Q. 411 - Intermediate Algebra

Question

In the following exercises, solve each system of equations. Use either substitution or elimination.

$\left\{\begin{array}{l}x+y-z=-1\\ 2x-y+2z=8\\ -3x+2y+z=-9\end{array}\right.$

Step-by-Step Solution

Answer

The solution for system of linear equation is (2,-2,1)

Step 1. Given

The system of equations 

$\left\{\begin{array}{l}x+y-z=-1\\ 2x-y+2z=8\\ -3x+2y+z=-9\end{array}\right.$

To find the solution of the system of equations.

Step 2. Solve the equations

Consider the system of equations: 

$\begin{array}{l}x+y-z=-1 ...........\left(1\right)\\ 2x-y+2z=8 ...........\left(2\right)\\ -3x+2y+z=-9 ......\left(3\right)\end{array}$

Multiply equation (1) with 2

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

From equation (2) and (4):

$$\begin{gathered} \frac{2x-y+2z=8 \\ 2x+2y-2z=-2}{4x+y=6} \\ 4x+y=6 .........\left(5\right) \end{gathered}$$

From equation (1) and (3):

$$\begin{gathered} \frac{x+y-z=-1 \\ -3x+2y+z=-9}{-2x+3y=-10} \\ -2x+3y=-10 ..........\left(6\right) \end{gathered}$$

Step 3. Solve for x

Multiply equation (5) by -3

$$\begin{gathered} -3(4x+y)=-3\left(6\right) \\ -\text{'}12x-3y=-18 ..........\left(7\right) \end{gathered}$$

from equation (6) and (7)

$$\begin{gathered} \frac{-12x-3y=-18 \\ -2x+3y=-10}{14x=-28} \\ x=2 \end{gathered}$$

Step 4. Solve for y

Put the value of $x=2$ in $-2x+3y=-10$

$$\begin{gathered} -2x+3y=-10 \\ -4+3y=-10 \\ 3y=-6 \\ y=-2 \end{gathered}$$

Step 5. Solve for z

Put $x=2,y=-2$ in the equation $$\begin{gathered} x+y-z=-1 \\ 2-2-z=-1 \\ z=1 \end{gathered}$$

Step 6. Checking the values

Put the values of $x=2,y=-2,z=1$ in the equation $x+y-z=-1$

$$\begin{gathered} x+y-z=-1 \\ 2-2-1=-1 \\ -1=-1 \end{gathered}$$

which holds true.