Solved Exercises on Solving Equations with Fractions in Grade 8

Master solving equations with fractions: single fractions, multiple fractions, complex fractions, equations with variables in denominators, and word problems through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Single Fraction Equation
Exercise 1
Solve for \(x\):
\(\frac{x}{4} = 3\)
Definition:

Single Fraction Equation: An equation with one fraction containing the variable

Method for Single Fractions:
  1. Multiply both sides by the denominator to eliminate the fraction
  2. Simplify to solve for the variable
  3. Verify the solution
Original Equation
\(\frac{x}{4} = 3\)
Multiply both sides by 4
\(x = 12\)
Step 1: Eliminate the fraction

Multiply both sides by the denominator (4): \(\frac{x}{4} \times 4 = 3 \times 4\)

Step 2: Simplify

Left side: \(\frac{x}{4} \times 4 = x\)

Right side: \(3 \times 4 = 12\)

So: \(x = 12\)

Step 3: Verify the solution

Substitute \(x = 12\) back into the original equation: \(\frac{12}{4} = 3\) ✓

\(x = 12\)
Final answer:

\(x = 12\)

Applied rules:

• Eliminate Fractions: Multiply both sides by the denominator

• Balanced Equation: Perform same operation on both sides

• Verification: Check solution in original equation

2 Multiple Fractions
Exercise 2
Solve for \(y\):
\(\frac{y}{3} + \frac{y}{6} = 4\)
Definition:

Multiple Fractions: An equation with multiple fractions containing the same variable

Original Equation
\(\frac{y}{3} + \frac{y}{6} = 4\)
Find LCD (6)
Multiply by 6
Multiply each term by 6
\(2y + y = 24\)
Combine like terms
\(3y = 24\)
Divide by 3
\(y = 8\)
Step 1: Find the LCD

The denominators are 3 and 6. The LCD is 6.

Step 2: Multiply every term by the LCD

\(6 \times \frac{y}{3} + 6 \times \frac{y}{6} = 6 \times 4\)

So: \(2y + y = 24\)

Step 3: Combine like terms

\(2y + y = 3y\)

So: \(3y = 24\)

Step 4: Solve for the variable

Divide both sides by 3: \(\frac{3y}{3} = \frac{24}{3}\)

So: \(y = 8\)

Step 5: Verify the solution

Substitute \(y = 8\) back into the original equation:

\(\frac{8}{3} + \frac{8}{6} = \frac{8}{3} + \frac{4}{3} = \frac{12}{3} = 4\) ✓

\(y = 8\)
Final answer:

\(y = 8\)

Applied rules:

• Least Common Denominator: Multiply every term by LCD to eliminate fractions

• Combine Like Terms: Add coefficients of like terms

• Balanced Equation: Perform same operation on both sides

3 Complex Fraction Equation
Exercise 3
Solve for \(z\):
\(\frac{z + 2}{5} = \frac{z - 1}{3}\)
Definition:

Cross Multiplication: When two fractions are equal, cross multiply: \(\frac{a}{b} = \frac{c}{d} \Rightarrow ad = bc\)

Original Equation
\(\frac{z + 2}{5} = \frac{z - 1}{3}\)
Cross Multiply
\(3(z + 2) = 5(z - 1)\)
Distribute
\(3z + 6 = 5z - 5\)
Move variables to one side
\(6 + 5 = 5z - 3z\)
Simplify
\(11 = 2z\)
Divide by 2
\(z = \frac{11}{2}\)
Step 1: Cross multiply

Multiply the numerator of the left fraction by the denominator of the right fraction and vice versa:

\((z + 2) \times 3 = (z - 1) \times 5\)

So: \(3(z + 2) = 5(z - 1)\)

Step 2: Apply distributive property

\(3(z + 2) = 3z + 6\) and \(5(z - 1) = 5z - 5\)

So: \(3z + 6 = 5z - 5\)

Step 3: Move variables to one side, constants to the other

Subtract \(3z\) from both sides and add 5 to both sides:

\(6 + 5 = 5z - 3z\)

So: \(11 = 2z\)

Step 4: Solve for the variable

Divide both sides by 2: \(\frac{11}{2} = \frac{2z}{2}\)

So: \(z = \frac{11}{2}\)

Step 5: Verify the solution

Substitute \(z = \frac{11}{2}\) back into the original equation:

Left side: \(\frac{\frac{11}{2} + 2}{5} = \frac{\frac{15}{2}}{5} = \frac{15}{10} = \frac{3}{2}\)

Right side: \(\frac{\frac{11}{2} - 1}{3} = \frac{\frac{9}{2}}{3} = \frac{9}{6} = \frac{3}{2}\)

Both sides equal \(\frac{3}{2}\) ✓

\(z = \frac{11}{2}\)
Final answer:

\(z = \frac{11}{2}\)

Applied rules:

• Cross Multiplication: \(\frac{a}{b} = \frac{c}{d} \Rightarrow ad = bc\)

• Distributive Property: \(a(b + c) = ab + ac\)

• Move Variables: Get all variable terms on one side

Solving Equations with Fractions Rules and Methods
\(\frac{x}{a} = b \Rightarrow x = ab\)
Single Fraction
Single Fraction
\(\frac{x}{a} = b \Rightarrow x = ab\)
Multiply both sides by the denominator
Multiple Fractions
\(\frac{x}{a} + \frac{x}{b} = c \Rightarrow x = \frac{abc}{a+b}\)
Multiply by LCD to eliminate fractions
Cross Multiplication
\(\frac{a}{b} = \frac{c}{d} \Rightarrow ad = bc\)
When two fractions are equal, cross multiply
LCD Strategy
Multiply every term by LCD
Eliminates all fractions
Variables in Denominator
\(\frac{a}{x} = b \Rightarrow x = \frac{a}{b}\)
Multiply both sides by the variable
Verification
Always check in original equation
Ensure solution is valid
Key Concepts: When solving equations with fractions, eliminate fractions first by multiplying by the LCD. Always verify your solution in the original equation.
Special Cases: When variables appear in denominators, be aware of restrictions (denominator cannot equal zero).
Tip 1: Always multiply every term by the LCD to eliminate all fractions.
Tip 2: For equations with variables in denominators, multiply both sides by the variable expression.
Tip 3: Always check your solution by substituting back into the original equation.
Solution: Exercises 4 to 5
4 Variable in Denominator
Exercise 4
Solve for \(w\):
\(\frac{12}{w} = 4\)
Definition:

Variable in Denominator: When the variable is in the denominator, multiply both sides by the variable expression

Original Equation
\(\frac{12}{w} = 4\)
Multiply both sides by w
\(12 = 4w\)
Divide both sides by 4
\(w = 3\)
Step 1: Eliminate the fraction

Multiply both sides by the variable expression (w): \(\frac{12}{w} \times w = 4 \times w\)

Step 2: Simplify

Left side: \(\frac{12}{w} \times w = 12\)

Right side: \(4 \times w = 4w\)

So: \(12 = 4w\)

Step 3: Solve for the variable

Divide both sides by 4: \(\frac{12}{4} = \frac{4w}{4}\)

So: \(w = 3\)

Step 4: Verify the solution

Substitute \(w = 3\) back into the original equation: \(\frac{12}{3} = 4\) ✓

Step 5: Check restrictions

Ensure the denominator is not zero: \(w = 3 \neq 0\) ✓

\(w = 3\)
Final answer:

\(w = 3\)

Applied rules:

• Multiply by Variable: When variable is in denominator, multiply both sides by the variable

• Restriction Check: Ensure the solution doesn't make denominator zero

• Balanced Equation: Perform same operation on both sides

5 Word Problem
Exercise 5
A number divided by 4 is equal to the same number divided by 6 plus 2. Find the number.
Definition:

Word Problem Translation: Convert the verbal statement into a mathematical equation

Let the number be x
\(\frac{x}{4} = \frac{x}{6} + 2\)
Find LCD (12)
Multiply by 12
Multiply each term by 12
\(3x = 2x + 24\)
Subtract 2x from both sides
\(x = 24\)
Step 1: Define the variable

Let \(x\) be the unknown number.

Step 2: Translate the problem into an equation

"A number divided by 4" translates to \(\frac{x}{4}\)

"is equal to" translates to \(=\)

"the same number divided by 6" translates to \(\frac{x}{6}\)

"plus 2" translates to \(+ 2\)

So: \(\frac{x}{4} = \frac{x}{6} + 2\)

Step 3: Find the LCD and eliminate fractions

The denominators are 4 and 6. The LCD is 12.

Multiply every term by 12: \(12 \times \frac{x}{4} = 12 \times \frac{x}{6} + 12 \times 2\)

So: \(3x = 2x + 24\)

Step 4: Solve for the variable

Subtract \(2x\) from both sides: \(3x - 2x = 2x - 2x + 24\)

So: \(x = 24\)

Step 5: Verify the solution

Substitute \(x = 24\) back into the original equation:

Left side: \(\frac{24}{4} = 6\)

Right side: \(\frac{24}{6} + 2 = 4 + 2 = 6\)

Both sides equal 6 ✓

The number is 24
Final answer:

The number is 24

Applied rules:

• Word Problem Translation: Convert verbal statements to mathematical expressions

• Least Common Denominator: Multiply every term by LCD to eliminate fractions

• Balanced Equation: Perform same operation on both sides

Complete Guide: Solving Equations with Fractions, Rules, Methods, and Applications
\(\frac{x}{a} + \frac{x}{b} = c \Rightarrow x = \frac{abc}{a+b}\)
Multiple Fractions
Key definitions:

Fraction: A number representing a part of a whole, written as \(\frac{a}{b}\) where \(a\) is the numerator and \(b\) is the denominator

Least Common Denominator (LCD): The smallest common multiple of the denominators in a set of fractions

Cross Multiplication: When \(\frac{a}{b} = \frac{c}{d}\), then \(ad = bc\)

Restrictions: Values that make the denominator equal to zero are not allowed

Complete methodology:
  1. Identify the type: Determine which method to use based on the equation structure
  2. Eliminate fractions: Use LCD multiplication or cross multiplication as appropriate
  3. Solve: Use standard equation-solving techniques
  4. Verify: Check solution in original equation
  5. Check restrictions: Ensure solution doesn't violate domain restrictions
Tip 1: When you have multiple fractions, always multiply every term by the LCD to eliminate all fractions at once.
Tip 2: For equations with variables in denominators, multiply both sides by the variable expression.
Tip 3: Always check that your solution doesn't make any denominator equal to zero.
Tip 4: Always verify your solution by substituting it back into the original equation.
Common errors: Forgetting to multiply every term by the LCD, making sign errors when distributing, not checking domain restrictions, incorrectly applying cross multiplication.
Real-World Applications: Solving rate problems, mixture problems, and many other practical scenarios that involve fractional relationships.
Essential rules to memorize:

• Single Fraction: \(\frac{x}{a} = b \Rightarrow x = ab\)

• LCD Strategy: Multiply every term by LCD to eliminate fractions

• Cross Multiplication: \(\frac{a}{b} = \frac{c}{d} \Rightarrow ad = bc\)

• Variables in Denominator: Multiply both sides by the variable expression

• Domain Restrictions: Denominators cannot equal zero

• Verification: Always check solution in original equation

Exercise with Visualization: Equations with Fractions Solutions
Exercise 6: Fraction Equation Solutions Comparison
Consider the following equations and their solutions:
\(f_1(x) = \frac{x}{4} = 3 \Rightarrow x = 12\)
\(f_2(y) = \frac{y}{3} + \frac{y}{6} = 4 \Rightarrow y = 8\)
\(f_3(z) = \frac{z + 2}{5} = \frac{z - 1}{3} \Rightarrow z = \frac{11}{2}\)

Analysis: The chart shows how different fraction equations yield different solutions.

  • \(\frac{x}{4} = 3\) (solution: \(x = 12\))
  • \(\frac{y}{3} + \frac{y}{6} = 4\) (solution: \(y = 8\))
  • \(\frac{z + 2}{5} = \frac{z - 1}{3}\) (solution: \(z = \frac{11}{2} = 5.5\))

Questions & Answers

Question: Why do I multiply by the LCD to eliminate fractions? Can't I just work with the fractions?

Answer: While you can solve equations with fractions without eliminating them, multiplying by the LCD makes the process much simpler:

Without LCD: \(\frac{x}{3} + \frac{x}{6} = 4\)

Finding a common denominator: \(\frac{2x}{6} + \frac{x}{6} = 4\), so \(\frac{3x}{6} = 4\), then \(\frac{x}{2} = 4\), so \(x = 8\)

With LCD: \(\frac{x}{3} + \frac{x}{6} = 4\)

Multiply by 6: \(2x + x = 24\), so \(3x = 24\), so \(x = 8\)

Both methods give the same answer, but the LCD method eliminates fractions early, reducing complexity and potential for errors.

The LCD method transforms the equation into one with integers, which are generally easier to work with.

Question: What if I have a variable in the denominator like \(\frac{5}{x} = 2\)? How do I solve that?

Answer: When you have a variable in the denominator, multiply both sides by that variable:

For \(\frac{5}{x} = 2\):

  1. Multiply both sides by \(x\): \(\frac{5}{x} \times x = 2 \times x\)
  2. Simplify: \(5 = 2x\)
  3. Divide by 2: \(\frac{5}{2} = x\)
  4. So: \(x = \frac{5}{2}\)

Important restriction: The variable cannot equal zero because division by zero is undefined.

In our example, \(x = \frac{5}{2} \neq 0\), so the solution is valid.

Always check that your solution doesn't make any denominator zero!

Question: How do I find the LCD when I have multiple fractions? What if the denominators are big numbers?

Answer: To find the LCD of multiple fractions, find the LCM (Least Common Multiple) of the denominators:

Example: For denominators 4, 6, and 8

  1. Prime factorization: \(4 = 2^2\), \(6 = 2 \times 3\), \(8 = 2^3\)
  2. Take the highest power of each prime: \(2^3 \times 3 = 8 \times 3 = 24\)
  3. So LCD = 24

Alternative method: List multiples of the largest denominator until you find one that all denominators divide evenly into:

Multiples of 8: 8, 16, 24, 32, ...

Does 4 divide 8? Yes. Does 6 divide 8? No.

Does 4 divide 16? Yes. Does 6 divide 16? No.

Does 4 divide 24? Yes. Does 6 divide 24? Yes. So LCD = 24.

For large numbers, use the prime factorization method as it's more systematic.

Question: What happens if I get a solution that makes the denominator zero? What does that mean?

Answer: If your solution makes any denominator zero, it's called an extraneous solution and must be rejected:

Example: Solve \(\frac{x}{x-2} = \frac{4}{x-2}\)

  1. Since denominators are equal, numerators must be equal: \(x = 4\)
  2. Check: If \(x = 4\), then denominators become \(4 - 2 = 2 \neq 0\) ✓

Another example: Solve \(\frac{x}{x-3} = \frac{3}{x-3}\)

  1. Setting numerators equal: \(x = 3\)
  2. Check: If \(x = 3\), then denominators become \(3 - 3 = 0\)
  3. Since division by zero is undefined, \(x = 3\) is not a valid solution
  4. The equation has no solution

Always check your solution by substituting it back into the original equation to ensure it doesn't make any denominator zero!

Question: How do I know when to use cross multiplication vs. multiplying by the LCD?

Answer: Use these guidelines to choose the right method:

Use Cross Multiplication when:

  • You have two fractions set equal to each other: \(\frac{a}{b} = \frac{c}{d}\)
  • There are no other terms in the equation
  • Example: \(\frac{x+1}{3} = \frac{x-2}{4}\)

Use LCD Multiplication when:

  • You have multiple fractions on one or both sides
  • There are other terms besides fractions
  • Example: \(\frac{x}{2} + \frac{x}{3} = 5\) or \(\frac{x}{4} + 2 = \frac{x}{6}\)

For the example \(\frac{x}{4} + 2 = \frac{x}{6}\):

This has extra terms (the "+2"), so use LCD method. LCD of 4 and 6 is 12:

\(12 \times \frac{x}{4} + 12 \times 2 = 12 \times \frac{x}{6}\)

\(3x + 24 = 2x\), so \(x = -24\)

Cross multiplication would only work if both sides were single fractions!