Question 127480



Start with the given system of equations:


{{{1x-2y=-2}}}

{{{1x+2y=6}}}





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-2y=-2}}} Start with the given equation



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



{{{-2y=-x-2}}} Rearrange the equation



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



{{{y=(-1/-2)x+(-2)/(-2)}}} Break up the fraction



{{{y=(1/2)x+1}}} Reduce



Now lets graph {{{y=(1/2)x+1}}} (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, (1/2)x+1) }}} Graph of {{{y=(1/2)x+1}}}




So let's solve for y on the second equation


{{{1x+2y=6}}} Start with the given equation



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



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



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



{{{y=(-1/2)x+(6)/(2)}}} Break up the fraction



{{{y=(-1/2)x+3}}} Reduce




Now lets add the graph of {{{y=(-1/2)x+3}}} to our first plot to get:


{{{ graph( 600, 600, -10, 10, -10, 10, (1/2)x+1,(-1/2)x+3) }}} Graph of {{{y=(1/2)x+1}}}(red) and {{{y=(-1/2)x+3}}}(green)


From the graph, we can see that the two lines intersect at the point ({{{2}}},{{{2}}})