Q90. - Algebra 1

Question

Use elimination to solve each system of equations.

$$\begin{gathered} \mathbf{4}\mathit{x}\mathbf{-}\mathbf{3}\mathit{y}\mathbf{=}\mathbf{12} \\ \mathbf{ }\mathbf{ }\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{14} \end{gathered}$$

Step-by-Step Solution

Answer

The solution of the given system of equations is $\left(6,4\right)$.

Step 1. Multiply.

Multiply each side $x+2y=14$ by 4.

$$\begin{gathered} 4\left(x+2y\right)=4\left(14\right) \\ 4x+8y=56 \end{gathered}$$

Step 2. Subtract.

Subtract $4x-3y=12$ from $4x+8y=56$. Simplify to obtain the value of y.

$$\begin{gathered} \left(4x+8y\right)-\left(4x-3y\right)=56-12 \\ 11y=44 \\ \frac{11y}{11}=\frac{44}{11} \\ y=4 \end{gathered}$$

Step 3. Substitute.

Substitute 4 for y into $x+2y=14$ and simplify.

$$\begin{gathered} x+2\left(4\right)=14 \\ x+8-8=14-8 \\ x=6 \end{gathered}$$ 

 

Therefore, the solution of the given system of equations is $\left(6,4\right)$.