Question 34056
<pre><b>The second car travels at 15+x
THe first car travels at x
Since its 1 hours less it's
d/(x)=d/(x+15)+1
and d=180
Equation:
*[tex \frac{180}{x}-\frac{180}{x+15}=1]
Cross mutliply:
180[(15+x)-(x)]=(x+15)(x)
{{{180(15)=x^2+15x}}}
{{{x^2+15x-2700}}}
a=1, b=15, c=-2700 in quardatic formula:
*[tex x=\frac{-15\frac{+}{-}sqrt(15^2-4*1*-2700)}{2*1}]
simplfy:
x=45 or x=-60
remove the negative and x=45.
<font color = green>Hence, the speed of the first car was 45mph and the second car was 60mph.
</font>Paul.