Question

Solving linear systems with 3 variables

Solve for x y and z using substitution

5x+2y-2z=-57
3x-10y-z=-11
y=-2

Answer 1

x=-9

y=-2

z=4

Explanation:

5x+2y-2z=-57

3x-10y-z=-11

y=-2

5x+2(-2)-2z=-57

3x-10(-2)-z=-11

y = -2

5x-4-2z=-57

3x+20-z=-11

y = -2

5x-2z=-53

3x-z=-31

y = -2

5x-2z=-53

3x+31=z

y = -2

5x-2(3x+31)=-53

3x+31=z

y = -2

5x-6x-64=-53

3x+31=z

y = -2

-x=9

3x+31=z

y = -2

x=-9

3(-9)+31=z

y = -2

x=-9

-27+31=z

y = -2

x=-9

z=4

y = -2