Q52. - Algebra 2

Question

Solve each system of equations.

 

 $\begin{array}{c}6x-2y-3z=-10\\ -6x+y+9z=3\\ 8x-3y=-16\end{array}$

Step-by-Step Solution

Answer

The solution of the system of equations:  $\begin{array}{c}6x-2y-3z=-10\\ -6x+y+9z=3\\ 8x-3y=-16\end{array}$ is $\left(\frac{1}{4},6,-\frac{1}{6}\right)$. 

Step 1 - Use elimination method

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.

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

Step 2 - Solve the system of equations in two variables

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.

$\begin{array}{l}24x-9y=-48\\ \underset{¯}{-24x-10y=-54}\\ y=6\end{array}$

Step 3 - Solve for variable x

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

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

Step 4 - Solve for variable z

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

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