Question 140987
{{{5y - 8x = 40}}} Start with the second equation



{{{- 8x + 5y  = 40}}} Rearrange the terms







So we now have the system of equations:


{{{5x-8y=-40}}}

{{{-8x+5y=40}}}





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


{{{5x-8y=-40}}} Start with the given equation



{{{-8y=-40-5x}}}  Subtract {{{5 x}}} from both sides



{{{-8y=-5x-40}}} Rearrange the equation



{{{y=(-5x-40)/(-8)}}} Divide both sides by {{{-8}}}



{{{y=(-5/-8)x+(-40)/(-8)}}} Break up the fraction



{{{y=(5/8)x+5}}} Reduce



Now lets graph {{{y=(5/8)x+5}}} (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, (5/8)x+5) }}} Graph of {{{y=(5/8)x+5}}}




So let's solve for y on the second equation


{{{-8x+5y=40}}} Start with the given equation



{{{5y=40+8x}}} Add {{{8 x}}} to both sides



{{{5y=+8x+40}}} Rearrange the equation



{{{y=(+8x+40)/(5)}}} Divide both sides by {{{5}}}



{{{y=(+8/5)x+(40)/(5)}}} Break up the fraction



{{{y=(8/5)x+8}}} Reduce




Now lets add the graph of {{{y=(8/5)x+8}}} to our first plot to get:


{{{ graph( 600, 600, -10, 10, -10, 10, (5/8)x+5,(8/5)x+8) }}} Graph of {{{y=(5/8)x+5}}}(red) and {{{y=(8/5)x+8}}}(green)


With help from a graphing calculator, we can see that the two lines intersect at the point ({{{-40/13}}},{{{40/13}}}) 




This means that system is independent and consistent. So you are correct.