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Pearson Algebra 2 Common Core, 2011 View details

5. Systems With Three Variables

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Exercise 39 Page 172

When using the Substitution Method to solve a system of equations, it is necessary to isolate a variable.

(2,4,6)

Practice makes perfect

The given system consists of equations of planes. We will use the Substitution Method to solve this system. When using this method, it is necessary to isolate a variable. In the third equation, it will be easiest to isolate z.

6x+y-4z=-8 & (I) y4- z6=0 & (II) 2x-z=-2 & (III)

SubEqn

(III): LHS-2x=RHS-2x

6x+y-4z=-8 y4- z6=0 - z=-2-2x

MultEqn

(III):LHS * (-1)=RHS* (-1)

6x+y-4z=-8 y4- z6=0 z=2+2x

With a variable isolated in one of the equations, we can substitute its equivalent expression into the remaining equations. In the final step of the simplification of these substitutions, our goal is to have yet another variable isolated.

6x+y-4z=-8 y4- z6=0 z=2+2x

(I), (II): z= 2+2x

6x+y-4( 2+2x)=-8 y4- 2+2x6=0 z=2+2x

▼

(I), (II): Simplify

Distr

(I): Distribute (-4)

6x+y-8-8x=-8 y4- 2+2x6=0 z=2+2x

MultEqn

(II): LHS * 12=RHS* 12

6x+y-8-8x=-8 12( y4)-12( 2+2x6)=0 z=2+2x

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