Question 811699
To solve a system by substitution, you start by finding an equation that is a formula to calculate one variable from the other. It would start with
{{{y}}}{{{"=..."}}}
or {{{x}}}{{{"=..."}}}
If there is no equation like that, you "make" one by solving for {{{y}}} or {{{x}}} in one of the equations.
 
Your system has an equation like that. (In fact, it has two of those).
If {{{y = (2/3)x - 1}}} , you can substitute the expression {{{(2/3)x - 1}}} for {{{y}}} in the other equation to get
{{{(2/3)x - 1= -x+4}}}
Now, you just solve for {{{x}}} , to start.
Adding {{{x}}} to both sides, you get
{{{x+(2/3)x - 1=x-x+4}}}
{{{x+(2/3)x - 1=4}}}
Adding {{{1}}} to both sides, you get
{{{x+(2/3)x - 1=4}}}
{{{x+(2/3)x - 1+1=4+1}}}
{{{x+(2/3)x=5}}}
{{{(1+2/3)x=5}}}
{{{(5/3)x=5}}}
Now, multiplying both sides times {{{3/5}}} ,
or dividing both sides by {{{5/3}}} (same thing), you find that
{{{(3/5)(5/3)x=5(3/5)}}}
{{{x=5(3/5)}}}
{{{highlight(x=3)}}}
Now you go back to the equation that had {{{y}}}{{{"=..."}}} (any of the two) and use it to calculate {{{y}}} from the {{{x=3}}} you just found.
Substituting {{{3}}} for {{{x}}} in {{{y=-x+4}}} you find
{{{y=-3+4}}}
{{{highlight(y=1)}}}