Question 124846
Start with the given system

{{{2x+1y=7}}}
{{{y=x+1}}}




{{{2x+1(x+1)=7}}}  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=7}}} Distribute



{{{3x+1=7}}} Combine like terms on the left side



{{{3x=7-1}}}Subtract 1 from both sides



{{{3x=6}}} Combine like terms on the right side



{{{x=(6)/(3)}}} Divide both sides by 3 to isolate x




{{{x=2}}} Divide





Now that we know that {{{x=2}}}, we can plug this into {{{y=x+1}}} to find {{{y}}}




{{{y=(2)+1}}} Substitute {{{2}}} for each {{{x}}}



{{{y=3}}} Simplify



So our answer is {{{x=2}}} and {{{y=3}}} which also looks like *[Tex \LARGE \left(2,3\right)]




Notice if we graph the two equations, we can see that their intersection is at *[Tex \LARGE \left(2,3\right)]. So this verifies our answer.



{{{ graph( 500, 500, -5, 5, -5, 5, 7-2x, x+1) }}} Graph of {{{2x+1y=7}}} (red) and {{{y=x+1}}} (green)