Question 127865
I'll do the first two to get you started



# 1





Start with the given system of equations:


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




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+y=1}}} Start with the first equation



{{{y=1-x}}}  Subtract {{{x}}} from both sides



{{{y=-x+1}}} Rearrange the equation



{{{y=(-x+1)/(1)}}} Divide both sides by {{{1}}}



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



{{{y=-x+1}}} Reduce




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


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




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




{{{2x+x-1=-2}}} Distribute the negative



{{{3x-1=-2}}} Combine like terms on the left side



{{{3x=-2+1}}}Add 1 to both sides



{{{3x=-1}}} Combine like terms on the right side



{{{x=(-1)/(3)}}} Divide both sides by 3 to isolate x




{{{x=-1/3}}} Reduce






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



So the first part of our answer is: {{{x=-1/3}}}










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




{{{y=-x+1}}} Start with the equation where {{{y}}} was previously isolated.



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



{{{y=1/3+1}}} Multiply



{{{y=4/3}}} Combine like terms  (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=4/3}}}










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


So our answers are:


{{{x=-1/3}}} and {{{y=4/3}}}


which form the point *[Tex \LARGE \left(-\frac{1}{3},\frac{4}{3}\right)] 





<hr>




# 2






Start with the given system of equations:


{{{system(5x-3y=-11,x-2y=2)}}}




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-3y=-11}}} Start with the first equation



{{{-3y=-11-5x}}}  Subtract {{{5x}}} from both sides



{{{-3y=-5x-11}}} Rearrange the equation



{{{y=(-5x-11)/(-3)}}} Divide both sides by {{{-3}}}



{{{y=((-5)/(-3))x+(-11)/(-3)}}} Break up the fraction



{{{y=(5/3)x+11/3}}} Reduce




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


Since {{{y=(5/3)x+11/3}}}, we can now replace each {{{y}}} in the second equation with {{{(5/3)x+11/3}}} to solve for {{{x}}}




{{{x-2highlight(((5/3)x+11/3))=2}}} Plug in {{{y=(5/3)x+11/3}}} into the first equation. In other words, replace each {{{y}}} with {{{(5/3)x+11/3}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{x+(-2)(5/3)x+(-2)(11/3)=2}}} Distribute {{{-2}}} to {{{(5/3)x+11/3}}}



{{{x-(10/3)x-22/3=2}}} Multiply



{{{(3)(1x-(10/3)x-22/3)=(3)(2)}}} 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>)




{{{3x-10x-22=6}}} Distribute and multiply the LCM to each side




{{{-7x-22=6}}} Combine like terms on the left side



{{{-7x=6+22}}}Add 22 to both sides



{{{-7x=28}}} Combine like terms on the right side



{{{x=(28)/(-7)}}} Divide both sides by -7 to isolate x




{{{x=-4}}} Divide






-----------------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=(5/3)x+11/3}}} (remember we previously solved for {{{y}}} in the first equation).




{{{y=(5/3)x+11/3}}} Start with the equation where {{{y}}} was previously isolated.



{{{y=(5/3)(-4)+11/3}}} Plug in {{{x=-4}}}



{{{y=-20/3+11/3}}} Multiply



{{{y=-3}}} 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=-3}}}










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


So our answers are:


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


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





{{{
drawing(500, 500, -10,10,-10,10,
  graph(500, 500, -10,10,-10,10, (-11-5*x)/(-3), (2-1*x)/(-2) ),
  blue(circle(-4,-3,0.1)),
  blue(circle(-4,-3,0.12)),
  blue(circle(-4,-3,0.15))
)
}}} graph of {{{5x-3y=-11}}} (red) and {{{x-2y=2}}} (green)  and the intersection of the lines (blue circle).