Solving Systems of Two Linear Equations by Substitution

TL;DR

You'll learn to solve two equations with two unknowns by isolating one variable and plugging it into the other equation. This method works every time when the system has exactly one solution. You'll get both variables' values in just a few algebraic steps.

1. The Mental Model

You have two equations, each describing a relationship between x and y. The substitution method lets you eliminate one variable by expressing it in terms of the other. You solve for one variable, then back-substitute to find the second. That's the whole idea.

2. The Core Material

The Substitution Process

Start with any system like this:
- Equation 1: 2x + 3y = 11
- Equation 2: x - y = 1

Pick the easier equation to solve for one variable. Here, equation 2 is simpler, so solve for x:
x - y = 1
x = y + 1

Now you've got x in terms of y. Substitute this expression for x into the first equation:
2(y + 1) + 3y = 11
2y + 2 + 3y = 11
5y + 2 = 11
5y = 9
y = 9/5

Finally, substitute y = 9/5 back into x = y + 1:
x = 9/5 + 1 = 9/5 + 5/5 = 14/5

Your solution is x = 14/5, y = 9/5.

Choosing Which Variable to Isolate

Look for coefficients of 1 or -1. If you see x = something or y = something already, use that. If not, pick the variable with the smallest coefficient to minimize fractions. In 3x + y = 7 and 2x - 5y = 4, isolate y from the first equation since its coefficient is 1.

Sometimes you'll have fractional coefficients like 1/2 or 3/4. Don't panic—just work through the algebra carefully. You can always multiply the entire equation by the denominator first to clear fractions if that feels easier.

Verification Strategy

Always check your answer in both original equations. If x = 14/5 and y = 9/5:
- Check equation 1: 2(14/5) + 3(9/5) = 28/5 + 27/5 = 55/5 = 11 ✓
- Check equation 2: 14/5 - 9/5 = 5/5 = 1 ✓

If either check fails, you made an arithmetic error somewhere. Don't skip this step—it catches mistakes before you move on.

flowchart TD
    A["Two equations with x and y"] --> B["Pick easier equation"]
    B --> C["Solve for one variable"]
    C --> D["Substitute into other equation"]
    D --> E["Solve for remaining variable"]
    E --> F["Back-substitute to find first variable"]
    F --> G["Check both values in original equations"]

Special Cases You'll Encounter

Sometimes you'll get 0 = 0 after substitution. This means infinitely many solutions—the equations represent the same line. If you get something like 5 = 0, there's no solution because the lines are parallel.

On the SAT, most problems have exactly one solution, but knowing these special cases prevents confusion if you encounter them.

3. Worked Example

Let's solve this system:
- 3x + 2y = 16
- x + y = 6

Step 1: Choose which variable to isolate. The second equation has coefficients of 1, so I'll solve for x:
x + y = 6
x = 6 - y

Step 2: Substitute x = 6 - y into the first equation:
3(6 - y) + 2y = 16
18 - 3y + 2y = 16
18 - y = 16
-y = 16 - 18
-y = -2
y = 2

Step 3: Find x by substituting y = 2 back into x = 6 - y:
x = 6 - 2 = 4

Step 4: Verify in both original equations:
- First equation: 3(4) + 2(2) = 12 + 4 = 16 ✓
- Second equation: 4 + 2 = 6 ✓

The solution is x = 4, y = 2.

4. Examiner's Breakdown

4.1 What Examiners Actually Reward

• Correct isolation: Show x = [expression in y] or y = [expression in x] clearly
• Proper substitution: Write the substituted equation completely before simplifying
• Combined like terms: Show 3y + 2y = 5y explicitly
• Back-substitution work: Don't just write the final answer—show how you found the second variable
• Verification step: Check your solution in at least one original equation
• Solution format: Write your final answer as an ordered pair (x, y) or clearly state both values

4.2 Trapdoor Mistakes

• Sign errors during substitution: When substituting x = y + 1 into 3x - 2y = 5, students write 3(y + 1) - 2y but then get the signs wrong. Distribute carefully: 3y + 3 - 2y.
• Forgetting to back-substitute: Finding y = 3 but forgetting to plug it back to find x. Always find both variables.
• Arithmetic errors with fractions: When y = 7/3, substituting into x = 2y gives x = 14/3, not x = 9/3. Multiply the numerators.
• Using the wrong equation for verification: Checking your answer in the modified equations instead of the original two equations you started with.

4.3 Score-Boosting Comparisons

5. Now Try It

Solve this system using substitution: 2x - y = 5 and x + 3y = 7. Work through each step—isolate a variable, substitute, solve, back-substitute, and verify. Take about 15 minutes and show all your work.

What success looks like: You should get specific numerical values for both x and y, and when you plug them back into both original equations, you get true statements.