Question 145383
Well it's hard to tell what you got if you don't post your solutions =)



I'll do one of each to help you in the right direction.



<a name="one">
# 1

<a href="#three">Jump to problem #3</a>


Start with the given system of equations:


{{{system(x+y=15,4x+3y=38)}}}




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=15}}} Start with the first equation



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



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



{{{y=(-x+15)/(1)}}} Divide both sides by {{{1}}}



{{{y=((-1)/(1))x+(15)/(1)}}} Break up the fraction



{{{y=-x+15}}} Reduce




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


Since {{{y=-x+15}}}, we can now replace each {{{y}}} in the second equation with {{{-x+15}}} to solve for {{{x}}}




{{{4x+3highlight((-x+15))=38}}} Plug in {{{y=-x+15}}} into the first equation. In other words, replace each {{{y}}} with {{{-x+15}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{4x+(3)(-1)x+(3)(15)=38}}} Distribute {{{3}}} to {{{-x+15}}}



{{{4x-3x+45=38}}} Multiply



{{{x+45=38}}} Combine like terms on the left side



{{{x=38-45}}}Subtract 45 from both sides



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






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



So the first part of our answer is: {{{x=-7}}}










Since we know that {{{x=-7}}} we can plug it into the equation {{{y=-x+15}}} (remember we previously solved for {{{y}}} in the first equation).




{{{y=-x+15}}} Start with the equation where {{{y}}} was previously isolated.



{{{y=-(-7)+15}}} Plug in {{{x=-7}}}



{{{y=7+15}}} Multiply



{{{y=22}}} Combine like terms 




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



So the second part of our answer is: {{{y=22}}}










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


So our answers are:


{{{x=-7}}} and {{{y=22}}}


which form the point *[Tex \LARGE \left(-7,22\right)] 








<hr>





<a name="three">
# 3

<a href="#one">Jump to problem #1</a>






Start with the given system of equations:


{{{system(5x-y=12,3x+y=4)}}}




Now in order to solve this system by using elimination/addition, we need to solve (or isolate) one variable. I'm going to solve for y.






In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for {{{y}}}, we would have to eliminate {{{x}}} (or vice versa).



So lets eliminate {{{x}}}. In order to do that, we need to have both {{{x}}} coefficients that are equal in magnitude but have opposite signs (for instance 2 and -2 are equal in magnitude but have opposite signs). This way they will add to zero. By adding to zero, they can be eliminated.




So to make the {{{x}}} coefficients equal in magnitude but opposite in sign, we need to multiply both {{{x}}} coefficients by some number to get them to an common number. So if we wanted to get {{{5}}} and {{{3}}} to some equal number, we could try to get them to the LCM.




Since the LCM of {{{5}}} and {{{3}}} is {{{15}}}, we need to multiply both sides of the top equation by {{{3}}} and multiply both sides of the bottom equation by {{{-5}}} like this:





{{{3(5x-y)=3(12)}}}  Multiply the top equation (both sides) by {{{3}}}
{{{-5(3x+y)=-5(4)}}}  Multiply the bottom equation (both sides) by {{{-5}}}





Distribute and multiply


{{{15x-3y=36}}}
{{{-15x-5y=-20}}}



Now add the equations together. In order to add 2 equations, group like terms and combine them


{{{(15x-15x)+(-3y-5y)=36-20}}}


Combine like terms and simplify




{{{cross(15x-15x)-8y=16}}} Notice how the x terms cancel out





{{{-8y=16}}} Simplify





{{{y=16/-8}}} Divide both sides by {{{-8}}} to isolate y





{{{y=-2}}} Reduce




Now plug this answer into the top equation {{{5x-y=12}}} to solve for x


{{{5x-y=12}}} Start with the first equation




{{{5x-(-2)=12}}} Plug in {{{y=-2}}}





{{{5x+2=12}}} Multiply




{{{5x=12-2}}}Subtract 2 from both sides



{{{5x=10}}} Combine like terms on the right side



{{{x=(10)/(5)}}} Divide both sides by 5 to isolate x




{{{x=2}}} Divide





So our answer is

{{{x=2}}} and {{{y=-2}}}




which also looks like *[Tex \LARGE \left(2,-2\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,-2\right)]. This visually verifies our answer.





{{{
drawing(500, 500, -10,10,-10,10,
  graph(500, 500, -10,10,-10,10, (12-5*x)/(-1), (4-3*x)/(1) ),
  blue(circle(2,-2,0.1)),
  blue(circle(2,-2,0.12)),
  blue(circle(2,-2,0.15))
)
}}} graph of {{{5x-y=12}}} (red) and {{{3x+y=4}}} (green)  and the intersection of the lines (blue circle).