# another word problem

#### Rutgers24

##### New member
Hi, another word problem from my algebra self-study book I'm stumped on:

By lowering the price of apples one cent a dozen a dealer sells for sixty cents sixty apples more than before. What was the original price per dozen?

Are there 3 unknowns here? P = price of apples, C = cost of apples per dozen, and X = number of apples? From my last question posted I assume P = C * X / 12, but I think I'm having trouble forming the right equation to solve here because I'm not sure how many unknowns I should use. Please help!

#### Dr.Peterson

##### Elite Member
You can use as many variables as you like; some people like starting with several, others prefer as few as possible. That's a matter of style.

Minimally, you want a variable for the quantity you are asked for, the original price per dozen. You could use only that; additional variables may make it easier for you to write equations.

The trouble with your three variables is just that they need more details. Maybe try this:
• C = original price for a dozen apple
• X = original number of apples
Then you can use your equation as an aid in writing the equations you need. (You don't need P as a variable, because that's known: 60 cents.) With two variables, you might write one equation for the original situation, and one after the price is lowered.

Give that a try.

#### Rutgers24

##### New member
ok if P is 60 cents then I write:

60 = (C - 1)(X + 60) / 12

Not sure this is right, also not sure how to write the other "original" situation into an equation that would allow me to substitute either C or X in this.

#### Dr.Peterson

##### Elite Member

(C - 1)(X + 60) / 12 = 60​
If the price per dozen is decreased by 1 cent and the number of apples is increased by 60, the price is 60 cents.​

That sounds good.

What you're missing is the situation before the price change, which is the easy one (I would have written this one first!):

CX/12 = 60​
At the original price per dozen and number of apples, the price is 60​

Try solving this system of equations. Then we can discuss other ways to solve the problem.

#### Rutgers24

##### New member
I see! I think what tripped me up was thinking I couldn't use 60 as the price in the 1st situation, before the change. Thank you for your help Dr. Peterson!