Question 165085
I'm going to use substitution to solve the system.



Start with the given system

{{{2y-x=3}}}
{{{x=3y-5}}}




{{{2y-(3y-5)=3}}}  Plug in {{{x=3y-5}}} into the first equation. In other words, replace each {{{x}}} with {{{3y-5}}}. Notice we've eliminated the {{{x}}} variables. So we now have a simple equation with one unknown.



{{{2y-3y+5=3}}} Distribute



{{{-y+5=3}}} Combine like terms on the left side



{{{-y=3-5}}}Subtract 5 from both sides



{{{-y=-2}}} Combine like terms on the right side



{{{y=(-2)/(-1)}}} Divide both sides by -1 to isolate y




{{{y=2}}} Divide





Now that we know that {{{y=2}}}, we can plug this into {{{x=3y-5}}} to find {{{x}}}




{{{x=3(2)-5}}} Substitute {{{2}}} for each {{{y}}}



{{{x=1}}} Simplify



So our answer is {{{x=1}}} and {{{y=2}}} which also looks like *[Tex \LARGE \left(1,2\right)]




Notice if we graph the two equations (you have to solve for "y" for each equation first), we can see that their intersection is at *[Tex \LARGE \left(1,2\right)]. So this verifies our answer.



{{{ drawing(500, 500, -10, 10, -10, 10,
    grid(1),
    graph( 500, 500, -10, 10, -10, 10, (3+x)/2, (x+5)/3) 
)}}} Graph of {{{2y-x=3}}} (red) and {{{x=3y-5}}} (green)