Q. 46 - Intermediate Algebra

Question

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

$$\begin{gathered} 5x+2y=2 \\ -3x-y=0 \end{gathered}$$

Step-by-Step Solution

Answer

The solution for the system of linear equation is $\left(-2,6\right)$.

Step 1. Given Information

The system of linear equations is:

$$\begin{gathered} 5x+2y=2 ..........\left(1\right) \\ -3x-y=0 ..........\left(2\right) \end{gathered}$$

Step 2. Calculation

Here both equations are in standard form. To get opposite coefficients of y multiply second equation with 2.

$$\begin{gathered} 2\left(-3x-y\right)=0 \\ -6x-2y=0 ..........\left(3\right) \end{gathered}$$

Solve for x:

From equations $\left(1\right)$and $\left(3\right)$:

$$\begin{gathered} 5x+2y=2 \\ \underline{-6x-2y=0} \\ -x =2 \\ x=-2 \end{gathered}$$

Solve for y:

Substitute$x=-2$ 

in the equation $5x+2y=2$ and solve for y.

$$\begin{gathered} 5x+2y=2 \\ 5\left(-2\right)+2y=2 \\ -10+2y=2 \\ 2y=10+2 \\ 2y=12 \\ y=\frac{12}{2} \\ y=6 \end{gathered}$$