Question 128750
Let's graph the first equation  {{{x-2y=0}}} 


Let's find a point using {{{x=0}}}


{{{x-2y=0}}} Start with the first equation


{{{0-2y=0}}} Plug in {{{x=0}}}


{{{-2y=0}}} Simplify


{{{y=0/-2}}} Divide both sides by -2 to isolate y


{{{y=0}}} Reduce


So when {{{x=0}}}, {{{y=0}}}. So our first point is (0,0)






Let's find another point using {{{x=1}}}


{{{x-2y=0}}} Start with the first equation


{{{1-2y=0}}} Plug in {{{x=1}}}


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


{{{y=-1/-2}}} Divide both sides by -2 to isolate y


{{{y=1/2}}} Reduce



So when {{{x=1}}}, {{{y=1/2}}}. So our second point is *[Tex \LARGE \left(1,\frac{1}{2}\right)]




Now plot the points *[Tex \LARGE \left(0,0\right)] and *[Tex \LARGE \left(1,\frac{1}{2}\right)]




{{{drawing(500,500,-9,11,-9,11,
graph(500,500,-9,11,-9,11,0),
grid(1),
circle(0,0,0.1),
circle(0,0,0.12),
circle(0,0,0.15),
circle(1,1/2,0.1),
circle(1,1/2,0.12),
circle(1,1/2,0.15)
) }}}




Now draw a straight line through the two points. This line is the graph of {{{x-2y=0}}}


{{{drawing(500,500,-9,11,-9,11,
graph(500,500,-9,11,-9,11,(1/2)x),
grid(1),
circle(0,0,0.1),
circle(0,0,0.12),
circle(0,0,0.15),
circle(1,1/2,0.1),
circle(1,1/2,0.12),
circle(1,1/2,0.15)
) }}} Graph of {{{x-2y=0}}} through the two points *[Tex \LARGE \left(0,0\right)] and *[Tex \LARGE \left(1,\frac{1}{2}\right)]





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Let's graph the second equation  {{{2x-y= -3}}} 


Let's find a point using {{{x=0}}}


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


{{{2(0)-y= -3}}} Plug in {{{x=0}}}


{{{-y=-3}}} Simplify


{{{y=-3/-1}}} Divide both sides by -1 to isolate y


{{{y=3}}} Reduce


So when {{{x=0}}}, {{{y=3}}}. So our first point is (0,3)






Let's find another point using {{{x=1}}}


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


{{{2(1)-y= -3}}} Plug in {{{x=1}}}



{{{2-y= -3}}} Multiply



{{{-y=-5}}} Subtract 2 from both sides


{{{y=-5/-1}}} Divide both sides by -1 to isolate y


{{{y=5}}} Reduce



So when {{{x=1}}}, {{{y=5}}}. So our second point is (1,5)





Now plot the points *[Tex \LARGE \left(0,3\right)] and *[Tex \LARGE \left(1,5\right)]




{{{drawing(500,500,-9,11,-9,11,
graph(500,500,-9,11,-9,11,0),
grid(1),
circle(0,3,0.1),
circle(0,3,0.12),
circle(0,3,0.15),
circle(1,5,0.1),
circle(1,5,0.12),
circle(1,5,0.15)
) }}}




Now draw a straight line through the two points. This line is the graph of {{{2x-y= -3}}}


{{{drawing(500,500,-9,11,-9,11,
graph(500,500,-9,11,-9,11,(-3-2x)/(-1)),
grid(1),
circle(0,3,0.1),
circle(0,3,0.12),
circle(0,3,0.15),
circle(1,5,0.1),
circle(1,5,0.12),
circle(1,5,0.15)
) }}} Graph of {{{2x-y= -3}}} through the two points *[Tex \LARGE \left(0,3\right)] and *[Tex \LARGE \left(1,5\right)]




Now simply graph these two line together to get 


{{{drawing(500,500,-9,11,-9,11,
graph(500,500,-9,11,-9,11,(-3-2x)/(-1),(1/2)x),
grid(1),
circle(0,3,0.1),
circle(0,3,0.12),
circle(0,3,0.15),
circle(1,5,0.1),
circle(1,5,0.12),
circle(1,5,0.15),
circle(0,0,0.1),
circle(0,0,0.12),
circle(0,0,0.15),
circle(1,1/2,0.1),
circle(1,1/2,0.12),
circle(1,1/2,0.15),
blue(circle(-2,-1,0.1)),
blue(circle(-2,-1,0.12)),
blue(circle(-2,-1,0.15))
) }}} Graph of {{{x-2y=0}}} (green) and {{{2x-y= -3}}} (red)



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