 = 3a

a = 5 / 3

Exercise 4.3

Question 4: The taxi fare in a city is as follows: For the first kilometre, the fares is Rs 8 and for the

subsequent distance it is Rs 5 per km. Taking the distance covered as x km and total fare as Rs y, write a linear equation for this information, and draw its graph.

Answer :Total distance covered = x km Fare for 1st kilometre = Rs 8 Fare for the rest of the distance = Rs (x − 1) 5 Total fare = Rs [8 + (x − 1) 5] y = 8 + 5x − 5 y = 5x + 3 5x − y + 3 = 0 It can be observed that point (0, 3) and (-3 / 5, 0) satisfies the above equation. Therefore, these are the solutions of this equation.

The graph of this equation is constructed as follows.

x 0 -3/5

y 3 0

Here, it can be seen that variable x and y are representing the distance covered and the fare paid for that distance respectively and these quantities may not be negative. Hence, only those values of x andy which are lying in the 1st quadrant will be considered.

Exercise 4.3

Question 7: Yamini and Fatima, two students of Class IX of a school, together

contributed Rs 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as Rs x and Rs y.) Draw the graph of the same.

Answer : Let the amount that Yamini and Fatima contributed

be x and y respectively towards the Prime Minister’s Relief fund. Amount contributed by Yamini + Amount contributed by Fatima = 100 x + y = 100 It can be observed that (100, 0) and (0, 100) satisfy the above equation. Therefore, these are the solutions of the above equation. The graph is constructed as follows.

Exercise 4.4

Question 2: Give the geometric representations of 2x + 9 = 0 as an equation (1) in one variable (2) in two variables Answer :

(1) In one variable, 2x + 9 = 0 represents a point  x = - 9/2 = - 4.5 as shown in the following figure. (2) In two variables, 2x + 9 = 0 represents a straight line passing through point 

(−4.5, 0) and parallel to  y-axis. It is a collection of all points of the plane, having their x-coordinate as 4.5.

Let`s Practice

Something Interesting

Let`s Try To Do …..

Q1. Determine the point on the graph of the linear equation x + y=6, whose ordinate is twice its abscissa.

Q2. How many solution(s) of the equation 3x+2=2x-3 are there on the i) Number Line ii) Cartesian plane

Q3. Draw the graph of the equation represented by the straight line which is parallel to the x-axis and 3 units above it.

Q4. Find the solutions of the linear equation x+2y=8, which represents a point on i) x axis ii) y-axis

Q5. For what values of c, the linear equation 2x+cy=8 has equal values of x and y as its solution.

Q6. Give the geometrical interpretations of 5x+3=3x-7 as an equation i) in one variable ii) In two variables

Q7. Draw the graph of the equation 3x+4y=6. At what points, the graph cut the x-axis and the y-axis.

Q8. At what point does the graph of equation 2x+3y=9 meet a line which is parallel to y -axis at a distance 4 units from the origin and on the right side of the y-axis.

Q9. Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case: (i) -2x + 3y = 6 (ii) x = 3y (iii) 2x = -5y 

Q10. Find the value of k if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Q11. If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a?

Q12. (i) Draw the graph of the linear equation using given Celsius for x-axis and Fahrenheit for y-axis. F =(9/5)C + 32(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?(iii) If the temperature is 95°F, what is the temperature in Celsius?(iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

Summary An equation in the form of ax + by + c = 0, where a, b and c are real

numbers and a and b are not both zeros is called a linear equation in two variables x and y. A linear equation in two variables has infinitely many solutions.

The solutions of a linear equation can be obtained by substituting different values for x in the equation to find the corresponding values of y.

The values of x and y are represented as an order pair. To plot the graph of this equation, its solutions is found algebraically and then plot the points on the graph.

Any linear equation of the form 'ax + by + c = 0' represents a straight line on the graph. Every point that satisfies the linear equation lies on the line. Every point that lies on the line is a solution of the linear equation. A point that does not lie on the line is not solution of the linear equation.

The equation of x-axis is y = 0. Any line parallel to x-axis is of the form y = k.

The equation of y-axis is x = 0. Any line  parallel to y-axis is of the form x = k.