Question 113282
Start with the given system

{{{4x-y=19}}}
{{{y=x-1}}}




{{{4x-(x-1)=19}}}  Plug in {{{y=x-1}}} into the first equation. In other words, replace each {{{y}}} with {{{x-1}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.



{{{4x-x+1=19}}} Distribute the negative



{{{3x+1=19}}} Combine like terms on the left side



{{{3x=19-1}}}Subtract 1 from both sides



{{{3x=18}}} Combine like terms on the right side



{{{x=(18)/(3)}}} Divide both sides by 3 to isolate x




{{{x=6}}} Divide





Now that we know that {{{x=6}}}, we can plug this into {{{y=x-1}}} to find {{{y}}}




{{{y=(6)-1}}} Substitute {{{6}}} for each {{{x}}}



{{{y=5}}} Simplify



So our answer is {{{x=6}}} and {{{y=5}}}