laundry attendant
Resume:
*** ******* *. ******* ** LINEAR EQUATIONS
*.* 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. The substitution
method is fairly straightforward to use. First, you solve either equation for
either variable, then substitute the result into the other equation. The result
is an equation in a single variable. Solve that equation, then substitute the
result into any of the other equations to find the remaining unknown variable.
You Try It!
Solve the following system of EXAMPLE 1. Solve the following system of equations:
equations:
2x 5y = 8 (4.7)
9x + 2y = 19 y = 3x 1 (4.8)
y = 13 + 3x
Solution: Equation (4.8) is already solved for y . Substitute equation (4.8)
into equation (4.7). This means we will substitute 3x 1 for y in equation (4.7).
2x 5y = 8 Equation (4.7).
2x 5(3x 1) = 8 Substitute 3x 1 for y in (4.7).
Now solve for x.
2x 15x + 5 = 8 Distribute 5.
13x + 5 = 8 Simplify.
13x = 13 Subtract 5 from both sides.
y x=1 Divide both sides by 13.
5
2x 5y = 8
As we saw in Solving Systems by Graphing, the solution to the system is the
point of intersection of the two lines represented by the equations in the system.
This means that we can substitute the answer x = 1 into either equation to find
(1, 2)
the corresponding value of y . We choose to substitute 1 for x in equation (4.8),
x
5 5 then solve for y, but you will get exactly the same result if you substitute 1 for
x in equation (4.7).
5
y = 3x 1 Equation (4.8).
y = 3x 1
y = 3(1) 1 Substitute 1 for x.
Figure 4.27: 2x 5y = 8
y=2 Simplify.
and y = x 3 intersect at
(1, 2).
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 (4.7) and (4.8).
263
4.2. SOLVING SYSTEMS BY SUBSTITUTION
Substitute (x, y ) = (1, 2) in equa- Substitute (x, y ) = (1, 2) in equa-
tion (4.7): tion (4.8):
2x 5y = 8 y = 3x 1
2(1) 5(2) = 8 2 = 3(1) 1
2 10 = 8 2=3 1
8 = 8 2=2
Thus, (1, 2) satisfies equation (4.7). Thus, (1, 2) satisfies equation (4.8).
Because (x, y ) = (1, 2) satisfies both equations, it is a solution of the system. Answer: ( 3, 4)
Substitution method. The substitution method involves these steps:
1. Solve either equation for either variable.
2. Substitute the result from step one into the other equation. Solve the
resulting equation.
3. Substitute the result from step two into either of the original system equa-
tions or the resulting equation from step one (whichever seems easiest),
then solve to find the remaining unknown variable.
You Try It!
EXAMPLE 2. Solve the following system of equations: Solve the following system of
equations:
5x 2y = 12 (4.9)
x 2y = 13
4x + y = 6 (4.10)
4x 3y = 26
Solution: The first step is to solve either equation for either variable. This
means that we can solve the first equation for x or y, but it also means that we
could first solve the second equation for x or y . Of these four possible choices,
solving the second equation (4.10) for y seems the easiest way to start.
4x + y = 6 Equation (4.10).
y = 6 4x Subtract 4x from both sides.
264 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
Next, substitute 6 4x for y in equation (4.9).
5x 2y = 12 Equation (4.9).
5x 2(6 4x) = 12 Substitute 6 4x for y in (4.9).
5x 12 + 8x = 12 Distribute 2.
13x 12 = 12 Simplify.
13x = 24 Add 12 to both sides.
24
x= Divide both sides by 13.
13
Finally, to find the y -value, substitute 24/13 for x in the equation y = 6 4x
(you can also substitute 24/13 for x in equations (4.9) or (4.10)).
y = 6 4x
y 24
5x 2y = 12
y =6 4 Substitute 24/13 for x in y = 6 4x.
5 13
78 96
y= Multiply, then make equivalent fractions.
13 13
18
y= Simplify.
x
13
5 5
(24/13, 18/13)
Hence, (x, y ) = (24/13, 18/13) is the solution of the system.
Check: Let’s use the graphing calculator to check the solution. First, we store
5 24/13 in X with the following keystrokes (see the result in Figure 4.29).
4x + y = 6
Figure 4.28: 5x 2y = 12 X, T, θ, n
2 4 1 3 STO ENTER
and 4x + y = 6 intersect at
(24/13, 18/13).
Next, we store 18/13 in the variable Y with the following keystrokes (see the
result in Figure 4.29). 1 8 1 3 1
STO ALPHA ENTER
Now, clear the calculator screen by pressing the CLEAR button, then enter
the left-hand side of equation 4.9 with the following keystrokes (see the result
in Figure 4.30).
X, T, θ, n
5 2 1
ALPHA ENTER
Now enter the left-hand side of equation 4.10 with the following keystrokes
(see the result in Figure 4.30). Note that each left-hand side produces the
number on the right-hand sides of equations (4.9) and (4.10). Thus, the solution
(x, y ) = (24/13, 18/13) checks.
265
4.2. SOLVING SYSTEMS BY SUBSTITUTION
X, T, θ, n +
4 1
ALPHA ENTER
Figure 4.29: Storing 24/13 and Figure 4.30: Checking equa-
18/13 in X and Y. tions (4.9) and (4.10).
Answer: (13/5, 26/5)
You Try It!
EXAMPLE 3. Solve the following system of equations: Solve the following system of
equations:
3x 2y = 6 (4.11) 3x 5y = 3
4x + 5y = 20 (4.12) 5x 6y = 2
Solution: Dividing by 2 gives easier fractions to deal with than dividing by
3, 4, or 5, so let’s start by solving equation (4.11) for y .
3x 2y = 6 Equation (4.11).
2y = 6 3x Subtract 3x from both sides.
6 3x
y= Divide both sides by 2.
2
3
y = 3 + x Divide both 6 and 3x by 2
2 using distributive property.
266 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
Substitute 3 + 3 x for y in equation (4.12).
2
4x + 5y = 20 Equation (4.12).
3 3
4x + 5 3 + x = 20 Substitute 3 + x for y .
2 2
15
4x 15 + x = 20 Distribute the 5.
2
8x 30 + 15x = 40 Clear fractions by multiplying
y
4x + 5y = 20
both sides by 2.
5
23x = 70 Simplify. Add 30 to both sides.
70
x= Divide both sides by 23.
(70/23, 36/23)
23
x
5 5 To find y, substitute 70/23 for x into equation y = 3 + 3 x. You could also
2
substitute 70/23 for x in equations (4.11) or (4.12) and get the same result.
5 3
3x 2y = 6 y = 3 + x
2
Figure 4.31: 3x 2y = 6 3 70
y = 3 + Substitute 70/23 for x.
and 4x + 5y = 20 intersect at 2 23
(70/23, 36/23). 69 105
y= + Multiply. Make equivalent fractions.
23 23
36
y= Simplify.
23
Hence, (x, y ) = (70/23, 36/23) is the solution of the system.
Check: To check this solution, let’s use the graphing calculator to find the
solution of the system. We already know that 3x 2y = 6 is equivalent to
y = 3 + 3 x. Let’s also solve equation (4.12) for y .
2
4x + 5y = 20 Equation (4.12).
5y = 20 4x Subtract 4x from both sides.
20 4x
y= Divide both sides by 5.
5
4
y =4 x Divide both 20 and 4x by 5
5 using the distributive property.
Enter y = 3+ 3 x and y = 4 4 x into the Y= menu of the graphing calculator
2 5
(see Figure 4.32). Press the ZOOM button and select 6:ZStandard. Press
2ND CALC to open the CALCULATE menu, select 5:intersect, then press
the ENTER key three times in succession to enter “Yes” to the queries “First
curve,” “Second curve,” and “Guess.” The result is shown in Figure 4.33.
At the bottom of the viewing window in Figure 4.33, note how the coordinates
of the point of intersection are stored in the variables X and Y. Without
267
4.2. SOLVING SYSTEMS BY SUBSTITUTION
Figure 4.32: Enter y 3 + 3 x and Figure 4.33: Use 5:intersect on
2
y = 4 4 x in Y1 and Y2, respec- the CALC menu to calculate the
5
tively. point of intersection.
moving the cursor, (the variables X and Y contain the coordinates of the
cursor), quit the viewing window by pressing 2ND QUIT, which is located
above the MODE key. Then press the CLEAR button to clear the calculator
screen.
Now press the X,T,θ,n key, then the MATH button on your calculator:
X, T, θ, n MATH
This will open the MATH menu on your calculator (see Figure 4.34). Select
1: Frac, then press the ENTER key to produce the fractional equivalent of
the decimal content of the variable X (see Figure 4.35).
Figure 4.34: The MATH menu. Figure 4.35: Changing the contents
of the variables X and Y to frac-
tions.
Repeat the procedure for the variable Y. Enter:
1
ALPHA MATH
Select 1: Frac, then press the ENTER key to produce the fractional equivalent
of the decimal content of the variable Y (see Figure 4.35). Note that the
fractional equivalents for X and Y are 70/23 and 36/23, precisely the same
answers we got with the substitution method above. Answer: ( 8/7, 9/7)
268 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
Exceptional Cases Revisited
It is entirely possible that you might apply the substitution method to a system
of equations that either have an infinite number of solutions or no solutions at
all. Let’s see what happens should you do that.
You Try It!
Solve the following system of EXAMPLE 4. Solve the following system of equations:
equations:
2x + 3y = 6 (4.13)
4 2
x= y 7 y = x+4 (4.14)
3 3
6x 8y = 3
Solution: Equation (4.14) is already solved for y, so let’s substitute 2 x + 4
3
for y in equation (4.13).
2x + 3y = 6 Equation (4.13).
2 2
2x + 3 x + 4 =6 Substitute x + 4 for y .
3 3
2x 2x + 12 = 6 Distribute the 3.
12 = 6 Simplify.
Goodness! What happened to the x? How are we supposed to solve for x in
this situation? However, note that the resulting statement, 12 = 6, is false, no
y matter what we use for x and y . This should give us a clue that there are no
2
y = 3x + 4
solutions. Perhaps we are dealing with parallel lines?
5
Let’s solve equation (4.13) for y, putting the equation into slope-intercept
form, to help determine the situation.
2x + 3y = 6 Equation (4.13).
x
5 5 3y = 2x + 6 Subtract 2x from both sides.
2x + 3y = 6
2
y = x+2 Divide both sides by 3.
3
5
Thus, our system is equivalent to the following two equations.
Figure 4.36: 2x + 3y = 6 and
2
y = 3 x + 4 are parallel. No 2
y = x+2 (4.15)
solution. 3
2
y = x+4 (4.16)
3
These lines have the same slope 2/3, but different y -intercepts (one has y -
intercept (0, 2), the other has y -intercept (0, 4)). Hence, these are two distinct
Answer: no solution parallel lines and the system has no solution.
269
4.2. SOLVING SYSTEMS BY SUBSTITUTION
You Try It!
EXAMPLE 5. Solve the following system of equations: Solve the following system of
equations:
2x 6y = 8 (4.17)
x = 3y 4 (4.18) 28x + 14y = 126
y = 2x 9
Solution: Equation (4.18) is already solved for x, so let’s substitute 3y 4 for
x in equation (4.17).
2x 6y = 8 Equation (4.17).
y
2(3y 4) 6y = 8 Substitute 3y 4 for x.
5
6y 8 6y = 8 Distribute the 2.
x = 3y 4
8 = 8 Simplify.
( 1, 1)
Goodness! What happened to the x? How are we supposed to solve for x in (2, 2)
( 4, 0)
this situation? However, note that the resulting statement, 8 = 8, is a true x
5 5
statement this time. Perhaps this is an indication that we are dealing with the
2x 6y = 8
same line?
Let’s put both equations (4.17) and (4.18) into slope-intercept form so that
we can compare them. 5
Solve equation (4.17) for y : Solve equation (4.18) for y : Figure 4.37: 2x 6y = 8
and x = 3y 4 are the same
2x 6y = 8 x = 3y 4
line. Infinite number of solu-
6y = 2x 8 x + 4 = 3y tions.
2x 8 x+4
y= =y
6 3
1 4
1 4
y = x+
y = x+
3 3
3 3
Answer: There are an
Hence, the lines have the same slope and the same y -intercept and they are infinite number of solutions.
exactly the same lines. Thus, there are an infinite number of solutions. Indeed, Examples of solution points
any point on either line is a solution. Examples of solution points are ( 4, 0), are (0, 9), (5, 1), and
( 1, 1), and (2, 2). ( 3, 15).
Tip. When you substitute one equation into another and the variable disap-
pears, consider:
1. If the resulting statement is false, then you have two distinct parallel lines
and there is no solution.
2. If the resulting statement is true, then you have the same lines and there
are an infinite number of solutions.
270 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
Exercises
lll lll
In Exercises 1-8, use the substitution method to solve each of the following systems. Check your answer
manually, without the use of a calculator.
1. 5.
7x + 7y = 63 x = 5 2y
y = 6 2x 2x 6y = 18
2. 6.
3x 8y = 27 x = 15 + 6y
y = 4 7x 9x + 3y = 21
3. 7.
x = 19 + 7y 6x 8y = 24
3x 3y = 3 y = 15 + 3x
4. 8.
x = 39 + 8y 9x + 8y = 45
9x + 2y = 71 y = 15 8x
In Exercises 9-28, use the substitution method to solve each of the following systems.
9. 18.
x + 9y = 46 6x 6y = 102
7x 4y = 27 9x y = 63
10. 19.
x + 9y = 12 4x 8y = 4
4x + 7y = 38 y = 2y 4
11. 20.
x + 4y = 22 3x + 6y = 2
8x + 7y = 20 y = 2y + 2
12. 21.
x 2y = 15 2x 2y = 26
3x 9y = 15 7x + y = 19
13. 22.
x + 2y = 4 2x 8y = 30
6x 4y = 56 6x + y = 10
14. 23.
x + 8y = 79 3x 4y = 43
4x + 6y = 82 3x + y = 22
15. 24.
x + 6y = 49 2x + 8y = 14
3x + 4y = 7 8x + y = 43
16. 25.
x 4y = 33 8x 4y = 24
4x + 7y = 6 9x y = 71
17. 26.
2x + 8y = 50 8x 2y = 14
9x y = 3 6x y = 9
271
4.2. SOLVING SYSTEMS BY SUBSTITUTION
8x 7y = 2 9x + 4y = 3
27. 28.
8 9
y = x+9 y = x+6
7 4
In Exercises 29-36, use the substitution method to solve each of the following systems. Use your
graphing calculator to check your solution.
29. 33.
3x 5y = 3 3x + 8y = 6
5x 7y = 2 2x + 7y = 2
30. 34.
4x 5y = 4 3x 7y = 6
3x 2y = 1 2x 3y = 1
31. 35.
4x + 3y = 8 4x + 5y = 4
3x + 4y = 2 3x 2y = 1
32. 36.
3x + 8y = 3 5x + 4y = 5
4x 9y = 2 4x + 5y = 2
In Exercises 37-48, use the substitution method to determine how many solutions each of the following
linear systems has.
9x + 6y = 9
37. 43. y = 7y + 18
3 9x 63y = 162
y = x 8
2
44. y = 4y 9
3x 5y = 9
38.
10x + 40y = 90
3
y = x+6
45.
5 x = 2y + 3
39. y = 2x 16 4x + 8y = 4
14x 7y = 112
46. x = 2y + 4
40. y = 12x + 12
3x + 6y = 5
120x + 10y = 120
47. 9x + 4y = 73
41. x = 16 5y
y = 3 2x
4x + 2y = 24
42. 48.
x = 18 4y 6x + 9y = 27
7x 7y = 49 y = 16 5x
272 CHAPTER 4. SYSTEMS OF LINEAR EQUATIONS
Answers
lll lll
1. ( 1, 8) 25. ( 7, 8)
3. ( 2, 3) 27. No solution
5. (3, 4) 29. ( 11/4, 9/4)
7. ( 8, 9) 31. (26/7, 16/7)
9. ( 1, 5) 33. (58/5, 18/5)
11. ( 6, 4) 35. ( 13/7, 16/7)
13. ( 8, 2) 37. No solutions
15. ( 7, 7) 39. Infinite number of solutions
17. (1, 6) 41. One solution
19. 43. Infinite number of solutions
21. ( 4, 9) 45. No solutions
23. ( 5, 7) 47. One solution
Contact this candidate