Question

Solve the system of equations with three variablesusing elimination.43 + 4y + z = 242x – 4y + z = 05x – 4y – 5z = 12

Answer 1

$\begin{gathered} \text{Let:} \\ 4x+4y+z=24\text{ (1)} \\ 2x-4y+z=0\text{ (2)} \\ 5x-4y-5z=12\text{ (3)} \\ (1)+(2) \\ 4x+2x+4y-4y+z+z=24+0 \\ 6x+2z=24\text{ (4)} \end{gathered}$$\begin{gathered} (2)-(3) \\ 2x-5x-4y-(-4y)+z-(-5z)=0-12 \\ -3x+6z=-12\text{ (5)} \end{gathered}$$\begin{gathered} (4)+2(5) \\ 6x-6x+2z+12z=24-24 \\ 14z=0 \\ z=0 \end{gathered}$

Replace z into (4)

$\begin{gathered} 6x+2z=24 \\ 6x+2(0)=24 \\ 6x+0=24 \\ x=\frac{24}{6} \\ x=4 \end{gathered}$

Replace x and z into (2):

$\begin{gathered} 2x-4y+z=0 \\ 2(4)-4y+0=0 \\ 8-4y=0 \\ 4y=8 \\ y=\frac{8}{4} \\ y=2 \end{gathered}$