91Ó°ÊÓ

Let the equations be:

$6x-2y-3z=-10$ … (i)

$-6x+y+9z=3$ … (ii)

$8x-3y=-16$ … (iii)

Use elimination to make a system of two equations (i) and (ii) in two variables. Multiply first equation by 3 and then add it with equation (ii) to eliminate variable z.

$\begin{array}{l}18x−6y−9z=−30\\ \underset{¯}{−6x+y+9z=3}\\ \text{12}x−5y=−27\end{array}$

Solve the system of equations $8x-3y=-16$and $12x-5y=-27$. Multiply first equation by 3 and second equation by 2 then subtract the two equations.

$\begin{array}{l}\text{ â¶Ä‰â¶Ä‰â¶Ä‰}24x−9y=−48\\ \underset{¯}{−\text{ â¶Ä‰â¶Ä‰â¶Ä‰}24x−10y=−54\text{ â¶Ä‰â¶Ä‰}}\\ \text{ â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â¶Ä‰}y=6\end{array}$

Now substitute 6 for y into the equation $8x-3y=-16$.

$\begin{array}{c}8x−3\left(6\right)=−16\\ 8x−18+18=−16+18\\ 8x=2\\ x=\frac{1}{4}\end{array}$

Now substitute $\frac{1}{4}$ for x and 6 for y into the equation $6x-2y-3z=-10$.

$\begin{array}{c}6\left(\frac{1}{4}\right)−2\left(6\right)−3z=−10\\ \frac{3}{2}−12−3z=−10\\ −\frac{21}{2}−3z+\frac{21}{2}=−10+\frac{21}{2}\\ −3z=\frac{−20+21}{2}\\ −3z=\frac{1}{2}\\ z=−\frac{1}{6}\end{array}$