Q 4.63 - Intermediate Algebra

Question

Solve the system by elimination: $$\begin{gathered} 3x+y-z=2 \\ 2x-3y-2z=1 \\ 4x-y-3z=0 \end{gathered}$$

Step-by-Step Solution

Answer

The solution of given system of equations is $(2,-1,3)$. 

Step 1. Given information

We are given system of linear equations 

$$\begin{gathered} 3x+y-z=2 -\left(1\right) \\ 2x-3y-2z=1 -\left(2\right) \\ 4x-y-3z=0 -\left(3\right) \end{gathered}$$

Step 2. Solving equation

Adding first and third equation, we get

$$\begin{gathered} 4x+3x+y-y-z-3z=2+0 \\ 7x-4z=2 -\left(4\right) \end{gathered}$$

Now, Multiplying by $3$ in first equation, we get  $9x+3y-3z=6 -\left(5\right)$

Adding fifth and second equation, we get

$$\begin{gathered} 9x+2x+3y-3y-3z-2z=6+1 \\ 11x-5z=7 -\left(6\right) \end{gathered}$$

Now, Multiplying $11$ to fourth equation and $7$ to sixth equation, we get

$$\begin{gathered} 77x-44z=22 \\ 77x-35z=49 \end{gathered}$$

Now, subtracting the above given equation,

$$\begin{gathered} 9z=27 \\ z=\frac{27}{9} \\ z=3 \end{gathered}$$

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

$$\begin{gathered} 7x-4\left(3\right)=2 \\ 7x-12=2 \\ 7x=12+2 \\ 7x=14 \\ x=2 \end{gathered}$$

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

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

Hence the solution is $(2,-1,3)$.

Step 3. Checking the solution

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

$$\begin{gathered} 3x+y-z=2 \\ 3\left(2\right)-1-3=2 \\ 6-4=2 \\ 2=2 \\ 2x-3y-2z=1 \\ 2\left(2\right)-3(-1)-2\left(3\right)=1 \\ 4+3-6=1 \\ 1=1 \\ 4x-y-3z=0 \\ 4\left(2\right)-(-1)-3\left(3\right)=0 \\ 8+1-9=0 \\ 0=0 \end{gathered}$$

This is true, hence the solution is correct.