Question 1014933
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The perimeter of rectangle is 80 dm. The width is 1 more than one half the length. Find the length and width of the rectangle.
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You need to solve this system of two equations

x + y = 40,
y - 1 = {{{x/2}}},

where x is the length and y is the width.

Substitute y = 40 -x from (1) into (2).

You will get 

(40 - x) - 1 = {{{x/2}}}.

Multiply this equation by 2. You will get

80 - 2x - 2 = x,   or

78 + x + 2x = 3x.

Hence, x = {{{78/3}}} = 26 dm. It is the length.

The width is 40 - 26 = 14 dm.
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