Solved Exercises on Equations with Variables on Both Sides in Integrated Math 1

Master equations with variables on both sides: simple to complex examples through these 5 detailed exercises with comprehensive solutions.

Solution: Exercises 1 to 3
1 Basic variables on both sides
Exercise 1
Solve for x:
\(4x + 3 = 2x + 9\)
Definition:

Equation with variables on both sides: An equation where the variable appears on both sides of the equals sign

Goal: Collect all variable terms on one side and all constant terms on the other side

Strategy: Move variable terms to one side and constants to the other

Variables on both sides method:

To solve equations with variables on both sides:

  1. Collect variables: Move all variable terms to one side using inverse operations
  2. Collect constants: Move all constant terms to the other side using inverse operations
  3. Isolate the variable: Divide by the coefficient
  4. Verify: Substitute solution back into original equation
Original
\(4x + 3 = 2x + 9\)
Subtract 2x
\(2x + 3 = 9\)
Subtract 3
\(2x = 6\)
Divide by 2
\(x = 3\)
Step 1: Write the equation

\(4x + 3 = 2x + 9\)

Step 2: Move variables to the left side by subtracting 2x from both sides

\(4x + 3 - 2x = 2x + 9 - 2x\)

\(2x + 3 = 9\)

Step 3: Move constants to the right side by subtracting 3 from both sides

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

\(2x = 6\)

Step 4: Isolate the variable by dividing both sides by 2

\(\frac{2x}{2} = \frac{6}{2}\)

\(x = 3\)

Step 5: Verify the solution

Substitute \(x = 3\) into original:

Left side: \(4(3) + 3 = 12 + 3 = 15\)

Right side: \(2(3) + 9 = 6 + 9 = 15\)

Since both sides equal 15, the solution is correct ✓

\(x = 3\)
Final answer:

The solution is \(x = 3\)

Applied rules:

• Variable Collection: Move all variable terms to one side by using inverse operations

• Constant Collection: Move all constant terms to the other side by using inverse operations

• Addition/Subtraction Property: Adding or subtracting the same number from both sides maintains equality

• Multiplication/Division Property: Dividing both sides by the same non-zero number maintains equality

Tip: It doesn't matter which side you move variables to - just be consistent!
Tip: Always verify by substituting back into the original equation to check both sides equal.
2 Variables on both sides with subtraction
Exercise 2
Solve for x:
\(5x - 2 = 3x + 8\)
Definition:

Variables on both sides with subtraction: An equation with variables and constants on both sides, including subtraction

Key concept: Treat subtraction as adding a negative number

Original
\(5x - 2 = 3x + 8\)
Subtract 3x
\(2x - 2 = 8\)
Add 2
\(2x = 10\)
Divide by 2
\(x = 5\)
Step 1: Write the equation

\(5x - 2 = 3x + 8\)

Step 2: Move variables to the left side by subtracting 3x from both sides

\(5x - 2 - 3x = 3x + 8 - 3x\)

\(2x - 2 = 8\)

Step 3: Move constants to the right side by adding 2 to both sides

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

\(2x = 10\)

Step 4: Isolate the variable by dividing both sides by 2

\(\frac{2x}{2} = \frac{10}{2}\)

\(x = 5\)

Step 5: Verify the solution

Substitute \(x = 5\) into original:

Left side: \(5(5) - 2 = 25 - 2 = 23\)

Right side: \(3(5) + 8 = 15 + 8 = 23\)

Since both sides equal 23, the solution is correct ✓

\(x = 5\)
Final answer:

The solution is \(x = 5\)

Applied rules:

• Variable Collection: Move all variable terms to one side

• Constant Collection: Move all constant terms to the other side

• Addition/Subtraction Property: Maintaining equality when performing operations

• Sign Awareness: Be careful with negative constants

Tip: When moving a negative term, it becomes positive when added to both sides!
3 Variables on both sides with negative coefficients
Exercise 3
Solve for x:
\(2x + 7 = -3x + 12\)
Definition:

Variables on both sides with negative coefficients: An equation where one side has a negative variable coefficient

Key concept: Adding a positive term to both sides eliminates the negative term on one side

Original
\(2x + 7 = -3x + 12\)
Add 3x
\(5x + 7 = 12\)
Subtract 7
\(5x = 5\)
Divide by 5
\(x = 1\)
Step 1: Write the equation

\(2x + 7 = -3x + 12\)

Step 2: Move variables to the left side by adding 3x to both sides

\(2x + 7 + 3x = -3x + 12 + 3x\)

\(5x + 7 = 12\)

Step 3: Move constants to the right side by subtracting 7 from both sides

\(5x + 7 - 7 = 12 - 7\)

\(5x = 5\)

Step 4: Isolate the variable by dividing both sides by 5

\(\frac{5x}{5} = \frac{5}{5}\)

\(x = 1\)

Step 5: Verify the solution

Substitute \(x = 1\) into original:

Left side: \(2(1) + 7 = 2 + 7 = 9\)

Right side: \(-3(1) + 12 = -3 + 12 = 9\)

Since both sides equal 9, the solution is correct ✓

\(x = 1\)
Final answer:

The solution is \(x = 1\)

Applied rules:

• Variable Collection: Move all variable terms to one side

• Sign Rules: Adding a positive number to eliminate a negative term

• Addition/Subtraction Property: Maintaining equality when performing operations

• Verification: Always check with negative coefficients

Tip: When adding a positive number to eliminate a negative term, remember: \(-3x + 3x = 0\)
Tip: Negative coefficients don't change the solving method - just be careful with signs!
Rules and methods, laws,...
\(ax + b = cx + d \Rightarrow x = \frac{d - b}{a - c}\)
General Form
\(ax + b = cx + d \Rightarrow (a-c)x = d - b\)
Collection Form
Basic
\(ax + b = cx + d\)
Collect: \((a-c)x = d - b\)
Negative Coeff
\(ax + b = -cx + d\)
Collect: \((a+c)x = d - b\)
Both Negative
\(-ax + b = -cx + d\)
Collect: \((-a+c)x = d - b\)
Complex
\(a(x + b) = c(x + d)\)
Distribute first, then collect
Equality Property: Both sides of an equation must remain equal when performing operations.
Collection Strategy: Move all variable terms to one side, all constants to the other.
Solution: Exercises 4 to 5
4 Variables on both sides with distribution
Exercise 4
Solve for x:
\(3(x + 2) = 2x + 14\)
Definition:

Variables on both sides with distribution: An equation requiring distribution before collecting variables

Key concept: Distribute first, then collect variables and constants

Original
\(3(x + 2) = 2x + 14\)
Distribute
\(3x + 6 = 2x + 14\)
Subtract 2x
\(x + 6 = 14\)
Subtract 6
\(x = 8\)
Step 1: Write the equation

\(3(x + 2) = 2x + 14\)

Step 2: Distribute on the left side

\(3(x + 2) = 3x + 6\), so: \(3x + 6 = 2x + 14\)

Step 3: Move variables to the left side by subtracting 2x from both sides

\(3x + 6 - 2x = 2x + 14 - 2x\)

\(x + 6 = 14\)

Step 4: Move constants to the right side by subtracting 6 from both sides

\(x + 6 - 6 = 14 - 6\)

\(x = 8\)

Step 5: Verify the solution

Substitute \(x = 8\) into original:

Left side: \(3(8 + 2) = 3(10) = 30\)

Right side: \(2(8) + 14 = 16 + 14 = 30\)

Since both sides equal 30, the solution is correct ✓

\(x = 8\)
Final answer:

The solution is \(x = 8\)

Applied rules:

• Distributive Property: Multiply the outside term by each term inside parentheses

• Variable Collection: Move all variable terms to one side

• Constant Collection: Move all constant terms to the other side

• Order of Operations: Distribute before collecting terms

Tip: Always distribute before collecting variables and constants!
5 Complex variables on both sides
Exercise 5
Solve for x:
\(4x - 3(x - 2) = 2(x + 1) + x\)
Definition:

Complex variables on both sides: An equation with multiple distributions and variables on both sides

Key concept: Distribute completely first, then collect like terms and variables

Original
\(4x - 3(x - 2) = 2(x + 1) + x\)
Distribute all
\(4x - 3x + 6 = 2x + 2 + x\)
Combine like terms
\(x + 6 = 3x + 2\)
Subtract x
\(6 = 2x + 2\)
Subtract 2
\(4 = 2x\)
Divide by 2
\(x = 2\)
Step 1: Write the equation

\(4x - 3(x - 2) = 2(x + 1) + x\)

Step 2: Distribute on the left side

\(-3(x - 2) = -3x + 6\), so: \(4x - 3x + 6 = 2(x + 1) + x\)

Simplified: \(x + 6 = 2(x + 1) + x\)

Step 3: Distribute on the right side

\(2(x + 1) = 2x + 2\), so: \(x + 6 = 2x + 2 + x\)

Step 4: Combine like terms on the right side

\(2x + x = 3x\), so: \(x + 6 = 3x + 2\)

Step 5: Move variables to the right side by subtracting x from both sides

\(x + 6 - x = 3x + 2 - x\)

\(6 = 2x + 2\)

Step 6: Move constants to the left side by subtracting 2 from both sides

\(6 - 2 = 2x + 2 - 2\)

\(4 = 2x\)

Step 7: Isolate the variable by dividing both sides by 2

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

\(2 = x\)

Or: \(x = 2\)

Step 8: Verify the solution

Substitute \(x = 2\) into original:

Left side: \(4(2) - 3(2 - 2) = 8 - 3(0) = 8 - 0 = 8\)

Right side: \(2(2 + 1) + 2 = 2(3) + 2 = 6 + 2 = 8\)

Since both sides equal 8, the solution is correct ✓

\(x = 2\)
Final answer:

The solution is \(x = 2\)

Applied rules:

• Multiple Distributions: Apply distributive property to every set of parentheses

• Sign Awareness: Pay attention to negative signs when distributing

• Systematic Approach: Distribute completely before collecting terms

• Variable Collection: Move all variables to one side

• Verification: Complex equations require careful checking

Tip: For complex equations, work slowly and check each step to avoid calculation errors.
Tip: When subtracting a group like \(-3(x - 2)\), distribute the negative: \(-3x + 6\).
Comprehensive Guide to Variables on Both Sides
\(ax + b = cx + d \Rightarrow (a-c)x = d - b \Rightarrow x = \frac{d - b}{a - c}\)
Systematic Solution
Key definitions:

Variables on both sides: An equation where the variable appears on both sides of the equals sign

Variable collection: Moving all variable terms to one side of the equation

Constant collection: Moving all constant terms to the other side of the equation

Systematic approach: Following a consistent order of operations to solve efficiently

Complete methodology:
  1. Distribute if necessary: Apply distributive property to eliminate parentheses
  2. Combine like terms: Simplify each side by combining similar terms
  3. Collect variables: Move all variable terms to one side using inverse operations
  4. Collect constants: Move all constant terms to the other side using inverse operations
  5. Isolate the variable: Divide by the coefficient
  6. Verify: Substitute solution back into original equation
Tip 1: It doesn't matter which side you move variables to - pick the side that seems simpler!
Tip 2: When moving terms, remember: what you do to one side, you must do to the other side.
Tip 3: Keep your work organized - each step should flow logically to the next.
Tip 4: Always verify your solution by substituting back into the original equation.
Tip 5: If you get 0 = 0, the equation has infinitely many solutions. If you get a contradiction like 5 = 3, there's no solution.
Common errors: Forgetting to move all variable terms to one side, making sign errors, not verifying solutions, performing operations on only one side.
Exam preparation: Practice equations with distribution, negative coefficients, and complex combinations. Focus on organization and verification.
Formulas to know by heart:

• General form: \(ax + b = cx + d \Rightarrow x = \frac{d - b}{a - c}\)

• Variable collection: Move all \(x\) terms to one side

• Constant collection: Move all numbers to the other side

• Sign distribution: \(-a(b + c) = -ab - ac\)

• Verification: Always substitute back to check

Variables on Both Sides Workflow

📊
Solving Process
1
Distribute
2
Combine Like Terms
3
Collect Variables
4
Collect Constants
5
Isolate Variable
6
Verify
Variables on Both Sides Forms
Basic: ax + b = cx + d → collect variables and constants
With distribution: a(x + b) = c(x + d) → distribute first
With negatives: ax + b = -cx + d → be careful with signs
Complex: multiple operations → systematic approach
Master This Systematic Approach to Excel in Algebra!

Questions & Answers

Question: When I have an equation like \(5x + 3 = 2x + 9\), how do I decide which side to move the variables to?

Answer: It actually doesn't matter which side you move the variables to - the final answer will be the same! However, here are some strategic considerations:

Move to the side with the larger coefficient: In \(5x + 3 = 2x + 9\), since 5 > 2, move variables to the left side to get a positive coefficient: \(5x - 2x = 3x\).

Move to the side that seems simpler: If one side has fewer terms, that might be easier.

For example, if you moved variables to the right side in \(5x + 3 = 2x + 9\):

  • Subtract 5x from both sides: \(3 = 2x - 5x + 9\)
  • Simplify: \(3 = -3x + 9\)
  • Subtract 9: \(-6 = -3x\)
  • Divide by -3: \(x = 2\)

Same answer, but you had to deal with negative coefficients!

Question: What happens if I get an equation like \(2x + 5 = 2x + 3\)? It seems impossible to solve!

Answer: You're right to notice something unusual! Let's solve it step by step:

Starting with: \(2x + 5 = 2x + 3\)

  • Subtract \(2x\) from both sides: \(2x + 5 - 2x = 2x + 3 - 2x\)
  • Simplify: \(5 = 3\)

This is a contradiction! Since \(5 \neq 3\), this equation has no solution.

This happens when the coefficients of the variable are the same on both sides, but the constants are different. The lines represented by these expressions are parallel and never intersect.

On the other hand, if you got something like \(2x + 5 = 2x + 5\), you'd get \(0 = 0\), meaning infinitely many solutions (the expressions are identical).

Question: Why do we need to verify our solution when we follow the correct steps?

Answer: Verification serves several important purposes:

  1. Error detection: Even when following correct steps, calculation errors can occur. Verification catches these mistakes.
  2. Understanding: Substituting back helps confirm that the solution actually works.
  3. Special cases: Verification helps identify when equations have no solution or infinite solutions.
  4. Confidence: Seeing both sides equal confirms your answer is correct.

For example, if you solved \(4x + 3 = 2x + 9\) and got \(x = 2\), verification would show:

  • Left side: \(4(2) + 3 = 8 + 3 = 11\)
  • Right side: \(2(2) + 9 = 4 + 9 = 13\)

Since \(11 \neq 13\), you know you made an error and need to go back and check your work.

Verification is an essential step that shouldn't be skipped!