Question 146408



Start with the given system of equations:


{{{2x-6y=42}}}

{{{3x-9y=-21}}}





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


{{{2x-6y=42}}} Start with the given equation



{{{-6y=42-2x}}}  Subtract {{{2 x}}} from both sides



{{{-6y=-2x+42}}} Rearrange the equation



{{{y=(-2x+42)/(-6)}}} Divide both sides by {{{-6}}}



{{{y=(-2/-6)x+(42)/(-6)}}} Break up the fraction



{{{y=(1/3)x-7}}} Reduce



Now lets graph {{{y=(1/3)x-7}}} (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, (1/3)x-7) }}} Graph of {{{y=(1/3)x-7}}}




So let's solve for y on the second equation


{{{3x-9y=-21}}} Start with the given equation



{{{-9y=-21-3x}}}  Subtract {{{3 x}}} from both sides



{{{-9y=-3x-21}}} Rearrange the equation



{{{y=(-3x-21)/(-9)}}} Divide both sides by {{{-9}}}



{{{y=(-3/-9)x+(-21)/(-9)}}} Break up the fraction



{{{y=(1/3)x+7/3}}} Reduce




Now lets add the graph of {{{y=(1/3)x+7/3}}} to our first plot to get:


{{{ graph( 600, 600, -10, 10, -10, 10, (1/3)x-7,(1/3)x+7/3) }}} Graph of {{{y=(1/3)x-7}}}(red) and {{{y=(1/3)x+7/3}}}(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.