Question 134322
I'm assuming that the first two are the first system of equations. Do you want to use substitution?


# 1





Start with the given system of equations:


{{{system(3x+2y=12,5x-2y=4)}}}




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+2y=12}}} Start with the first equation



{{{2y=12-3x}}}  Subtract {{{3x}}} from both sides



{{{2y=-3x+12}}} Rearrange the equation



{{{y=(-3x+12)/(2)}}} Divide both sides by {{{2}}}



{{{y=((-3)/(2))x+(12)/(2)}}} Break up the fraction



{{{y=(-3/2)x+6}}} Reduce




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


Since {{{y=(-3/2)x+6}}}, we can now replace each {{{y}}} in the second equation with {{{(-3/2)x+6}}} to solve for {{{x}}}




{{{5x-2highlight(((-3/2)x+6))=4}}} Plug in {{{y=(-3/2)x+6}}} into the first equation. In other words, replace each {{{y}}} with {{{(-3/2)x+6}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{5x+(-2)(-3/2)x+(-2)(6)=4}}} Distribute {{{-2}}} to {{{(-3/2)x+6}}}



{{{5x+(6/2)x-12=4}}} Multiply



{{{(2)(5x+(6/2)x-12)=(2)(4)}}} Multiply both sides by the LCM of 2. 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>)




{{{10x+6x-24=8}}} Distribute and multiply the LCM to each side




{{{16x-24=8}}} Combine like terms on the left side



{{{16x=8+24}}}Add 24 to both sides



{{{16x=32}}} Combine like terms on the right side



{{{x=(32)/(16)}}} Divide both sides by 16 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=(-3/2)x+6}}} (remember we previously solved for {{{y}}} in the first equation).




{{{y=(-3/2)x+6}}} Start with the equation where {{{y}}} was previously isolated.



{{{y=(-3/2)(2)+6}}} Plug in {{{x=2}}}



{{{y=-6/2+6}}} 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=2}}} and {{{y=3}}}


which form the point *[Tex \LARGE \left(2,3\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(2,3\right)]. This visually verifies our answer.





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




<hr>


# 2






Start with the given system of equations:


{{{system(3x+5y=26,2x+3y=16)}}}




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+5y=26}}} Start with the first equation



{{{5y=26-3x}}}  Subtract {{{3x}}} from both sides



{{{5y=-3x+26}}} Rearrange the equation



{{{y=(-3x+26)/(5)}}} Divide both sides by {{{5}}}



{{{y=((-3)/(5))x+(26)/(5)}}} Break up the fraction



{{{y=(-3/5)x+26/5}}} Reduce




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


Since {{{y=(-3/5)x+26/5}}}, we can now replace each {{{y}}} in the second equation with {{{(-3/5)x+26/5}}} to solve for {{{x}}}




{{{2x+3highlight(((-3/5)x+26/5))=16}}} Plug in {{{y=(-3/5)x+26/5}}} into the first equation. In other words, replace each {{{y}}} with {{{(-3/5)x+26/5}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{2x+(3)(-3/5)x+(3)(26/5)=16}}} Distribute {{{3}}} to {{{(-3/5)x+26/5}}}



{{{2x-(9/5)x+78/5=16}}} Multiply



{{{(5)(2x-(9/5)x+78/5)=(5)(16)}}} Multiply both sides by the LCM of 5. 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>)




{{{10x-9x+78=80}}} Distribute and multiply the LCM to each side




{{{x+78=80}}} Combine like terms on the left side



{{{x=80-78}}}Subtract 78 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=(-3/5)x+26/5}}} (remember we previously solved for {{{y}}} in the first equation).




{{{y=(-3/5)x+26/5}}} Start with the equation where {{{y}}} was previously isolated.



{{{y=(-3/5)(2)+26/5}}} Plug in {{{x=2}}}



{{{y=-6/5+26/5}}} Multiply



{{{y=4}}} 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=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, (26-3*x)/(5), (16-2*x)/(3) ),
  blue(circle(2,4,0.1)),
  blue(circle(2,4,0.12)),
  blue(circle(2,4,0.15))
)
}}} graph of {{{3x+5y=26}}} (red) and {{{2x+3y=16}}} (green)  and the intersection of the lines (blue circle).