Question 286193
Write 2 equations, 1 for each case:
{{{d = r*t}}}
case #1
{{{280 = rt}}}
{{{280 = (r + 5)*(t - 1)}}}
{{{280 = rt + 5t - r - 5}}}
By substitution:
{{{280 = 280 + 5t - r - 5}}}
{{{0 = 5t - r - 5}}}
{{{r = 5t - 5}}}
Substituting again:
{{{280 = (5t - 5)*t}}}
{{{280 = 5t^2 - 5t}}}
{{{t^2 - t - 56 = 0}}}
{{{(t + 7)(t - 8) = 0}}}
{{{t}}} can't be negative, so {{{t = 8}}}
and
{{{280 = rt}}}
{{{280 = r*8}}}
{{{r = 35}}} mi/hr
check answer:
{{{280 = (r + 5)*(t - 1)}}}
{{{280 = (35 + 5)*(t - 1)}}}
{{{280 = 40t - 40}}}
{{{320 = 40t}}}
{{{t = 8}}}
OK