Systems of Equations: Substitution Method

The substitution method solves a system of equations by rearranging one equation for a single variable and substituting it into the other. This lesson works through the steps, including what happens when a system has no solution or infinitely many.

The substitution method solves a system by:

1. Solving one equation for one variable (usually the easiest one, isolated or with coefficient 1)
2. Substituting that expression into the other equation
3. Solving the resulting one-variable equation
4. Substituting the value back into either original equation to find the second variable
5. Writing the solution as (x, y) and checking in both equations

Example system:

$\begin{cases} y = 3x - 2 \\ 2x + y = 7 \end{cases}$

1. First equation already solves for y: y = 3x − 2
2. Substitute into second: 2x + (3x − 2) = 7
3. 5x − 2 = 7 → 5x = 9 → x = 9/5 = 1.8
4. y = 3(1.8) − 2 = 5.4 − 2 = 3.4

Solution: (9/5, 17/5) or (1.8, 3.4)

Special cases:

• No solution: substitution leads to contradiction (e.g., 0 = 5)
• Infinite solutions: substitution leads to identity (e.g., 0 = 0)

Solve:

$\begin{cases} 3x + 2y = 13 \\ x = y + 1 \end{cases}$

1. Second equation already solved for x: x = y + 1
2. Substitute into first: 3(y + 1) + 2y = 13
3. 3y + 3 + 2y = 13 → 5y + 3 = 13 → 5y = 10 → y = 2
4. x = 2 + 1 = 3

Solution: (3, 2)

Check:

• 3(3) + 2(2) = 9 + 4 = 13 ✓
• x = 3, y + 1 = 2 + 1 = 3 ✓

Both true.

Another (no solution case):

$\begin{cases} 2x + 3y = 5 \\ 4x + 6y = 12 \end{cases}$

1. Solve first for x: 2x = 5 − 3y → x = (5 − 3y)/2
2. Substitute: 4((5 − 3y)/2) + 6y = 12
3. 2(5 − 3y) + 6y = 12 → 10 − 6y + 6y = 12 → 10 = 12 (contradiction)

No solution (parallel lines).

Substitution is great for systems where one variable is easy to isolate:

• Two phone plans: Plan A: 30 + 0.10t = total, Plan B: 45 + 0.05t = total. Set equal: 30 + 0.10t = 45 + 0.05t → solve for t (texts when equal)
• Two jobs: Job A: 20h + 50 bonus, Job B: 18h + 120 bonus. 20h + 50 = 18h + 120 → substitution finds h (hours when pay matches)
• Mixture: Blend 5 dollars per pound coffee with 8 dollars per pound to get 6.50 dollars per pound. Set up system and substitute to find amounts.

This method helps compare options, find balance points, or mix solutions.