Using the substitution method, solve the simultaneous equation

Question

$$ (x+y)/3=(x-y)/5=2x-3y+5 $$

I tried using the elimination method to get rid of the fractions.

And ended up with $$ 5x+5y=3x-3y=30x-45y+75 $$ After getting this, I’m lost and don’t know what to do next.

Answer 1

First of all you can manipulare the first two equation as follow: $$5x+5y=3x-3y\implies 2x=-8y\implies \color{blue}{y=-\frac 14x}\tag{1}$$ Then you have to solve the second inequality: $$3x-3y=30x-45y+75\implies 27x-42y+75=0\implies \color{green}{y=\frac{75+27x}{42}}\tag{2}$$

Now you have to match $(1)$ and $(2)$ in order to obtain the value for $x$: $$-\frac 14x=\frac{75+27x}{42}\implies x=-2$$ Substituting it into $(1)$ you get: $$y=\frac 12$$

This means that the solution of your system is the point $\color{red}{\big(-2,\dfrac 12\big)}$.

Answer 2

Geometrically, we can take this triple equation as 3 bundle lines intersecting at one point, meanwhile 3 below family of lines: $$(1) \ 5x+5y=k_1\\(2)\ 3x-3y=k_2\\(3) \ 30x-45y+75=k_3$$from $5x+5y=3x-3y$ we get $x=-4y$ and by substituting in $5x+5y=30x-45y+75$ we obtain $x=-2,y=-\frac{1}{2}$. A aketch of this solution is as following: