Question

Solve the system by substitution 5x-2y+3z=-32x-4y-3z=0X+6y-8z=-37

Answer 1

The given system is

`\begin{gathered} 5x-2y+3z=-3\rightarrow(1) \\ 2x-4y-3z=0\rightarrow(2) \\ x+6y-8z=-37\rightarrow(3) \end{gathered}`

Use equation (3) to find x in terms of y and z

`x=-6y+8z-37\rightarrow(4)`

Substitute x in (1) and (2) by (4)

`\begin{gathered} 5(-6y+8z-37)-2y+3z=-3 \\ -30y+40z-185-2y+3z=-3 \\ -32y+43z=182\rightarrow(5) \end{gathered}`

`\begin{gathered} 2(-6y+8z-37)-4y-3z=0 \\ -12y+16z-74-4y-3z=0 \\ -16y+13z=74\rightarrow(6) \end{gathered}`

Multiply (6) by 2

`\begin{gathered} -32y+26z=148 \\ -32y+26z-26z=148-26z \\ -32y=-26z+148\rightarrow(7) \end{gathered}`

Substitute -32y in (5) by (7)

`\begin{gathered} -26z+148+43z=182 \\ 17z=182-148 \\ 17z=34 \\ z=(34)/(17) \\ z=2 \end{gathered}`

Substitute the values of z in equation (7) to find y

`\begin{gathered} -32y=-26(2)+148 \\ -32y=-52+148 \\ -32y=96 \\ y=(96)/(-32) \\ y=-3 \end{gathered}`

Substitute the values of z and y in equation (4) to find x

`\begin{gathered} x=-6(-3)+8(2)-37 \\ x=18+16-37 \\ x=-3 \end{gathered}`

The solution is x = -3, y = -3, z = 2