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 13 Page 209

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

(- 3, - 3)

Practice makes perfect

When solving a system of equations using the Substitution 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.

It looks like it will be simplest to isolate y in the first equation. y-x=0 ⇔ y=x Now that we have isolated y, we can solve the system by substitution.

y=x & (I) 2x-5y=9 & (II)
y=x 2x-5 x=9
y=x - 3x=9
y=x - 3x- 3= 9- 3
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(II): Calculate quotient
y=x - 3- 3x= 9- 3
y=x - 3- 3x=- 9 3
y=x 1x=-3
y=x x=-3

Great! Now, to find the value of y, we need to substitute x=-3 into the first equation.

y=x x=-3
y=-3 x=-3

The solution, or point of intersection, to this system of equations is the point (-3,-3).

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=0 & (I) 2x-5y=9 & (II)

(I), (II): x= -3, y= -3

-3-( -3)? =0 2( -3)-5( -3)? =9
-3-(-3)? =0 -2(3)-5(-3)? =9
-3-(-3)? =0 -2(3)+5(3)? =9
-3-(-3)? =0 -6+15? =9
-3+3? =0 -6+15? =9

(I), (II): Add terms

0=0 ✓ 9=9 ✓

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