Q 167. - Intermediate Algebra

Question

In the following exercises, solve the system of equations.

$\left\{\begin{array}{l}6x-5y+2z=3\\ 2x+y-4z=5\\ 3x-3y+z=-1\end{array}\right.$

Step-by-Step Solution

Answer

The solution is $(4,5,2)$.

Step 1. Given Information

We are given system of linear equations,

$\left\{\begin{array}{l}6x-5y+2z=3 -\left(1\right)\\ 2x+y-4z=5 -\left(2\right)\\ 3x-3y+z=-1 -\left(3\right)\end{array}\right.$

Step 2. Solving the equations

Multiplying third equation by $2$, we get

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

Now, Subtracting fourth and first equation, we get

$$\begin{gathered} 6x-6x-5y-(-6y)+2z-2z=3-(-2) \\ -5y+6y=3+2 \\ y=5 \end{gathered}$$

Now, multiplying by $4$ in third equation, we get

$$\begin{gathered} 4(3x-3y+z)=4\times -1 \\ 12x-12y+4z=-4 -\left(5\right) \end{gathered}$$

Now, adding fifth and second equation, we get

$$\begin{gathered} 12x+2x-12y+y+4z-4z=-4+5 \\ 14x-11y=1 -\left(6\right) \end{gathered}$$

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

$$\begin{gathered} 14x-11y=1 \\ 14x-11\left(5\right)=1 \\ 14x-55=1 \\ 14x=56 \\ x=\frac{56}{14} \\ x=4 \end{gathered}$$

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

$$\begin{gathered} 6x-5y+2z=3 \\ 6\left(4\right)-5\left(5\right)+2z=3 \\ 24-25+2z=3 \\ 2z=3+1 \\ z=\frac{4}{2} \\ z=2 \end{gathered}$$

Step 3. Checking the solution

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

$$\begin{gathered} 6x-5y+2z=3 \\ 6\left(4\right)-5\left(5\right)+2\left(2\right)=3 \\ 24-25+4=3 \\ 3=3 \\ 2x+y-4z=5 \\ 2\left(4\right)+5-4\left(2\right)=5 \\ 8+5-8=5 \\ 5=5 \\ 3x-3y+z=-1 \\ 3\left(4\right)-3\left(5\right)+2=-1 \\ 12-15+2=-1 \\ -1=-1 \end{gathered}$$

This is true, hence the solution is correct.