Q 4.65 - Intermediate Algebra

Question

Solve :

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

Step-by-Step Solution

Answer

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

Step 1. Given information

We are given system of linear equations,

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

Step 2. Solving the equations

Multiplying first equation by $3$ and second equation by $4$, we get

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

Now, adding the above equations, we get

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

Now, multiplying $2$ in sixth equation and $9$ in third equation, we get

$$\begin{gathered} 18x+16y=10 -\left(7\right) \\ 18x+27y=54 -\left(8\right) \end{gathered}$$

Now, subtracting the above equations, we get

$$\begin{gathered} 18x-18x+27y-16y=54-10 \\ 11y=44 \\ y=\frac{44}{11} \\ y=4 \end{gathered}$$

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

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

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

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

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

Step 3. Checking the solution

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

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

This is true, hence the solution is correct.