a photograph of a child has a width 2 inches longer than its length.

we let L = length and W = width.

we get:

L = W + 2

that we hold for future reference.

the proud parents decide the photo is cute enough that they want a larger version. the picture is made bigger by increasing both its length and its width by 2 inches, its area increases by 20 square inches.

we get (L + 2) * (W + 2) = A + 20

original Area is given by the equation:

L * W = A

what are the dimensions of the original picture?

we have 3 equations to work with:

equation 1: L = W + 2

equation 2: (L + 2) * (W + 2) = A + 20

equation 3: (L * W) = A

we can use these relationships to reduce the number of unknowns down to 1.

that would be W.

first we replace A with (L * W).

equation 2 becomes: (L + 2) * (W + 2) = (L * W) + 20

next we replace L with W + 2.

equation 2 becomes: (W + 2 + 2) * (W + 2) = ( (W + 2) * W ) + 20

we now solve for W.

equation 2 start out as:

(W + 2 + 2) * (W + 2) = ( (W + 2) * W ) + 20

this becomes:

(W + 4) * (W + 2) = ( (W + 2) * W ) + 20

multiplying the factors out we get:

W^2 + 6W + 8 = W^2 + 2W + 20

if we subtract W^2 + 2W from both sides of the equation, we get:

4W + 8 = 20

if we subtract 8 = from both sides of the equation, we get:

4W = 12

divide both sides by 4 to get:

W = 3

since L = W + 2, we get:

L = 5

original dimensions of the photograph are

Length = 5 and width = 3

new photo becomes length = 7 and width = 5

area of the original photo is 5 * 3 = 15 square inches.

area of the larger photo is 7 * 5 = 35 square inches.

area has increased by 20 square inches.

all requirements have been satisfied so the answer is:

length of the original photo is 5 inches.

width of the original photo is 3 inches.