Question 170179
I'll do the first two to get you started



# 1



Start with the given system of equations:


{{{system(3x+y=7,4x-y=21)}}}




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}}}




{{{4x-highlight((-3x+7))=21}}} 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.




{{{4x+3x-7=21}}} Distribute the negative



{{{7x-7=21}}} Combine like terms on the left side



{{{7x=21+7}}}Add 7 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=-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(4)+7}}} Plug in {{{x=4}}}



{{{y=-12+7}}} Multiply



{{{y=-5}}} Combine like terms 




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



So the second part of our answer is: {{{y=-5}}}










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


So our answers are:


{{{x=4}}} and {{{y=-5}}}


which form the point *[Tex \LARGE \left(4,-5\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,-5\right)]. This visually verifies our answer.





{{{
drawing(500, 500, -10,10,-10,10,
  grid(1),
  graph(500, 500, -10,10,-10,10, (7-3*x)/(1), (21-4*x)/(-1) ),
  blue(circle(4,-5,0.1)),
  blue(circle(4,-5,0.12)),
  blue(circle(4,-5,0.15))
)
}}} graph of {{{3x+y=7}}} (red) and {{{4x-y=21}}} (green)  and the intersection of the lines (blue circle).




<hr>



# 2





Start with the given system of equations:

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



{{{-2(3x+5y)=-2(5)}}} Multiply the both sides of the second equation by -2.



{{{-6x-10y=-10}}} Distribute and multiply.



So we have the new system of equations:

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



Now add the equations together. You can do this by simply adding the two left sides and the two right sides separately like this:



{{{(6x+2y)+(-6x-10y)=(2)+(-10)}}}



{{{(6x+-6x)+(2y+-10y)=2+-10}}} Group like terms.



{{{0x+-8y=-8}}} Combine like terms. Notice how the x terms cancel out.



{{{-8y=-8}}} Simplify.



{{{y=(-8)/(-8)}}} Divide both sides by {{{-8}}} to isolate {{{y}}}.



{{{y=1}}} Reduce.



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



{{{6x+2y=2}}} Now go back to the first equation.



{{{6x+2(1)=2}}} Plug in {{{y=1}}}.



{{{6x+2=2}}} Multiply.



{{{6x=2-2}}} Subtract {{{2}}} from both sides.



{{{6x=0}}} Combine like terms on the right side.



{{{x=(0)/(6)}}} Divide both sides by {{{6}}} to isolate {{{x}}}.



{{{x=0}}} Reduce.



So our answer is {{{x=0}}} and {{{y=1}}}.



Which form the ordered pair *[Tex \LARGE \left(0,1\right)].



This means that the system is consistent and independent.



Notice when we graph the equations, we see that they intersect at *[Tex \LARGE \left(0,1\right)]. So this visually verifies our answer.



{{{drawing(500,500,-10,10,-9,11,
grid(1),
graph(500,500,-10,10,-9,11,(2-6x)/(2),(5-3x)/(5)),
circle(0,1,0.05),
circle(0,1,0.08),
circle(0,1,0.10)
)}}} Graph of {{{6x+2y=2}}} (red) and {{{3x+5y=5}}} (green)