Q25. - Algebra 2

Question

Solve each equation.

$\left[\begin{array}{cc}3x-5& x+y\\ 12& 9z\end{array}\right]\mathbf{=}\left[\begin{array}{cc}10& 8\\ 12& 3x+y\end{array}\right]$

Step-by-Step Solution

Answer

The solution is $\left(\mathbf{5}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{2}\right)$.

Step 1 – Description of step.

As the matrices are equal, the corresponding elements are equal. Write the three linear equations to show this equality.

$\begin{array}{c}3x-5=10\\ x+y=8\\ 3x+y=9z\end{array}$

Step 2 – Description of step.

Add 5 on both the sides of $3x-5=10$ and simplify.

$\begin{array}{c}3x-5+5=10+5\\ 3x=15\end{array}$

Step 3 – Description of step.

Divide each side of $3x=15$ by 3 to find the value of x.

 $\begin{array}{c}\frac{3x}{3}=\frac{15}{3}\\ x=5\end{array}$

Step 4 – Description of step.

Substitute 5 for x into $x+y=8$ and simplify.

 $\begin{array}{c}5+y=8\\ y=3\end{array}$

Step 5 – Description of step.

Substitute 5 for x and 3 for y into $3x+y=9z$ and simplify.

 $\begin{array}{c}3\left(5\right)+3=9z\\ 15+3=9z\\ 18=9z\end{array}$

Step 6 – Description of step.

Divide each side of $18=9z$ by 9 to find the value of z.

 $\begin{array}{c}\frac{18}{9}=\frac{9z}{9}\\ z=2\end{array}$

Therefore, the solution is $\left(5,3,2\right)$.