Question 126093


Start with the given system of equations:


{{{2x+y=4}}}

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





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+y=4}}} Start with the given equation



{{{1y=4-2x}}}  Subtract {{{2 x}}} from both sides



{{{1y=-2x+4}}} Rearrange the equation



{{{y=(-2x+4)/(1)}}} Divide both sides by {{{1}}}



{{{y=(-2/1)x+(4)/(1)}}} Break up the fraction



{{{y=-2x+4}}} Reduce



Now lets graph {{{y=-2x+4}}} (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, -2x+4) }}} Graph of {{{y=-2x+4}}}




So let's solve for y on the second 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 add the graph of {{{y=-x+3}}} to our first plot to get:


{{{ graph( 600, 600, -10, 10, -10, 10, -2x+4,-x+3) }}} Graph of {{{y=-2x+4}}}(red) and {{{y=-x+3}}}(green)


From the graph, we can see that the two lines intersect at the point ({{{1}}},{{{2}}})