Example

Solving a Two-Number Word Problem Using a System of Equations by Substitution

Apply the seven-step problem-solving strategy for systems of linear equations to an abstract number problem, using substitution to solve the resulting system.

Problem: The sum of two numbers is zero. One number is nine less than the other. Find the numbers.

  1. Read the problem.
  2. Identify what to find: two numbers.
  3. Name the unknowns: Let nn = the first number and mm = the second number.
  4. Translate into a system of equations. "The sum of two numbers is zero" gives n+m=0n + m = 0. "One number is nine less than the other" gives n=m9n = m - 9. The system is:

{n+m=0n=m9\left\{\begin{array}{l} n + m = 0 \\ n = m - 9 \end{array}\right.

  1. Solve using substitution. Because the second equation is already solved for nn, substitute m9m - 9 for nn in the first equation:

m9+m=0m - 9 + m = 0

Combine like terms: 2m9=02m - 9 = 0. Add 99 to both sides: 2m=92m = 9. Divide both sides by 22: m=92m = \frac{9}{2}.

Substitute m=92m = \frac{9}{2} into the second equation to find nn:

n=929=92182=92n = \frac{9}{2} - 9 = \frac{9}{2} - \frac{18}{2} = -\frac{9}{2}

  1. Check: Does 92+(92)=0\frac{9}{2} + \left(-\frac{9}{2}\right) = 0? Yes ✓. Is 92-\frac{9}{2} nine less than 92\frac{9}{2}? 929=92\frac{9}{2} - 9 = -\frac{9}{2} ✓.
  2. Answer: The numbers are 92\frac{9}{2} and 92-\frac{9}{2}.

This example demonstrates how using two variables and a system of equations can simplify the setup of a word problem. Each verbal statement translates directly into one equation — one for the sum constraint and one for the relationship between the numbers — without needing to express both unknowns through a single variable. The substitution method is ideal here because the second equation already expresses nn in terms of mm, so substitution can begin immediately. The fractional solution reinforces the practical advantage of substitution over graphing for non-integer answers.

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Updated 2026-04-21

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