Question 189187
I can write two equations, one for Brooke:
(1) {{{d[a] = r[a]*t[a]}}}
and one for Bart,
(2) {{{d[b] = r[b]*t[b]}}}
The key to the problem is that Bart catches Brooke.
The distance that each drives is the same
The time Bart spends driving must equal the time
Brooke spends driving

given:
{{{d[a] = d[b]}}} mi
{{{r[a] = 35}}} mi/hr
{{{r[b] = 50}}} mi/hr
{{{t[a] = t[b] + .75}}} hr
---------------------------
(1) {{{d[a] = r[a]*t[a]}}}
(1) {{{d[a] = 35*(t[b] + .75)}}}
(1) {{{d(a) = 35t[b] + 26.25}}}
and
(2) {{{d[b] = r[b]*t[b]}}}
(2) {{{d[b] = 50t[b]}}}
And, since {{{d[a] = d[b]}}}
(2) {{{d[a] = 50t[b]}}}
Subtract (1) from (2)
(2) {{{d[a] = 50t[b]}}}
(1) {{{d(a) = 35t[b] + 26.25}}}
(3) {{{0 = 15t[b] - 26.25}}}
(3) {{{15t[b] = 26.25}}}
(3) {{{t[b] = 1.75}}}
Bart's time driving is 1.75 hrs
If you add his delays,
{{{1.75 + .75 = 2.5}}}
So, 2.5 hrs after Brooke left, Bart caught her
which was 7 AM + 2.5 = 9:30 AM
check answer:
Brooke's driving time was 2.5 hrs
(1) {{{d[a] = 35*2.5}}}
(1) {{{d(a) = 87.5}}} mi
and
(2) {{{d[a] = 50t[b]}}}
(2) {{{87.5 = 50t[b]}}}
(2) {{{t[b] = 1.75}}} hr
This is what I said Bart's driving time was, so
it checks