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MathematicsGrades 8–11Intermediate

How to Solve Systems of Equations: Substitution & Elimination Step-by-Step

Master solving linear systems using substitution, elimination, and graphing with clear worked examples.

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Sparky Academic TeamCurriculum Specialist
9 min readUpdated 2026-10-17
Quick Summary / TL;DR

A system of equations consists of two or more equations with the same variables (x and y). You can solve them using Substitution (isolate one variable and plug its expression into the other equation) or Elimination (add or subtract equations to cancel a variable). The final solution is the coordinate point (x, y) where the lines intersect!

Core Equation / Rule:
Substitution: y = mx + b → Plug into Eq 2 | Elimination: (Ax + By) + (Cx - By) = Cancel y
💡
🎯 Isolated variable? Substitute! Columns lined up? Eliminate!
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The 4-Step Solving Process

1

Choose the Best Method

y = 2x + 1 ⟹ Substitution | 3x + 2y = 12 and 5x - 2y = 4 ⟹ Elimination

Inspect the equations: if one variable already has a coefficient of 1 or -1, use Substitution. If terms line up in standard form, use Elimination.

💡 Pro Tip: Both methods always give the exact same answer—pick whichever requires fewer fraction steps!
2

Eliminate One Variable

2x + (2x + 1) = 9 ⟹ 4x + 1 = 9

In substitution, replace the variable with its expression. In elimination, multiply one or both equations so coefficients cancel when added.

💡 Pro Tip: Make sure you distribute minus signs carefully when substituting parentheses into the second equation.
3

Solve for the Remaining Variable

4x = 8 ⟹ x = 2

You now have a simple one-variable linear equation. Solve for x or y.

💡 Pro Tip: You're only halfway done! A system solution must contain values for BOTH x and y.
4

Back-Substitute & Write as (x, y)

y = 2(2) + 1 = 5 ⟹ Solution: (2, 5)

Plug your found value back into either original equation to find the second variable, then check your answer.

💡 Pro Tip: Plug (x, y) into BOTH equations to verify that it makes both true!

What is a System of Equations?

A system of linear equations is a set of two or more equations containing the same variables (usually x and y).

Graphically, each equation forms a straight line on a coordinate grid. The solution to the system is the single point (x, y) where the two lines cross (intersect)!


When to Use Substitution vs. Elimination

ConditionRecommended MethodWhy?
One variable is already isolated (e.g. y = 3x - 4)SubstitutionPlug directly in with zero rearrangement
One variable has a coefficient of 1 or -1 (e.g. x + 4y = 9)SubstitutionEasy to isolate x = 9 - 4y without creating fractions
Both equations are in standard form (Ax + By = C)EliminationCoefficients can easily cancel out by adding or subtracting
Matching or opposite coefficients (e.g. +2y and -2y)EliminationAdding the equations eliminates the variable immediately

Method 1: The Substitution Method

The Step-by-Step Game Plan:

  1. Isolate one variable in whichever equation is simplest.
  2. Substitute that expression into the other equation (use parentheses!).
  3. Solve the resulting single-variable equation.
  4. Back-substitute that number to find the second variable.

Worked Example 1: Solving with Substitution

Problem:

Solve the system:

Equation 1: y = 3x - 5

Equation 2: 2x + y = 10

Step 1: Substitute Equation 1 into Equation 2

Since Equation 1 already states y = (3x - 5), replace y in Equation 2:

2x + (3x - 5) = 10

Step 2: Combine like terms and solve for x

5x - 5 = 10

Add 5 to both sides:

5x = 15

Divide by 5:

x = 3

Step 3: Find y using back-substitution

Plug x = 3 into Equation 1:

y = 3(3) - 5 = 9 - 5 = 4

Step 4: Write as an ordered pair and check

Solution: (3, 4)

Verification Check:

  • •Eq 1: 4 = 3(3) - 5 ⟹ 4 = 4 (True)
  • •Eq 2: 2(3) + 4 = 10 ⟹ 6 + 4 = 10 (True)

Interactive Practice Check
Given the system: $y = 2x - 1$ and $3x + 2y = 12$. Step 1: Substitute $(2x - 1)$ for $y$ into the second equation: $3x + 2(2x - 1) = 12 ⟹ 3x + 4x - 2 = 12 ⟹ 7x = 14$. What is the value of $x$?

Method 2: The Elimination Method (Addition Method)

In the elimination method, you add or subtract the two equations together so that one variable adds to zero and disappears!


Worked Example 2: Solving with Elimination

Problem:

Solve the system:

Equation 1: 3x + 2y = 16

Equation 2: 5x - 2y = 0

Step 1: Notice matching opposite coefficients

The y terms are +2y and -2y. Adding them together will cancel out y completely!

Step 2: Add the two equations vertically

\begin{aligned} 3x + 2y &= 16 \\ + 5x - 2y &= 0 \\ \hline 8x + 0y &= 16 \end{aligned}
8x = 16 ⟹ x = 2

Step 3: Back-substitute to find y

Substitute x = 2 into Equation 2:

5(2) - 2y = 0
10 - 2y = 0 ⟹ 2y = 10 ⟹ y = 5

Final Solution: The lines intersect at (2, 5).


What If Coefficients Don't Match? (Multiplying First)

If neither variable matches, multiply one or both equations by a constant number first:

\begin{aligned} 2x + 3y &= 12 \\ x - y &= 1 \end{aligned}

Multiply the second equation by 3:

3(x - y = 1) ⟹ 3x - 3y = 3

Now add to the first equation:

(2x + 3y) + (3x - 3y) = 12 + 3 ⟹ 5x = 15 ⟹ x = 3

Special Cases: No Solution vs. Infinitely Many Solutions

Not all systems have exactly one solution:

  • •No Solution (Parallel Lines): The lines never cross. When solving, variables cancel out and leave an impossible statement like 0 = 7.
  • •Infinitely Many Solutions (Same Line): Both equations represent the exact same line. Variables cancel out leaving a universal truth like 0 = 0.
⚠️WATCH OUT

Always Distribute Minus Signs Carefully!

When substituting an expression like y = (2x - 3) into 5x - y = 8, remember the minus sign applies to both terms inside:

5x - (2x - 3) = 5x - 2x + 3 = 8. Dropping the negative on the second term is the most common test mistake!

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