Q. 52 - Intermediate Algebra

Question

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

$$\begin{gathered} 3x+8y=-3 \\ 2x+5y=-3 \end{gathered}$$

Step-by-Step Solution

Answer

The solution for the system of linear equation is $\left(-9,3\right)$.

Step 1. Given Information

System of linear equations:

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

Step 2. Calculation

Here both equations are in standard form. To get opposite coefficients of y multiply first equation with $-5$.

$$\begin{gathered} -5\left(3x+8y\right)=-5\left(-3\right) \\ -15x-40y=15 ..........\left(3\right) \end{gathered}$$ 

Also multiply second equation with 8.

$$\begin{gathered} 8\left(2x+5y\right)=8\left(-3\right) \\ 16x+40y=-24 ..........\left(4\right) \end{gathered}$$

Solve for x:

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

$$\begin{gathered} -15x-40y=15 \\ \underline{16x+40y=-24 } \\ x =-9 \\ x=-9 \end{gathered}$$

Solve for y:

Substitute $x=-9$

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

$$\begin{gathered} 3x+8y=-3 \\ 3\left(-9\right)+8y=-3 \\ -27+8y=-3 \\ 8y=27-3 \\ 8y=24 \\ y=\frac{24}{8} \\ y=3 \end{gathered}$$

The solution for the system of linear equation is $\left(-9,3\right)$.