Envision Math 2.0: Grade 8, Volume 2
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Envision Math 2.0: Grade 8, Volume 2 View details
3. Solve Systems by Substitution
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Exercise 13 Page 276

Practice makes perfect
Consider the given system of linear equations. x=8y-4 & (I) x+8y=6 & (II) We will solve this system by using the Substitution Method. Since the variable x is isolated in the first equation, we can substitute its equivalent expression into the second equation.

x=8y-4 & (I) x+8y=6 & (II)
x=8y-4 8y-4+8y=6
x=8y-4 16y-4=6
x=8y-4 16y=10
x=8y-4 y=10/16
x=8y-4 y=5/8

We will now substitute y= 58 into one of the equations to find the value of x. Let's use the first equation.

x=8y-4 y=5/8
x=8( 5/8)-4 y=5/8
x=5-4 y=5/8
x=1 y=5/8

The solution to the given system of linear equations is the point (1, 58).

Note that the variable x has already been isolated in the first equation. Therefore, it is easier to substitute the expression equivalent to x in the first equation into the second equation in order to solve the system. x= 8y-4 & (I) x+8y=6 & (II) [1.5em] substitution ↓ [0.2em] x= 8y-4 & (I) 8y-4+8y=6 & (II)

Extra

Steps of the Substitution Method
Let's review the steps that we need to follow when solving a system of equations by the Substitution Method.

What is done?
Step I Isolate one variable in any of the equations.
Step II Substitute the equivalent expression of the isolated variable into the other equation. This will result in one equation in one variable.
Step III Solve the obtained equation in one variable.
Step IV Substitute the result into one of the original equations to find the other variable.