Q50. - Algebra 2

Question

Solve each equation.50 

$\left[\begin{array}{c}3\text{x}+2\\ 15\end{array}\right]=\left[\begin{array}{c}23\\ -4\text{y}-1\end{array}\right]$

Step-by-Step Solution

Answer

The answer is $(7,-4)$  that is$\text{x}=7$   and $\text{y}=-4$ .

Step 1 - State the definition of a matrix.

A rectangular array of numbers arranged in rows and columns is called a matrix.

Step 2 - Solve the equation.

Using the definition of equal matices is used to solve the equation.

 If the dimension of the matrices are same and corresponding elements are equal then the matrices are equal.

 Here the matrices are:

$\left[\begin{array}{c}3x+2\\ 15\end{array}\right]=\left[\begin{array}{c}23\\ -4y-1\end{array}\right]$

Since, the matrices are equal then corresponding elements are also equal.

 Then, the two linear equations formed are:

$\begin{array}{l}3x+2=23\\ -4y-1=15\end{array}$

Solve the equations to obtain values of x  and y .

$\begin{array}{c}3x+2=23\\ 3x+2-2=23-2\\ 3x=21\\ x=7\end{array}$

and $\begin{array}{c}-4y-1=15\\ -4y-1+1=15+1\\ -4y=16\\ y=-4\end{array}$

Step 3 - State the conclusion.

Therefore, $x=7$  and $y=-4$ . The solution is  $(7,-4)$