Word Problem Help?

[Hanajima]

Set up an equation and solve the following problem.
Felipe jogs for 9 miles and then walks another 9 miles. He jogs 1.5 miles per hour faster than he walks, and the entire distance of 18 miles takes 9 hours. Find the rate at which he walks and the rate at which he jogs.

Equation:

9/x+1.5 + 9/x = 9

Multiply both sides by x(x+1.5) to clear denominators, get:

9x + 9x +13.5 = 9x^2 +13.5

Set it to zero, to get:

9x^2 -18=0

x= 0,2

x cannot be zero so x=2

so walking rate=2mph
and jogging= 3.5mph?

I typed this in but my program says I'm wrong

 

[soroban]

Hello, Hanajima!

A small oversight . . .

Felipe jogs for 9 miles and then walks another 9 miles.
He jogs 1.5 miles per hour faster than he walks, and the entire distance of 18 miles takes 9 hours.
Find the rate at which he walks and the rate at which he jogs.


\(\displaystyle \text{Equation: }\:\frac{9}{x+1.5} + \frac{9}{x} \;=\; 9\)


\(\displaystyle \text{Multiply both sides by }x(x+1.5)\text{ to clear denominators, and get:}\)

. . \(\displaystyle 9x + 9(x + 1.5) \;=\;9x(x+1.5)\)

. . \(\displaystyle 9x + 9x +13.5 = 9x^2 +13.5x\)
. . . . . . . . . . = . . . = . . . . . \(\displaystyle \Uparrow\)

 

[lookagain]

\(\displaystyle > \ > \ >\)Set up an equation and solve the following problem.\(\displaystyle < \ < \ <\)

Part of solving this is to define your variables.

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Let x = the number of mph at which he walks
Let x + 1.5 = the number of mph at which he jogs


9/(x + 1.5) + 9/x = 9 \(\displaystyle \ . \ . \ . \ . \ . You \ must \ have \ grouping \ symbols \ around \ the \ denominator, \ x + \ 1.5\)


\(\displaystyle \frac{9}{x + 1.5} + \frac{9}{x} \ = \ 9\)


A suggestion is to next multiply all terms on each side by \(\displaystyle \frac{1}{9}\) to get:


\(\displaystyle \frac{1}{x + 1.5} + \frac{1}{x} \ = \ 1\)


After you multiply each side by the lcd of x(x + 1.5), you would have:

\(\displaystyle x + 1.5 + x = x^2 + 1.5x\)


\(\displaystyle 2x + 1.5 = x^2 + 1.5x\)


This is already a quadratic equation. Combine like terms, change it to a
trinomial, and set it to zero:


\(\displaystyle x^2 - 0.5x - 1.5 \ = \ 0\)


Multiply by 2 to get integer coefficients:


\(\displaystyle 2x^2 - x - 3 \ = \ 0\)


\(\displaystyle (2x - 3)(x + 1) \ = \ 0\)


\(\displaystyle Then \ x \ = \ \frac{3}{2} \ = \ 1.5, \ ignore \ x \ = \ -1\)


\(\displaystyle And \ x + 1.5 \ = \ 3\)


The walking speed is 1.5 mph.

The jogging speed is 3 mph.

 

[Denis]

9/x+1.5 + 9/x = 9
Multiply both sides by x(x+1.5) to clear denominators, get:
9x + 9x +13.5 = 9x^2 +13.5

CAREFUL! Last term should be 13.5x