Example

Solving a Three-Sided Fencing Problem Using a System of Equations by Substitution

Apply the seven-step problem-solving strategy for systems of linear equations to a real-world geometry problem where only three sides of a rectangle need fencing (because the fourth side is an existing wall), using substitution to solve the resulting system.

Problem: Randall has 125 feet of fencing to enclose the rectangular part of his backyard adjacent to his house. He will only need to fence around three sides, because the fourth side will be the wall of the house. He wants the length of the fenced yard (parallel to the house wall) to be 5 feet more than four times the width. Find the length and the width.

  1. Read the problem and draw a rectangle with one side against a house wall.
  2. Identify: The length and width of the fenced yard.
  3. Name: Let LL = the length of the fenced yard and WW = the width of the fenced yard.
  4. Translate into a system of equations. Because only three sides require fencing — one length and two widths — the fencing constraint is L+2W=125L + 2W = 125 (not the standard perimeter formula 2L+2W2L + 2W). The relationship between the dimensions gives the second equation: L=4W+5L = 4W + 5. The system is:

{L+2W=125L=4W+5\left\{\begin{array}{l} L + 2W = 125 \\ L = 4W + 5 \end{array}\right.

  1. Solve using substitution. Because the second equation is already solved for LL, substitute 4W+54W + 5 for LL in the first equation:

4W+5+2W=1254W + 5 + 2W = 125

Combine like terms: 6W+5=1256W + 5 = 125. Subtract 5: 6W=1206W = 120. Divide by 6: W=20W = 20.

Substitute W=20W = 20 into the second equation: L=4(20)+5=80+5=85L = 4(20) + 5 = 80 + 5 = 85.

  1. Check: Three fenced sides: 20+85+20=12520 + 85 + 20 = 125 ✓. Is the length 5 more than four times the width? 420+5=854 \cdot 20 + 5 = 85 ✓.
  2. Answer: The length is 85 feet and the width is 20 feet.

This example differs from standard rectangle perimeter problems in one important way: because one side of the rectangle is a house wall that does not need fencing, the fencing equation is L+2W=125L + 2W = 125 rather than 2L+2W=P2L + 2W = P. Recognizing which sides require fencing — and writing the correct constraint equation from the physical setup — is the key translation step. Once the system is formed, the solving process (substitution, combining like terms, back-substituting) follows the same pattern as a full-perimeter problem.

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Updated 2026-04-21

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