8 MODULE 3. EQUATIONS 3b Solving Systems of Two Equations Algebraically Solving Systems by Substitution In this section we introduce an algebraic technique for solving systems of two equations in two unknowns called the substitution method. Substitution method. The substitution method involves two consecutive steps: 1. Solve either equation for either variable. 2. Substitute the result from step one into the other equation. You Try It! EXAMPLE 1. Solve the following system of equations: 2x - 5y = -8 (3.5) y =3x - 1 (3.6) Solution: Equation (3.30) is already solved for y. Substitute equation (3.30) into equation (3.29). 2x - 5y = -8 Equation (3.29). 2x - 5(3x - 1) = -8 Substitute (3.30) in (3.29). Now solve for x. 2x - 15x +5= -8 Distribute -5. -13x +5= -8 Simplify. -13x = -13 Subtract 5 from both sides. x =1 Divide both sides by -13. Substitute x = 1 in equation (3.30), then solve for y. y =3x - 1 Equation (3.30). y = 3(1) - 1 Substitute 1 for x. y =2 Simplify. Hence, (x, y) = (1, 2) is the solution of the system. Check: To show that the solution (x, y) = (1, 2) is a solution of the system, we need to show that (x, y) = (1, 2) satisfies both equations (3.29) and (3.30).
Transcript
8 MODULE 3. EQUATIONS
3b Solving Systems of Two Equations Algebraically
Solving Systems by Substitution
In this section we introduce an algebraic technique for solving systems of twoequations in two unknowns called the substitution method.
Substitution method. The substitution method involves two consecutivesteps:
1. Solve either equation for either variable.
2. Substitute the result from step one into the other equation.
You Try It!
EXAMPLE 1. Solve the following system of equations:
2x! 5y = !8 (3.5)
y = 3x! 1 (3.6)
Solution: Equation (3.30) is already solved for y. Substitute equation (3.30)into equation (3.29).
2x! 5y = !8 Equation (3.29).
2x! 5(3x! 1) = !8 Substitute (3.30) in (3.29).
Now solve for x.
2x! 15x+ 5 = !8 Distribute !5.
!13x+ 5 = !8 Simplify.
!13x = !13 Subtract 5 from both sides.
x = 1 Divide both sides by !13.
Substitute x = 1 in equation (3.30), then solve for y.
y = 3x! 1 Equation (3.30).
y = 3(1)! 1 Substitute 1 for x.
y = 2 Simplify.
Hence, (x, y) = (1, 2) is the solution of the system.
Check: To show that the solution (x, y) = (1, 2) is a solution of the system,we need to show that (x, y) = (1, 2) satisfies both equations (3.29) and (3.30).
3B. SOLVING SYSTEMS OF TWO EQUATIONS ALGEBRAICALLY 9
Substitute (x, y) = (1, 2) in equa-tion (3.29):
2x! 5y = !8
2(1)! 5(2) = !8
2! 10 = !8
!8 = !8
Thus, (1, 2) satisfies equa-tion (3.29).
Substitute (x, y) = (1, 2) in equa-tion (3.30):
y = 3x! 1
2 = 3(1)! 1
2 = 3! 1
2 = 2
Thus, (1, 2) satisfies equa-tion (3.30).
!
You Try It!
EXAMPLE 2. Solve the following system of equations:
5x! 2y = 12 (3.7)
4x+ y = 6 (3.8)
Solution: The first step is to solve either equation for either variable. Thismeans that we can solve the first equation for x or y, but it also means that wecould first solve the second equation for x or y. Of these four possible choices,solving equation (3.32) for y seems the easiest way to start.
4x+ y = 6 Equation (3.32).
y = 6! 4x Subtract 4x from both sides.
Next, substitute y = 6! 4x for y in equation (3.31).
5x! 2y = 12 Equation (3.31).
5x! 2(6! 4x) = 12 Substitute y = 6! 4x for y.
5x! 12 + 8x = 12 Distribute !2.
13x! 12 = 12 Simplify.
13x = 24 Add 12 to both sides.
x =24
13Divide both sides by 13.
Finally, to find the y-value, substitute x = 24/13 into the equation y =
10 MODULE 3. EQUATIONS
6! 4x (you can also substitute x = 24/13 into equation (3.31) or (3.32)).
y = 6! 4x
y = 6! 4
!
24
13
"
Substitute x = 24/13 in y = 6! 4x.
y =78
13!
96
13Multiply, then make equivalent fractions.
y = !
18
13Simplify.
Hence, (x, y) = (24/13,!18/13) is the solution of the system.
!
You Try It!
EXAMPLE 3. Solve the following system of equations:
3x! 2y = 6 (3.9)
4x+ 5y = 20 (3.10)
Solution: Dividing by 2 gives easier fractions to deal with than dividing by3, 4, or 5, so let’s start by solving equation (3.33) for y.
3x! 2y = 6 Equation (3.33).
!2y = 6! 3x Subtract 3x from both sides.
y =6! 3x
!2Divide both sides by !2.
y = !3 +3
2x Divide both terms by !2
using distributive property.
Substitute y = !3 + 3
2x for y in equation (3.34).
4x+ 5y = 20 Equation (3.34).
4x+ 5
!
!3 +3
2x
"
= 20 Substitute !3 +3
2x for y.
4x! 15 +15
2x = 20 Distribute the 5.
8x! 30 + 15x = 40 Clear fractions by multiplyingboth sides by 2.
23x = 70 Simplify. Add 30 to both sides.
x =70
23Divide both sides by 23.
3B. SOLVING SYSTEMS OF TWO EQUATIONS ALGEBRAICALLY 11
To find y, substitute x = 70/23 for x into equation y = !3 + 3
2x.
y = !3 +3
2x
y = !3 +3
2
!
70
23
"
Substitute 70/23 for x.
y = !
69
23+
105
23Multiply. Make equivalent fractions.
y =36
23Simplify.
Hence, (x, y) = (70/23, 36/23) is the solution of the system.
!
Exceptional Cases Revisited
It is entirely possible that you might apply the substitution method to a systemof equations that either have an infinite number of solutions or no solutions atall. Let’s see what happens should you do that.
You Try It!
EXAMPLE 4. Solve the following system of equations:
2x+ 3y = 6 (3.11)
y = !
2
3x+ 4 (3.12)
Solution: Equation (3.36) is already solved for y, so let’s substitute y =!
2
3x+ 4 for y in equation 3.35.
2x+ 3y = 6 Equation (3.35).
2x+ 3
!
!
2
3x+ 4
"
= 6 Substitute !
2
3x+ 4 for y.
2x! 2x+ 12 = 6 Distribute the 3.
12 = 6 Simplify.
Goodness! What happened to the x? How are we supposed to solve for x inthis situation?
However, the resulting statement, 12 = 6, is false, no matter what we usefor x and y. This should give us a clue that there are no solutions. Perhaps weare dealing with parallel lines?
Let’s solve equation (3.35) to determine the situation.
2x+ 3y = 6 Equation (3.35).
3y = !2x+ 6 Subtract 2x from both sides.
y = !
2
3x+ 2 Divide both sides by 3.
12 MODULE 3. EQUATIONS
Thus, our system is equivalent to the following two equations.
y = !
2
3x+ 2 (3.13)
y = !
2
3x+ 4 (3.14)
These lines have the same slope !2/3, but di!erent y-intercepts (one has y-intercept (0, 2), the other has y-intercept (0, 4)). Hence, these are two distinctparallel lines and the system has no solution.
!
You Try It!
EXAMPLE 5. Solve the following system of equations:
2x! 6y = !8 (3.15)
x = 3y ! 4 (3.16)
Solution: Equation (3.40) is already solved for x, so let’s substitute x = 3y!4for x in equation (3.39).
2x! 6y = !8 Equation (3.39).
2(3y ! 4)! 6y = !8 Substitute 3y ! 4 for x.
6y ! 8! 6y = !8 Distribute the 2.
!8 = !8 Simplify.
Goodness! What happened to the x? How are we supposed to solve for x inthis situation? However, note that the resulting statement, !8 = !8, is a truestatement this time. Perhaps this is an indication that we are dealing with thesame line?
Let’s put both equations (3.39) and (3.40) into slope-intercept form so thatwe can compare them.
Solve equation (3.39) for y:
2x! 6y = !8
!6y = !2x! 8
y =!2x! 8
!6
y =1
3x+
4
3
Solve equation (3.40) for y:
x = 3y ! 4
x+ 4 = 3y
x+ 4
3= y
y =1
3x+
4
3
Hence, the lines have the same slope and the same y-intercept and they areexactly the same lines. Thus, there are an infinite number of solutions. Indeed,
3B. SOLVING SYSTEMS OF TWO EQUATIONS ALGEBRAICALLY 13
any point on either line is a solution. Examples of solution points are (0, 4/3),(!4, 0), and (!1, 1).
!
Tip. When you substitute one equation into another and the variable disap-pears, consider:
1. If the resulting statement is false, suspect that the lines are parallel linesand there is no solution.
2. If the resulting statement is true, suspect that you have the same linesand there are an infinite number of solutions.
Solving Systems by Elimination
When both equations of a system are in standard form Ax + By = C, then aprocess called elimination is the best procedure to use to find the solution ofthe system. Elimination is based on two simple ideas, the first of which shouldbe familiar.
1. Multiplying both sides of an equation by a non-zero number does notchange its solutions. Thus, the equation
x+ 3y = 7 (3.17)
will have the same solutions (it’s the same line) as the equation obtainedby multiplying equation (3.17) by 2.
2x+ 6y = 14 (3.18)
2. Adding two true equations produces another true equation. For example,consider what happens when you add 4 = 4 to 5 = 5.
4 = 45 = 59 = 9
Even more importantly, consider what happens when you add two equa-tions that have (2, 1) as a solution. The result is a third equation whosegraph also passes through the solution.
14 MODULE 3. EQUATIONS
x + y = 3x ! y = 1
2x = 4x = 2
!5 5
!5
5
x
y
x = 2x! y = 1
x+ y = 3
Elimination. Adding a multiple of an equation to a second equation producesand equation that passes through the same solution as the first two equations.
Let’s use these ideas to solve a system of equations.
3B. SOLVING SYSTEMS OF TWO EQUATIONS ALGEBRAICALLY 15
You Try It!
EXAMPLE 6. Solve the following system of equations.
x+ 2y = !5 (3.19)
2x! y = !5 (3.20)
Solution: Our focus will be on eliminating the variable x. We note that if wemultiply equation (3.19) by !2, then add the result to equation (3.20), the xterms will be eliminated.
!2x ! 4y = 10 Multiply equation (3.19) by !2.2x ! y = !5 Equation (3.20).
! 5y = 5 Add the equations.
Thus, y = !1.To find the corresponding value of x, substitute y = !1 in equation (3.19)
(or equation (3.20)) and solve for x.
x+ 2y = !5 Equation (3.19)
x+ 2(!1) = !5 Substitute y = !1.
x = !3 Solve for x.
Check: To check, we need to show that the point (x, y) = (!3, 1) satisfiesboth equations.
Substitute (x, y) = (!3,!1) intoequation (3.19).
x+ 2y = !5
!3 + 2(!1) = !5
!5 = !5
Substitute (x, y) = (!3,!1) intoequation (3.20).
2x! y = !5
2(!3)! (!1) = !5
!5 = !5
Thus, the point (x, y) = (!3,!1) satisfies both equations and is therefore thesolution of the system.
!
To show that you have the option of which variable you choose to eliminate,let’s try Example 6 a second time, this time eliminating y instead of x.
You Try It!
EXAMPLE 7. Solve the following system of equations.
x+ 2y = !5 (3.21)
2x! y = !5 (3.22)
16 MODULE 3. EQUATIONS
Solution: This time we focus on eliminating the variable y. We note that ifwe multiply equation (3.22) by 2, then add the result to equation (3.21), the yterms will be eliminated.
x + 2y = !5 Equation (3.21).4x ! 2y = !10 Multiply equation (3.22) by 2.
5x = !15 Add the equations.
Thus, x = !3.To find the corresponding value of y, substitute x = !3 in equation (3.21)
(or equation (3.22)) and solve for y.
x+ 2y = !5 Equation (3.21)
!3 + 2y = !5 Substitute x = !3.
2y = !2 Add 3 to both sides.
y = !1 Divide both sides by 2.
Hence, (x, y) = (!3,!1), just as in Example 6, is the solution of the system.
!
Sometimes the elimination method requires a process similar to that offinding a common denominator.
You Try It!
EXAMPLE 8. Solve the following system of equations.
3x+ 4y = 12 (3.23)
2x! 5y = 10 (3.24)
Solution: Let’s focus on eliminating the x-terms. Note that if we multiplyequation (3.23) by 2, then multiply equation (3.24) by !3, the x-terms willbe eliminated when we add the resulting equations. Note that 6 is the leastcommon multiple of 3 and 2.
6x + 8y = 24 Multiply equation (3.23) by 2.!6x + 15y = !30 Multiply equation (3.24) by !3.
23y = !6 Add the equations.
Hence, y = !6/23.At this point, we could substitute y = !6/23 in either equation, then solve
the result for x. However, working with y = !6/23 is a bit daunting, partic-ularly in the light of elimination being easier. So let’s use elimination again,this time focusing on eliminating y. Similarly, if we multiply equation (3.23)by 5, then multiply equation (3.24) by 4, when we add the results, the y-termswill be eliminated.
3B. SOLVING SYSTEMS OF TWO EQUATIONS ALGEBRAICALLY 17
15x + 20y = 60 Multiply equation (3.23) by 5.8x ! 20y = 40 Multiply equation (3.24) by 4.
23x = 100 Add the equations.
Thus, x = 100/23, and the system of the system is (x, y) = (100/23,!6/23).
!
Exceptional Cases
In the previous section, we saw that if the substitution method led to a falsestatement, we should suspect that we have parallel lines. The same thing canhappen with the elimination method of this section.
You Try It!
EXAMPLE 9. Solve the following system of equations.
x+ y = 3 (3.25)
2x+ 2y = !6 (3.26)
Solution: Let’s focus on eliminating the x-terms. Note that if we multiplyequation (3.25) by!2, the x-terms will be eliminated when we add the resultingequations.
Because of our experience with this solving this exceptional case with substi-tution, the fact that both variables have disappeared should not be completelysurprising. Note that this last statement is false, regardless of the values of xand y. Hence, the system has no solution.
Indeed, if we find the intercepts of each equation and plot them, then wecan easily see that the lines of this system are parallel (see Figure 3.8). Parallellines never intersect, so the system has no solutions.
!
You Try It!
EXAMPLE 10. Solve the following system of equations.
x! 7y = 4 (3.27)
!3x+ 21y = !12 (3.28)
18 MODULE 3. EQUATIONS
!5 5
!5
5
x
y
(3, 0)
(0, 3)
(!3, 0)
(0,!3)
x+ y = 3
2x+ 2y = !6
Figure 3.8: The lines x+ y = 3 and 2x+ 2y = !6 are parallel.
Solution: We might recognize that the equations 3.27 and (3.28) are identi-cal. But it’s also conceivable that we don’t see that right away and begin theelimination method. Let’s multiply the first equation by 3, then add. This willeliminate the x-terms.
The equations are identical! Hence, there are an infinite number of points ofintersection. Indeed, any point on either line is a solution. Example points ofsolution are (4, 0), (0,!4/7), and (18, 2).
!
3B. SOLVING SYSTEMS OF TWO EQUATIONS ALGEBRAICALLY 19
Solving Systems by Substitution
In this section we introduce an algebraic technique for solving systems of twoequations in two unknowns called the substitution method.
Substitution method. The substitution method involves two consecutivesteps:
1. Solve either equation for either variable.
2. Substitute the result from step one into the other equation.
You Try It!
EXAMPLE 11. Solve the following system of equations:
2x! 5y = !8 (3.29)
y = 3x! 1 (3.30)
Solution: Equation (3.30) is already solved for y. Substitute equation (3.30)into equation (3.29).
2x! 5y = !8 Equation (3.29).
2x! 5(3x! 1) = !8 Substitute (3.30) in (3.29).
Now solve for x.
2x! 15x+ 5 = !8 Distribute !5.
!13x+ 5 = !8 Simplify.
!13x = !13 Subtract 5 from both sides.
x = 1 Divide both sides by !13.
Substitute x = 1 in equation (3.30), then solve for y.
y = 3x! 1 Equation (3.30).
y = 3(1)! 1 Substitute 1 for x.
y = 2 Simplify.
Hence, (x, y) = (1, 2) is the solution of the system.
Check: To show that the solution (x, y) = (1, 2) is a solution of the system,we need to show that (x, y) = (1, 2) satisfies both equations (3.29) and (3.30).
20 MODULE 3. EQUATIONS
Substitute (x, y) = (1, 2) in equa-tion (3.29):
2x! 5y = !8
2(1)! 5(2) = !8
2! 10 = !8
!8 = !8
Thus, (1, 2) satisfies equa-tion (3.29).
Substitute (x, y) = (1, 2) in equa-tion (3.30):
y = 3x! 1
2 = 3(1)! 1
2 = 3! 1
2 = 2
Thus, (1, 2) satisfies equa-tion (3.30).
!
You Try It!
EXAMPLE 12. Solve the following system of equations:
5x! 2y = 12 (3.31)
4x+ y = 6 (3.32)
Solution: The first step is to solve either equation for either variable. Thismeans that we can solve the first equation for x or y, but it also means that wecould first solve the second equation for x or y. Of these four possible choices,solving equation (3.32) for y seems the easiest way to start.
4x+ y = 6 Equation (3.32).
y = 6! 4x Subtract 4x from both sides.
Next, substitute y = 6! 4x for y in equation (3.31).
5x! 2y = 12 Equation (3.31).
5x! 2(6! 4x) = 12 Substitute y = 6! 4x for y.
5x! 12 + 8x = 12 Distribute !2.
13x! 12 = 12 Simplify.
13x = 24 Add 12 to both sides.
x =24
13Divide both sides by 13.
Finally, to find the y-value, substitute x = 24/13 into the equation y = 6! 4x
3B. SOLVING SYSTEMS OF TWO EQUATIONS ALGEBRAICALLY 21
(you can also substitute x = 24/13 into equation (3.31) or (3.32)).
y = 6! 4x
y = 6! 4
!
24
13
"
Substitute x = 24/13 in y = 6! 4x.
y =78
13!
96
13Multiply, then make equivalent fractions.
y = !
18
13Simplify.
Hence, (x, y) = (24/13,!18/13) is the solution of the system.
!
You Try It!
EXAMPLE 13. Solve the following system of equations:
3x! 2y = 6 (3.33)
4x+ 5y = 20 (3.34)
Solution: Dividing by 2 gives easier fractions to deal with than dividing by3, 4, or 5, so let’s start by solving equation (3.33) for y.
3x! 2y = 6 Equation (3.33).
!2y = 6! 3x Subtract 3x from both sides.
y =6! 3x
!2Divide both sides by !2.
y = !3 +3
2x Divide both terms by !2
using distributive property.
Substitute y = !3 + 3
2x for y in equation (3.34).
4x+ 5y = 20 Equation (3.34).
4x+ 5
!
!3 +3
2x
"
= 20 Substitute !3 +3
2x for y.
4x! 15 +15
2x = 20 Distribute the 5.
8x! 30 + 15x = 40 Clear fractions by multiplyingboth sides by 2.
23x = 70 Simplify. Add 30 to both sides.
x =70
23Divide both sides by 23.
22 MODULE 3. EQUATIONS
To find y, substitute x = 70/23 for x into equation y = !3 + 3
2x.
y = !3 +3
2x
y = !3 +3
2
!
70
23
"
Substitute 70/23 for x.
y = !
69
23+
105
23Multiply. Make equivalent fractions.
y =36
23Simplify.
Hence, (x, y) = (70/23, 36/23) is the solution of the system.
!
Exceptional Cases Revisited
It is entirely possible that you might apply the substitution method to a systemof equations that either have an infinite number of solutions or no solutions atall. Let’s see what happens should you do that.
You Try It!
EXAMPLE 14. Solve the following system of equations:
2x+ 3y = 6 (3.35)
y = !
2
3x+ 4 (3.36)
Solution: Equation (3.36) is already solved for y, so let’s substitute y =!
2
3x+ 4 for y in equation 3.35.
2x+ 3y = 6 Equation (3.35).
2x+ 3
!
!
2
3x+ 4
"
= 6 Substitute !
2
3x+ 4 for y.
2x! 2x+ 12 = 6 Distribute the 3.
12 = 6 Simplify.
Goodness! What happened to the x? How are we supposed to solve for x inthis situation?
However, the resulting statement, 12 = 6, is false, no matter what we usefor x and y. This should give us a clue that there are no solutions. Perhaps weare dealing with parallel lines?
Let’s solve equation (3.35) to determine the situation.
2x+ 3y = 6 Equation (3.35).
3y = !2x+ 6 Subtract 2x from both sides.
y = !
2
3x+ 2 Divide both sides by 3.
3B. SOLVING SYSTEMS OF TWO EQUATIONS ALGEBRAICALLY 23
Thus, our system is equivalent to the following two equations.
y = !
2
3x+ 2 (3.37)
y = !
2
3x+ 4 (3.38)
These lines have the same slope !2/3, but di!erent y-intercepts (one has y-intercept (0, 2), the other has y-intercept (0, 4)). Hence, these are two distinctparallel lines and the system has no solution.
!
You Try It!
EXAMPLE 15. Solve the following system of equations:
2x! 6y = !8 (3.39)
x = 3y ! 4 (3.40)
Solution: Equation (3.40) is already solved for x, so let’s substitute x = 3y!4for x in equation (3.39).
2x! 6y = !8 Equation (3.39).
2(3y ! 4)! 6y = !8 Substitute 3y ! 4 for x.
6y ! 8! 6y = !8 Distribute the 2.
!8 = !8 Simplify.
Goodness! What happened to the x? How are we supposed to solve for x inthis situation? However, note that the resulting statement, !8 = !8, is a truestatement this time. Perhaps this is an indication that we are dealing with thesame line?
Let’s put both equations (3.39) and (3.40) into slope-intercept form so thatwe can compare them.
Solve equation (3.39) for y:
2x! 6y = !8
!6y = !2x! 8
y =!2x! 8
!6
y =1
3x+
4
3
Solve equation (3.40) for y:
x = 3y ! 4
x+ 4 = 3y
x+ 4
3= y
y =1
3x+
4
3
Hence, the lines have the same slope and the same y-intercept and they areexactly the same lines. Thus, there are an infinite number of solutions. Indeed,
24 MODULE 3. EQUATIONS
any point on either line is a solution. Examples of solution points are (0, 4/3),(!4, 0), and (!1, 1).
Tip. When you substitute one equation into another and the variable disap-pears, consider:
1. If the resulting statement is false, suspect that the lines are parallel linesand there is no solution.
2. If the resulting statement is true, suspect that you have the same linesand there are an infinite number of solutions.
