Solved Exercises on Solving Systems of Linear Equations by Elimination in Integrated Math 1

Master elimination method: adding/subtracting equations, multiplying equations, and solving through these 5 detailed exercises with comprehensive solutions.

Solution: Exercises 1 to 3
1 Addition Elimination
Exercise 1
Solve the system of equations by elimination:
\(x + y = 7\) and \(x - y = 1\)
Verify your solution.
Definition:

Elimination Method: Adding or subtracting equations to eliminate one variable and solve for the other

Method for elimination:
  1. Align equations so like terms are in columns
  2. Determine which variable to eliminate
  3. Add or subtract equations to eliminate that variable
  4. Solve the resulting single-variable equation
  5. Substitute back to find the other variable
  6. Verify the solution in both original equations
System
x + y = 7, x - y = 1
Addition
2x = 8
Solution
(4, 3)
Step 1: Align the equations

\(x + y = 7\)

\(x - y = 1\)

Step 2: Identify which variable to eliminate

The y-coefficients are 1 and -1, so adding the equations will eliminate y

Step 3: Add the equations

\((x + y) + (x - y) = 7 + 1\)

\(x + y + x - y = 8\)

\(2x = 8\)

Step 4: Solve for x

\(x = \frac{8}{2} = 4\)

Step 5: Substitute x back to find y

Using the first equation: \(x + y = 7\)

\(4 + y = 7\)

\(y = 7 - 4 = 3\)

Step 6: Verify the solution

First equation: \(x + y = 7\)

\(4 + 3 = 7\) ✓

Second equation: \(x - y = 1\)

\(4 - 3 = 1\) ✓

Solution: (4, 3)
Final answer:

The solution to the system is (4, 3).

Applied rules:

• Elimination Principle: Adding equations preserves equality

• Variable Elimination: Opposite coefficients cancel when added

• Verification: Check solution in both original equations

2 Subtraction Elimination
Exercise 2
Solve the system of equations by elimination:
\(2x + 3y = 13\) and \(2x - y = 5\)
Show all steps clearly.
Definition:

Subtraction Elimination: When coefficients of the same variable are equal, subtracting equations eliminates that variable

System
2x + 3y = 13, 2x - y = 5
Subtraction
4y = 8
Solution
(4, 1)
Step 1: Align the equations

\(2x + 3y = 13\)

\(2x - y = 5\)

Step 2: Identify which variable to eliminate

The x-coefficients are both 2, so subtracting the second equation from the first will eliminate x

Step 3: Subtract the second equation from the first

\((2x + 3y) - (2x - y) = 13 - 5\)

\(2x + 3y - 2x + y = 8\)

\(4y = 8\)

Step 4: Solve for y

\(y = \frac{8}{4} = 2\)

Step 5: Substitute y back to find x

Using the second equation: \(2x - y = 5\)

\(2x - 2 = 5\)

\(2x = 7\)

\(x = \frac{7}{2} = 3.5\)

Step 6: Verify the solution

First equation: \(2x + 3y = 13\)

\(2(3.5) + 3(2) = 7 + 6 = 13\) ✓

Second equation: \(2x - y = 5\)

\(2(3.5) - 2 = 7 - 2 = 5\) ✓

Solution: (3.5, 2)
Final answer:

The solution to the system is (3.5, 2).

Applied rules:

• Subtraction Elimination: Equal coefficients cancel when subtracted

• Distribution: When subtracting expressions, distribute the negative sign

• Verification: Always check solution in both original equations

3 Multiplication Before Elimination
Exercise 3
Solve the system of equations by elimination:
\(3x + 2y = 16\) and \(x + 3y = 13\)
Express your answer as a mixed number if needed.
Definition:

Multiplication Elimination: When coefficients are not equal/opposite, multiply equations to make them so

System
3x + 2y = 16, x + 3y = 13
Multiplication
Multiply 2nd by 3
Solution
(2, 4)
Step 1: Identify coefficients

First equation: \(3x + 2y = 16\) (coefficients: 3, 2)

Second equation: \(x + 3y = 13\) (coefficients: 1, 3)

Neither x nor y coefficients are equal or opposite

Step 2: Choose which variable to eliminate

To eliminate x: multiply the second equation by 3 to get coefficient 3

New system: \(3x + 2y = 16\) and \(3x + 9y = 39\)

Step 3: Subtract to eliminate x

\((3x + 9y) - (3x + 2y) = 39 - 16\)

\(3x + 9y - 3x - 2y = 23\)

\(7y = 23\)

Step 4: Solve for y

\(y = \frac{23}{7}\)

Step 5: Substitute y back to find x

Using the second original equation: \(x + 3y = 13\)

\(x + 3(\frac{23}{7}) = 13\)

\(x + \frac{69}{7} = 13\)

\(x = 13 - \frac{69}{7} = \frac{91}{7} - \frac{69}{7} = \frac{22}{7}\)

Step 6: Verify the solution

First equation: \(3x + 2y = 16\)

\(3(\frac{22}{7}) + 2(\frac{23}{7}) = \frac{66}{7} + \frac{46}{7} = \frac{112}{7} = 16\) ✓

Second equation: \(x + 3y = 13\)

\(\frac{22}{7} + 3(\frac{23}{7}) = \frac{22}{7} + \frac{69}{7} = \frac{91}{7} = 13\) ✓

Solution: (22/7, 23/7)
Final answer:

The solution to the system is \(\left(\frac{22}{7}, \frac{23}{7}\right)\) or approximately (3.14, 3.29).

Applied rules:

• Multiplication Property: Multiply both sides of an equation by the same value

• Common Multiples: Find LCM to make coefficients equal

• Fraction Arithmetic: Be careful with operations on fractions

Elimination Method Rules and Methods
\(\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}\)
System of Linear Equations
Step 1
Align equations
Line up like terms
Step 2
Choose variable
Decide which to eliminate
Step 3
Make coefficients equal/opposite
Multiply if needed
Key definitions:

Elimination Method: An algebraic technique for solving systems by adding or subtracting equations to eliminate one variable

System of Linear Equations: A set of two or more linear equations with the same variables

Solution to System: An ordered pair that makes all equations in the system true simultaneously

Complete methodology:
  1. Align Equations: Write equations in standard form with like terms aligned
  2. Choose Variable: Determine which variable is easier to eliminate
  3. Make Coefficients Equal/Opposite: Multiply equations by constants if needed
  4. Combine Equations: Add or subtract to eliminate the chosen variable
  5. Solve Single Variable: Solve the resulting one-variable equation
  6. Find Other Variable: Substitute back into one of the original equations
  7. Verify Solution: Check in both original equations
Tip 1: Look for variables with coefficients of 1 or -1 to make elimination easier.
Tip 2: When multiplying equations, multiply every term by the same constant.
Tip 3: Always verify your solution by substituting into both original equations.
Tip 4: If coefficients are already opposites, add the equations; if equal, subtract them.
Common errors: Forgetting to multiply all terms when scaling equations, sign errors when subtracting, not verifying the solution, arithmetic mistakes with fractions.
Exam preparation: Practice with different coefficient types, master fraction arithmetic, work with word problems.
Formulas to know by heart:

• Addition Elimination: If coefficients are opposites, add equations

• Subtraction Elimination: If coefficients are equal, subtract equations

• Multiplication Property: Multiply entire equation by same value preserves equality

• Solution Verification: Substitute solution into both original equations

Solution: Exercises 4 to 5
4 Word Problem Application
Exercise 4
A store sells pens for $2 each and notebooks for $3 each. A customer buys 8 items for a total of $19. How many pens and notebooks did the customer buy? Use elimination to solve.
Definition:

Word Problem Modeling: Translating real-world situations into mathematical equations

Variables
x=pens, y=notebooks
System
x + y = 8, 2x + 3y = 19
Solution
(5, 3)
Step 1: Define variables and set up equations

Let \(x\) = number of pens, \(y\) = number of notebooks

Total items: \(x + y = 8\)

Total cost: \(2x + 3y = 19\)

Step 2: Choose variable to eliminate

Eliminate x by multiplying the first equation by 2: \(2x + 2y = 16\)

Step 3: Set up for elimination

New system: \(\begin{cases} 2x + 2y = 16 \\ 2x + 3y = 19 \end{cases}\)

Step 4: Subtract equations to eliminate x

\((2x + 3y) - (2x + 2y) = 19 - 16\)

\(y = 3\)

Step 5: Substitute y back to find x

Using the first original equation: \(x + y = 8\)

\(x + 3 = 8\)

\(x = 5\)

Step 6: Verify the solution

Total items: \(x + y = 5 + 3 = 8\) ✓

Total cost: \(2x + 3y = 2(5) + 3(3) = 10 + 9 = 19\) ✓

Solution: (5, 3) - 5 pens, 3 notebooks
Final answer:

The customer bought 5 pens and 3 notebooks.

Applied rules:

• Word Problem Translation: Convert verbal constraints to equations

• Elimination Method: Multiply to make coefficients equal, then subtract

• Verification: Check solution against original problem constraints

5 Age Problem
Exercise 5
Sarah is twice as old as her brother Tom. The sum of their ages is 30 years. How old are Sarah and Tom? Use elimination to solve.
Definition:

Age Problems: Using algebraic methods to solve problems involving current or future ages

Variables
s=Sarah's age, t=Tom's age
System
s - 2t = 0, s + t = 30
Solution
(20, 10)
Step 1: Define variables and set up equations

Let \(s\) = Sarah's current age, \(t\) = Tom's current age

Sarah is twice as old as Tom: \(s = 2t\) or \(s - 2t = 0\)

Sum of ages is 30: \(s + t = 30\)

Step 2: Align equations for elimination

\(\begin{cases} s - 2t = 0 \\ s + t = 30 \end{cases}\)

Step 3: Choose variable to eliminate

Eliminate s by subtracting the first equation from the second

Step 4: Subtract equations to eliminate s

\((s + t) - (s - 2t) = 30 - 0\)

\(s + t - s + 2t = 30\)

\(3t = 30\)

\(t = 10\)

Step 5: Find s using either equation

Using the second equation: \(s + t = 30\)

\(s + 10 = 30\)

\(s = 20\)

Step 6: Verify the solution

Sarah is twice as old as Tom: \(20 = 2(10) = 20\) ✓

Sum of ages: \(s + t = 20 + 10 = 30\) ✓

Solution: (20, 10) - Sarah 20, Tom 10
Final answer:

Sarah is 20 years old and Tom is 10 years old.

Applied rules:

• Age Relationship Modeling: Translate verbal relationships to equations

• Elimination Strategy: Choose the variable that's easier to eliminate

• Verification: Check solution against all original conditions

Elimination Method Fundamentals
\(\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}\)
System of Linear Equations
Key definitions:

Elimination Method: An algebraic technique for solving systems by adding or subtracting equations to eliminate one variable, then solving for the remaining variable

Variable Elimination: The process of removing one variable from a system by combining equations

Back-Substitution: Using a known value to find the remaining variable in a system

Complete methodology:
  1. Standard Form: Write equations in standard form (ax + by = c)
  2. Choose Variable: Select the variable that will be easier to eliminate
  3. Make Coefficients Equal/Opposite: Multiply equations by constants if necessary
  4. Combine Equations: Add or subtract to eliminate the chosen variable
  5. Solve Single Variable: Solve the resulting one-variable equation
  6. Find Other Variable: Substitute the known value into one of the original equations
  7. Verify Solution: Check the solution in both original equations
Tip 1: When coefficients are opposites (like 3 and -3), add the equations.
Tip 2: When coefficients are the same (like 2 and 2), subtract the equations.
Tip 3: When multiplying equations, remember to multiply every term on both sides.
Tip 4: Always verify your solution to catch arithmetic errors.
Applications: Economics (cost/profit analysis), physics (motion problems), business (break-even analysis), chemistry (mixture problems), engineering (circuit analysis).
Method Advantages: Works well when coefficients are already equal/opposite, provides exact solutions, efficient for systems with integer coefficients.
Essential formulas:

• Addition Elimination: If coefficients are opposites, add equations

• Subtraction Elimination: If coefficients are equal, subtract equations

• Multiplication Property: \(a(bx + cy) = abx + acy\)

• Solution verification: Check \((x, y)\) in both original equations

Elimination Method Visualization
Exercise 6: Elimination Process
Visualize the elimination process for:
x + y = 7 and x - y = 1
Show how addition eliminates y.

Analysis: The chart shows how elimination method combines equations to eliminate one variable.

  • Original system: Two equations with two variables
  • Elimination: Adding equations eliminates y (y + (-y) = 0)
  • Result: Single equation 2x = 8 in variable x

Questions & Answers

Question: When should I use elimination vs. substitution?

Answer: Choose the method based on the system's structure:

  • Use elimination when: Coefficients of one variable are the same or opposites, or can easily be made the same/opposite
  • Use substitution when: One equation is already solved for a variable (like y = 2x + 3)
  • General rule: If both equations are in standard form, elimination is often easier

For example, with the system x + y = 5 and 2x - y = 4, elimination is ideal because y-coefficients are opposites (1 and -1).

With the system y = 3x - 2 and 2x + y = 8, substitution is ideal because the first equation is already solved for y.

Both methods will give the same solution, but one is usually more efficient than the other.

Question: I keep making sign errors when subtracting equations. Any tips?

Answer: Sign errors are common when subtracting equations. Here are strategies:

  • Think of subtraction as adding the opposite: Instead of "subtracting equation 2 from equation 1," think of it as "adding equation 1 to -1 times equation 2"
  • Distribute the negative sign: When subtracting (a + b) - (c + d), it becomes a + b - c - d
  • Be systematic: Subtract term by term: x-terms, y-terms, constants separately

For example, if you have (2x + 3y = 7) - (2x - y = 5), rewrite as:

(2x + 3y = 7) + (-2x + y = -5)

This gives: 2x - 2x + 3y + y = 7 + (-5), which simplifies to 4y = 2

The key is to apply the negative sign to every term in the equation being subtracted.

Question: What happens if I get a contradiction or identity when solving by elimination?

Answer: These special cases indicate the nature of the system:

If you get a contradiction (like 0 = 5):

  • The system is inconsistent (no solution)
  • The lines are parallel and never intersect
  • Example: Solving x + y = 3 and x + y = 8 leads to 0 = 5 (contradiction)

If you get an identity (like 0 = 0):

  • The system is dependent (infinitely many solutions)
  • The equations represent the same line
  • Example: Solving x + y = 5 and 2x + 2y = 10 leads to 0 = 0 (identity)

These outcomes are mathematically valid and tell you about the relationship between the equations in the system.

Question: How do I choose which variable to eliminate?

Answer: Choose the variable that will be easiest to eliminate:

  • Look for coefficients that are already equal or opposite - this avoids multiplication
  • Choose the variable with smaller coefficients - this makes arithmetic simpler
  • Select the variable that requires less multiplication - fewer steps, fewer chances for errors

For example, in the system 2x + 3y = 7 and 4x + y = 5:

  • To eliminate x: multiply first equation by 2 → 4x + 6y = 14, then subtract second equation
  • To eliminate y: multiply second equation by 3 → 12x + 3y = 15, then subtract first equation

Eliminating y would be easier since it only requires multiplying by 3, while eliminating x requires multiplying by 2 and then subtracting.

Both approaches work, but choose the one that minimizes the number of operations.