a 5 in. by 7in. picture is enlarged so that the length is 8 in. more than its width Find the length and width.

Answer 1

20x28

Current difference is 2in, needs to be 4x that so just multiply current measurements by 4

Answer 2

Here's the way to figure it out:

To enlarge a picture, you would multiply the length and the width by the same amount (call this value x).

So, the length would be 7 times x (length = 7x) and the width would be 5 times x (width = 5x).

Now, we are trying to find where the length is 8 more than the width, or (length - width = 8)

So, replacing the values we got above: 7x - 5x = 8. This can be simplified to make it 2x = 8. Now, we can solve for x by dividing both sides of the equation by 2, making the answer x=4. So, you take length = 7x = 7 * 4 = 28, and width = 5x = 5 * 4 = 28.

Answer 3

The width is 20 and the length is 28.

This is how you can find it:

5(width) x (width)
----------- = --------------
7(length) x+8(length)

cross multiply and get

5x + 40 = 7x
40 = 2x
20 = x

if you look above x was the variable set for the width; therefore, the width is 20.

x+8 was the variable set for the length; therefore, the length is 28

Therefore, the picture was enlarged 4x.

Answer 4

20 x 28

Answer 5

use ratio

let x = width and x + 8 = length

5 is to 7 as is x is to (x +8)

multiply extremes 5 (x+8) = 5x + 40

multiply 7 times x = 7x

7x = 5x + 40

2x = 40

x = 20

therefore x + 8 = 20 + 8 = 28

so answer is 20 and 28

Answer 6

Assuming that the aspect ratio is kept the same ...
Currently the length is 2 inches more than the width.
To make that difference 8 inches, you would have to enlarge the picture 4 times its original size.
So, 5x7 => 20x28

Answer 7

are you cheating on homework?

Answer 8

20 by 28 duh