Question

Solve by elimination: 3x + 4y = 0 5x – 3y = -58

Answer 1

Given the equation system

$\begin{cases}3x+4y=0 \\ 5x-3y=-58\end{cases}$

To solve the equation system by using the elimination method, the first step is to multiply one or both equations by a number, so that either the x-terms or y-terms of the equations have the same coefficient.

I will multiply the first equation by 5 and the second equation by 3

First equation:

$\begin{gathered} 5(3x+4y)=5\cdot0 \\ 5\cdot3x+5\cdot4y=0 \\ 15x+20y=0 \end{gathered}$

Second equation

$\begin{gathered} 3(5x-3y)=3\cdot(-58) \\ 3\cdot5x-3\cdot3y=3\cdot(-58) \\ 15x-9y=-174 \end{gathered}$

Now you have to subtract both expressions:

After subtracting both equations we have determined an expression with "y" as the only unknown. The next step is to solve the expression for y.

To do so, you have to divide both sides of the expression by 29

$\begin{gathered} 29y=174 \\ \frac{29y}{29}=\frac{174}{29} \\ y=6 \end{gathered}$

The value of y is 6.

Now that we have determined the value of y, you can replace it in either one of the original equations to determine the value of x:

$\begin{gathered} 3x+4y=0 \\ 3x+4\cdot6=0 \\ 3x+24=0 \\ 3x+24-24=0-24 \\ 3x=-24 \\ \frac{3x}{3}=-\frac{24}{3} \\ x=-8 \end{gathered}$

The solution of this equation system is x=-8 and y=6