Solving $5x=6y-4$ and $2y=3x+4$ system of linear equations.

Question

I'm given to solve the following system of equations.

$5x=6y-4$,

$2y=3x+4$

Here's what I'm doing. Rearranging the equations with variables on left.

$5x-6y=-4$ and $2y-3x=4$

I then solve for $y$ in the second equation.

$2y-3x=4 \therefore 2y=4+3x \therefore y = \frac{4+3x}{2}$

I plug this $y$ back into the second equation and solve for $x$

$2\left(\frac{4+3x}{2}\right)-3x=4 \therefore 4+3x-3x=4 \therefore 4=4$

It seems like my $x$ variable simply cancelled out, what does this mean?

Answer 1

You got $y=\frac{4+3x}2$ from the second equation, which is correct. And now, there's nothing more that you can get from that equation. Plug $y=\frac{4+3x}2$ in the first equation now.

Answer 2

You can write the system in the form $$5x-6y=-4$$ $$-3x+2y=4$$ Multiplying the second equation by $3$ and adding to the first we get $$-4x=8$$ so $$x=-2$$