Full solution

Q. Solve the system of equations by elimination.

\newline

3x - y - z = 7

\newline

-2x + y + 3z = -7

\newline

2x - 2y + z = 14

Eliminate y by adding equations: First, let's add the first and second equations to eliminate y. $\newline$ $(3x - y - z) + (-2x + y + 3z) = 7 + (-7)$ $\newline$ $3x - y - z - 2x + y + 3z = 0$ $\newline$ $x + 2z = 0$
Simplify third equation: Now, let's multiply the third equation by $\frac{1}{2}$ to simplify it. $\newline$ $\left(\frac{1}{2}\right)(2x - 2y + z) = \left(\frac{1}{2}\right)(14)$ $\newline$ $x - y + \left(\frac{1}{2}\right)z = 7$
Eliminate y by adding equations: Next, we'll add the modified third equation to the first equation to eliminate y. $\newline$ $(x - y + (1/2)z) + (3x - y - z) = 7 + 7$ $\newline$ $x - y + (1/2)z + 3x - y - z = 14$ $\newline$ $4x - 2y - (1/2)z = 14$
Eliminate z by adding equations: We can now add the new equation from the previous step to the modified second equation to eliminate z. $\newline$ $(4x - 2y - \frac{1}{2}z) + (x + 2z) = 14 + 0$ $\newline$ $4x - 2y - \frac{1}{2}z + x + 2z = 14$ $\newline$ $5x - 2y + \frac{3}{2}z = 14$
Get rid of fraction: Let's multiply the new equation by $2$ to get rid of the fraction. $\newline$ $2(5x - 2y + \frac{3}{2}z) = 2(14)$ $\newline$ $10x - 4y + 3z = 28$
Eliminate y by adding equations: Now, we'll add the modified third equation to the new equation to eliminate y. $\newline$ $(10x - 4y + 3z) + (x - y + \frac{1}{2}z) = 28 + 7$ $\newline$ $10x - 4y + 3z + x - y + \frac{1}{2}z = 35$ $\newline$ $11x - 5y + \frac{7}{2}z = 35$
Solve for x: We can now solve for x by adding the first and third equations. $\newline$ $(3x - y - z) + (11x - 5y + \frac{7}{2}z) = 7 + 35$ $\newline$ $3x - y - z + 11x - 5y + \frac{7}{2}z = 42$ $\newline$ $14x - 6y + \frac{5}{2}z = 42$
Divide to solve for x: Divide the new equation by $14$ to solve for $x$ .
$(14x - 6y + (5/2)z) / 14 = 42 / 14$
$x - (6/14)y + (5/28)z = 3$
$x = 3 + (6/14)y - (5/28)z$