Question 197168
Start with the given system of equations:


{{{system(3x+y=7,2x-5y=-1)}}}




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+y=7}}} Start with the first equation



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



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




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


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




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




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



{{{2x+15x-35=-1}}} Multiply



{{{17x-35=-1}}} Combine like terms on the left side



{{{17x=-1+35}}}Add 35 to both sides



{{{17x=34}}} Combine like terms on the right side



{{{x=(34)/(17)}}} Divide both sides by 17 to isolate x




{{{x=2}}} Divide






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



So the first part of our answer is: {{{x=2}}}










Since we know that {{{x=2}}} we can plug it into the equation {{{y=-3x+7}}} (remember we previously solved for {{{y}}} in the first equation).




{{{y=-3x+7}}} Start with the equation where {{{y}}} was previously isolated.



{{{y=-3(2)+7}}} Plug in {{{x=2}}}



{{{y=-6+7}}} Multiply



{{{y=1}}} Combine like terms 




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



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










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


So our answers are:


{{{x=2}}} and {{{y=1}}}


which form the point *[Tex \LARGE \left(2,1\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(2,1\right)]. This visually verifies our answer.





{{{
drawing(500, 500, -10,10,-10,10,
  graph(500, 500, -10,10,-10,10, (7-3*x)/(1), (-1-2*x)/(-5) ),
  blue(circle(2,1,0.1)),
  blue(circle(2,1,0.12)),
  blue(circle(2,1,0.15))
)
}}} 


Graph of {{{3x+y=7}}} (red) and {{{2x-5y=-1}}} (green)  and the intersection of the lines (blue circle).