Little Indices Problem

• Jul 21st 2010, 05:23 AM
Lites
Little Indices Problem
Hello again guys,

I have five problems to solve but found myself stuck by this question other thought to be very simple but I am unfortunately not. :( Could you guys explain it how to solve this problem? I don't know how to make the base number the same because both a and b is variable and not number. Thank you very much for your help. :)

Solve the following simultaneous equations:

$75 = ab^2, 275 = ab^3$

My first time using the Latex software, so please forgive me if I've made a mistake.
• Jul 21st 2010, 05:29 AM
Quote:

Originally Posted by Lites
Hello again guys,

I have five problems to solve but found myself stuck by this question other thought to be very simple but I am unfortunately not. :( Could you guys explain it how to solve this problem? I don't know how to make the base number the same because both a and b is variable and not number. Thank you very much for your help. :)

Solve the following simultaneous equations:

$75 = ab^2, 275 = ab^3$

My first time using the Latex software, so please forgive me if I've made a mistake.

$b\left(ab^2\right)=ab^3\ \Rightarrow\ b=\frac{275}{75}=\frac{11}{3}$

Then $a=\frac{75}{b^2}$
• Jul 21st 2010, 05:52 AM
Lites
Quote:

$b\left(ab^2\right)=ab^3\ \Rightarrow\ b=\frac{275}{75}=\frac{11}{3}$

Then $a=\frac{75}{b^2}$

Thank you for your time Archie Meade, you've always come to help. But when I check it again, the key of the book says Do you know how do they get those answers?

a = 3, b = 5?

I'm sorry if I don't understand you fully.
• Jul 21st 2010, 05:52 AM
HallsofIvy
What I would do is "divide one equation by the other":
$\frac{275}{75}= \frac{ab^3}{ab^2}= b$

$\frac{275}{75}= \frac{11(25)}{3(25)}= \frac{11}{3}$ as Archie Mead said.

Once you know b, replace b in either of the equations by that and solve for a.
• Jul 21st 2010, 06:09 AM
Lites
Quote:

Originally Posted by HallsofIvy
What I would do is "divide one equation by the other":
$\frac{275}{75}= \frac{ab^3}{ab^2}= b$

$\frac{275}{75}= \frac{11(25)}{3(25)}= \frac{11}{3}$ as Archie Mead said.

Once you know b, replace b in either of the equations by that and solve for a.

I just want to clarify several things,

What would you get after dividing both equation with each other? To know the B right? And the B is $\frac{11}{3}$
• Jul 21st 2010, 06:27 AM
eumyang
Quote:

Originally Posted by Lites
Thank you for your time Archie Meade, you've always come to help. But when I check it again, the key of the book says Do you know how do they get those answers?

a = 3, b = 5?

I'm sorry if I don't understand you fully.

These answers can't be the solutions to the system of equations
$75 = ab^2$
$275 = ab^3$
... because if you plug in a and b into the 2nd equation you will get 375 on the right side, not 275. So maybe the original problem should have been
$75 = ab^2$
$375 = ab^3$
... and there was a typo?
• Jul 22nd 2010, 05:07 AM
Lites
There was a typo indeed, I'm so sorry, it should be 375. Thank you for your kind and patient help guys.