Question 147154

{{{(3/4)x+5y=20}}} Start with the first equation.



{{{4((3/cross(4))x+5y)=4(20)}}} Multiply both sides by the LCD {{{4}}} to clear any fractions.



{{{3x+20y=80}}} Distribute and multiply.



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




{{{(2/3)x-(1/2)y=12}}} Move onto the next equation.



{{{6((2/cross(3))x-(1/cross(2))y)=6(12)}}} Multiply both sides by the LCD {{{6}}} to clear any fractions.



{{{4x-3y=72}}} Distribute and multiply.





So we have the given system of equations:

{{{system(3x+20y=80,4x-3y=72)}}}



{{{-4(3x+20y)=-4(80)}}} Multiply the both sides of the first equation by -4.



{{{-12x-80y=-320}}} Distribute and multiply.



{{{3(4x-3y)=3(72)}}} Multiply the both sides of the second equation by 3.



{{{12x-9y=216}}} Distribute and multiply.



So we have the new system of equations:

{{{system(-12x-80y=-320,12x-9y=216)}}}



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-80y)+(12x-9y)=(-320)+(216)}}}



{{{(-12x+12x)+(-80y+-9y)=-320+216}}} Group like terms.



{{{0x+-89y=-104}}} Combine like terms. Notice how the x terms cancel out.



{{{-89y=-104}}} Simplify.



{{{y=(-104)/(-89)}}} Divide both sides by {{{-89}}} to isolate {{{y}}}.



{{{y=104/89}}} Reduce.



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



{{{-12x-80y=-320}}} Now go back to the first equation.



{{{-12x-80(104/89)=-320}}} Plug in {{{y=104/89}}}.



{{{-12x-8320/89=-320}}} Multiply.



{{{89(-12x-8320/cross(89))=89(-320)}}} Multiply both sides by the LCD {{{89}}} to clear any fractions.



{{{-1068x-8320=-28480}}} Distribute and multiply.



{{{-1068x=-28480+8320}}} Add {{{8320}}} to both sides.



{{{-1068x=-20160}}} Combine like terms on the right side.



{{{x=(-20160)/(-1068)}}} Divide both sides by {{{-1068}}} to isolate {{{x}}}.



{{{x=1680/89}}} Reduce.



So our answer is {{{x=1680/89}}} and {{{y=104/89}}}.



Which form the ordered pair *[Tex \LARGE \left(\frac{1680}{89},\frac{104}{89}\right)].



This means that the two equations are consistent and independent.