Q. 54 - Intermediate Algebra

Question

In the following exercises, solve the systems of equations by elimination.

$$\begin{gathered} 3x+8y=67 \\ 5x+3y=60 \end{gathered}$$

Step-by-Step Solution

Answer

The solution for the system of linear equation is $\left(9,5\right)$.

Step 1. Given Information

The system of linear equations:

$$\begin{gathered} 3x+8y=67 ........\left(1\right) \\ 5x+3y=60 ........\left(2\right) \end{gathered}$$

Step 2. Calculation

Here both equations are in standard form. To get opposite coefficients of y multiply first equation with 3.

$$\begin{gathered} 3\left(3x+8y\right)=3\left(67\right) \\ 9x+24y=201 ........\left(3\right) \end{gathered}$$

Also multiply second equation with $-8$.

$$\begin{gathered} -8\left(5x+3y\right)=-8\left(60\right) \\ -40x-24y=-480 ........\left(4\right) \end{gathered}$$

Solve for x:

From equations $\left(3\right)$and $\left(4\right)$:

$$\begin{gathered} 9x+24y=201 \\ -40x-24y=-480 \\ -31x=-279 \\ x=\frac{-279}{-31} \\ x=9 \end{gathered}$$

Solve for y:

Substitute $x=9$

in the equation $3x+8y=67$and solve for y.

$$\begin{gathered} 3x+8y=67 \\ 3\left(9\right)+8y=67 \\ 27+8y=67 \\ 8y=67-27 \\ 8y=40 \\ y=\frac{40}{8} \\ y=5 \end{gathered}$$

The solution for the system of linear equation is $\left(9,5\right)$.