Q 172. - Intermediate Algebra

Question

In the following exercises, solve the system of equations. 

$\left\{\begin{array}{l}3x+8y+2z=-5\\ 2x+5y-3z=0\\ x+2y-2z=-1\end{array}\right.$

Step-by-Step Solution

Answer

The solution is $(-9,3,-1)$.

Step 1. Given Information

We are given system of linear equations,  

$\left\{\begin{array}{l}3x+8y+2z=-5 -\left(1\right)\\ 2x+5y-3z=0 -\left(2\right)\\ x+2y-2z=-1 -\left(3\right)\end{array}\right.$

Step 2. Eliminating the same variable

Multiplying by $2$ in third equation, we get

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

Now, subtracting fourth and second equation, we get

$$\begin{gathered} 2x-2x+5y-4y-3z+4z=0+2 \\ y+z=2 -\left(5\right) \end{gathered}$$

Multiplying by $3$ in third equation, we get

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

Now, subtracting sixth and first equation, we get

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

Step 3. Solving the new system of equation

Now, subtracting the seventh and fifth equation, we get

$$\begin{gathered} y-y+4z-z=-1-2 \\ 3z=-3 \\ z=\frac{-3}{3} \\ z=-1 \end{gathered}$$

Putting the value of $z$ in fifth equation, we get

$$\begin{gathered} y+z=2 \\ y-1=2 \\ y=2+1 \\ y=3 \end{gathered}$$

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

$$\begin{gathered} 3 x+8 y+2 z=-5 \\ 3 x+8\left(3\right)+2(-1)=-5 \\ 3 x+24-2=-5 \\ 3 x+22=-5 \\ 3x=-5-22 \\ 3x=-27 \\ x=\frac{-27}{3} \\ x=-9 \end{gathered}$$

Step 4. Checking the solution

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

$$\begin{gathered} 3 x+8 y+2 z =-5 \\ 3(-9)+8\left(3\right)+2(-1) =-5 \\ -27+24-2 =-5 \\ -5 =-5 \\ 2 x+5 y-3 z=0 \\ 2(-9)+5\left(3\right)-3(-1)=0 \\ -18+15+3=0 \\ 0=0 \\ x+2 y-2 z=-1 \\ -9+2\left(3\right)-2(-1)=-1 \\ -9+6+2=-1 \\ -1=-1 \end{gathered}$$

Hence the solution is correct.