Question 133040
{{{8y-5x=40}}} Start with the second equation



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






Start with the given system of equations:


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

{{{-5x+8y=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


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



{{{8y=40+5x}}} Add {{{5 x}}} to 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 add the graph of {{{y=(5/8)x+5}}} to our first plot to get:


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


From the graph, we can see that the two lines are parallel and will never intersect. So there are no solutions and the system is inconsistent.