Question 134113



Start with the given system of equations:


{{{system(x+y=20,30x+25y=550)}}}




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=20}}} Start with the first equation



{{{y=20-x}}}  Subtract {{{x}}} from both sides



{{{y=-x+20}}} Rearrange the equation



{{{y=(-x+20)/(1)}}} Divide both sides by {{{1}}}



{{{y=((-1)/(1))x+(20)/(1)}}} Break up the fraction



{{{y=-x+20}}} Reduce




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


Since {{{y=-x+20}}}, we can now replace each {{{y}}} in the second equation with {{{-x+20}}} to solve for {{{x}}}




{{{30x+25highlight((-x+20))=550}}} Plug in {{{y=-x+20}}} into the first equation. In other words, replace each {{{y}}} with {{{-x+20}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{30x+(25)(-1)x+(25)(20)=550}}} Distribute {{{25}}} to {{{-x+20}}}



{{{30x-25x+500=550}}} Multiply



{{{5x+500=550}}} Combine like terms on the left side



{{{5x=550-500}}}Subtract 500 from both sides



{{{5x=50}}} Combine like terms on the right side



{{{x=(50)/(5)}}} Divide both sides by 5 to isolate x




{{{x=10}}} Divide






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



So the first part of our answer is: {{{x=10}}}










Since we know that {{{x=10}}} we can plug it into the equation {{{y=-x+20}}} (remember we previously solved for {{{y}}} in the first equation).




{{{y=-x+20}}} Start with the equation where {{{y}}} was previously isolated.



{{{y=-(10)+20}}} Plug in {{{x=10}}}



{{{y=-10+20}}} Multiply



{{{y=10}}} Combine like terms 




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



So the second part of our answer is: {{{y=10}}}










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


So our answers are:


{{{x=10}}} and {{{y=10}}}


which form the point *[Tex \LARGE \left(10,10\right)] 









Now let's graph the two equations (if you need help with graphing, check out this <a href=http://www.algebra.com/algebra/homework/Linear-equations/graphing-linear-equations.solver>solver</a>)



From the graph, we can see that the two equations intersect at *[Tex \LARGE \left(10,10\right)]. This visually verifies our answer.





{{{
drawing(500, 500, -15,15,-15,15,
  graph(500, 500, -15,15,-15,15, (20-1*x)/(1), (550-30*x)/(25) ),
  blue(circle(10,10,0.1)),
  blue(circle(10,10,0.12)),
  blue(circle(10,10,0.15))
)
}}} graph of {{{x+y=20}}} (red) and {{{30x+25y=550}}} (green)  and the intersection of the lines (blue circle).