Q 4.64 - Intermediate Algebra

Question

Solve the system by elimination: $$\begin{gathered} 4x+y+z=-1 \\ -2x-2y+z=2 \\ 2x+3y-z=1 \end{gathered}$$

Step-by-Step Solution

Answer

The solution of given system of equation is $(-2,3,4)$.

Step 1. Given information

We are given system of linear equations

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

Step 2. Solving equations

Subtracting the first and second equation, we get

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

Now, adding second and third equation, we get

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

Putting the value of $y$ in fourth equation, we get

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

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

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

Hence the solution is $(-2,3,4)$.

Step 3. Checking the solution

Checking the solution by putting the value of $x,y,z$ in the equations, we get

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

This is true, hence the solution is correct.