Question 142855


Start with the given system of equations:


{{{system(6x+2y=10,5x-3y=27)}}}




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


{{{6x+2y=10}}} Start with the first equation



{{{2y=10-6x}}}  Subtract {{{6x}}} from both sides



{{{2y=-6x+10}}} Rearrange the equation



{{{y=(-6x+10)/(2)}}} Divide both sides by {{{2}}}



{{{y=((-6)/(2))x+(10)/(2)}}} Break up the fraction



{{{y=-3x+5}}} Reduce




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


Since {{{y=-3x+5}}}, we can now replace each {{{y}}} in the second equation with {{{-3x+5}}} to solve for {{{x}}}




{{{5x-3highlight((-3x+5))=27}}} Plug in {{{y=-3x+5}}} into the first equation. In other words, replace each {{{y}}} with {{{-3x+5}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{5x+(-3)(-3)x+(-3)(5)=27}}} Distribute {{{-3}}} to {{{-3x+5}}}



{{{5x+9x-15=27}}} Multiply



{{{14x-15=27}}} Combine like terms on the left side



{{{14x=27+15}}}Add 15 to both sides



{{{14x=42}}} Combine like terms on the right side



{{{x=(42)/(14)}}} Divide both sides by 14 to isolate x




{{{x=3}}} Divide






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



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










Since we know that {{{x=3}}} we can plug it into the equation {{{y=-3x+5}}} (remember we previously solved for {{{y}}} in the first equation).




{{{y=-3x+5}}} Start with the equation where {{{y}}} was previously isolated.



{{{y=-3(3)+5}}} Plug in {{{x=3}}}



{{{y=-9+5}}} Multiply



{{{y=-4}}} Combine like terms 




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



So the second part of our answer is: {{{y=-4}}}










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


So our answers are:


{{{x=3}}} and {{{y=-4}}}


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





{{{
drawing(500, 500, -10,10,-10,10,
  graph(500, 500, -10,10,-10,10, (10-6*x)/(2), (27-5*x)/(-3) ),
  blue(circle(3,-4,0.1)),
  blue(circle(3,-4,0.12)),
  blue(circle(3,-4,0.15))
)
}}} graph of {{{6x+2y=10}}} (red) and {{{5x-3y=27}}} (green)  and the intersection of the lines (blue circle).