Pearson Algebra 1 Common Core, 2011
PA
Pearson Algebra 1 Common Core, 2011 View details
2. Multiplying Powers With the Same Base
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Exercise 61 Page 431

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

(4,7)

Practice makes perfect

When solving a system of equations using substitution, 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.

Observing the given equations, it looks like it will be simplest to isolate y in the second equation.

2x+y=15 - 12x+y=5
2x+y=15 y=5+ 12x

Now that we've isolated y, we can solve the system by substitution.

2x+y=15 y=5+ 12x
2x+ 5+ 12x=15 y=5+ 12x
â–¼
(I):Solve for x
2x+ 12x=10 y=5+ 12x
42x+ 12x=15 y=5+ 12x
52x=10 y=5+ 12x
52x* 25=10* 25 y=5+ 12x
x* 52 * 25=10* 25 y=5+ 12x
x* 1=10 * 25 y=5+ 12x
x=10 * 25 y=5+ 12x
x= 205 y=5+ 12x
x=4 y=5+ 12x

Great! Now, to find the value of y, we need to substitute x=4 into the second equation.

x=4 y=5+ 12x
x=4 y=5+ 12( 4)
x=4 y=5+ 42
x=4 y=5+2
x=4 y=7

The solution, or point of intersection, to this system of equations is the point (4,7).