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.
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The original area \( A_{\text{original}} \) of the screen can be expressed as: \[ A_{\text{original}} = l \times w = (w + 5) \times w = w^2 + 5w \]
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The new dimensions will be:
- New length: \( 2l = 2(w + 5) = 2w + 10 \)
- New width: \( w - 1 \)
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The new area \( A_{\text{new}} \) will then be: \[ A_{\text{new}} = (2w + 10)(w - 1) \]
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Expanding \( A_{\text{new}} \): \[ A_{\text{new}} = (2w + 10)(w - 1) = 2w^2 - 2w + 10w - 10 = 2w^2 + 8w - 10 \]
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According to the problem, the new area is 170 square inches greater than the original area: \[ A_{\text{new}} = A_{\text{original}} + 170 \] Therefore: \[ 2w^2 + 8w - 10 = w^2 + 5w + 170 \]
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Rearranging this equation will give: \[ 2w^2 + 8w - 10 - w^2 - 5w - 170 = 0 \] \[ w^2 + 3w - 180 = 0 \]
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We can factor this quadratic equation: \[ (w + 15)(w - 12) = 0 \]
So, the possible solutions for \( w \) are: \[ w + 15 = 0 \Rightarrow w = -15 \quad (\text{not valid since width cannot be negative}) \] \[ w - 12 = 0 \Rightarrow w = 12 \]
- 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:
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Original area: \[ A_{\text{original}} = l \times w = 17 \times 12 = 204 \text{ in}^2 \]
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New dimensions: \[ \text{New length} = 2l = 34 \text{ in}, \quad \text{New width} = w - 1 = 11 \text{ in} \]
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New area: \[ A_{\text{new}} = (34)(11) = 374 \text{ in}^2 \]
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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