Question 129270
Start with the given system

{{{2x-3y=12}}}
{{{x=4y+1}}}




{{{2(4y+1)-3y=12}}}  Plug in {{{x=4y+1}}} into the first equation. In other words, replace each {{{x}}} with {{{4y+1}}}. Notice we've eliminated the {{{x}}} variables. So we now have a simple equation with one unknown.



{{{8y+2-3y=12}}} Distribute



{{{5y+2=12}}} Combine like terms on the left side



{{{5y=12-2}}}Subtract 2 from both sides



{{{5y=10}}} Combine like terms on the right side



{{{y=(10)/(5)}}} Divide both sides by 5 to isolate y




{{{y=2}}} Divide





Now that we know that {{{y=2}}}, we can plug this into {{{x=4y+1}}} to find {{{x}}}




{{{x=4(2)+1}}} Substitute {{{2}}} for each {{{y}}}



{{{x=9}}} Simplify



So our answer is {{{x=9}}} and {{{y=2}}} which also looks like *[Tex \LARGE \left(9,2\right)]




Notice if we graph the two equations, we can see that their intersection is at *[Tex \LARGE \left(9,2\right)]. So this verifies our answer.



{{{ graph( 500, 500, -10, 10, -10, 10, (12-2x)/(-3), (x-1)/(4)) }}} Graph of {{{2x-3y=12}}} (red) and {{{x=4y+1}}} (green)