Question 124951



Start with the given system of equations:


{{{3x-6y=9}}}

{{{1x-2y=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


{{{3x-6y=9}}} Start with the given equation



{{{-6y=9-3x}}}  Subtract {{{3 x}}} from both sides



{{{-6y=-3x+9}}} Rearrange the equation



{{{y=(-3x+9)/(-6)}}} Divide both sides by {{{-6}}}



{{{y=(-3/-6)x+(9)/(-6)}}} 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


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


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


From the graph, we can see that the two lines are identical (one lies perfectly on top of the other) and intersect at all points of both lines. So there are an infinite number of solutions and the system is dependent.