Solution

There is a common sense solution and an algebraic one. (Both admit quite a few variations.)

If a pen is twice more expensive than a pencil, 3 pencils and 4 pen cost as much as 3 + 2×4 = 11 pencils,markers,pens,pencils. By the same token, 4 pencils and 3 pens would cost as much as 2×3+4=10,2×3+4=10,3+2×4=11,3×4=12,3×3+2×2=13 pencils. The difference of $1 is then the price of 11 - 10 = 1 pencil. So a pencil goes for $1 and a pen which is twice as expensive costs $2.

You are welcome to check the result (and checking your results is a very good practice) by adding the costs of pens and pencils in the required quantities and verifying that the difference is indeed $1. We shall solve the problem algebraically and see if that leads to the same answer.

Let x be the price of a pencil and y the price of a pen. We know that

y = 2x.

The cost of 3 pencils and 4 pens is $1 more than the he cost of 4 pencils and 3 pens translates into

(3x + 4y) - (4x + 3y) = 1.

The equation is simplified to

y - x = 1.

It is important to understand that this is not the only way to compose the equations, and what we are going to do is not the only way to solve the equations. Here's just one possibility.

Recollect that y = 2x, and replace y in the equation with 2x to obtain

2x - x = x = 1.

So a pencil costs $1 and a pen $2 (which can be computed as either $1×2 or $1 + $1.)

The algebraic solution could have been shorten had we surmised at the outset that the conditions of the problem mean that in the second case the mom would have purchased 1 pen less but 1 pencil more than she actually bought. So it is quite immediate that $1 is the difference in the prices of a pen and a pencil:

y - x = 1.