Q 169. - Intermediate Algebra

Question

In the following exercises, solve the system of equations.

$\left\{\begin{array}{l}5x-3y+2z=-5\\ 2x-y-z=4\\ 3x-2y+2z=-7\end{array}\right.$

Step-by-Step Solution

Answer

The solution is $(7,12,-2)$.

Step 1. Given Information

We are given system of linear equations, 

$\left\{\begin{array}{l}5x-3y+2z=-5 -\left(1\right)\\ 2x-y-z=4 -\left(2\right)\\ 3x-2y+2z=-7 -\left(3\right)\end{array}\right.$

Step 2. Eliminating the same variable

Multiplying by $3$ in second equation, we get

$$\begin{gathered} 3(2x-y-z)=3\times 4 \\ 6x-3y-3z=12 -\left(4\right) \end{gathered}$$

Now, subtracting fourth and first equation, we get

$$\begin{gathered} 6x-5x-3y+3y-3z-2z=12+5 \\ x-5z=17 -\left(5\right) \end{gathered}$$

Now, multiplying $2$ in second equation, we get

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

Now, subtracting sixth and third equation, we get

$$\begin{gathered} 4x-3x-2y+2y-2z-2z=8+7 \\ x-4z=15 -\left(7\right) \end{gathered}$$

Step 3. Solving the new system of equation

Subtracting fifth and seventh equation, we get

$$\begin{gathered} x-x-5z+4z=17-15 \\ -z=2 \\ z=-2 \end{gathered}$$

Now, putting the value of $z$ in seventh equation, we get

$$\begin{gathered} x-4z=15 \\ x-4(-2)=15 \\ x+8=15 \\ x=15-8 \\ x=7 \end{gathered}$$

Now, putting the value of $x,z$ in second equation, we get

$$\begin{gathered} 2x-y-z=4 \\ 2\left(7\right)-y-(-2)=4 \\ 14-y+2=4 \\ y=16-4 \\ y=12 \end{gathered}$$

Step 4. Checking the solution

Putting the value of $x,y,z$ in the equations, we get

$$\begin{gathered} 5x-3y+2z=-5 \\ 5\left(7\right)-3\left(12\right)+2(-2)=-5 \\ 35-36-4=-5 \\ -5=-5 \\ 2x-y-z=4 \\ 2\left(7\right)-12-(-2)=4 \\ 14-12+2=4 \\ 4=4 \\ 3x-2y+2z=-7 \\ 3\left(7\right)-2\left(12\right)+2(-2)=-7 \\ 21-24-4=-7 \\ -7=-7 \end{gathered}$$

Hence the solution is correct.