SOLUTION: I can not use algebra. I painted a picture that is 4 inches longer than it is wide. If I put a 1 inch frame around my picture, the area increases by 48 square inches. What are t

Question 780636: I can not use algebra.
I painted a picture that is 4 inches longer than it is wide. If I put a 1 inch frame around my picture, the area increases by 48 square inches. What are the dimensions of my picture?

Answer by josgarithmetic(39625) (Show Source): You can put this solution on YOUR website!
You said, "I can not use algebra."
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Do you mean, you have too much difficulty trying to use algebra? An analysis of the exercise can be shown and explained, and will use algebra.
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Do you mean, the use of algebra is not permitted for this exercise? That prohibition is unnacceptable; algebra is very well suited and is expected to this exercise.

Painted picture, w for width, y for length.

Area for just the picture is

1 inche frame extends length and width to be now
width = w+1+1=
length = y+1+1=(4+w)+1+1=
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Think about those distance size changes very carefully before continuing. We add 2, not just 1. We extend that 1 inch in TWO opposite directions, so that the 1 inche distance frame-width goes all the way around the picture.

The extended area is

Given is that the area increases by 48 square inches. This means the difference between the two areas is 48 square inches.

That appears to be a quadratic equation and just needs simplification. Some help in starting through that is,...

In fact this will not remain as a quadratic equation, since w^2-w^2=0.
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and you can compute the y value yourself.
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