# WORD PROBLEM HELP PLEASE!You are going to hang a painting in your living room. If the painting's width is three inches shorter than the length and the frame, which is 4 inches wide (by that it...

WORD PROBLEM HELP PLEASE!

You are going to hang a painting in your living room. If the painting's width is three inches shorter than the length and the frame, which is 4 inches wide (by that it means the actual length of the wall to wall on the inside of the frame), and the total area of the painting and frame is 270 square inches, what are the dimensions of the painting and frame?

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First state what is known:

Large Area (A)= 270 sq in

I'm a little confused with the reference to the frame "from wall to wall" but think you mean that the frame itself is 4 inches wide all round which means that the length and width of the painting would be 8 inches smaller than the frame (as there is 4 in on left, on right, top and bottom).

The length of the painting is l and therefore the width is( l-3). So from the formula for area of a rectangle =l x b we deduce:

A = l(l-3)

The width of the overall painting plus frame will also be 3 inches narrower (shorter). Therefore:

270 = l(l-3)

270 = l^2 - 3l

0= l^2 - 3l - 270

0= (l+15)(l-18)

l=-15 (but l cannot equal -15)

or l = 18

It follows that the length of the painting plus frame is 18 in, the width is 15 in and as there is 4 in each side to deduct, the width of the painting is (15-8) 7in and the length is (18-8) 10 in as the frame is the same width all the way round.

**Painting: 10in by 7 in**

**Frame and painting: 18in by 15 in**