Question 401141
Using the distance equals rate times time, d = r*t, we can write an equation for each part of the trip:
{{{d[1] = r[1]*t[1]}}}
{{{d[2] = r[2]*t[2]}}}
with the subscripts describing the part of  the trip. Since we know the distances for each part we can substitute the numbers for those distaces:
{{{48 = r[1]*t[1]}}}
{{{19 = r[2]*t[2]}}}<br>
At this point we have two equations and <i>four</i> variables. ({{{r[1]}}} and {{{r[2]}}} count as separate variables. So do {{{t[1]}}} and {{{t[2]}}}.) So we need two more equations. We are told that the rate during the second part was 5 mph slower than the rate during the first part. So:
{{{r[2] = r[1] - 5}}}
We are also told that the total time was 3 hours. So:
{{{t[1] + t[2] = 3}}}
If we subtract {{{t[1]}}} from each side we can solve for {{{t[2]}}}:
{{{t[2] = 3 - t[1]}}}
Now we can use these last two equations and substitute for {{{r[2]}}} and {{{t[2]}}} in the {{{19 = r[2]*t[2]}}} equation:
{{{19 = (r[1]-5)*(3-t[1])}}}
which simplifies as follows:
{{{19 = r[1]*3-r[1]*t[1]-5*3-5*t[1]}}}
{{{19 = 3r[1]-r[1]*t[1]-15-5*t[1]}}}
From {{{48 = r[1]*t[1]}}} we know that {{{r[1]*t[1]}}} is 48. Replacing the {{{r[1]*t[1]}}} with 48 in the previous equation we get:
{{{19 = 3r[1]-48-15-5*t[1]}}}
or
{{{19 = 3r[1]-63-5*t[1]}}}
Adding 63 to each side we get:
{{{82 = 3r[1]-5*t[1]}}}
Next we can solve for {{{t[1]}}} in the equation {{{48 = r[1]*t[1]}}} by dividing both sides by {{{r[1]}}}:
{{{48/r[1] = t[1]}}}
Now we can substitute for {{{t[1]}}} in {{{82 = 3r[1]-5*t[1]}}}:
{{{82 = 3r[1]-5*(48/r[1])}}}
We finally have an equation with a single variable! We can now solve for {{{r[1]}}}. Multiplying both sides by {{{r[1]}}} (to eliminate the fraction) we get:
{{{82r[1] = 3r[1]^2-240}}}
This is a quadratic equation so we want one side to be zero. Subtracting {{{82r[1]}}} from each side we get:
{{{0 = 3r[1]^2-82r[1]-240}}}
There are just too many possible factors of 240 for me to want to try to factor this. So I will use the Quadratic Formula instead:
{{{r[1] = (-(-82) +- sqrt((-82)^2-4(3)(-240)))/2(3)}}}
which simplifies as follows:
{{{r[1] = (-(-82) +- sqrt(6724-4(3)(-240)))/2(3)}}}
{{{r[1] = (-(-82) +- sqrt(6724+2880))/2(3)}}}
{{{r[1] = (-(-82) +- sqrt(9604))/2(3)}}}
{{{r[1] = (82 +- sqrt(9604))/6}}}
{{{r[1] = (82 +- 98)/6}}}
In long form this is:
{{{r[1] = (82 + 98)/6}}} or {{{r[1] = (82 - 98)/6}}}
which simplify as:
{{{r[1] = (180)/6}}} or {{{r[1] = (-16)/6}}}
{{{r[1] = 30}}} or {{{r[1] = (-8)/3}}}
(BTW, this means the equation would have factored into:
{{{0 = (r[1]-30)(3r[1]+8)}}}<br>
The second solution for the rate is negative which makes no sense in the context of this word problem. So we will reject it.<br>
So {{{r[1]}}}, the rate during the first part of the trip, is 30 mph. The rate during the second part, {{{r[2]}}}, is 5 mpg slower so it must be 30-5 or 25 mph. (Since the problem only asks for the rates we are finished. We could use these values to find {{{t[1]}}} and {{{t[2]}}} also.)