Q9. - Algebra 2

Question

Solve each system of equations.

9. 

$\begin{array}{l}x+y+z=12\\ 6x-2y-z=16\\ 3x+4y+2z=28\end{array}$

Step-by-Step Solution

Answer

 The solution is $x=4,y=0,z=8$.

Step-1 –Using elimination to obtain two equations with two variables

Given system of equations are 

$\begin{array}{l}x+y+z=12\\ 6x-2y-z=16\\ 3x+4y+2z=28\end{array}$

Multiplying first equation by $2$ and adding to the second first equation, we get

$\begin{array}{l}2x+6x+2z-z=24+16\\ 8x+z=40\end{array}$

Again multiplying the second  equation by $2$ and adding to the third equation, we get

$\begin{array}{l}12x+3x-2z+2z=32+28\\ 15x=60\\ x=4\end{array}$

Step-2 –Putting the value of \[x\] in the equation with two variables

$\begin{array}{l}8x+z=40\\ 8\left(4\right)+z=40\\ 32+z=40\\ z=40-32\\ z=8\end{array}$

Step-3 –Substituting the value of x and z in the equation with three variables

$\begin{array}{l}x+y+z=12\\ 4+y+18=12\\ y=12-12\\ y=0\end{array}$