Question

Solve the problem. Check answers to be sure they are reasonable.
The dimensions of a rectangular monitor screen are such that its length is 5 in. more than its width. If the length were doubled and if the width were decreased by 1 ​in., the area would be increased by 170 in squared. What are the length and width of the​ screen?

Answer 1

Let the width of the monitor screen be \( w \) inches. Since the length is 5 inches more than the width, the length can be expressed as \( l = w + 5 \) inches.

Now, we can set up the equation based on the information that if the length is doubled and the width is decreased by 1 inch, the area increases by 170 square inches.

1. The original area \( A_{\text{original}} \) of the screen can be expressed as:
\[
A_{\text{original}} = l \times w = (w + 5) \times w = w^2 + 5w
\]

2. The new dimensions will be:
- New length: \( 2l = 2(w + 5) = 2w + 10 \)
- New width: \( w - 1 \)

3. The new area \( A_{\text{new}} \) will then be:
\[
A_{\text{new}} = (2w + 10)(w - 1)
\]

4. Expanding \( A_{\text{new}} \):
\[
A_{\text{new}} = (2w + 10)(w - 1) = 2w^2 - 2w + 10w - 10 = 2w^2 + 8w - 10
\]

5. According to the problem, the new area is 170 square inches greater than the original area:
\[
A_{\text{new}} = A_{\text{original}} + 170
\]
Therefore:
\[
2w^2 + 8w - 10 = w^2 + 5w + 170
\]

6. Rearranging this equation will give:
\[
2w^2 + 8w - 10 - w^2 - 5w - 170 = 0
\]
\[
w^2 + 3w - 180 = 0
\]

7. We can factor this quadratic equation:
\[
(w + 15)(w - 12) = 0
\]

So, the possible solutions for \( w \) are:
\[
w + 15 = 0 \Rightarrow w = -15 \quad (\text{not valid since width cannot be negative})
\]
\[
w - 12 = 0 \Rightarrow w = 12
\]

8. Now we can find the length:
\[
l = w + 5 = 12 + 5 = 17
\]

Thus, the width of the screen is \( 12 \) inches and the length is \( 17 \) inches.

### Conclusion:
- Width: \( 12 \) inches
- Length: \( 17 \) inches

### Verification:
1. Original area:
\[
A_{\text{original}} = l \times w = 17 \times 12 = 204 \text{ in}^2
\]

2. New dimensions:
\[
\text{New length} = 2l = 34 \text{ in}, \quad \text{New width} = w - 1 = 11 \text{ in}
\]

3. New area:
\[
A_{\text{new}} = (34)(11) = 374 \text{ in}^2
\]

4. Change in area:
\[
\Delta A = A_{\text{new}} - A_{\text{original}} = 374 - 204 = 170 \text{ in}^2
\]

The values check out, so the solution is confirmed:
- Width: \( 12 \) inches
- Length: \( 17 \) inches