Question 174547
I'll do the first two to get you started



a)



Start with the given system of equations:


{{{system(x+y=6,2x+y=8)}}}



Let's solve the system by substitution



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=6}}} Start with the first equation



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



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



{{{y=(-x+6)/(1)}}} Divide both sides by {{{1}}}



{{{y=((-1)/(1))x+(6)/(1)}}} Break up the fraction



{{{y=-x+6}}} Reduce




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


Since {{{y=-x+6}}}, we can now replace each {{{y}}} in the second equation with {{{-x+6}}} to solve for {{{x}}}




{{{2x+highlight((-x+6))=8}}} Plug in {{{y=-x+6}}} into the second equation. In other words, replace each {{{y}}} with {{{-x+6}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{x+6=8}}} Combine like terms on the left side



{{{x=8-6}}}Subtract 6 from both sides



{{{x=2}}} Combine like terms on the right side






-----------------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=-x+6}}} (remember we previously solved for {{{y}}} in the first equation).




{{{y=-x+6}}} Start with the equation where {{{y}}} was previously isolated.



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



{{{y=-2+6}}} Multiply



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




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



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










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


So our answers are:


{{{x=2}}} and {{{y=4}}}


which form the point *[Tex \LARGE \left(2,4\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(2,4\right)]. This visually verifies our answer.





{{{
drawing(500, 500, -10,10,-10,10,
  graph(500, 500, -10,10,-10,10, (6-1*x)/(1), (8-2*x)/(1) ),
  blue(circle(2,4,0.1)),
  blue(circle(2,4,0.12)),
  blue(circle(2,4,0.15))
)
}}} graph of {{{x+y=6}}} (red) and {{{2x+y=8}}} (green)  and the intersection of the lines (blue circle).





<hr>



b)



Start with the given system of equations:

{{{system(7x+3y=14,5x+9y=10)}}}



{{{-3(7x+3y)=-3(14)}}} Multiply the both sides of the first equation by -3.



{{{-21x-9y=-42}}} Distribute and multiply.



So we have the new system of equations:

{{{system(-21x-9y=-42,5x+9y=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:



{{{(-21x-9y)+(5x+9y)=(-42)+(10)}}}



{{{(-21x+5x)+(-9y+9y)=-42+10}}} Group like terms.



{{{-16x+0y=-32}}} Combine like terms. Notice how the y terms cancel out.



{{{-16x=-32}}} Simplify.



{{{x=(-32)/(-16)}}} Divide both sides by {{{-16}}} to isolate {{{x}}}.



{{{x=2}}} Reduce.



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



{{{-21x-9y=-42}}} Now go back to the first equation.



{{{-21(2)-9y=-42}}} Plug in {{{x=2}}}.



{{{-42-9y=-42}}} Multiply.



{{{-9y=-42+42}}} Add {{{42}}} to both sides.



{{{-9y=0}}} Combine like terms on the right side.



{{{y=(0)/(-9)}}} Divide both sides by {{{-9}}} to isolate {{{y}}}.



{{{y=0}}} Reduce.



So our answer is {{{x=2}}} and {{{y=0}}}.



Which form the ordered pair *[Tex \LARGE \left(2,0\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(2,0\right)]. So this visually verifies our answer.



{{{drawing(500,500,-8,12,-10,10,
grid(1),
graph(500,500,-8,12,-10,10,(14-7x)/(3),(10-5x)/(9)),
circle(2,0,0.05),
circle(2,0,0.08),
circle(2,0,0.10)
)}}} Graph of {{{7x+3y=14}}} (red) and {{{5x+9y=10}}} (green)