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6-2 Solving Systems by Substitution
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Warm Up
California Standards
Lesson Presentation
6-2 Solving Systems by Substitution
Warm Up
Solve each equation for x.
1. y = x + 3
2. y = 3x – 4
x=y–3
Simplify each expression.
3. 2(x – 5) 2x – 10
4. 12 – 3(x + 1) 9 – 3x
Evaluate each expression for the given value
of x.
5. 2 x + 8 for x = 6 12 6. 3(x – 7) for x = 10 9
3
6-2 Solving Systems by Substitution
California
Standards
9.0 Students solve a system of two
linear equations in two variables algebraically
and are able to interpret the answer graphically.
Student are able to solve a system of two linear
inequalities in two variables and to sketch the
solution sets.
6-2 Solving Systems by Substitution
Sometimes it is difficult to identify the
exact solution to a system by graphing. In
this case, you can use a method called
substitution.
Substitution is used to reduce the system
to one equation that has only one variable.
Then you can solve this equation by the
methods taught in Chapter 2.
6-2 Solving Systems by Substitution
Solving Systems of Equations by Substitution
Step 1
Solve for one variable in at least one equation,
if necessary.
Step 2
Substitute the resulting expression into the
other equation.
Step 3
Solve that equation to get the value of the
first variable.
Step 4
Substitute that value into one of the original
equations and solve for the other variable.
Step 5
Write the values from steps 3 and 4 as an
ordered pair, (x, y), and check.
6-2 Solving Systems by Substitution
Additional Example 1A: Solving a System of Linear
Equations by Substitution
Solve the system by substitution.
y = 3x
y=x–2
Step 1 y = 3x
Both equations are solved for y.
y=x–2
Step 2 y = x – 2 Substitute 3x for y in the second
3x = x – 2
equation.
Step 3 –x
–x
Now solve this equation for x.
2x =
–2
Subtract x from both sides and
2x = –2
then divide by 2.
2
2
x = –1
6-2 Solving Systems by Substitution
Additional Example 1A Continued
Solve the system by substitution.
Step 4
Step 5
y = 3x
y = 3(–1)
y = –3
(–1, –3)
Write one of the original
equations.
Substitute –1 for x.
Write the solution as an
ordered pair.
Check Substitute (–1, –3) into both equations in the
system.
y = 3x
y=x–2
–3 3(–1)
–3 –1 – 2
–3
–3 
–3
–3 
6-2 Solving Systems by Substitution
Helpful Hint
You can substitute the value of one variable into
either of the original equations to find the value
of the other variable.
6-2 Solving Systems by Substitution
Additional Example 1B: Solving a System of Linear
Equations by Substitution
Solve the system by substitution.
y=x+1
4x + y = 6
The first equation is solved for y.
Step 1 y = x + 1
Write the second equation.
Step 2 4x + y = 6
Substitute x + 1 for y in the
4x + (x + 1) = 6 second equation.
5x + 1 = 6
Simplify. Solve for x.
Step 3
–1 –1
Subtract 1 from both sides.
5x
= 5
5x = 5
Divide both sides by 5.
5
5
x=1
6-2 Solving Systems by Substitution
Additional Example 1B Continued
Solve the system by substitution.
Step 4
Step 5
y=x+1
y=1+1
y=2
(1, 2)
Write one of the original
equations.
Substitute 1 for x.
Write the solution as an
ordered pair.
6-2 Solving Systems by Substitution
Additional Example 1C: Solving a System of Linear
Equations by Substitution
Solve the system by substitution.
x + 2y = –1
x–y=5
Step 1 x + 2y = –1
Solve the first equation for x by
subtracting 2y from both sides.
−2y −2y
x = –2y – 1
Step 2 x – y = 5
(–2y – 1) – y = 5
–3y – 1 = 5
Substitute –2y – 1 for x in the
second equation.
Simplify.
6-2 Solving Systems by Substitution
Additional Example 1C Continued
Step 3 –3y – 1 = 5
+1 +1
–3y = 6
–3y = 6
–3 –3
y = –2
Step 4 x – y
x – (–2)
x+2
–2
x
=5
=5
=5
–2
=3
Step 5 (3, –2)
Solve for y.
Add 1 to both sides.
Divide both sides by –3.
Write one of the original
equations.
Substitute –2 for y.
Subtract 2 from both sides.
Write the solution as an
ordered pair.
6-2 Solving Systems by Substitution
Check It Out! Example 1a
Solve the system by substitution. Check your
answer.
y= x+3
y = 2x + 5
Step 1 y = x + 3
y = 2x + 5
Both equations are solved for y.
Step 2 y = x + 3
2x + 5 = x + 3
Substitute 2x + 5 for y in the first
equation.
Step 3 2x + 5 = x + 3 Solve for x. Subtract x and 5
from both sides.
–x – 5 –x – 5
x = –2
6-2 Solving Systems by Substitution
Check It Out! Example 1a Continued
Solve the system by substitution. Check your
answer.
Write one of the original
Step 4 y = x + 3
equations.
y = –2 + 3
Substitute –2 for x.
y=1
Step 5 (–2, 1)
Write the solution as an
ordered pair.
Check Substitute (–2, 1) into both equations in the
system.
y=x+3
y = 2x + 5
1 2(–2) + 5
1 (–2) + 3
1 –4 + 5
1
1
1 1
6-2 Solving Systems by Substitution
Check It Out! Example 1b
Solve the system by substitution. Check your
answer.
x = 2y – 4
x + 8y = 16
The first equation is solved for x.
Step 1 x = 2y – 4
Step 2 x + 8y = 16
(2y – 4) + 8y = 16
Substitute 2y – 4 for x in the
second equation.
Step 3 10y – 4 = 16
+4 +4
10y = 20
10y
20
=
10
10
y=2
Simplify. Then solve for y.
Add 4 to both sides.
Divide both sides by 10.
6-2 Solving Systems by Substitution
Check It Out! Example 1b Continued
Solve the system by substitution. Check your
answer.
Step 4
x + 8y = 16
x + 8(2) = 16
x + 16 = 16
– 16 –16
x
= 0
Step 5 (0, 2)
Write one of the original
equations.
Substitute 2 for y.
Simplify.
Subtract 16 from both
sides.
Write the solution as
an ordered pair.
6-2 Solving Systems by Substitution
Check It Out! Example 1b Continued
Solve the system by substitution. Check your
answer.
Check Substitute (0, 2) into both equations in the
system.
x = 2y – 4
x + 8y = 16
0 + 8(2) 16
0 2(2) – 4
0
0
16 16 
6-2 Solving Systems by Substitution
Check It Out! Example 1c
Solve the system by substitution. Check your
answer.
2x + y = –4
x + y = –7
Step 1
x + y = –7
–y –y
x
= –y – 7
Step 2
x = –y – 7
2(–y – 7) + y = –4
2(–y – 7) + y = –4
–2y – 14 + y = –4
Solve the second equation for x
by subtracting y from each
side.
Substitute –y – 7 for x in the
first equation.
Distribute 2.
6-2 Solving Systems by Substitution
Check It Out! Example 1c Continued
Solve the system by substitution. Check your
answer.
Combine like terms.
Step 3 –2y – 14 + y = –4
–y – 14 = –4
+14 +14
Add 14 to each side.
–y
= 10
y = –10
x + y = –7
Step 4
x + (–10) = –7
x – 10 = – 7
Write one of the original
equations.
Substitute –10 for y.
6-2 Solving Systems by Substitution
Check It Out! Example 1c Continued
Solve the system by substitution. Check your
answer.
Step 5 x – 10 = –7
Add 10 to both sides.
+10 +10
Step 6
x=3
(3, –10)
Write the solution as an
ordered pair.
Check Substitute (3, –10) into both equations in the
system.
x + y = –7
2x + y = –4
3 + (–10)
–7
–7
–7 
2(3) + (–10) –4
–4 –4 
6-2 Solving Systems by Substitution
Sometimes you substitute an
expression for a variable that has a
coefficient. When solving for the
second variable in this situation, you
can use the Distributive Property.
6-2 Solving Systems by Substitution
Caution
When you solve one equation for a variable, you
must substitute the value or expression into the
other original equation, not the one that had just
been solved.
6-2 Solving Systems by Substitution
Additional Example 2: Using the Distributive Property
Solve
y + 6x = 11
by substitution.
3x + 2y = –5
Solve the first equation for y
Step 1 y + 6x = 11
by subtracting 6x from each
– 6x – 6x
side.
y = –6x + 11
Step 2
3x + 2y = –5 Substitute –6x + 11 for y in the
second equation.
3x + 2(–6x + 11) = –5
3x + 2(–6x + 11) = –5 Distribute 2 to the expression
in parentheses.
6-2 Solving Systems by Substitution
Additional Example 2 Continued
Solve
y + 6x = 11
by substitution.
3x + 2y = –5
Simplify. Solve for x.
Step 3 3x + 2(–6x) + 2(11) = –5
3x – 12x + 22 = –5
–9x + 22 = –5
– 22 –22 Subtract 22 from
–9x = –27
both sides.
–9x = –27 Divide both sides
by –9.
–9
–9
x=3
6-2 Solving Systems by Substitution
Additional Example 2 Continued
Solve
Step 4
y + 6x = 11
by substitution.
3x + 2y = –5
y + 6x = 11
y + 6(3) = 11
y + 18 = 11
–18 –18
Write one of the original
equations.
Substitute 3 for x.
Simplify.
Subtract 18 from each side.
y = –7
Step 5
(3, –7)
Write the solution as an
ordered pair.
6-2 Solving Systems by Substitution
Check It Out! Example 2
Solve
–2x + y = 8
3x + 2y = 9
by substitution. Check
your answer.
Step 1 –2x + y = 8
+ 2x
+2x
y = 2x + 8
Solve the first equation for y
by adding 2x to each side.
Step 2
3x + 2y = 9
3x + 2(2x + 8) = 9
Substitute 2x + 8 for y in the
second equation.
3x + 2(2x + 8) = 9
Distribute 2 to the expression
in parentheses.
6-2 Solving Systems by Substitution
Check It Out! Example 2 Continued
Solve
–2x + y = 8
3x + 2y = 9
by substitution. Check
your answer.
Step 3 3x + 2(2x) + 2(8) = 9
3x + 4x + 16 = 9
7x + 16 = 9
–16 –16
7x = –7
7x = –7
7
7
x = –1
Simplify. Solve for x.
Subtract 16 from
both sides.
Divide both sides
by 7.
6-2 Solving Systems by Substitution
Check It Out! Example 2 Continued
Solve
Step 4
–2x + y = 8
3x + 2y = 9
–2x + y = 8
–2(–1) + y = 8
y+2=8
–2 –2
y
Step 5
by substitution. Check
your answer.
Write one of the original
equations.
Substitute –1 for x.
Simplify.
Subtract 2 from each side.
=6
(–1, 6)
Write the solution as an
ordered pair.
6-2 Solving Systems by Substitution
Check It Out! Example 2 Continued
Solve
–2x + y = 8
3x + 2y = 9
by substitution. Check
your answer.
Check Substitute (–1, 6) into both equations in the
system.
3x + 2y = 9
–2x + y = 8
3(–1) + 2(6)
–3 + 12
9
9
9
9
–2(–1) + (6)
2 + (6)
8
8
8
8
6-2 Solving Systems by Substitution
Additional Example 3: Consumer Economics
Application
Jenna is deciding between two cell-phone
plans. The first plan has a $50 sign-up fee and
costs $20 per month. The second plan has a
$30 sign-up fee and costs $25 per month. After
how many months will the total costs be the
same? What will the costs be? If Jenna has to
sign a one-year contract, which plan will be
cheaper? Explain.
Write an equation for each option. Let t represent
the total amount paid and m represent the number
of months.
6-2 Solving Systems by Substitution
Additional Example 3 Continued
Total
paid
is
sign-up
fee
plus
monthly
fee
times
months.
Option 1
t
=
$50
+
$20

m
Option 2
t
=
$30
+
$25

m
Step 1 t = 50 + 20m
t = 30 + 25m
Both equations are solved
for t.
Substitute 50 + 20m for t in
Step 2 50 + 20m = 30 + 25m the second equation.
6-2 Solving Systems by Substitution
Additional Example 3 Continued
Step 3 50 + 20m = 30 + 25m
–20m
– 20m
50
= 30 + 5m
–30
–30
20
=
5m
20 = 5m
5
5
m=4
Step 4 t = 30 + 25m
t = 30 + 25(4)
t = 30 + 100
t = 130
Solve for m. Subtract 20m
from both sides.
Subtract 30 from both
sides.
Divide both sides by 5.
Write one of the original
equations.
Substitute 4 for m.
Simplify.
6-2 Solving Systems by Substitution
Additional Example 3 Continued
Step 5
(4, 130)
Write the solution as an
ordered pair.
In 4 months, the total cost for each option would be
the same–$130.
If Jenna has to sign a one-year contract,
which plan will be cheaper? Explain.
Option 1: t = 50 + 20(12) = 290
Option 2: t = 30 + 25(12) = 330
Jenna should choose the first plan because it costs
$290 for the year and the second plan costs $330.
6-2 Solving Systems by Substitution
Check It Out! Example 3
One cable television provider has a $60 setup
fee and $80 per month, and the second has a
$160 equipment fee and $70 per month.
a. In how many months will the cost be the
same? What will that cost be.
Write an equation for each option. Let t
represent the total amount paid and m
represent the number of months.
6-2 Solving Systems by Substitution
Check It Out! Example 3 Continued
Total
paid
is
fee
plus
monthly
fee
times
months.
Option 1
t
=
$60
+
$80

m
Option 2
t
= $160
+
$70

m
Step 1 t = 60 + 80m
t = 160 + 70m
Both equations are solved
for t.
Substitute 60 + 80m for t in
Step 2 60 + 80m = 160 + 70m the second equation.
6-2 Solving Systems by Substitution
Check It Out! Example 3 Continued
Step 3 60 + 80m = 160 + 70m Solve for m. Subtract
–70m
–70m 70m from both sides.
60 + 10m = 160
Subtract 60 from both
–60
–60
sides.
10m = 100
Divide both sides by 10.
10
10
m = 10
Write one of the original
Step 4 t = 160 + 70m
equations.
t = 160 + 70(10)
Substitute 10 for m.
t = 160 + 700
Simplify.
t = 860
6-2 Solving Systems by Substitution
Check It Out! Example 3 Continued
Step 5 (10, 860) Write the solution as an ordered pair.
In 10 months, the total cost for each option
would be the same–$860.
b. If you plan to move in 6 months, which is
the cheaper option? Explain.
Option 1: t = 60 + 80(6) = 540
Option 2: t = 160 + 270(6) = 580
The first option is cheaper for the first six months.
6-2 Solving Systems by Substitution
Lesson Quiz: Part I
Solve each system by substitution.
1.
2.
3.
y = 2x
(–2, –4)
x = 6y – 11
3x – 2y = –1
–3x + y = –1
x–y=4
(1, 2)
6-2 Solving Systems by Substitution
Lesson Quiz: Part II
4. Plumber A charges $60 an hour. Plumber B
charges $40 to visit your home plus $55 for
each hour. For how many hours will the total
cost for each plumber be the same? How much
will that cost be? If a customer thinks they will
need a plumber for 5 hours, which plumber
should the customer hire? Explain.
8 hours; $480; plumber A: plumber A is
cheaper for less than 8 hours.