Q 51. - Intermediate Algebra

Question

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

$$\begin{gathered} 4x-3y=3 \\ 2x+5y=-31 \end{gathered}$$

Step-by-Step Solution

Answer

The solution for the system of linear equation is $\left(-3,-5\right)$.

Step 1. Given Information

System of linear equations: 

$$\begin{gathered} 4x-3y=3 .........\left(1\right) \\ 2x+5y=-31 ........\left(2\right) \end{gathered}$$

Step 2. Calculation

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

$$\begin{gathered} 5\left(4x-3y\right)=5\left(3\right) \\ 20x-15y=15 ........\left(3\right) \end{gathered}$$

Also multiply second equation with 3.

$$\begin{gathered} 3\left(2x+5y\right)=3\left(-31\right) \\ 6x+15y=-93 .........\left(4\right) \end{gathered}$$

Solve for x:

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

$$\begin{gathered} 20x-15y=15 \\ \underline{ 6x+15y=-93} \\ 26x =-78 \\ x=-\frac{78}{26} \\ x=-3 \end{gathered}$$

Solve for y:

Substitute $x=-3$

in the equation $4x-3y=3$and solve for y.

$$\begin{gathered} 4x-3y=3 \\ 4\left(-3\right)-3y=3 \\ -12-3y=3 \\ -3y=3+12 \\ -3y=15 \\ y=\frac{15}{-3} \\ y=-5 \end{gathered}$$

The solution for the system of linear equation is $\left(-3,-5\right)$.