Question 111723

Start with the given system

{{{2x+4y=16}}}
{{{x=3y-2}}}




{{{2(3y-2)+4y=16}}}  Plug in {{{x=3y-2}}} into the first equation. In other words, replace each {{{x}}} with {{{3y-2}}}. Notice we've eliminated the {{{x}}} variables. So we now have a simple equation with one unknown.



{{{6y-4+4y=16}}} Distribute



{{{10y-4=16}}} Combine like terms on the left side



{{{10y=16+4}}}Add 4 to both sides



{{{10y=20}}} Combine like terms on the right side



{{{y=(20)/(10)}}} Divide both sides by 10 to isolate y




{{{y=2}}} Divide





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




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



{{{x=4}}} Simplify



So our answer is {{{x=4}}} and {{{y=2}}}



Since a solution exists, and it's unique, the system is consistent and independent