Question 170002
# 1




Start with the given system of equations:

{{{system(3x-2y=8,-12x+8y=32)}}}



{{{4(3x-2y)=4(8)}}} Multiply the both sides of the first equation by 4.



{{{12x-8y=32}}} Distribute and multiply.



So we have the new system of equations:

{{{system(12x-8y=32,-12x+8y=32)}}}



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



{{{(12x-8y)+(-12x+8y)=(32)+(32)}}}



{{{(12x+-12x)+(-8y+8y)=32+32}}} Group like terms.



{{{0x+0y=64}}} Combine like terms. Notice how the x terms cancel out.



{{{0=64}}}Simplify.



Since {{{0=64}}} is <font size="4"><b>NEVER</b></font> true, this means that there are no solutions. So the system is inconsistent.




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# 2



Note: I've made the first equation {{{-4x+y=27}}} and the second equation {{{7x-5y=14}}}



Start with the given system of equations:


{{{system(-4x+y=27,7x-5y=14)}}}




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


{{{-4x+y=27}}} Start with the first equation



{{{y=27+4x}}} Add {{{4x}}} to both sides



{{{y=+4x+27}}} Rearrange the equation



{{{y=(+4x+27)/(1)}}} Divide both sides by {{{1}}}



{{{y=((+4)/(1))x+(27)/(1)}}} Break up the fraction



{{{y=4x+27}}} Reduce




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


Since {{{y=4x+27}}}, we can now replace each {{{y}}} in the second equation with {{{4x+27}}} to solve for {{{x}}}




{{{7x-5highlight((4x+27))=14}}} Plug in {{{y=4x+27}}} into the second equation. In other words, replace each {{{y}}} with {{{4x+27}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.




{{{7x+(-5)(4)x+(-5)(27)=14}}} Distribute {{{-5}}} to {{{4x+27}}}



{{{7x-20x-135=14}}} Multiply



{{{-13x-135=14}}} Combine like terms on the left side



{{{-13x=14+135}}}Add 135 to both sides



{{{-13x=149}}} Combine like terms on the right side



{{{x=(149)/(-13)}}} Divide both sides by -13 to isolate x




{{{x=-149/13}}} Reduce






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



So the first part of our answer is: {{{x=-149/13}}}










Since we know that {{{x=-149/13}}} we can plug it into the equation {{{y=4x+27}}} (remember we previously solved for {{{y}}} in the first equation).




{{{y=4x+27}}} Start with the equation where {{{y}}} was previously isolated.



{{{y=4(-149/13)+27}}} Plug in {{{x=-149/13}}}



{{{y=-596/13+27}}} Multiply



{{{y=-245/13}}} 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=-245/13}}}










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


So our answers are:


{{{x=-149/13}}} and {{{y=-245/13}}}


which form the ordered pair *[Tex \LARGE \left(-\frac{149}{13},-\frac{245}{13}\right)] 




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# 3





Start with the given system of equations:

{{{system(-4x+3y=5,12x-9y=-15)}}}



{{{3(-4x+3y)=3(5)}}} Multiply the both sides of the first equation by 3.



{{{-12x+9y=15}}} Distribute and multiply.



So we have the new system of equations:

{{{system(-12x+9y=15,12x-9y=-15)}}}



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



{{{(-12x+9y)+(12x-9y)=(15)+(-15)}}}



{{{(-12x+12x)+(9y+-9y)=15+-15}}} Group like terms.



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



{{{0=0}}}Simplify.



Since {{{0=0}}} is <font size="4"><b>ALWAYS</b></font> true, this means that there are an infinite number of solutions. So the system is consistent and dependent.