Q91. - Algebra 1

Question

Use elimination to solve each system of equations.

$$\begin{gathered} \mathbf{ }\mathbf{2}x-\mathbf{3}y=\mathbf{2} \\ \mathbf{5}\mathbf{x}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{28} \end{gathered}$$

Step-by-Step Solution

Answer

The solution of the given system of equations is $\left(4,2\right)$.

Step 1. Multiply.

Multiply each side $2x-3y=2$ by 5.

$$\begin{gathered} 5\left(2x-3y\right)=5\left(2\right) \\ 10x-15y=10 \end{gathered}$$

Step 2. Multiply.

Multiply each side $5x+4y=28$ by 2.

$$\begin{gathered} 2\left(5x+4y\right)=2\left(28\right) \\ 10x+8y=56 \end{gathered}$$

Step 3. Subtract.

Subtract $10x-15y=10$ from $10x+8y=56$. Simplify to obtain the value of y.

$$\begin{gathered} \left(10x+8y\right)-\left(10x-15y\right)=56-10 \\ 23y=46 \\ \frac{23y}{23}=\frac{46}{23} \\ y=2 \end{gathered}$$

Step 4. Substitute.

Substitute 2 for y into $2x-3y=2$ and simplify.

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

Therefore, the solution of the given system of equations is $\left(4,2\right)$.