Q26. - Algebra 2

Question

26: Solve each system of equations.

$\begin{array}{c}x+4y-z=6\\ 3x+2y+3z=16\\ 2x-y+z=3\end{array}$

Step-by-Step Solution

Answer

The solution is $x=1,y=2,z=3$.

Step-1 – Use elimination to make a system of two equations in two variables.

Given equations are-:

$\begin{array}{c}x+4y-z=\mathrm{6......}\left(1\right)\\ 3x+2y+3z=\mathrm{16.......}\left(2\right)\\ 2x-y+z=\mathrm{3.......}\left(3\right)\end{array}$

Multiplying the first equation by $3$ and subtracting with second equation, we get

$\begin{array}{c}12y-2y-3z-3z=18-16\\ 10y-6z=\mathrm{2........}\left(4\right)\end{array}$

Again, multiplying the first equation by $2$ and subtracting with third equation, we get

$\begin{array}{c}8y+y-2z-z=12-3\\ 9y-3z=\mathrm{9........}\left(5\right)\end{array}$

Step-2 – Solve the system of two equations.

$\begin{array}{c}10y-6z=\mathrm{2.......}\left(4\right)\\ 9y-3z=\mathrm{9........}\left(5\right)\end{array}$

Multiplying the fifth equation by $2$  and subtracting from fourth equation.

$\begin{array}{c}10y-18y=2-18\\ -8y=-16\\ y=2\end{array}$

Putting the value of $y$ in the fourth equation, we get

$\begin{array}{c}10y-6z=2\\ 10\left(2\right)-6z=2\\ -6z=2-20\\ -6z=-18\\ z=3\end{array}$

Step-3 – Substituting the value according to the coordinates in one of the equations with three variables.

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