Solving Systems of Equations
Use substitution and elimination methods
Solving Systems of Equations
What is a System of Equations?
A system of equations is a set of two or more equations with the same variables. The solution is the set of values that makes ALL equations true simultaneously.
Example system: y = 2x + 1 y = -x + 7
We need to find the (x, y) pair that satisfies BOTH equations.
What Does a Solution Look Like?
For a system of two linear equations in two variables:
One Solution: The lines intersect at one point (x, y) Example: y = 2x + 1 and y = -x + 7 intersect at (2, 5)
No Solution: The lines are parallel (never intersect) Example: y = 2x + 1 and y = 2x + 5 (same slope, different intercepts)
Infinitely Many Solutions: The lines are identical (overlap completely) Example: y = 2x + 1 and 2y = 4x + 2 (same line written differently)
Three Methods for Solving Systems
There are three algebraic methods to solve systems:
- Graphing (visual, approximate)
- Substitution (best when one equation is solved for a variable)
- Elimination (best when coefficients line up nicely)
Method 1: Substitution
When to use: One equation is already solved for a variable, or easily solved.
Steps:
- Solve one equation for one variable
- Substitute that expression into the other equation
- Solve for the remaining variable
- Substitute back to find the other variable
- 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:
- Line up equations with variables in columns
- Multiply one or both equations to make coefficients of one variable opposites
- Add equations to eliminate that variable
- Solve for the remaining variable
- Substitute back to find the other variable
- 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:
- Write each equation in slope-intercept form (y = mx + b)
- Graph both lines on the same axes
- Find the intersection point
- 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
-
Not substituting correctly Keep parentheses when substituting expressions
-
Arithmetic errors Be careful with negative signs
-
Stopping after finding one variable Must find both x and y!
-
Not checking the solution Always verify in both equations
-
Confusing methods Don't mix substitution and elimination mid-problem
Quick Reference
| Method | Best When | Steps | |--------|-----------|-------| | Substitution | One equation solved for variable | Substitute, solve, substitute back | | Elimination | Coefficients are opposites | Add/subtract equations | | Graphing | Visualization needed | Graph both, find intersection |
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
📚 Practice Problems
1Problem 1easy
❓ Question:
Solve the system using substitution: y = 2x + 1 y = -x + 7
💡 Show Solution
Step 1: Both equations are already solved for y, so set them equal: 2x + 1 = -x + 7
Step 2: Solve for x: 2x + x = 7 - 1 3x = 6 x = 2
Step 3: Substitute x = 2 into either original equation (using the first): y = 2(2) + 1 y = 4 + 1 y = 5
Step 4: Check in both equations: Equation 1: 5 = 2(2) + 1 → 5 = 5 ✓ Equation 2: 5 = -(2) + 7 → 5 = 5 ✓
Answer: (2, 5)
2Problem 2easy
❓ Question:
Solve using substitution:
💡 Show Solution
Since is already solved for , substitute into the second equation:
Step 1: Substitute
Step 2: Solve for
Step 3: Find using
Check: ✓
Answer:
3Problem 3easy
❓ Question:
Solve the system using elimination: 2x + 3y = 12 2x - y = 4
💡 Show Solution
Step 1: Notice that both equations have 2x, so we can eliminate x by subtracting: (2x + 3y) - (2x - y) = 12 - 4 2x + 3y - 2x + y = 8 4y = 8 y = 2
Step 2: Substitute y = 2 into either original equation (using the second): 2x - 2 = 4 2x = 6 x = 3
Step 3: Check in both equations: Equation 1: 2(3) + 3(2) = 6 + 6 = 12 ✓ Equation 2: 2(3) - 2 = 6 - 2 = 4 ✓
Answer: (3, 2)
4Problem 4medium
❓ Question:
Solve using elimination:
💡 Show Solution
Notice the coefficients are opposites, so we can add the equations:
Step 1: Add the equations
Step 2: Solve for
Step 3: Substitute into the first equation
Answer:
5Problem 5medium
❓ Question:
Solve the system: 3x + 2y = 16 x - y = 2
💡 Show Solution
Step 1: Use substitution. Solve the second equation for x: x = y + 2
Step 2: Substitute into the first equation: 3(y + 2) + 2y = 16 3y + 6 + 2y = 16 5y + 6 = 16 5y = 10 y = 2
Step 3: Substitute y = 2 back into x = y + 2: x = 2 + 2 x = 4
Step 4: Check in both equations: Equation 1: 3(4) + 2(2) = 12 + 4 = 16 ✓ Equation 2: 4 - 2 = 2 ✓
Answer: (4, 2)
6Problem 6medium
❓ Question:
Solve the system using elimination: 4x + 5y = 23 3x - 2y = 0
💡 Show Solution
Step 1: To eliminate y, multiply the first equation by 2 and the second by 5: 2(4x + 5y = 23) → 8x + 10y = 46 5(3x - 2y = 0) → 15x - 10y = 0
Step 2: Add the equations: 8x + 10y + 15x - 10y = 46 + 0 23x = 46 x = 2
Step 3: Substitute x = 2 into either original equation (using the second): 3(2) - 2y = 0 6 - 2y = 0 -2y = -6 y = 3
Step 4: Check in both equations: Equation 1: 4(2) + 5(3) = 8 + 15 = 23 ✓ Equation 2: 3(2) - 2(3) = 6 - 6 = 0 ✓
Answer: (2, 3)
7Problem 7hard
❓ Question:
Solve:
💡 Show Solution
Step 1: Notice that if we multiply the first equation by 2:
Step 2: Compare with the second equation:
This says the same expression equals two different numbers, which is impossible!
Answer: No solution (the lines are parallel and never intersect)
8Problem 8hard
❓ Question:
A theater sold 450 tickets. Adult tickets cost 8. If total sales were $4,800, how many of each type of ticket were sold?
💡 Show Solution
Step 1: Define variables: Let a = number of adult tickets Let c = number of child tickets
Step 2: Write equations from the problem: Total tickets: a + c = 450 Total sales: 12a + 8c = 4800
Step 3: Solve using substitution. From equation 1: c = 450 - a
Step 4: Substitute into equation 2: 12a + 8(450 - a) = 4800 12a + 3600 - 8a = 4800 4a + 3600 = 4800 4a = 1200 a = 300
Step 5: Find c: c = 450 - 300 = 150
Step 6: Check: Total tickets: 300 + 150 = 450 ✓ Total sales: 12(300) + 8(150) = 3600 + 1200 = 4800 ✓
Answer: 300 adult tickets and 150 child tickets
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