Q 374. - Intermediate Algebra

Question

Solve the system of equations.

$\left\{\begin{array}{l}-x-3y+2z=14\\ -x+2y-3z=-4\\ 3x+y-2z=6\end{array}\right.$

Step-by-Step Solution

Answer

The solution for the system of equations is,

$(8z+\frac{16}{7},\frac{11}{7}z-\frac{6}{7},z)$.

Step 1. Given the information.

The system of equations is,

$\left\{\begin{array}{l}-x-3y+2z=14...........\left(1\right)\\ -x+2y-3z=-4..........\left(2\right)\\ 3x+y-2z=6................\left(3\right)\end{array}\right.$

Step 2. Eliminating x from equations (1) and (2).

Eliminating $x$ from equations (1) and (2).

$$\begin{gathered} x+3y-2z=-14............\left(1\right)\times -1 \\ -x+2y-3z=-4................\left(2\right) \end{gathered}$$

Solving the equations, we get,

$5y-5z=-18.........\left(4\right)$

Step 3. Eliminating x from the equations (2) and (3).

Eliminating $x$ from the equations (2) and (3),

$$\begin{gathered} -3x+6y-9z=-12........\left(2\right)\times 3 \\ 3x+y-2z=6.................\left(3\right) \end{gathered}$$

Solving the equations, we get,

$7y-11z=-6 .........\left(5\right)$

Step 4. Showing how y is dependent on z .

Solving equation $\left(5\right), 7y-11z=-6$ for $y$ in terms of $z$,

$$\begin{gathered} 7y-11z=-6 \\ 7y=11z-6 \\ y=\frac{11}{7}z-\frac{6}{7} \end{gathered}$$

Step 5. Showing the value of x in terms of z .

Substituting $y=\frac{11}{2}z-\frac{6}{7}$ in equation (2),

$-x+2y-3z=-4$,

$$\begin{gathered} -x+2(\frac{11}{2}z-\frac{6}{7})-3z=-4 \\ -x+11z-\frac{12}{7}-3z=-4 \\ -x+8z=-4+\frac{12}{7} \\ -x+8z=\frac{-28+12}{7} \\ x=8z+\frac{16}{7} \end{gathered}$$