
Lesson 5.3: Solve Systems using Substitution
Presentation
•
Mathematics
•
8th Grade
•
Practice Problem
•
Medium
+1
Standards-aligned
Reed Carbone
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8 Slides • 4 Questions
1
By Reed Carbone
2
Definition:
Variable: A variable is a placeholder for a number we do not know yet.
In this lesson, we are going to use variables to hold the place for entire expressions.
Subject | Subject
Some text here about the topic of discussion
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4
Step 1: Solve one of your equations for a certain variable (either x or y it doesn't matter)
Step 2: Substitute the expression you find into your other equation.
Step 3: Combine like terms until you can solve for your remaining variable.
Step 4: Plug the value you find for your variable into your original equation to solve for the final variable.
5
x = 8y - 4
x + 8y = 6
6
Multiple Choice
What is the solution to the system of equations?
y = 3x - 10
3x + 2y = 16
(2,4)
(4,2)
(3,8)
(2,11)
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Pedro has 276 more hits than Ricky. Use substitution to solve the system of equations to find how many hits Pedro, p, and Ricky, r, have recorded. The total combined hits is 2,666.
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Multiple Choice
Jake buys a fruit smoothie and a protein bar for $5.90. Kobe buys 2 fruit smoothies and 4 protein bars for $16.80. Write a system of equations that can be used to find the price of each item individually. (Use s for smoothies and p for protein bars.)
s = 5.90 + p
p = 16.80 + s
s + p = 5.90
s + p = 16.80
s + p = 5.90
2s + 4p = 16.80
2s + 4p = 5.90
s + p = 16.80
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6x - 2y = -6
11 = y - 5x
Step 1: Solve an equation for a variable.
Step 2: Substitute that variable into your other equation.
Step 3: Solve the equation for your variable.
Step 4: Substitute your solution into the original equation.
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Multiple Choice
What is the solution to the system of equations?
5x + y = -24
x + 3y = -2
(-5,-1)
(5,1)
(-5,1)
(5,-1)
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Multiple Choice
Solve the previous system of equations to determine how much a protein bar and a smoothie cost.
s + p = 5.90
2s + 4p = 16.80
s = 3.40
p = 2.50
s= 2.50
p = 3.40
s = 5.20
p = 2.20
s = 2.20
p = 5.20
12
By Reed Carbone
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