Lesson 1: Systems of Linear Equations
Learning Objectives
- Understand what a system of linear equations represents and how to identify solutions
- Solve systems of linear equations by graphing and interpret the results geometrically
- Solve systems using the substitution method
- Solve systems using the elimination (addition) method
- Classify systems as consistent (one solution), inconsistent (no solution), or dependent (infinite solutions)
- Choose the most efficient method for solving a given system
- Set up and solve systems to model real-world applications
- Apply systems to mixture problems, distance problems, and cost problems
Introduction: What Are Systems of Equations?
System of Linear Equations: A collection of two or more linear equations involving the same set of variables.
For example:
2x + y = 7
x - y = 2
A solution to a system of equations is an ordered pair (x, y) that satisfies all equations in the system simultaneously. When we solve a system, we're looking for values of the variables that make all equations true at the same time.
Graphical Interpretation
Each linear equation in a system represents a line in the coordinate plane. The solution to the system corresponds to the point(s) where the lines intersect.
Types of Systems
Consistent System (One Solution): The lines intersect at exactly one point. The system has a unique solution.
Inconsistent System (No Solution): The lines are parallel and never intersect. There is no ordered pair that satisfies both equations.
Dependent System (Infinite Solutions): The equations represent the same line. Every point on the line is a solution, so there are infinitely many solutions.
Example 1: Verifying a Solution
Determine whether (3, 1) is a solution to the system:
2x + y = 7
x - y = 2
Solution:
We substitute x = 3 and y = 1 into both equations:
First equation: 2x + y = 7
2(3) + 1 = 6 + 1 = 7
Second equation: x - y = 2
3 - 1 = 2
Answer: Since (3, 1) satisfies both equations, it is a solution to the system.
Example 2: Checking a Non-Solution
Is (2, 3) a solution to the system?
x + y = 5
2x - y = 2
Solution:
Substitute x = 2 and y = 3:
First equation: x + y = 5
2 + 3 = 5
Second equation: 2x - y = 2
2(2) - 3 = 4 - 3 = 1 (not equal to 2)
Answer: Since (2, 3) does not satisfy both equations, it is NOT a solution to the system.
Section 1: Solving Systems by Graphing
The graphing method involves plotting both equations on the same coordinate plane and identifying the point of intersection.
Steps for Solving by Graphing
- Rewrite each equation in slope-intercept form (y = mx + b) if needed
- Graph both lines on the same coordinate plane
- Identify the point of intersection (if it exists)
- Verify the solution by substituting into both original equations
Example 3: System with One Solution (Consistent)
Solve the system by graphing:
y = 2x - 1
y = -x + 5
Solution:
Step 1: Both equations are already in slope-intercept form.
First equation: y = 2x - 1
- Slope: m = 2
- y-intercept: b = -1 (point: (0, -1))
- Another point: when x = 1, y = 2(1) - 1 = 1, so (1, 1)
Second equation: y = -x + 5
- Slope: m = -1
- y-intercept: b = 5 (point: (0, 5))
- Another point: when x = 2, y = -2 + 5 = 3, so (2, 3)
Step 2: Graphing these lines, they intersect at (2, 3).
Step 3: Verify:
First equation: y = 2(2) - 1 = 4 - 1 = 3
Second equation: y = -(2) + 5 = 3
Answer: The solution is (2, 3). This is a consistent system with one unique solution.
Example 4: Parallel Lines (Inconsistent System)
Solve the system by graphing:
y = 3x + 2
y = 3x - 4
Solution:
Analysis:
- Both lines have slope m = 3
- First line has y-intercept (0, 2)
- Second line has y-intercept (0, -4)
Since both lines have the same slope but different y-intercepts, they are parallel and will never intersect.
Answer: No solution. This is an inconsistent system. The lines are parallel.
Example 5: Same Line (Dependent System)
Solve the system by graphing:
2x + y = 4
4x + 2y = 8
Solution:
Step 1: Write both equations in slope-intercept form.
First equation: 2x + y = 4
y = -2x + 4
Second equation: 4x + 2y = 8
2y = -4x + 8
y = -2x + 4
Analysis: Both equations simplify to the same line: y = -2x + 4
Answer: Infinitely many solutions. This is a dependent system. Every point on the line y = -2x + 4 is a solution. We can write the solution as {(x, y) | y = -2x + 4} or {(x, -2x + 4)}.
Limitations of the Graphing Method
- May be difficult to read exact coordinates from a graph
- Solutions with fractions or decimals are hard to identify precisely
- Requires accurate graphing skills
- Best used when solutions are integers or when an approximate solution is acceptable
For exact solutions, we use algebraic methods: substitution and elimination.
Section 2: The Substitution Method
The substitution method is particularly useful when one variable is already isolated in one of the equations, or can be easily isolated.
Steps for the Substitution Method
- Solve one equation for one variable (choose the easiest)
- Substitute this expression into the other equation
- Solve the resulting equation for the remaining variable
- Substitute back to find the other variable
- Check your solution in both original equations
Example 6: Variable Already Isolated
Solve the system using substitution:
y = 3x - 5
2x + y = 10
Solution:
Step 1: The first equation already has y isolated: y = 3x - 5
Step 2: Substitute this expression for y into the second equation:
2x + (3x - 5) = 10
Step 3: Solve for x:
2x + 3x - 5 = 10
5x - 5 = 10
5x = 15
x = 3
Step 4: Substitute x = 3 back into y = 3x - 5:
y = 3(3) - 5 = 9 - 5 = 4
Step 5: Check the solution (3, 4):
First equation: y = 3(3) - 5 = 4
Second equation: 2(3) + 4 = 6 + 4 = 10
Answer: The solution is (3, 4).
Example 7: Solving for a Variable First
Solve the system using substitution:
x + 2y = 7
3x - y = 5
Solution:
Step 1: Solve the first equation for x (it has coefficient 1):
x + 2y = 7
x = 7 - 2y
Step 2: Substitute x = 7 - 2y into the second equation:
3(7 - 2y) - y = 5
Step 3: Solve for y:
21 - 6y - y = 5
21 - 7y = 5
-7y = -16
y = 16/7
Step 4: Substitute y = 16/7 into x = 7 - 2y:
x = 7 - 2(16/7) = 7 - 32/7 = 49/7 - 32/7 = 17/7
Answer: The solution is (17/7, 16/7) or approximately (2.43, 2.29).
Example 8: Substitution with Fractions
Solve the system using substitution:
y = (1/2)x + 3
x - 4y = -8
Solution:
Step 1: y is already isolated in the first equation.
Step 2: Substitute y = (1/2)x + 3 into the second equation:
x - 4[(1/2)x + 3] = -8
Step 3: Solve for x:
x - 2x - 12 = -8
-x - 12 = -8
-x = 4
x = -4
Step 4: Substitute x = -4 into y = (1/2)x + 3:
y = (1/2)(-4) + 3 = -2 + 3 = 1
Answer: The solution is (-4, 1).
Example 9: No Solution (Inconsistent System)
Solve the system using substitution:
y = 2x + 3
4x - 2y = 10
Solution:
Step 1: y is already isolated: y = 2x + 3
Step 2: Substitute into the second equation:
4x - 2(2x + 3) = 10
Step 3: Simplify:
4x - 4x - 6 = 10
-6 = 10
Analysis: We get a false statement (-6 = 10). This means there is no solution.
Answer: No solution. The system is inconsistent (parallel lines).
Example 10: Infinitely Many Solutions (Dependent System)
Solve the system using substitution:
x = 3y - 6
2x - 6y = -12
Solution:
Step 1: x is already isolated: x = 3y - 6
Step 2: Substitute into the second equation:
2(3y - 6) - 6y = -12
Step 3: Simplify:
6y - 12 - 6y = -12
-12 = -12
Analysis: We get a true statement (identity). This means the equations represent the same line.
Answer: Infinitely many solutions. The system is dependent. Solution: {(x, y) | x = 3y - 6} or {(3y - 6, y) | y is any real number}.
Example 11: Word Problem - Two Numbers
The sum of two numbers is 25, and their difference is 7. Find the two numbers.
Solution:
Step 1: Define variables. Let x = first number, y = second number.
Step 2: Write equations from the problem:
Sum is 25: x + y = 25
Difference is 7: x - y = 7
Step 3: Solve the second equation for x:
x = y + 7
Step 4: Substitute into the first equation:
(y + 7) + y = 25
2y + 7 = 25
2y = 18
y = 9
Step 5: Find x:
x = 9 + 7 = 16
Step 6: Check: 16 + 9 = 25 and 16 - 9 = 7
Answer: The two numbers are 16 and 9.
Section 3: The Elimination (Addition) Method
The elimination method involves adding or subtracting the equations to eliminate one variable. This method is especially efficient when coefficients of one variable are opposites or can be made opposites.
Steps for the Elimination Method
- Write both equations in standard form (Ax + By = C)
- If necessary, multiply one or both equations by constants to make the coefficients of one variable opposites
- Add the equations to eliminate one variable
- Solve for the remaining variable
- Substitute this value into either original equation to find the other variable
- Check your solution in both original equations
Example 12: Coefficients Already Opposite
Solve the system using elimination:
3x + y = 11
2x - y = 4
Solution:
Step 1: Notice that the y coefficients are already opposites (+1 and -1).
Step 2: Add the equations to eliminate y:
3x + y = 11
+ 2x - y = 4
___________
5x = 15
Step 3: Solve for x:
5x = 15
x = 3
Step 4: Substitute x = 3 into the first equation:
3(3) + y = 11
9 + y = 11
y = 2
Step 5: Check (3, 2):
First: 3(3) + 2 = 9 + 2 = 11
Second: 2(3) - 2 = 6 - 2 = 4
Answer: The solution is (3, 2).
Example 13: Multiply One Equation
Solve the system using elimination:
2x + 3y = 12
x + y = 5
Solution:
Step 1: To eliminate x, multiply the second equation by -2:
-2(x + y) = -2(5)
-2x - 2y = -10
Step 2: Add to the first equation:
2x + 3y = 12
-2x - 2y = -10
___________
y = 2
Step 3: Substitute y = 2 into x + y = 5:
x + 2 = 5
x = 3
Answer: The solution is (3, 2).
Example 14: Multiply Both Equations
Solve the system using elimination:
3x + 4y = 10
2x + 3y = 7
Solution:
Step 1: To eliminate x, multiply the first equation by 2 and the second by -3:
2(3x + 4y) = 2(10) → 6x + 8y = 20
-3(2x + 3y) = -3(7) → -6x - 9y = -21
Step 2: Add the equations:
6x + 8y = 20
-6x - 9y = -21
___________
-y = -1
Step 3: Solve for y:
y = 1
Step 4: Substitute y = 1 into 2x + 3y = 7:
2x + 3(1) = 7
2x + 3 = 7
2x = 4
x = 2
Answer: The solution is (2, 1).
Example 15: Eliminating the Other Variable
Solve the system using elimination (eliminate y this time):
5x - 2y = 14
3x + 4y = 4
Solution:
Step 1: To eliminate y, multiply the first equation by 2:
2(5x - 2y) = 2(14) → 10x - 4y = 28
Step 2: Add to the second equation:
10x - 4y = 28
3x + 4y = 4
___________
13x = 32
Step 3: Solve for x:
x = 32/13
Step 4: Substitute into 3x + 4y = 4:
3(32/13) + 4y = 4
96/13 + 4y = 4
4y = 4 - 96/13 = 52/13 - 96/13 = -44/13
y = -11/13
Answer: The solution is (32/13, -11/13).
Example 16: No Solution (Elimination Method)
Solve the system using elimination:
2x + 3y = 6
4x + 6y = 18
Solution:
Step 1: Multiply the first equation by -2:
-2(2x + 3y) = -2(6) → -4x - 6y = -12
Step 2: Add to the second equation:
-4x - 6y = -12
4x + 6y = 18
___________
0 = 6
Analysis: We get a false statement (0 = 6).
Answer: No solution. The system is inconsistent (parallel lines).
Example 17: Infinitely Many Solutions (Elimination Method)
Solve the system using elimination:
6x - 9y = 12
-2x + 3y = -4
Solution:
Step 1: Multiply the second equation by 3:
3(-2x + 3y) = 3(-4) → -6x + 9y = -12
Step 2: Add to the first equation:
6x - 9y = 12
-6x + 9y = -12
___________
0 = 0
Analysis: We get a true statement (identity: 0 = 0).
Answer: Infinitely many solutions. The system is dependent. Both equations represent the same line.
Section 4: Applications and Word Problems
Systems of linear equations are powerful tools for solving real-world problems. The key is to carefully define variables and translate the problem into equations.
Steps for Solving Word Problems
- Read the problem carefully and identify what you need to find
- Define variables for the unknowns
- Write two equations based on the given information
- Solve the system using substitution or elimination
- Answer the question in the context of the problem
- Check that your answer makes sense
Example 18: Number Problem
One number is 3 more than twice another number. The sum of the two numbers is 27. Find the numbers.
Solution:
Step 1: Define variables.
Let x = first number, y = second number
Step 2: Write equations.
"One number is 3 more than twice another": x = 2y + 3
"Sum is 27": x + y = 27
Step 3: Substitute x = 2y + 3 into x + y = 27:
(2y + 3) + y = 27
3y + 3 = 27
3y = 24
y = 8
Step 4: Find x:
x = 2(8) + 3 = 16 + 3 = 19
Step 5: Check: 19 = 2(8) + 3 and 19 + 8 = 27
Answer: The numbers are 19 and 8.
Example 19: Age Problem
Sarah is 4 years older than her brother Tom. In 5 years, the sum of their ages will be 38. How old are they now?
Solution:
Step 1: Define variables.
Let S = Sarah's current age, T = Tom's current age
Step 2: Write equations.
"Sarah is 4 years older than Tom": S = T + 4
"In 5 years, sum is 38": (S + 5) + (T + 5) = 38
Step 3: Simplify the second equation:
S + 5 + T + 5 = 38
S + T + 10 = 38
S + T = 28
Step 4: Substitute S = T + 4 into S + T = 28:
(T + 4) + T = 28
2T + 4 = 28
2T = 24
T = 12
Step 5: Find S:
S = 12 + 4 = 16
Step 6: Check: In 5 years, Sarah will be 21 and Tom will be 17. Their sum: 21 + 17 = 38
Answer: Sarah is 16 years old and Tom is 12 years old.
Example 20: Mixture Problem
A coffee shop wants to create a 50-pound blend of coffee by mixing a light roast that costs $8 per pound with a dark roast that costs $12 per pound. If the blend should cost $9.60 per pound, how many pounds of each type should be used?
Solution:
Step 1: Define variables.
Let L = pounds of light roast, D = pounds of dark roast
Step 2: Write equations.
"Total weight is 50 pounds": L + D = 50
"Total cost equation": 8L + 12D = 9.60(50)
Step 3: Simplify the second equation:
8L + 12D = 480
Step 4: Solve L + D = 50 for L:
L = 50 - D
Step 5: Substitute into the cost equation:
8(50 - D) + 12D = 480
400 - 8D + 12D = 480
400 + 4D = 480
4D = 80
D = 20
Step 6: Find L:
L = 50 - 20 = 30
Step 7: Check: 30 + 20 = 50
Cost: 8(30) + 12(20) = 240 + 240 = 480 = 9.60(50)
Answer: Use 30 pounds of light roast and 20 pounds of dark roast.
Example 21: Distance Problem
Two cars leave the same location traveling in opposite directions. One car travels at 60 mph and the other at 50 mph. After how many hours will they be 330 miles apart?
Solution:
Step 1: Define variables.
Let d₁ = distance traveled by first car
Let d₂ = distance traveled by second car
Let t = time (same for both cars)
Step 2: Write equations using distance = rate × time.
d₁ = 60t (first car)
d₂ = 50t (second car)
d₁ + d₂ = 330 (total distance apart)
Step 3: Substitute into the total distance equation:
60t + 50t = 330
110t = 330
t = 3
Step 4: Check: First car: 60(3) = 180 miles
Second car: 50(3) = 150 miles
Total: 180 + 150 = 330
Answer: They will be 330 miles apart after 3 hours.
Example 22: Cost Problem
A school is selling tickets to a play. Adult tickets cost $8 and student tickets cost $5. The school sold 350 tickets and collected $2,380. How many of each type of ticket were sold?
Solution:
Step 1: Define variables.
Let a = number of adult tickets, s = number of student tickets
Step 2: Write equations.
"Total tickets sold": a + s = 350
"Total revenue": 8a + 5s = 2380
Step 3: Use elimination. Multiply first equation by -5:
-5a - 5s = -1750
Step 4: Add to the second equation:
-5a - 5s = -1750
8a + 5s = 2380
___________
3a = 630
Step 5: Solve for a:
a = 210
Step 6: Substitute into a + s = 350:
210 + s = 350
s = 140
Step 7: Check: 210 + 140 = 350
Revenue: 8(210) + 5(140) = 1680 + 700 = 2380
Answer: 210 adult tickets and 140 student tickets were sold.
Check Your Understanding
1. Which ordered pair is a solution to the system: x + y = 5 and 2x - y = 4?
Answer: (3, 2)
Check: 3 + 2 = 5 and 2(3) - 2 = 6 - 2 = 4
2. What type of system is this: y = 2x + 1 and y = 2x - 3?
Answer: Inconsistent system (no solution)
Both lines have the same slope (m = 2) but different y-intercepts, so they are parallel and never intersect.
3. Solve by substitution: y = x - 1 and 2x + y = 8
Answer: (3, 2)
Substitute y = x - 1 into second equation:
2x + (x - 1) = 8
3x - 1 = 8
3x = 9
x = 3
Then y = 3 - 1 = 2
4. Solve by elimination: 3x + 2y = 16 and 5x - 2y = 8
Answer: (3, 3.5) or (3, 7/2)
Add the equations directly (y coefficients are opposites):
8x = 24, so x = 3
Substitute: 3(3) + 2y = 16
9 + 2y = 16
2y = 7
y = 3.5
5. Which method would be most efficient for this system: x = 4y - 1 and 3x + 2y = 13?
Answer: Substitution method
The first equation already has x isolated, making substitution the most efficient approach.
6. A restaurant sold 45 pizzas, with some cheese pizzas at $12 each and pepperoni pizzas at $15 each. Total sales were $600. Set up the system of equations.
Answer:
Let c = cheese pizzas, p = pepperoni pizzas
c + p = 45 (total pizzas)
12c + 15p = 600 (total sales)
7. What can you conclude if solving a system yields the result "0 = 5"?
Answer: The system has no solution (inconsistent system)
A false statement like 0 = 5 indicates the lines are parallel and never intersect.
8. What can you conclude if solving a system yields the result "0 = 0"?
Answer: The system has infinitely many solutions (dependent system)
An identity like 0 = 0 indicates the equations represent the same line.
9. Solve: 4x - 3y = 11 and 2x + y = 9
Answer: (4, 1.67) or (4, 5/3)
Using substitution: y = 9 - 2x
4x - 3(9 - 2x) = 11
4x - 27 + 6x = 11
10x = 38
x = 3.8... Actually, let me recalculate:
From second equation: y = 9 - 2x
Substitute: 4x - 3(9 - 2x) = 11
4x - 27 + 6x = 11
10x = 38
x = 19/5 or 3.8
Wait, checking again: Using elimination is cleaner.
Multiply second by 3: 6x + 3y = 27
Add to first: 10x = 38, x = 19/5
Then y = 9 - 2(19/5) = 45/5 - 38/5 = 7/5
Solution: (19/5, 7/5) or (3.8, 1.4)
10. Two numbers have a sum of 50 and a difference of 12. Find them using a system.
Answer: 31 and 19
System: x + y = 50 and x - y = 12
Add equations: 2x = 62, so x = 31
Then y = 50 - 31 = 19
Key Takeaways
- A system of linear equations can have one solution (consistent), no solution (inconsistent), or infinitely many solutions (dependent)
- Graphically, the solution is the point where the lines intersect
- The substitution method works well when a variable is isolated or easily isolated
- The elimination method works well when coefficients are opposites or can be made opposites
- A false statement (like 0 = 5) indicates no solution (parallel lines)
- An identity (like 0 = 0) indicates infinitely many solutions (same line)
- Always define variables clearly when solving word problems
- Common applications include mixture problems, distance problems, age problems, and cost problems
- Check your solutions by substituting back into both original equations
- Choose the most efficient method based on the form of the system