Question 166627
Note: I'm going to make the first equation {{{x+y = 80}}} and the second equation {{{4x+9y=395}}} (it is valid to relabel the equations)





Start with the given system of equations:


{{{system(x+y=80,4x+9y=395)}}}




Now in order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to solve for y.





So let's isolate y in the first equation


{{{x+y=80}}} Start with the first equation



{{{y=80-x}}}  Subtract {{{x}}} from both sides



{{{y=-x+80}}} Rearrange the equation



---------------------


Since {{{y=-x+80}}}, we can now replace each {{{y}}} in the second equation with {{{-x+80}}} to solve for {{{x}}}




{{{4x+9highlight((-x+80))=395}}} Plug in {{{y=-x+80}}} into the second equation. In other words, replace each {{{y}}} with {{{-x+80}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{4x+(9)(-1)x+(9)(80)=395}}} Distribute {{{9}}} to {{{-x+80}}}



{{{4x-9x+720=395}}} Multiply



{{{-5x+720=395}}} Combine like terms on the left side



{{{-5x=395-720}}}Subtract 720 from both sides



{{{-5x=-325}}} Combine like terms on the right side



{{{x=(-325)/(-5)}}} Divide both sides by -5 to isolate x




{{{x=65}}} Divide






-----------------First Answer------------------------------



So the first part of our answer is: {{{x=65}}}










Since we know that {{{x=65}}} we can plug it into the equation {{{y=-x+80}}} (remember we previously solved for {{{y}}} in the first equation).




{{{y=-x+80}}} Start with the equation where {{{y}}} was previously isolated.



{{{y=-(65)+80}}} Plug in {{{x=65}}}



{{{y=-65+80}}} Multiply



{{{y=15}}} Combine like terms 




-----------------Second Answer------------------------------



So the second part of our answer is: {{{y=15}}}










-----------------Summary------------------------------


So our answers are:


{{{x=65}}} and {{{y=15}}}


which form the ordered pair *[Tex \LARGE \left(65,15\right)]