Pearson Algebra 2 Common Core, 2011
PA
Pearson Algebra 2 Common Core, 2011 View details
5. Systems With Three Variables
Continue to next subchapter

Exercise 46 Page 173

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

3/2

Practice makes perfect

The given system consists of equations of planes. When using the Substitution Method to solve a system of equations, it is necessary to isolate a variable. In the first equation, y is isolated, so we will start by substituting it for its equivalent expression into the remaining equations.

y=-2x+10 - x+y-2z=-2 3x-2y+4z=7

(II), (III): y= -2x+10

y=-2x+10 - x+( -2x+10)-2z=-2 3x-2( -2x+10)+4z=7
â–¼
(II), (III): Simplify
y=-2x+10 - x-2x+10-2z=-2 3x-2(-2x+10)+4z=7
y=-2x+10 - x-2x+10-2z=-2 3x+4x-20+4z=7

(II), (III): Add and subtract terms

y=-2x+10 -3x+10-2z=-2 7x-20+4z=7
y=-2x+10 -3x+10=2z-2 7x-20+4z=7
y=-2x+10 -3x+12=2z 7x-20+4z=7
y=-2x+10 -6x+24=4z 7x-20+4z=7
y=-2x+10 4z=-6x+24 7x-20+4z=7
This time, 4z was isolated in the second equation. We can now substitute its equivalent expression into the third equation.

y=-2x+10 4z=-6x+24 7x-20+4z=7
â–¼
(III): Solve by substitution
y=-2x+10 4z=-6x+24 7x-20+( -6x+24)=7
y=-2x+10 4z=-6x+24 7x-20-6x+24=7
y=-2x+10 4z=-6x+24 x+4=7
y=-2x+10 4z=-6x+24 x=3

The value of x is 3. Substituting 3 for x into the first and the second equations, we can find the values of x and z.

y=-2x+10 4z=-6x+24 x=3

(I), (II): x= 3

y=-2( 3)+10 4z=-6( 3)+24 x=3
â–¼
(I), (II): Simplify

(I), (II): Multiply

y=-6+10 4z=-18+24 x=3

(I), (II): Add terms

y=4 4z=6 x=3
y=4 z= 64 x=3
y=4 z= 32 x=3

The value of z is 32.