Question 96328
You are taking the right approach!  Let's see how it goes:
{{{d[1] = r[1]t[1]}}} and...
{{{d[2] = r[2]t[2]}}} We are given that:
{{{r[1] = 60}}}
{{{r[2] = 80}}}
{{{d[1]+d[2] = 575}}} or {{{d[1] = 575-d[2]}}}
 and we have to find:
{{{t[1]}}} and {{{t[2]}}} but we also know that:
{{{t[1]+t[2] = 8}}} or {{{t[1] = 8-t[2]}}}
Let's put in some values.
{{{d[1] = 60t[1]}}} or {{{d[1] = 60(8-t[2])}}}
{{{d[2] = 80t[2]}}}
Now, since {{{d[1] = d[1]}}} we can write:
{{{575-d[2] = 60(8-t[2])}}} Solve this for {{{d[2]}}}
{{{d[2] = 575-60(8-t[2])}}}
{{{d[2] = 575-480+60t[2]}}}
{{{d[2] = 95+60t[2]}}} substitute {{{d[2] = 80t[2]}}}
{{{80t[2] = 95+60t[2]}}} Solve this for {{{t[2]}}}
{{{20t[2] = 95}}}
{{{t[2] = 95/20}}}
{{{t[2] = 4.75}}}hours.
{{{t[1] = 8-t[2]}}}
{{{t[1] = 8-4.75}}}
{{{t[1] = 3.25}}}hours.

Check:
{{{r[1]t[1]+r[2]t[2] = 575}}} The total trip distance.
{{{60(3.25)+80(4.75) = 195+380}}} = 575