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Explanation:
The old dimensions of the rectangle were 30 by 15.
The new dimensions are 30+x by 15+x, where x is some positive number.
The new rectangle dimensions multiply to 1000
(30+x)(15+x) = 1000
Use the FOIL rule to expand out the left side like so
(30+x)(15+x) = 1000
450+30x+15x+x^2 = 1000
450+45x+x^2 = 1000
x^2+45x+450
Then lets get everything to one side
x^2+45x+450 = 1000
x^2+45x+450-1000 = 0
x^2+45x-550 = 0
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From here we use the quadratic formula
Plug in a = 1, b = 45, and c = -550.
[tex]x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(45)\pm\sqrt{(45)^2-4(1)(-550)}}{2(1)}\\\\x = \frac{-45\pm\sqrt{4225}}{2}\\\\[/tex]
[tex]x = \frac{-45\pm65}{2}\\\\x = \frac{-45+65}{2} \ \text{ or } \ x = \frac{-45-65}{2}\\\\x = \frac{20}{2} \ \text{ or } \ x = \frac{-110}{2}\\\\x = 10 \ \text{ or } \ x = -55\\\\[/tex]
Earlier we defined x to be a positive number. This means we ignore x = -55. It makes no sense to add on a negative amount to each dimension.
The only practical answer is x = 10.
So each dimension is increased by 10 feet.
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If x = 10, then the old dimensions of 30 by 15 bump up to 30+10 = 40 by 15+10 = 25
Then note how 40*25 = 1000, which helps confirm we have the correct dimensions.
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Now go back to the question at hand: we want to find the new width. The old width was 30 feet. The new width is 30+x = 30+10 = 40 feet