Question 120606
Start with the given system

{{{2x+y=11}}}
{{{y=x+2}}}




{{{2x+x+2=11}}}  Plug in {{{y=x+2}}} into the first equation. In other words, replace each {{{y}}} with {{{x+2}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.





{{{3x+2=11}}} Combine like terms on the left side



{{{3x=11-2}}}Subtract 2 from both sides



{{{3x=9}}} Combine like terms on the right side



{{{x=(9)/(3)}}} Divide both sides by 3 to isolate x




{{{x=3}}} Divide





Now that we know that {{{x=3}}}, we can plug this into {{{y=x+2}}} to find {{{y}}}




{{{y=(3)+2}}} Substitute {{{3}}} for each {{{x}}}



{{{y=5}}} Simplify



So our answer is {{{x=3}}} and {{{y=5}}} which also looks like *[Tex \LARGE \left(3,5\right)]




Notice if we graph the two equations, we can see that their intersection is at *[Tex \LARGE \left(3,5\right)]. So this verifies our answer.



{{{ graph( 500, 500, -7, 7, -7, 7, 11-2x, x+2) }}} Graph of {{{2x+1y=11}}} (red) and {{{y=x+2}}} (green)