Question 127907

Start with the given system

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




{{{2x-1(x-1)=0}}}  Plug in {{{y=x-1}}} into the first equation. In other words, replace each {{{y}}} with {{{x-1}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.



{{{2x-1x+1=0}}} Distribute



{{{x+1=0}}} Combine like terms on the left side



{{{x=0-1}}}Subtract 1 from both sides



{{{x=-1}}} Combine like terms on the right side





Now that we know that {{{x=-1}}}, we can plug this into {{{y=x-1}}} to find {{{y}}}




{{{y=(-1)-1}}} Substitute {{{-1}}} for each {{{x}}}



{{{y=-2}}} Simplify



So our answer is {{{x=-1}}} and {{{y=-2}}} which also looks like *[Tex \LARGE \left(-1,-2\right)]




Notice if we graph the two equations, we can see that their intersection is at *[Tex \LARGE \left(-1,-2\right)]. So this verifies our answer.



{{{ graph( 500, 500, -5, 5, -5, 5, 2x, x-1) }}} Graph of {{{2x-1y=0}}} (red) and {{{y=x-1}}} (green)