# There is a rectangular picture frame with dimensions width=x+3 and length=3x-5, the picture inside the frame has dimensions of width=x-1 and length=8. the area of the shaded region is 48 inches...

There is a rectangular picture frame with dimensions width=x+3 and length=3x-5, the picture inside the frame has dimensions of width=x-1 and length=8. the area of the shaded region is 48 inches squared.

solve for x and give the dimension of the picture frame

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Assuming that the shaded area is the space between the frame and the picture inside, create an equationremembering that

A=l x b

The outer rectangle (the frame): A= (x + 3) x (3x-5)

The inner rectangle (picture): A = (x-1) x 8

The shaded area is the large rectangular space - small space = 48

`therefore48= (x+3)(3x-5) - 8(x-1)`

`therefore 48 = 3x^2 + 9x -5x -15 -8x +8`

Note that the symbol changes when -8 is times-ed by -1

Simplify and bring everythng together

`therefore 0 = 3x^2 - 4x - 55`

Now factorize using the factors of the first term (1`x` x 3`x` ) and those of the 3rd term (1 x 55 OR 5 x 11). The `1x times 3x` works with the 5 x 11 to create the middle term -4x

`therefore (x -5)(3x + 11) = 0`

`therefore x=5 or x= -11/3`

For our purposes only the value of **x=5 **is relevant as a length cannot be negative .If x=5 we can find the dimensions of the frame by substitution. The dimensions are: `(3x - 5) and (x+3)`

`= (3(5) - 5) and (5+3)`

**= 10 inches and 8 inches**

**x=5 and the dimensions are 10in by 8in**

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