Question

Solve the system of equations below using any method you learned in this unit. Show all work (even if you are using your calculator).

Answer 1

Given the system of equations

`\begin{gathered} x+4y-z=20-----1 \\ 3x+2y+z=8-----2 \\ 2x-3y+2z=-16-----3 \end{gathered}`

We can solve for x, y and z below.

Step-by-step explanation

Step 1: Find the value of z using the substitution method

`\begin{gathered} \begin{bmatrix}x+4y-z=20\\ 3x+2y+z=8\\ 2x-3y+2z=-16\end{bmatrix} \\ Isolate\text{ for x in equation 1} \\ x=20-4y+z \\ \mathrm{Substitute\:}x=20-4y+z\text{ in equation 2 and 3} \\ \begin{bmatrix}3\left(20-4y+z\right)+2y+z=8\\ 2\left(20-4y+z\right)-3y+2z=-16\end{bmatrix} \\ sinplify \\ \begin{bmatrix}-10y+4z+60=8 \\ -11y+4z+40=-16\end{bmatrix} \\ Isolate\text{ for y in}-10y+4z+60=8 \\ -10y=8-4z-60 \\ y=(8-4z-60)/(-10) \\ y=(-4z-52)/(-10) \\ y=(2\left(z+13\right))/(5) \\ \mathrm{Substitute\:}y=(2\left(z+13\right))/(5)\text{ in }-11y+4z+40=-16 \\ \begin{bmatrix}-11\cdot (2\left(z+13\right))/(5)+4z+40=-16\end{bmatrix} \\ simplify \\ \begin{bmatrix}(-2z-286)/(5)+40=-16\end{bmatrix} \\ multiply\text{ through by 5} \\ -2z-286+200=-80 \\ isolate\text{ for z} \\ -2z=-80-200+286 \\ -2z=6 \\ z=(6)/(-2) \\ z=-3 \end{gathered}`

Step 2: Find y

`\begin{gathered} \mathrm{Substitute\:}z=-3\text{ in}\mathrm{\:}y=(2\left(z+13\right))/(5) \\ y=(2(-3+13))/(5) \\ y=(2(10))/(5) \\ y=4 \end{gathered}`

Step 3: Find z

`\begin{gathered} \mathrm{Substitute\:}z=-3,\:y=4\text{ in }x=20-4y+z \\ x=20-4\cdot \:4-3 \\ x=1 \end{gathered}`

Answer: The solutions to the system of equations are

`x=1,\:z=-3,\:y=4`