Question 150378



Start with the given system of equations:


{{{system(x-3y=7,2x-5y=13)}}}




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-3y=7}}} Start with the first equation



{{{-3y=7-x}}}  Subtract {{{x}}} from both sides



{{{-3y=-x+7}}} Rearrange the equation



{{{y=(-x+7)/(-3)}}} Divide both sides by {{{-3}}}



{{{y=((-1)/(-3))x+(7)/(-3)}}} Break up the fraction



{{{y=(1/3)x-7/3}}} Reduce




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


Since {{{y=(1/3)x-7/3}}}, we can now replace each {{{y}}} in the second equation with {{{(1/3)x-7/3}}} to solve for {{{x}}}




{{{2x-5highlight(((1/3)x-7/3))=13}}} Plug in {{{y=(1/3)x-7/3}}} into the first equation. In other words, replace each {{{y}}} with {{{(1/3)x-7/3}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{2x+(-5)(1/3)x+(-5)(-7/3)=13}}} Distribute {{{-5}}} to {{{(1/3)x-7/3}}}



{{{2x-(5/3)x+35/3=13}}} Multiply



{{{(3)(2x-(5/3)x+35/3)=(3)(13)}}} Multiply both sides by the LCM of 3. This will eliminate the fractions  (note: if you need help with finding the LCM, check out this <a href=http://www.algebra.com/algebra/homework/divisibility/least-common-multiple.solver>solver</a>)




{{{6x-5x+35=39}}} Distribute and multiply the LCM to each side




{{{x+35=39}}} Combine like terms on the left side



{{{x=39-35}}}Subtract 35 from both sides



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






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



So the first part of our answer is: {{{x=4}}}










Since we know that {{{x=4}}} we can plug it into the equation {{{y=(1/3)x-7/3}}} (remember we previously solved for {{{y}}} in the first equation).




{{{y=(1/3)x-7/3}}} Start with the equation where {{{y}}} was previously isolated.



{{{y=(1/3)(4)-7/3}}} Plug in {{{x=4}}}



{{{y=4/3-7/3}}} Multiply



{{{y=-1}}} Combine like terms and reduce.  (note: if you need help with fractions, check out this <a href="http://www.algebra.com/algebra/homework/NumericFractions/fractions-solver.solver">solver</a>)




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



So the second part of our answer is: {{{y=-1}}}










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


So our answers are:


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


which form the point *[Tex \LARGE \left(4,-1\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(4,-1\right)]. This visually verifies our answer.





{{{
drawing(500, 500, -10,10,-10,10,
  graph(500, 500, -10,10,-10,10, (7-1*x)/(-3), (13-2*x)/(-5) ),
  blue(circle(4,-1,0.1)),
  blue(circle(4,-1,0.12)),
  blue(circle(4,-1,0.15))
)
}}} graph of {{{x-3y=7}}} (red) and {{{2x-5y=13}}} (green)  and the intersection of the lines (blue circle).