Word Problem/Equation

[mikosangeleyes]

This is from a distance=rate times time problem.

Amy drove the 200 miles to New Orleans at an average speed 10 miles per hour faster than her usual average spped. If she completed the trip in 1 hour less than usual, what is her usual driving speed, in miles per hour?

I got up to here and made this equation which according to the book is correct:

200 200
___ = ___ - 1 (These are fractions)
r+10 r


If you can't read it right this way this is another way to see it I guess:

200/r+10 = 200/r - 1


I would typically know how to solve this by cross multiplying but I don't know what to do with the negative one. Please give help and tell me how to do it.

Thanks

 

[jonboy]

This is what you mean.

\(\displaystyle \L\bold \frac{200}{r+10}=\frac{200}{r} -1\)

 

[mikosangeleyes]

No the negative one is seperate, it's not like r-1

This is the 2nd part after the = sign

200
___ - 1
r

 

[stapel]

If you can't read it right this way this is another way to see it I guess:

200/r+10 = 200/r - 1

Parentheses can be very helpful. You mean the above to be the following, as near as I can determine:

. . . . .200/(r + 10) = 200/r - 1

The parentheses indicate clearly that the "10" is part of the left-hand side's denominator, and that the "1" is not part of the right-hand side's denominator.

To "cross-multiply", you'll first need to combine the terms on the right-hand side. Hint: 1 = r/r.

Eliz.

 

[galactus]

Multiply by your LCD, r(r+10)

\(\displaystyle \L\\r\sout{(r+10)}\frac{200}{\sout{r+10}}=\sout{r}(r+10)\frac{200}{\sout{r}}-(1)r(r+10)\)

Now simplify and solve the resulting quadratic.

 

[mikosangeleyes]

I'm sorry I'm still really confused. I don't get that part of the LCD or changing the 1 to r/r?

What should I do with the -1?

Or can you give me more steps to how to do it?

 

[stapel]

What should I do with the -1?

Change it from "1" to "r/r". Subtract it from the other fraction. Simplify. Then cross-multiply.

Eliz.

 

[mikosangeleyes]

I changed the 1 to r/r and simplified it to (200-r) since 1r-1r = 0

After cross multiplying I got the quadratic equation:

r(squared) - 190r - 1800 = 0

Am I doing this right so far?

 

[jonboy]

No. But ill have a step by step explanation in about 6 more mins. this stuff is lengthy :shock:

 

[jonboy]

Hello mikosangeleyes

Solve: \(\displaystyle \L\bold \frac{200}{r+10}=\frac{200}{r} -1\)

This can also be written as \(\displaystyle \L\bold \frac{200}{r+10}=\frac{200}{r} +\frac{-1}{1}\)

First you need to solve everything on the right side of the equals sign. You need to have common denominaters, which means you need to find the LCM of our denominators \(\displaystyle (r & 1)\) with\(\displaystyle \L\bold \frac{200}{r} +\frac{-1}{1}\)
You should come up w/the LCM being \(\displaystyle \L\bold r\)

So know we have: \(\displaystyle \L\bold \frac{200}{r+10}=\frac{200}{r} +\frac{-1r}{r}\)

Solve the right side of the equation: \(\displaystyle \L\bold \frac{200}{r} +\frac{-1r}{r}\to\frac{200-1r}{r}\)

So now we have: \(\displaystyle \L\bold \frac{200}{r+10}=\frac{200-r}{r}\)

Cross Multiply: \(\displaystyle \L\bold 200r+2000-r^2-10r=200r\)

Bunch of basic simplifying gets you: \(\displaystyle \L\bold r^2+10r-2000=0\)

Now use the Quadratic Formula:

You will get a negative answer and a postive answer. The negative answer you just eradicate cuz you cant have a negative rate, time, nor distance.

Glad to help.

 

[mikosangeleyes]

I love you.


THANK YOU SO MUCH!!!!!!!!!!!!!!!!!!

That really meant a lot and I just needed to know how to solve that and it was bugging me so bad and you helped me and now I get it. You are my best math friend ever.

Thanks again wonderful

 

[jonboy]

Thank you I love you to :smile: You should get:

\(\displaystyle - b \pm \frac{{\sqrt {b^2 - 4ac} }}{\begin{array}{l}
2a \\
\\
\end{array}}=r\)

\(\displaystyle - 10 \pm \frac{{\sqrt {10^2 ( - 4)(1)( - 2000)} }}{\begin{array}{l}
2(1) \\
\\
\end{array}}=r\)

\(\displaystyle - 10 \pm \frac{{\sqrt {100 + 8000} }}{\begin{array}{l}
2(1) \\
\\
\end{array}}=r\)


\(\displaystyle - 10 \pm \frac{{\sqrt {8100} }}{\begin{array}{l}
2 \\
\\
\end{array}}=r\)

\(\displaystyle r=40 & -50\)

Remember to ignore the \(\displaystyle -50\) cuz rate is not negative.

So our answer is \(\displaystyle 40\).

Hope you concurred this as well.

 

[mikosangeleyes]

I don't mean to sound annoying, but I think you made a mistake only on a little part.

-4(-2000) = +8000 + 100 = 8100

The square root of that is 90 so -10 +90 = 80 / 2 = 40

That's the answer the book gave too.

 

[jonboy]

Yeah your right thanks for pointing that out. Well sry about that. I edited my post, (nothing ever happenen) :wink: .

 

[galactus]

Jonboy idid a fine job of explaining to you, but just an aside,

instead of the quad formula you could also just factor.

\(\displaystyle r^{2}+10r-2000\)

\(\displaystyle (r-40)(r+50)\)

 

[jonboy]

Yeah I know but I like the Quadratic Formula better I guess.

 

[mcrae]

Hope you concurred this as well.

lol, look up the definition of concurred


and

Jonboy idid a fine job of explaining to you, but just an aside,

instead of the quad formula you could also just factor.

\(\displaystyle r^{2}+10r-2000\)

\(\displaystyle (r-40)(r+50)\)

lollol.