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6-2 Solving Systems by Substitution
Activating Prior Knowledge
Simplify each expression.
1. 2(x – 5)
2x – 10
2. 12 – 3(x + 1)
9 – 3x
What is the difference between
simplifying and solving?
Tie to LO
6-2 Solving Systems by Substitution
Learning Objective
Today, we will solve
systems of
equations by
substitution.
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6-2 Solving Systems by Substitution
Concept Development Review
– Notes #1
A system of linear equations is a set
of two or more linear equations
containing two or more variables.
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6-2 Solving Systems by Substitution
Concept Development Review
– Notes #2 & 3
2.A solution of a system of linear
equations with two variables is an
ordered pair that satisfies each
equation in the system.
3. So, if an ordered pair is a solution,
it will make both equations true.
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6-2 Solving Systems by Substitution
Concept Development Review
Notes #4 & 5
4. Sometimes it is difficult to identify
the exact solution to a system by
graphing.
5. In this case, you can use a method
called substitution.
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6-2 Solving Systems by Substitution
Concept Development
Solving Systems of Equations by Substitution
Step 1
Step 2
Step 3
Step 4
Step 5
Solve for one variable in at least one
equation, if necessary.
Substitute the resulting expression into
the other equation.
Solve that equation to get the value of
the first variable.
Substitute that value into one of the original
equations and solve for the other variable.
Write the values from steps 3 and 4 as
an ordered pair, (x, y), and check.
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6-2 Solving Systems by Substitution
Concept Development
Helpful Hint
You can substitute the value of one
variable into either of the original
equations to find the value of the other
variable.
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6-2 Solving Systems by Substitution
Skill Development – Notes #6
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
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6-2 Solving Systems by Substitution
Skill Development – Cont. Notes #6
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 
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6-2 Solving Systems by Substitution
Skill Development – Notes #7
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
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6-2 Solving Systems by Substitution
Skill Development – Cont. Notes #7
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.
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6-2 Solving Systems by Substitution
Skill Development – Notes #8
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.
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6-2 Solving Systems by Substitution
Skill Development – Cont. Notes #8
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.
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6-2 Solving Systems by Substitution
Concept Development – Notes #9 & 10
9. Sometimes you substitute an
expression for a variable that has a
coefficient.
10. When solving for the second
variable in this situation, you can use
the Distributive Property.
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6-2 Solving Systems by Substitution
Concept Development
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
Skill Development – Notes #11
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.
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6-2 Solving Systems by Substitution
Skill Development – Cont. Notes #11
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
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6-2 Solving Systems by Substitution
Skill Development – Cont. Notes #11
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.
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6-2 Solving Systems by Substitution
Skill Development – Notes #12
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.
CFU
6-2 Solving Systems by Substitution
Skill Development – Cont. Notes #12
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.
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6-2 Solving Systems by Substitution
Skill Development – Cont. Notes #12
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.
CFU
6-2 Solving Systems by Substitution
Skill Development – Cont. Notes #12
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
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6-2 Solving Systems by Substitution
Closure
1. What did we learn today?
2. Why is this important to you?
3. How can you tell whether an ordered pair is a
solution of a given system?
4. Solve the system by substitution.
x = 6y – 11
3x – 2y = –1
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