4.3Solving Systems by Substitution
Isolate one variable, substitute its expression into the other equation, and solve algebraically for an exact solution.
Why This Matters
Substitution is a powerful algebraic technique used far beyond systems — it's the same idea behind u-substitution in Calculus and variable replacement in Physics. Mastering it now pays dividends in every advanced math course.
Workbook
Lesson, vocabulary, worked examples, and practice problems.
Essential Question
How does replacing one variable with an equivalent expression allow us to solve a system of equations algebraically, and when is substitution more efficient than graphing?
Lesson Overview
The substitution method solves a system algebraically by replacing one variable with an equivalent expression. This reduces two equations with two unknowns to a single equation with one unknown — which we already know how to solve. Substitution gives an exact answer, unlike graphing which can be imprecise. It works best when one variable is already isolated or has a coefficient of 1.
5-step substitution method
System:
y = 2x + 1
3x + y = 16
Substitute y = 2x+1 into Eq 2:
3x + 2x + 1 = 16
5x + 1 = 16
5x = 15
x = 3
Back-substitute x = 3:
y = 2(3) + 1 = 7
Solution: (3, 7)
Color-coded substitution path
Substitution finds the exact intersection (3, 7)
Substitution Steps
- Isolate one variable (y = …)
- Substitute the expression into the other equation
- Solve the single-variable equation
- Back-substitute to find the other variable
- Verify in both original equations
Before → After Substitution
Before
y = x + 3
2x + y = 9
After
2x+(x+3)=9
3x = 6
x = 2
One equation, one variable — easy to solve!
Substitution reduces two equations to one
Equation Balance: y = x + 4
Replace y with (x + 4) in the other equation:
3x + 4 = 10
x = 2, y = 6
Substitution keeps the equation balanced
Graphing vs. Substitution — when to use each
Graphing vs. Substitution — when to use each
Worked Examples
Step 1 — Isolate
y = 2x + 1
Variable is alone on one side
Step 2 — Substitute
3x + (2x+1) = 16
Replace y with the expression
Step 3 — Solve
5x + 1 = 16 → x = 3
One variable, one equation
Solve the system by substitution: y = 2x + 1 and 3x + y = 16
Equation 1 already has y isolated: y = 2x + 1.
Substitute into Equation 2: 3x + (2x + 1) = 16.
Simplify: 5x + 1 = 16 → 5x = 15 → x = 3.
Back-substitute: y = 2(3) + 1 = 7.
Verify: Eq 1: 7 = 2(3)+1 = 7 ✓ Eq 2: 3(3)+7 = 9+7 = 16 ✓
Graph confirms solution (3, 7)
Verification: Check (3, 7)
Equation 1
y = 2x + 1
7 = 2(3)+1
7 = 7 ✓
Equation 2
3x+y=16
3(3)+7=16
16=16 ✓
Both equations satisfied → (3, 7) is correct!
Always verify in both original equations
Solve the system by substitution: y = x + 4 and 2x + y = 10
Equation 1 already has y isolated: y = x + 4.
Substitute into Equation 2: 2x + (x + 4) = 10.
Simplify: 3x + 4 = 10 → 3x = 6 → x = 2.
Back-substitute: y = 2 + 4 = 6.
Verify: Eq 1: 6 = 2+4 ✓ Eq 2: 2(2)+6 = 10 ✓
Graph confirms solution (2, 6)
Solve the system by substitution: x + 2y = 8 and 3x − y = 3
Isolate x in Equation 1: x = 8 − 2y.
Substitute into Equation 2: 3(8 − 2y) − y = 3.
Expand: 24 − 6y − y = 3 → 24 − 7y = 3 → −7y = −21 → y = 3.
Back-substitute: x = 8 − 2(3) = 8 − 6 = 2.
Verify: Eq 1: 2+2(3) = 8 ✓ Eq 2: 3(2)−3 = 3 ✓
Graph confirms solution (2, 3)
Solve the system by substitution: y = 2x + 3 and 4x − 2y = −6
Equation 1 already has y isolated: y = 2x + 3.
Substitute into Equation 2: 4x − 2(2x + 3) = −6.
Expand: 4x − 4x − 6 = −6 → −6 = −6.
This is always true (identity) → the equations are the same line.
Infinitely Many Solutions
Solve the system by substitution: y = 2x + 3 and y = 2x − 1
Both equations have y isolated.
Set them equal: 2x + 3 = 2x − 1.
Subtract 2x from both sides: 3 = −1.
This is a contradiction → no solution.
No Solution — Parallel Lines
Guided Practice
Guided Practice Video: Solving Systems by Substitution
Watch the guided practice walkthrough for solving systems by substitution, then complete the problems below.
Video by Sang Real Math
Watch on YouTube ↗Solve by substitution: y = 3x − 2 and x + y = 6.
Hint: Equation 1 already has y isolated. Substitute (3x − 2) for y in Equation 2, then solve for x.
Solve by substitution: 2x + y = 8 and y = x − 1.
Hint: Equation 2 has y isolated. Substitute (x − 1) for y in Equation 1.
Solve by substitution: x + 3y = 9 and x = 2y.
Hint: Equation 2 has x isolated. Substitute 2y for x in Equation 1.
Solve by substitution: y = 4x + 1 and 8x − 2y = −2. How many solutions?
Hint: Substitute y = 4x+1 into the second equation. What happens when you simplify?
Solve by substitution: 3x − y = 5 and y = 3x + 2. How many solutions?
Hint: Substitute y = 3x+2 into the first equation. Do you get a true or false statement?
Key Vocabulary
Substitution Method
An algebraic technique where one variable is replaced by an equivalent expression derived from another equation.
Isolate a Variable
Rewrite an equation so one variable is alone on one side: y = … or x = …
Back-Substitution
After finding the value of one variable, substitute it back into an original equation to find the other variable.
Equivalent Expression
An expression equal in value to another; used to replace a variable during substitution.
Algebraic Solution
An exact answer found through equation manipulation, as opposed to a graphical estimate.
Contradiction
A false statement (e.g., 0 = 5) that results from substitution, indicating no solution.
Identity
A statement that is always true (e.g., 0 = 0) that results from substitution, indicating infinitely many solutions.
Coefficient
The numerical factor of a variable term. Choosing the variable with coefficient 1 makes isolation easiest.
Interactive Practice — 5 Questions
Solve by substitution: y = 3x − 1 and x + y = 7.
When solving by substitution you get 0 = 5. What does this mean?
Which system is best solved by substitution?
A student substitutes y = x + 4 into 2x + y = 10 and writes 2x + x = 10. What error did they make?
Solve by substitution: y = 2x + 1 and 4x − 2y = −2.
Independent Practice
Independent Practice
Solve by substitution: y = x + 5 and 2x + y = 11.
Solve by substitution: y = 3x and x + y = 8.
Solve by substitution: y = −x + 6 and 2x + y = 7.
Solve by substitution: x = 2y − 1 and 3x + y = 12.
Solve by substitution: 2x + y = 10 and y = x − 1.
Common Mistakes
Substituting the expression into the same equation it came from — creating a circular identity like 0 = 0.
Always substitute into the other equation — the one you did not use to isolate the variable.
Forgetting to distribute when substituting — e.g., substituting y = 2x + 3 into x + y = 9 as x + 2x + 3 = 9 but writing x + 2x + 3 without parentheses and making sign errors.
Use parentheses when substituting: x + (2x + 3) = 9. This prevents sign errors during distribution.
Stopping after finding x and forgetting to solve for y.
After finding x, substitute back into one of the original equations to find y. Write the answer as (x, y).
Choosing the harder equation to isolate a variable when an easier one is available.
Look for a variable with a coefficient of 1 or −1 — it's the easiest to isolate without fractions.
Math Tips
Pick the variable with a coefficient of 1 or −1 to isolate — this avoids fractions in your substitution.
Always use parentheses when substituting an expression: replace y with (x + 3), not just x + 3.
If you get a false statement like 5 = 0, the system has no solution (parallel lines).
If you get a true statement like 0 = 0, the system has infinitely many solutions (same line).
After finding x, always back-substitute to find y — and verify in both original equations.