# Thread: distance problem and finding Y

1. ## distance problem and finding Y

The distance (in miles) traveled when driving at a certain speed for hours,
then driving miles/hour faster for another hour.
Express the distance in terms of .

the distance traveled when driving at mph for hours is the expression = ?

ADDED to that distance is driving 1 hour at mph FASTER,
so the expression to be added is = ?

The final expression = ?

my second question is..

1/3y + 9 = 1/9y Solve for Y.

Thanks for any help..

2. let the distance be X..

$X = S \frac{miles}{hour} \times 13 \;hours \;+ (S+5)\frac{miles}{hour} \times 1\; hours\; = ...$

finish...

2nd question:

$\dfrac{1}{3y}+9 = \dfrac{1}{9y}$

$\dfrac{1}{3y}-\dfrac{1}{9y}=-9$

$\dfrac{1}{3y}\times\dfrac{3}{3}-\dfrac{1}{9y}=-9$

$\dfrac{3}{9y}-\dfrac{1}{9y}=-9$

finish...