Question 124408



Start with the given system of equations:


{{{4x+3y=12}}}

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





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


{{{4x+3y=12}}} Start with the given equation



{{{3y=12-4x}}}  Subtract {{{4 x}}} from both sides



{{{3y=-4x+12}}} Rearrange the equation



{{{y=(-4x+12)/(3)}}} Divide both sides by {{{3}}}



{{{y=(-4/3)x+(12)/(3)}}} Break up the fraction



{{{y=(-4/3)x+4}}} Reduce



Now lets graph {{{y=(-4/3)x+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, (-4/3)x+4) }}} Graph of {{{y=(-4/3)x+4}}}




So let's solve for y on the second equation


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



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



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



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



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



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




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


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


From the graph, we can see that the two lines intersect at the point ({{{6}}},{{{-4}}}) (note: you might have to adjust the window to see the intersection)