Question 111841
Start with the given system

{{{x-y=48}}}
{{{y=5x-8}}}




{{{x-(5x-8)=48}}}  Plug in {{{y=5x-8}}} into the first equation. In other words, replace each {{{y}}} with {{{5x-8}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.



{{{x-5x+8=48}}} Distribute the negative



{{{-4x+8=48}}} Combine like terms on the left side



{{{-4x=48-8}}}Subtract 8 from both sides



{{{-4x=40}}} Combine like terms on the right side



{{{x=(40)/(-4)}}} Divide both sides by -4 to isolate x




{{{x=-10}}} Divide





Now that we know that {{{x=-10}}}, we can plug this into {{{y=5x-8}}} to find {{{y}}}




{{{y=5(-10)-8}}} Substitute {{{-10}}} for each {{{x}}}



{{{y=-58}}} Simplify



So our answer is {{{x=-10}}} and {{{y=-58}}}


Check:


{{{x-y=48}}} Start with the first equation



{{{-10-(-58)=48}}} Plug in {{{x=-10}}} and {{{y=-58}}}



{{{48=48}}} works