Big Ideas Math: Modeling Real Life, Grade 8
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Big Ideas Math: Modeling Real Life, Grade 8 View details
Chapter Review
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Exercise 15 Page 228

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

Solution: (3,0)
Explanation: See solution.

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 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. For this exercise, y is already isolated in one equation, so we can skip straight to solving!
    3y+9=3x & (I) y=- 13x+1 & (II)
    3( - 13x+1 )+9=3x y=- 13x+1
    â–Ľ
    (I): Solve for x
    3(- x3+1 )+9=3x y=- 13x+1
    - 3* x3+3* 1 +9=3x y=- 13x+1
    - x+3* 1 +9=3x y=- 13x+1
    - x+3 +9=3x y=- 13x+1
    - x+3 +9+x=3x+x y=- 13x+1
    12=4x y=- 13x+1
    124= 4x4 y=- 13x+1
    124= 4x4 y=- 13x+1
    124=x y=- 13x+1
    3=x y=- 13x+1
    x=3 y=- 13x+1
    Great! Now, to find the value of y, we need to substitute x=3 into the second equation.
    x=3 & (I) y=- 13x+1 & (II)
    x=3 y=- 13( 3)+1
    x=3 y=- 1+1
    x=3 y=0
    The solution, or point of intersection, to this system of equations is the point (3,0).