Question 145582

Start with the given system

{{{3x+5y=3}}}
{{{x=8-4y}}}




{{{3(8-4y)+5y=3}}}  Plug in {{{x=8-4y}}} into the first equation. In other words, replace each {{{x}}} with {{{8-4y}}}. Notice we've eliminated the {{{x}}} variables. So we now have a simple equation with one unknown.



{{{24-12y+5y=3}}} Distribute



{{{-7y+24=3}}} Combine like terms on the left side



{{{-7y=3-24}}}Subtract 24 from both sides



{{{-7y=-21}}} Combine like terms on the right side



{{{y=(-21)/(-7)}}} Divide both sides by -7 to isolate y




{{{y=3}}} Divide





Now that we know that {{{y=3}}}, we can plug this into {{{x=8-4y}}} to find {{{x}}}




{{{x=8-4(3)}}} Substitute {{{3}}} for each {{{y}}}



{{{x=-4}}} Simplify



So our answer is {{{x=-4}}} and {{{y=3}}} which also looks like *[Tex \LARGE \left(-4,3\right)]




Notice if we graph the two equations, we can see that their intersection is at *[Tex \LARGE \left(-4,3\right)]. So this verifies our answer.



{{{ graph( 500, 500, -5, 5, -5, 5, (3-3x)/5, (x-8)/(-4)) }}} Graph of {{{3x+5y=3}}} (red) and {{{x=8-4y}}} (green)