John traveled 400 km from Manila to a nearby province. If in returning home he reduced his speed by 9 km and able to reach Manila 2 hours more time than going to the province, what was his original speed?

okay, I had a typo, and it messed everything up.

here we go again:

let be the original speed and let be the original time.
So we have and

They both equal 400 because that is the distance he traveled both to and from Manilla.
So we set them equal to each other and get:

[math] \Rightarrow /, t cdot s = t \cdot s -9t + 2s -18 [\math]
[math] \Rightarrow /, 0 = -9t + 2s -18 [\math]
[math] \Rightarrow /, 9t = 2s -18 [\math]
[math] \Rightarrow /, t = \frac{1}{9}(2s -18) [\math]

So now we have in terms of , and we just plug that into the original equation and get:

[math] \frac{1}{9}(2s -18) \cdot s = 400 [\math]
[math] \Rightarrow \, (2s -18) \cdot s = 3600 [\math]
[math] \Rightarrow \, 2s^2 -18 s = 3600 [\math]
[math] \Rightarrow \, s^2 -9 s = 1800 [\math]
[math] \Rightarrow \, s^2 -9 s - 1800 = 0 [\math]

Now we apply the quadratic formula and get:










Now since is negative and we can't go a negative speed, are only viable answer is



I hope this helps.

wow, I just can't get anything right today :(

your answer is , If you can wade through my horrible proofreading skills above, I explained it.