Question 146610


Start with the given system of equations:


{{{8x-6y=36}}}

{{{6y-8x=-36}}}









In order to graph these equations, we need to solve for y for each equation.




So let's solve for y on the first equation


{{{8x-6y=36}}} Start with the given equation



{{{-6y=36-8x}}}  Subtract {{{8 x}}} from both sides



{{{-6y=-8x+36}}} Rearrange the equation



{{{y=(-8x+36)/(-6)}}} Divide both sides by {{{-6}}}



{{{y=(-8/-6)x+(36)/(-6)}}} Break up the fraction



{{{y=(4/3)x-6}}} Reduce



Now lets graph {{{y=(4/3)x-6}}} (note: 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>)



{{{ graph( 600, 600, -10, 10, -10, 10, (4/3)x-6) }}} Graph of {{{y=(4/3)x-6}}}




{{{6y-8x=-36}}} Start with the second equation.



{{{6y=-36+8x}}} Add {{{8x}}} to both sides.



{{{y=(-36+8x)/(6)}}} Divide both sides by {{{6}}} to isolate {{{y}}}.




{{{y=(4/3)x-6}}} Break up the fraction and rearrange the terms





Now lets add the graph of {{{y=(3/4)x+9/2}}} to our first plot to get:


{{{ graph( 600, 600, -10, 20, -10, 20, (4/3)x-6,(4/3)x-6) }}} Graph of {{{y=(4/3)x-6}}}(red) and {{{y=(4/3)x-6}}}(green)


From the graph, we can see that the two lines are the same. So there are an infinite number of solutions



So the system is consistent and dependent.