Question

How do you solve the system by elimination $5x+4y=12$ and $3x-3y=18$?

Answer 1

Evaluating the given equations:

The given equations are:

$5x+4y=12......\left(1\right)$

$3x-3y=18......\left(2\right)$

Multiply equation $\left(1\right)$ with $3$ and equation $\left(2\right)$ with $-5$ and adding the equation $\left(2\right)$ and equation $\left(1\right)$ we get

$$\begin{gathered} \Rightarrow 5x+4y=12\left[\times 3\right] \\ \Rightarrow 3x-3y=18\left[\times -5\right] \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ \Rightarrow 15x+12y=36 \\ \Rightarrow -15x+15y=-90 \\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ \Rightarrow 27y=-54 \\ \Rightarrow y=\frac{-54}{27} \\ \Rightarrow y=-2 \end{gathered}$$

Substitute the value $y=-2$ in the equation $\left(1\right)$ we get

$$\begin{gathered} \Rightarrow 5x+4y=12 \\ \Rightarrow 5x+4\left(-2\right)=12 \\ \Rightarrow 5x-8=12 \\ \Rightarrow 5x=12+8 \\ \Rightarrow 5x=20 \\ \Rightarrow x=\frac{20}{5} \\ \Rightarrow x=4 \end{gathered}$$

Hence, the value of $x=4$ and $y=-2$.