Method 1: Substitution

When to use: One equation is already solved for a variable, or easily solved.

Steps:

1. Solve one equation for one variable
2. Substitute that expression into the other equation
3. Solve for the remaining variable
4. Substitute back to find the other variable
5. Check the solution in both equations

Example 1: Already Solved Solve: y = 2x + 1 y = -x + 7

Step 1: First equation already solved for y

Step 2: Substitute y = 2x + 1 into second equation 2x + 1 = -x + 7

Step 3: Solve for x 3x + 1 = 7 3x = 6 x = 2

Step 4: Substitute x = 2 back into either equation y = 2(2) + 1 = 5

Solution: (2, 5)

Step 5: Check in both equations ✓ y = 2(2) + 1 → 5 = 5 ✓ y = -(2) + 7 → 5 = 5

Example 2: Solve First Solve: x + y = 10 2x + 3y = 26

Step 1: Solve first equation for y y = 10 - x

Step 2: Substitute into second equation 2x + 3(10 - x) = 26

Step 3: Solve for x 2x + 30 - 3x = 26 -x + 30 = 26 -x = -4 x = 4

Step 4: Find y y = 10 - 4 = 6

Solution: (4, 6)

Check: ✓ 4 + 6 = 10 ✓ 2(4) + 3(6) = 8 + 18 = 26

Example 3: Requires Solving Solve: 2x + y = 8 3x - 2y = 5

Step 1: Solve first equation for y y = 8 - 2x

Step 2: Substitute into second equation 3x - 2(8 - 2x) = 5

Step 3: Solve 3x - 16 + 4x = 5 7x - 16 = 5 7x = 21 x = 3

Step 4: Find y y = 8 - 2(3) = 8 - 6 = 2

Solution: (3, 2)

Method 2: Elimination (Addition/Subtraction)

When to use: Coefficients of one variable are opposites or can be made opposites.

Steps:

1. Line up equations with variables in columns
2. Multiply one or both equations to make coefficients of one variable opposites
3. Add equations to eliminate that variable
4. Solve for the remaining variable
5. Substitute back to find the other variable
6. Check the solution

Example 1: Already Opposites Solve: 3x + 2y = 16 3x - 2y = 8

Step 1: Notice 2y and -2y are opposites

Step 2: Add equations 3x + 2y = 16

• 3x - 2y = 8 6x = 24

Step 3: Solve x = 4

Step 4: Substitute into first equation 3(4) + 2y = 16 12 + 2y = 16 2y = 4 y = 2

Solution: (4, 2)

Example 2: Create Opposites Solve: 2x + 3y = 12 x + 2y = 7

Step 1: Make x coefficients opposites Multiply second equation by -2: 2x + 3y = 12 -2x - 4y = -14

Step 2: Add equations 2x + 3y = 12

• -2x - 4y = -14 -y = -2

Step 3: Solve y = 2

Step 4: Substitute into second original equation x + 2(2) = 7 x + 4 = 7 x = 3

Solution: (3, 2)

Example 3: Both Need Multiplying Solve: 3x + 4y = 10 2x + 3y = 7

Step 1: Eliminate x by making coefficients opposites Multiply first equation by 2: 6x + 8y = 20 Multiply second equation by -3: -6x - 9y = -21

Step 2: Add 6x + 8y = 20

• -6x - 9y = -21 -y = -1

Step 3: Solve y = 1

Step 4: Substitute 3x + 4(1) = 10 3x + 4 = 10 3x = 6 x = 2

Solution: (2, 1)

Special Cases

No Solution (Parallel Lines)

Example: y = 2x + 3 y = 2x + 5

Using substitution: 2x + 3 = 2x + 5 3 = 5 (FALSE!)

This means no solution (parallel lines).

Alternative form: 2x - y = -3 2x - y = -5

Using elimination (subtract): 0 = 2 (FALSE!)

No solution.

Infinitely Many Solutions (Same Line)

Example: x + y = 5 2x + 2y = 10

Using elimination (multiply first by -2): -2x - 2y = -10 2x + 2y = 10

Add: 0 = 0 (TRUE for all values!)

This means infinitely many solutions (same line).

Choosing the Best Method

Use Substitution when:

• One equation is already solved for a variable (y = ...)
• One variable has a coefficient of 1
• Equations are in slope-intercept form

Use Elimination when:

• Coefficients are already opposites or nearly so
• Both equations are in standard form
• No variable has coefficient 1

Use Graphing when:

• You want to visualize the solution
• An approximate answer is acceptable
• Using technology (graphing calculator)

Solving by Graphing (Overview)

While less precise, graphing helps visualize systems.

Steps:

1. Write each equation in slope-intercept form (y = mx + b)
2. Graph both lines on the same axes
3. Find the intersection point
4. Check the point in both equations

Example: y = x + 1 y = -x + 5

Graph both lines, they intersect at (2, 3)

Check: 3 = 2 + 1 ✓ and 3 = -2 + 5 ✓

Systems with Fractions

Strategy: Clear fractions by multiplying by LCD before solving.

Example: x/2 + y/3 = 5 x/4 + y = 7

Multiply first equation by 6: 3x + 2y = 30 Multiply second equation by 4: x + 4y = 28

Now solve using elimination or substitution: From second: x = 28 - 4y Substitute: 3(28 - 4y) + 2y = 30 84 - 12y + 2y = 30 -10y = -54 y = 5.4

x = 28 - 4(5.4) = 6.4

Solution: (6.4, 5.4)

Systems with Decimals

Strategy: Clear decimals by multiplying by powers of 10.

Example: 0.5x + 0.3y = 1.9 0.2x + 0.4y = 1.4

Multiply both by 10: 5x + 3y = 19 2x + 4y = 14

Now solve normally.

Checking Solutions

Always check by substituting into BOTH original equations.

Solution (3, 4) for: 2x + y = 10 x - y = -1

Check: 2(3) + 4 = 6 + 4 = 10 ✓ 3 - 4 = -1 ✓

Real-World Applications

Systems model situations with multiple constraints.

Example 1: Money Problem: You have 15 coins worth $1.80, all nickels and dimes. How many of each?

Let n = nickels, d = dimes

Number equation: n + d = 15 Value equation: 0.05n + 0.10d = 1.80

Clear decimals in second: 5n + 10d = 180

From first: n = 15 - d Substitute: 5(15 - d) + 10d = 180 75 - 5d + 10d = 180 5d = 105 d = 21... wait, this is impossible (can't have 21 dimes if total is 15 coins)

Let me recalculate: 5n + 10d = 180 Simplify: n + 2d = 36

System: n + d = 15 n + 2d = 36

Subtract: -d = -21, so d = 21... error!

Correct approach: n + d = 15 5n + 10d = 180 → divide by 5 → n + 2d = 36

These are incompatible. Let me verify the original problem...

Actually for $1.80 with 15 coins: n + d = 15 0.05n + 0.10d = 1.80

Multiply second by 20: n + 2d = 36 Subtract from first: -d = -21...

This suggests error in problem setup. Correct version:

Example 1 (Corrected): 12 coins worth $1.80: n + d = 12 5n + 10d = 180 → n + 2d = 36

Subtract: -d = -24, d = 24 (still wrong)

Let's use n + d = 12, so n = 12 - d 5(12-d) + 10d = 180 60 - 5d + 10d = 180 5d = 120 d = 24 (impossible)

Correctly working example: You have 20 coins worth $3.00, all nickels and dimes.

n + d = 20 5n + 10d = 300 → n + 2d = 60

Subtract: -d = -40, so d = 40 (still impossible!)

Working Example: 15 coins worth $1.20: n + d = 15 5n + 10d = 120 → n + 2d = 24

System: n + d = 15 n + 2d = 24

Subtract: -d = -9, so d = 9 Then n = 6

Answer: 6 nickels, 9 dimes Check: 6(0.05) + 9(0.10) = 0.30 + 0.90 = $1.20 ✓

Common Mistakes

1. Not substituting correctly Keep parentheses when substituting expressions

2. Arithmetic errors Be careful with negative signs

3. Stopping after finding one variable Must find both x and y!

4. Not checking the solution Always verify in both equations

5. Confusing methods Don't mix substitution and elimination mid-problem

Quick Reference

Practice Tips

• Master both substitution and elimination
• Identify which method is easier before starting
• Show all work clearly
• Always check your solution
• Practice special cases (no solution, infinite solutions)
• Clear fractions and decimals early
• Keep equations organized in columns

❓ Question:

Step 2: Solve for $x$ $x = 4$

Step 3: Substitute $x = 4$ into the first equation $3(4) + 2y = 16$ $12 + 2y = 16$ $2y = 4$ $y = 2$

Answer: $(4, 2)$

Step 2: Compare with the second equation: $4x + 6y = 14$ $4x + 6y = 10$

This says the same expression equals two different numbers, which is impossible!

Answer: No solution (the lines are parallel and never intersect)