Question 126120



Start with the given system of equations:


{{{1x+y=3}}}

{{{1x+y=-1}}}





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


{{{1x+y=3}}} Start with the given equation



{{{1y=3-x}}}  Subtract {{{ x}}} from both sides



{{{1y=-x+3}}} Rearrange the equation



{{{y=(-x+3)/(1)}}} Divide both sides by {{{1}}}



{{{y=(-1/1)x+(3)/(1)}}} Break up the fraction



{{{y=-x+3}}} Reduce



Now lets graph {{{y=-x+3}}} (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, -x+3) }}} Graph of {{{y=-x+3}}}




So let's solve for y on the second equation


{{{1x+y=-1}}} Start with the given equation



{{{1y=-1-x}}}  Subtract {{{ x}}} from both sides



{{{1y=-x-1}}} Rearrange the equation



{{{y=(-x-1)/(1)}}} Divide both sides by {{{1}}}



{{{y=(-1/1)x+(-1)/(1)}}} Break up the fraction



{{{y=-x-1}}} Reduce




Now lets add the graph of {{{y=-x-1}}} to our first plot to get:


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