a plane goes to an airport that is 4200km away
on the way back (4200km) it goes 100km/h faster than before
if the total time taken (there and back) is 13 hours, how fast was the airplane going originally (x)

2. Use the formula

$r_{avg} = \frac{2r_1r_2}{r_1 + r_2}$

to obtain

$\frac{2(4200)}{13} = \frac{2x(x + 100)}{x + (x + 100)}$

3. and where do you go from there??

+ we never learned that equation

4. If you would like, I can derive the equation. You would have to derive it (in some sense) in order to solve the problem.

Continuing on:

$\frac{8400}{13} = \frac{2x^2 + 200x}{2x + 100}$

$\frac{8400(2x + 100)}{13} = 2x^2 + 200x$

$8400(2x + 100) = 13(2x^2 + 200x)$

$16800x + 8400 = 26x^2 + 2600x$

$26x^2 - 14200x - 8400 = 0$

$x = \frac{14200 \pm \sqrt{14200^2 + 4(26)(8400)}}{52}$

Recognizing that the negative answer for x makes no sense, we have

$x = \frac{14200 + \sqrt{14200^2 + 4(26)(8400)}}{52}$

5. Originally Posted by attack4u
a plane goes to an airport that is 4200km away
on the way back (4200km) it goes 100km/hr faster than before
if the total time taken (there and back) is 13 hours, how fast was the airplane going originally (x)
distance = rate *time,
so, time = distance /rate -----**

time in going 4200km away = 4200/x hrs.
time in going back = 4200 /(x +100) hrs.

4200/x +4200/(x+100) = 13
Clear the fractions, multiply both sides by x(x+100),
4200(x+100) +4200(x) = 13(x)(x+100)
4200x +420,000 +4200x = 13x^2 +1300x
13x^2 -7100x -420,000 = 0