# Profit Help

• April 14th 2014, 06:07 AM
awfulatmath
Profit Help
I'm having a little difficulty solving growth/decay facts, I suppose this question is about that but I'm having a little difficulty understanding what it means

Question: Keller industries' profit were up $50,000$ this year over last year. This was an increase of $12.5%$

Let T represent the profit this year and L the profit from last year and write a system of equations that can be used to determine the profits?
• April 14th 2014, 06:42 AM
earboth
Re: Profit Help
Quote:

Originally Posted by awfulatmath
...
Question: Keller industries' profit were up $50,000$ this year over last year. This was an increase of $12.5%$

Let T represent the profit this year and L the profit from last year and write a system of equations that can be used to determine the profits?

Hello,

you only have to extract 2 equations from the text:

$\begin{array}{r}L + \frac{12.5}{100} \cdot L = T \\ L + 50,000 = T \end{array}$

Solve for L and T.
• April 14th 2014, 09:29 AM
awfulatmath
Re: Profit Help
Thanks so much, for the equations.

Now when solving is it possible to use the Elimination or Substitution method when solving? Since I can solve for both L and T variables?
• April 14th 2014, 09:49 AM
romsek
Re: Profit Help
Quote:

Originally Posted by awfulatmath
Thanks so much, for the equations.

Now when solving is it possible to use the Elimination or Substitution method when solving? Since I can solve for both L and T variables?

either will work. Since you have two expressions for T you might as well just set them equal and immediately eliminate T from the mix. You'd get

$L+\dfrac{12.5}{100}\cdot L = L + 50,000$

and this can be solved for $L$

$L$ can then be used to find $T$ from either equation.
• April 14th 2014, 03:35 PM
awfulatmath
Re: Profit Help
I solved for $L$ and received $.125$

Or is that completely wrong? Would I take the value for $L$ and plug into $T$?
• April 14th 2014, 04:16 PM
HallsofIvy
Re: Profit Help
What do you think? 50,000 was an increase of 12.5 (% although you didn't say that) of last year's profit. Is 12.5% of .125 equal to 50,000?
You are really saying that .125L= 50000.

(Note the "0.125", NOT 12.5!)
• April 15th 2014, 04:54 PM
romsek
Re: Profit Help
Quote:

Originally Posted by awfulatmath
I solved for $L$ and received $.125$

Or is that completely wrong? Would I take the value for $L$ and plug into $T$?

it's wrong.

subtract $L$ from both sides and you are left with

$0.125 L = 50000$

surely you can solve it from here.

Then $T=L + 50000=$