Introductory Algebra for College Students (7th Edition)

The width of the rectangular lot is $70$ yards and the length is $130$ yards.
The formula for the perimeter of a rectangle is given by the following formula: $$P = 2l + 2w$$ From the problem, we know that the perimeter of the rectangular lot is $400$ yards and that the length $l$ of the lot is $2w - 10$. We can now devise an equation that incorporates all of this information: $$400 = 2(2w - 10) + 2w$$ First, we distribut what is in the parentheses: $$400 = 4w - 20 + 2w$$ Simplify by combining like terms: $$400 = 6w - 20$$ Add $20$ to each side of the equation to isolate the constants on one side of the equation: $$6w = 420$$ Divide each side of the equation by $6$ to isolate $w$: $$w = 70$$ We can now substitute $70$ in for $w$ to find the length $l$: $$l = 2(70) - 10$$ Multiply first, according to order of operations: $$l = 140 - 10$$ Subtract to solve for $l$: $$l = 130$$ The width of the rectangular lot is $70$ yards and the length is $130$ yards.