Question 124907



Start with the given system of equations:


{{{system(3x-4y=8,6x-2y=10)}}}




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


{{{3x-4y=8}}} Start with the first equation



{{{-4y=8-3x}}}  Subtract {{{3x}}} from both sides



{{{-4y=-3x+8}}} Rearrange the equation



{{{y=(-3x+8)/(-4)}}} Divide both sides by {{{-4}}}



{{{y=((-3)/(-4))x+(8)/(-4)}}} Break up the fraction



{{{y=(3/4)x-2}}} Reduce




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


Since {{{y=(3/4)x-2}}}, we can now replace each {{{y}}} in the second equation with {{{(3/4)x-2}}} to solve for {{{x}}}




{{{6x-2highlight(((3/4)x-2))=10}}} Plug in {{{y=(3/4)x-2}}} into the first equation. In other words, replace each {{{y}}} with {{{(3/4)x-2}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{6x+(-2)(3/4)x+(-2)(-2)=10}}} Distribute {{{-2}}} to {{{(3/4)x-2}}}



{{{6x-(6/4)x+4=10}}} Multiply



{{{(4)(6x-(6/4)x+4)=(4)(10)}}} Multiply both sides by the LCM of 4. 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>)




{{{24x-6x+16=40}}} Distribute and multiply the LCM to each side




{{{18x+16=40}}} Combine like terms on the left side



{{{18x=40-16}}}Subtract 16 from both sides



{{{18x=24}}} Combine like terms on the right side



{{{x=(24)/(18)}}} Divide both sides by 18 to isolate x




{{{x=4/3}}} Reduce






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



So the first part of our answer is: {{{x=4/3}}}










Since we know that {{{x=4/3}}} we can plug it into the equation {{{y=(3/4)x-2}}} (remember we previously solved for {{{y}}} in the first equation).




{{{y=(3/4)x-2}}} Start with the equation where {{{y}}} was previously isolated.



{{{y=(3/4)(4/3)-2}}} Plug in {{{x=4/3}}}



{{{y=12/12-2}}} 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=4/3}}} and {{{y=-1}}}


which form the point *[Tex \LARGE \left(\frac{4}{3},-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(\frac{4}{3},-1\right)]. This visually verifies our answer.





{{{
drawing(500, 500, -10,10,-10,10,
  graph(500, 500, -10,10,-10,10, (8-3*x)/(-4), (10-6*x)/(-2) ),
  blue(circle(4/3,-1,0.1)),
  blue(circle(4/3,-1,0.12)),
  blue(circle(4/3,-1,0.15))
)
}}} graph of {{{3x-4y=8}}} (red) and {{{6x-2y=10}}} (green)  and the intersection of the lines (blue circle).