Question 147010
Start with the given system

{{{2x+y-8=0}}}
{{{y=x+2/3}}}




{{{2x+x+2/3-8=0}}}  Plug in {{{y=x+2/3}}} into the first equation. In other words, replace each {{{y}}} with {{{x+2/3}}}. Notice we've eliminated the {{{y}}} variables. So we now have a simple equation with one unknown.



{{{(3)(2x+x+2/3-8)=(3)(0)}}} Multiply both sides by the LCM of 3. This will eliminate the fractions  (note: if you need help with finding the LCM, check out this <a href=http://www.algebra.com/algebra/homework/divisibility/least-common-multiple.solver>solver</a>)




{{{6x+3x+2-24=0}}} Distribute and multiply the LCM to each side




{{{6x+3x+2-24=0}}} Distribute



{{{9x-22=0}}} Combine like terms on the left side



{{{9x=0+22}}}Add 22 to both sides



{{{9x=22}}} Combine like terms on the right side



{{{x=(22)/(9)}}} Divide both sides by 9 to isolate x




Now that we know that {{{x=22/9}}}, we can plug this into {{{y=x+2/3}}} to find {{{y}}}




{{{y=(22/9)+2/3}}} Substitute {{{22/9}}} for each {{{x}}}



{{{y=28/9}}} Simplify



So our answer is {{{x=22/9}}} and {{{y=28/9}}} which forms the ordered pair *[Tex \LARGE \left(\frac{22}{9},\frac{28}{9}\right)]