System of Equations

**ss103** · March 29th 2009, 04:34 PM

Please help with the following problem. Thank You!

Two planes leave Pittsburgh and Philadelphia at the same time. Each going to the other city. One plane flies 25mph faster than the other plane. Find the air speed of each plane if the cities are 275 miles apart and the planes pass each other after 40 minutes of flying time.

**skeeter** · March 29th 2009, 04:45 PM

Originally Posted by ss103

Please help with the following problem. Thank You!

Two planes leave Pittsburgh and Philadelphia at the same time. Each going to the other city. One plane flies 25mph faster than the other plane. Find the air speed of each plane if the cities are 275 miles apart and the planes pass each other after 40 minutes of flying time.

let $v$ = speed of slower plane in mph

$v+25$ = speed of the faster plane in mph

note that $40$ min = $\frac{2}{3}$ hr

(speed)(time) = distance

$v\left(\frac{2}{3}\right) + (v+25)\left(\frac{2}{3}\right) = 275$

solve for $v$

**ss103** · March 29th 2009, 04:52 PM

thanks! but we have to find the speed of each plane, separately. v is just the speed of one plane

**Shyam** · March 29th 2009, 04:58 PM

Originally Posted by ss103

Please help with the following problem. Thank You!

Two planes leave Pittsburgh and Philadelphia at the same time. Each going to the other city. One plane flies 25mph faster than the other plane. Find the air speed of each plane if the cities are 275 miles apart and the planes pass each other after 40 minutes of flying time.

Let the speed of first plane (which leaves Pittsburgh) = x mph
Let the speed of second plane = (x + 25) mph

Let the two planes pass each other at distance "y" from Pittsburgh, after 40 min.

So, distance travelled by first plane in 40 min = y
Speed = x
Time = 40 min = 40/60 h = 2/3 h

Distance = speed . Time

so, $y = x \times \frac{2}{3}$

$y = \frac{2}{3}\;x$

3y = 2x .....................................(1)

Now, distance travelled by second plane in 40 min = (275 - y)
Speed = (x + 25)
Time = 40 min = 40/60 h = 2/3 h

Distance = speed . Time

so, $275 - y = (x+25) \times \frac{2}{3}$

$275 - y = \frac{2}{3}\;(x+25)$

$3(275 - y) = 2(x+25)$

825 - 3y = 2x + 50

2x + 3y = 775 .................................(2)

Now, solve these two eqns (1) and (2). FINISH it.
The speeds of two planes will be 193.75 mph and 218.75 mph.

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