Question

In the following exercises, solve the system of equations.

$\left\{\begin{array}{l}2x-5y+3z=8\\ 3x-y+4z=7\\ x+3y+2z=-3\end{array}\right.$

Short answer

The solution is $(6,-1,-3)$.

Step-by-step solution

Step 1. Given Information   

We are given system of linear equations,

$\left\{\begin{array}{l}2x-5y+3z=8\\ 3x-y+4z=7\\ x+3y+2z=-3\end{array}\right.$

Step 2. Eliminating the same variable

Multiplying by $2$in third equation, we get

role="math" localid="1644655865067" $$\begin{gathered} 2(x+3y+2z)=2\times -3 \\ 2x+6y+4z=-6-\left(4\right) \end{gathered}$$

Subtracting first and fourth equation, we get

$$\begin{gathered} 2x-2x+6y+5y+4z-3z=-6-8 \\ 11y+z=-14-\left(5\right) \end{gathered}$$

Multiplying $3$in third equation, we get

role="math" localid="1644656147135" $$\begin{gathered} 3(x+3y+2z)=3\times -3 \\ 3x+9y+6z=-9-\left(6\right) \end{gathered}$$

Now, subtracting sixth and second equation,

role="math" localid="1644656533866" $$\begin{gathered} 3x-3x+9y+y+6z-4z=-9-7 \\ 10y+2z=-16 \\ 5y+z=-8-\left(7\right) \end{gathered}$$

Step 3. Solving the new system of equation

Subtracting seventh and fifth equation, we get

$$\begin{gathered} 11y-5y+z-z=-14+8 \\ 6y=-6 \\ y=\frac{-6}{6} \\ y=-1 \end{gathered}$$

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

$$\begin{gathered} 5y+z=-8 \\ 5(-1)+z=-8 \\ z=-8+5 \\ z=-3 \end{gathered}$$

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

$$\begin{gathered} x+3y+2z=-3 \\ x+3(-1)+2(-3)=-3 \\ x-3-6=-3 \\ x=6 \end{gathered}$$

Step 4. Checking the solution  

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

$$\begin{gathered} 2x-5y+3z=8 \\ 2\left(6\right)-5(-1)+3(-3)=8 \\ 12+5-9=8 \\ 8=8 \\ 3x-y+4z=7 \\ 3\left(6\right)-(-1)+4(-3)=7 \\ 18+1-12=7 \\ 7=7 \\ x+3y+2z=-3 \\ 6+3(-1)+2(-3)=-3 \\ 6-3-6=-3 \\ -3=-3 \end{gathered}$$

Hence the solution is correct.