Question 122873
If "The sum of two numbers is 38", then the first equation is {{{x+y=38}}}




If "Their difference is 12", then the second equation is {{{x-y=12}}}




So we have the system:



{{{system(x+y=38,x-y=12)}}}




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=38}}} Start with the first equation



{{{y=38-x}}}  Subtract {{{x}}} from both sides



{{{y=-x+38}}} Rearrange the equation





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Since {{{y=-x+38}}}, we can now replace each {{{y}}} in the second equation with {{{-x+38}}} to solve for {{{x}}}




{{{x-highlight((-x+38))=12}}} Plug in {{{y=-x+38}}} into the first equation. In other words, replace each {{{y}}} with {{{-x+38}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{x+x-38=12}}} Distribute the negative



{{{2x-38=12}}} Combine like terms on the left side



{{{2x=12+38}}}Add 38 to both sides



{{{2x=50}}} Combine like terms on the right side



{{{x=(50)/(2)}}} Divide both sides by 2 to isolate x




{{{x=25}}} Divide






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



So the first part of our answer is: {{{x=25}}}










Since we know that {{{x=25}}} we can plug it into the equation {{{y=-x+38}}} (remember we previously solved for {{{y}}} in the first equation).




{{{y=-x+38}}} Start with the equation where {{{y}}} was previously isolated.



{{{y=-(25)+38}}} Plug in {{{x=25}}}



{{{y=-25+38}}} Multiply



{{{y=13}}} Combine like terms 




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



So the second part of our answer is: {{{y=13}}}










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


So our answers are:


{{{x=25}}} and {{{y=13}}}




Answer:



So the two numbers are 25 and 13