1. What is the difference between simplifying and solving?
Activating Prior Knowledge -
Simplify each expression.
9 – 3x
1. 2(x – 5)
2x – 10
2. 12 – 3(x + 1)
What is the difference between simplifying and solving?
Tie to LO

2. Today, we will solve systems of equations by substitution.
Learning Objective
Today, we will solve systems of equations by substitution.
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3. 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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4. 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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5. 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. Solving Systems of Equations by Substitution
Concept Development
Solving Systems of Equations by Substitution
Step 2
Step 3
Step 4
Step 5
Solve for one variable in at least one equation, if necessary.
Step 1
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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7. Concept Development CFU
You can substitute the value of one variable into either of the original equations to find the value of the other variable.
Helpful Hint
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8. 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
3x = x – 2
Substitute 3x for y in the second equation.
Step 3
–x –x
2x = –2
2x = –2
x = –1
Now solve this equation for x. Subtract x from both sides and then divide by 2.
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9. Skill Development – Cont. Notes #6
Solve the system by substitution.
Write one of the original equations.
Step 4
y = 3x
y = 3(–1)
y = –3
Substitute –1 for x.
Write the solution as an ordered pair.
Step 5
(–1, –3)
Check Substitute (–1, –3) into both equations in the system.
y = 3x
–3 3(–1)
–3 –3 
y = x – 2
–3 –1 – 2
–3 –3 
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10. 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 second equation.
4x + (x + 1) = 6
5x + 1 = 6
Simplify. Solve for x.
Step 3
–1 –1
5x = 5
Subtract 1 from both sides.
5x = 5
Divide both sides by 5.
x = 1
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11. Skill Development – Cont. Notes #7
Solve the system by substitution.
Write one of the original equations.
Step 4
y = x + 1
y = 1 + 1
y = 2
Substitute 1 for x.
Step 5
(1, 2)
Write the solution as an ordered pair.
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12. 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
Substitute –2y – 1 for x in the second equation.
–3y – 1 = 5
Simplify.
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13. Skill Development – Cont. Notes #8
Step 3
–3y – 1 = 5
Solve for y.
+1 +1
–3y = 6
Add 1 to both sides.
–3y = 6
–3 –3
y = –2
Divide both sides by –3.
Step 4
x – y = 5
Write one of the original equations.
x – (–2) = 5
x + 2 = 5
Substitute –2 for y.
–2 –2
x = 3
Subtract 2 from both sides.
Write the solution as an ordered pair.
Step 5 (3, –2)
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14. 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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15. Concept Development
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.
Caution

16. Skill Development – Notes #11
y + 6x = 11
Solve by substitution.
3x + 2y = –5
Solve the first equation for y by subtracting 6x from each side.
Step 1 y + 6x = 11
– 6x – 6x
y = –6x + 11
3x + 2(–6x + 11) = –5
3x + 2y = –5
Step 2
Substitute –6x + 11 for y in the second equation.
3x + 2(–6x + 11) = –5
Distribute 2 to the expression in parentheses.
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17. Skill Development – Cont. Notes #11
y + 6x = 11
Solve by substitution.
3x + 2y = –5
Step 3
3x + 2(–6x) + 2(11) = –5
Simplify. Solve for x.
3x – 12x + 22 = –5
–9x + 22 = –5
–9x = –27
– 22 –22
Subtract 22 from both sides.
–9x = –27
– –9
Divide both sides by –9.
x = 3
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18. Skill Development – Cont. Notes #11
y + 6x = 11
Solve by substitution.
3x + 2y = –5
Write one of the original equations.
Step 4
y + 6x = 11
y + 6(3) = 11
Substitute 3 for x.
y + 18 = 11
Simplify.
–18 –18
y = –7
Subtract 18 from each side.
Step 5
(3, –7)
Write the solution as an ordered pair.
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19. Skill Development – Notes #12
–2x + y = 8
Solve by substitution. Check your answer.
3x + 2y = 9
Step 1 –2x + y = 8
Solve the first equation for y by adding 2x to each side.
+ 2x x
y = 2x + 8
3x + 2(2x + 8) = 9
3x + 2y = 9
Step 2
Substitute 2x + 8 for y in the second equation.
3x + 2(2x + 8) = 9
Distribute 2 to the expression in parentheses.
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20. Skill Development – Cont. Notes #12
–2x + y = 8
Solve by substitution. Check your answer.
3x + 2y = 9
Step 3
3x + 2(2x) + 2(8) = 9
Simplify. Solve for x.
3x + 4x = 9
7x = 9
7x = –7
–16 –16
Subtract 16 from both sides.
7x = –7
Divide both sides by 7.
x = –1
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21. Skill Development – Cont. Notes #12
–2x + y = 8
Solve by substitution. Check your answer.
3x + 2y = 9
Write one of the original equations.
Step 4
–2x + y = 8
–2(–1) + y = 8
Substitute –1 for x.
y + 2 = 8
Simplify.
–2 –2
y = 6
Subtract 2 from each side.
Step 5
(–1, 6)
Write the solution as an ordered pair.
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22. Skill Development – Cont. Notes #12
–2x + y = 8
Solve by substitution. Check your answer.
3x + 2y = 9
Check Substitute (–1, 6) into both equations in the system.
3x + 2y = 9
3(–1) + 2(6) 9  –
–2x + y = 8
–2(–1) + (6) 8
8 8 
2 + (6) 8
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23. Closure CFU 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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