1. ## 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:

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

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

2. 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:

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

My first time using the Latex software, so please forgive me if I've made a mistake.
$\displaystyle b\left(ab^2\right)=ab^3\ \Rightarrow\ b=\frac{275}{75}=\frac{11}{3}$

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

3. Originally Posted by Archie Meade
$\displaystyle b\left(ab^2\right)=ab^3\ \Rightarrow\ b=\frac{275}{75}=\frac{11}{3}$

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

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

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

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

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

$\displaystyle \frac{275}{75}= \frac{11(25)}{3(25)}= \frac{11}{3}$ 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 $\displaystyle \frac{11}{3}$

6. 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
$\displaystyle 75 = ab^2$
$\displaystyle 275 = ab^3$
... 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
$\displaystyle 75 = ab^2$
$\displaystyle 375 = ab^3$
... and there was a typo?

7. There was a typo indeed, I'm so sorry, it should be 375. Thank you for your kind and patient help guys.