Big Ideas Math: Modeling Real Life, Grade 8
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2. Solving Systems of Linear Equations by Substitution
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Exercise 11 Page 207

Does either of the equations have an isolated variable in it?

(1,2)

Practice makes perfect

In this system of equations, at least one of the variables has a coefficient of 1. Therefore, we will approach its solution with the Substitution Method. When solving a system of equations using this method, there are three steps.

  1. Isolate a variable in one of the equations.
  2. Substitute the expression for that variable into the other equation and solve.
  3. Substitute this solution into one of the equations and solve for the value of the other variable.

Since the expression equal to y in the second equation is simpler, let's use that for our initial substitution.

y=- x+3 & (I) y=2x & (II)
2x=- x+3 y=2x
2x+x=- x+3+x y=2x
3x=3 y=2x
3x3= 33 y=2x
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(I): Calculate quotient
33x= 33 y=2x
1x=1 y=2x
x=1 y=2x

Great! Now, to find the value of y, we need to substitute x=1 into equation (I) in the given system. Let's use the second equation.

x=1 y=2x
x=1 y=2( 1)
x=1 y=2

The solution, or point of intersection, to this system of equations is the point (1,2).

Checking Our Answer

To check our answer, we will substitute our solution into both equations. If doing so results in true statements, then our solution is correct.

y=- x+3 & (I) y=2x & (II)

(I), (II): x= 1, y= 2

2? =- 1+3 2? =2( 1)
2? =- 1+3 2=2
2=2 ✓ 2=2 ✓

Because both equations are true statements, we know that our solution is correct.