Question 118052
Let x=speed of first car, y=speed of second car


You're on the right track since you have the correct equations set up




Start with the given system

{{{2x+2y=208}}}
{{{x=8+y}}}




{{{2(8+y)+2y=208}}}  Plug in {{{x=8+y}}} into the first equation. In other words, replace each {{{x}}} with {{{8+y}}}. Notice we've eliminated the {{{x}}} variables. So we now have a simple equation with one unknown.



{{{16+2y+2y=208}}} Distribute



{{{4y+16=208}}} Combine like terms on the left side



{{{4y=208-16}}}Subtract 16 from both sides



{{{4y=192}}} Combine like terms on the right side



{{{y=(192)/(4)}}} Divide both sides by 4 to isolate y




{{{y=48}}} Divide





Now that we know that {{{y=48}}}, we can plug this into {{{x=8+y}}} to find {{{x}}}




{{{x=8+(48)}}} Substitute {{{48}}} for each {{{y}}}



{{{x=56}}} Simplify



So our answer is {{{x=56}}} and {{{y=48}}}



So the first car was going 56 mph and the second car was going 48 mph