Solving Systems by Substitution

TipLearning Objectives

By the end of this lesson, you’ll be able to:

  • Recognize when substitution is the most efficient solving method.
  • Substitute expressions correctly into the other equation.
  • Solve for both variables and check your answers.

Key Ideas

Substitution is often the cleanest method when:

  • One equation is already isolated (e.g., \(y = 3x - 1\)).
  • One equation can be isolated quickly with one step.
  • A variable has coefficient 1 or –1, making isolation simple.

General plan:

  1. Isolate one variable in either equation.
  2. Substitute that expression into the other equation.
  3. Solve the resulting single-variable equation.
  4. Substitute back to find the second variable.
  5. Check the solution in both equations.

Common Problem Types

When a variable is already alone

Ideal form: \(y = \dots\) or \(x = \dots\).

When a variable is easy to isolate

Example: \(x - 2y = 7\) → \(x = 7 + 2y\).

When substitution avoids fractions

Sometimes isolating the “simplest-looking” variable reduces complexity.

When substitution is cleaner than elimination

Especially when the coefficients don’t align nicely.

Strategies

  • Pick the equation that is easier to isolate—avoid fractions when possible.
  • Substitute carefully using parentheses, especially with negatives.
  • Write the substituted equation on a new line to keep steps clean.
  • Always solve the simpler equation first to avoid messy algebra.
  • After finding both variables, do a quick check in the original system.

Worked Examples

Example 1 — Variable Already Isolated

Solve: \[ \begin{cases} y = 2x + 1 \\ 3x + y = 10 \end{cases} \]

Substitute the expression for \(y\): \[ 3x + (2x + 1) = 10 \]

Simplify: \[ 5x + 1 = 10 \Rightarrow 5x = 9 \Rightarrow x = \frac{9}{5} \]

Now plug back: \[ y = 2\left(\frac{9}{5}\right) + 1 = \frac{23}{5} \]

Solution:
\[ \left(\frac{9}{5}, \frac{23}{5}\right) \]

Example 2 — Isolate First, Then Substitute

Solve: \[ \begin{cases} x + 2y = 11 \\ 3x - y = 4 \end{cases} \]

Isolate \(x\) in the first equation: \[ x = 11 - 2y \]

Substitute into the second: \[ 3(11 - 2y) - y = 4 \]

Simplify: \[ 33 - 6y - y = 4 \] \[ 33 - 7y = 4 \] \[ -7y = -29 \Rightarrow y = \frac{29}{7} \]

Then substitute back to find \(x\): \[ x = 11 - 2\left(\frac{29}{7}\right) \]

(You can leave in fractional form.)

WarningCommon Mistakes
  • Substituting into the wrong equation.
  • Forgetting parentheses when substituting expressions like \(-x + 5\) or \(3y - 2\).
  • Solving for one variable but never computing the second.
  • Dropping negatives during substitution.

Practice Problems

  1. \[ \begin{cases} y = 4x \\ x + y = 15 \end{cases} \]

  2. \[ \begin{cases} x = 3y - 2 \\ 2x + y = 13 \end{cases} \]

  3. \[ \begin{cases} y = -x + 5 \\ 2x + y = 4 \end{cases} \]

1.
Substitute into \(x + y = 15\):
\(x + 4x = 15\)
\(5x = 15\) → \(x = 3\)
\(y = 4(3) = 12\)
Answer: \((3, 12)\)

2.
\(2(3y - 2) + y = 13\)
\(6y - 4 + y = 13\)
\(7y = 17\) → \(y = \frac{17}{7}\)
\(x = 3y - 2 = \frac{23}{7}\)
Answer: \(\left(\frac{23}{7}, \frac{17}{7}\right)\)

3.
Substitute:
\(2x + (-x + 5) = 4\)
\(x + 5 = 4\) → \(x = -1\)
\(y = -(-1) + 5 = 6\)
Answer: \((-1, 6)\)

Summary

  • Substitution works best when a variable is already isolated or easy to isolate.
  • Always substitute carefully using parentheses.
  • After solving for one variable, substitute back to get the other.
  • Check solutions in both equations for accuracy.
  • Choose the equation that is easiest to isolate.
  • Parentheses are your friend—use them when substituting expressions.
  • If substitution gets messy, consider switching to elimination.
  • Always find both variables before declaring a solution.