Question 131038
 


Start with the given system of equations:


{{{system(5x+15y=10,2x+8y=6)}}}




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


{{{5x+15y=10}}} Start with the first equation



{{{15y=10-5x}}}  Subtract {{{5x}}} from both sides



{{{15y=-5x+10}}} Rearrange the equation



{{{y=(-5x+10)/(15)}}} Divide both sides by {{{15}}}



{{{y=((-5)/(15))x+(10)/(15)}}} Break up the fraction



{{{y=(-1/3)x+2/3}}} Reduce




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


Since {{{y=(-1/3)x+2/3}}}, we can now replace each {{{y}}} in the second equation with {{{(-1/3)x+2/3}}} to solve for {{{x}}}




{{{2x+8highlight(((-1/3)x+2/3))=6}}} Plug in {{{y=(-1/3)x+2/3}}} into the first equation. In other words, replace each {{{y}}} with {{{(-1/3)x+2/3}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{2x+(8)(-1/3)x+(8)(2/3)=6}}} Distribute {{{8}}} to {{{(-1/3)x+2/3}}}



{{{2x-(8/3)x+16/3=6}}} Multiply



{{{(3)(2x-(8/3)x+16/3)=(3)(6)}}} 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-8x+16=18}}} Distribute and multiply the LCM to each side




{{{-2x+16=18}}} Combine like terms on the left side



{{{-2x=18-16}}}Subtract 16 from both sides



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



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




{{{x=-1}}} Divide






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



So the first part of our answer is: {{{x=-1}}}










Since we know that {{{x=-1}}} we can plug it into the equation {{{y=(-1/3)x+2/3}}} (remember we previously solved for {{{y}}} in the first equation).




{{{y=(-1/3)x+2/3}}} Start with the equation where {{{y}}} was previously isolated.



{{{y=(-1/3)(-1)+2/3}}} Plug in {{{x=-1}}}



{{{y=1/3+2/3}}} 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=-1}}} and {{{y=1}}}


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





{{{
drawing(500, 500, -10,10,-10,10,
  graph(500, 500, -10,10,-10,10, (10-5*x)/(15), (6-2*x)/(8) ),
  blue(circle(-1,1,0.1)),
  blue(circle(-1,1,0.12)),
  blue(circle(-1,1,0.15))
)
}}} graph of {{{5x+15y=10}}} (red) and {{{2x+8y=6}}} (green)  and the intersection of the lines (blue circle).