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The perimeter of a rectangle is 88. The length is five more than twice the width. Find the length and the width.
ALength 13 and width 31
BLength 44 and width 22
CLength 29 and width 15
DLength 31 and width 13
Answer & Solution
Correct answer: D. Length 31 and width 13
1. Let L be the length and W be the width, giving the system 2L + 2W = 88 and L = 2W + 5.
2. Substitute 2W + 5 for L in the perimeter equation: 2(2W plus 5) plus 2W equals 88.
3. Distribute and simplify: 4W plus 10 plus 2W equals 88, so 6W equals 78, giving W equals 13.
4. Substitute W = 13 into L = 2W + 5 to get L equals 31, and checking confirms 2(31) plus 2(13) equals 88.
_Source: OpenStax Elementary Algebra (CC BY 4.0), Ch 5 "Systems of Linear Equations", section 5.2 Solve Systems of Equations by Substitution_
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