Question 126101
Note: these are not inequalities. These are equations. 



Start with the given system of equations:


{{{1x+2y=3}}}

{{{2x-3y=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=3}}} Start with the given equation



{{{2y=3-x}}}  Subtract {{{ x}}} from both sides



{{{2y=-x+3}}} Rearrange the equation



{{{y=(-x+3)/(2)}}} Divide both sides by {{{2}}}



{{{y=(-1/2)x+(3)/(2)}}} Break up the fraction



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



Now lets graph {{{y=(-1/2)x+3/2}}} (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+3/2) }}} Graph of {{{y=(-1/2)x+3/2}}}




So let's solve for y on the second equation


{{{2x-3y=6}}} Start with the given equation



{{{-3y=6-2x}}}  Subtract {{{2 x}}} from both sides



{{{-3y=-2x+6}}} Rearrange the equation



{{{y=(-2x+6)/(-3)}}} Divide both sides by {{{-3}}}



{{{y=(-2/-3)x+(6)/(-3)}}} Break up the fraction



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




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


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


From the graph, we can see that the two lines intersect at the point ({{{3}}},{{{0}}})