Question 171699
{{{d = r*t}}} for both of them
For Brenda:
(1) {{{d[b] = r[b]*t[b]}}}
For Randy:
(2) {{{d[r] = r[r]*t[r]}}}
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Given is:
{{{d[b] = 30}}}
{{{d[r] = 60}}}
{{{r[r] = r[b] + 5}}}
{{{t[r] = t[b] + 1}}}
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Now I can rewrite (1) and (2)
(1) {{{30 = r[b]*t{b]}}}
{{{t[b] = 30/r[b]}}}
(2) {{{60 = (r[b] + 5)*(t[b] + 1)}}}
(2) {{{60 = (r[b] + 5)((30/r[b]) + 1)}}}
(2) {{{60 = 30 + 150/r[b] + r[b] + 5}}}
Multiply both sides by {{{r[b]}}}
(2) {{{60r[b] = 35r[b] + 150 + r[b]^2}}}
(2) {{{r[b]^2 - 25r[b] + 150 = 0}}}
Looking at this, I see {{{(-15)*(-10) = 150}}} and {{{-10 -15 = -25}}}
(2) {{{(r[b] - 15)(r[b] - 10) = 0}}}
This is true if either {{{r[b] = 10}}} or {{{r[b] = 15}}}
If {{{r[b] = 10}}}, then
{{{r[r] = r[b] + 5}}}
{{{r[r] = 10 + 5}}}
{{{r[r] = 15}}}
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{{{t[b] = 30/r[b]}}}
{{{t[b] = 30/10}}}
{{{t[b] = 3}}}
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{{{t[r] = t[b] + 1}}}
{{{t[r] = 3 + 1}}}
{{{t[r] = 4}}}
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Putting this together,
For Brenda:
(1) {{{30 = 10*3}}}
For Randy:
(2) {{{60 = 15*4}}}
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These answers work, but what if {{{r[b] = 15}}}?
{{{t[b] = 30/r[b]}}}
{{{t[b] = 30/15}}}
{{{t[b] = 2}}}
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{{{r[r] = r[b] + 5}}}
{{{r[r] = 15 + 5}}}
{{{r[r] = 20}}}
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{{{t[r] = t[b] + 1}}}
{{{t[r] = 2 + 1}}}
{{{t[r] = 3}}}
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Putting these together,
(1) {{{30 = 15*2}}}
For Randy:
(2) {{{60 = 20*3}}}
These answers work too, so
The speeds of the cyclists were either
(1) Brenda went 10 mi/hr and Randy went 15 mi/hr
(2) Brenda went 15 mi/hr and Randy went 20 mi/hr