Rational Equations

Grade 9 · algebra · 71 practice problems · read aloud

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Rational Equations: Solving Fractional Problems

What Are Rational Equations? 🤔

A rational equation is an equation containing at least one fraction whose numerator and denominator are polynomials. You see these when working with rates, proportions, or scaling problems. For example, 1/x + 1/3 = 1/2 could help you figure out how long it takes two people to complete a job together.

How to Solve Rational Equations

  1. Find the LCD: Identify the Least Common Denominator of all fractions in the equation.
  2. Multiply & Clear: Multiply every term on both sides by the LCD. This eliminates the denominators!
  3. Solve: Solve the resulting polynomial equation (usually linear or quadratic).
  4. Check for Extraneous Solutions: Substitute your answers back into the original equation. If any solution makes a denominator zero, it's extraneous and must be rejected.

Worked Examples

Example 1: Solve 2/x = 5

  1. LCD is x
  2. Multiply both sides by x: 2 = 5x
  3. Solve: x = 2/5
  4. Check: 2 ÷ (2/5) = 5 ✓ (Valid solution)

Example 2: Solve 3/(x+1) = 2/(x-1)

  1. LCD is (x+1)(x-1)
  2. Multiply both sides by LCD: 3(x-1) = 2(x+1)
  3. Solve: 3x - 3 = 2x + 2 → x = 5
  4. Check: 3/(5+1) = 2/(5-1) → 3/6 = 2/4 → 1/2 = 1/2 ✓ (Valid solution)

Common Mistakes to Avoid 🚫

Forgetting to check for extraneous solutions: Always verify your answers don't make any denominator zero.

Incorrectly finding the LCD: The LCD must contain every factor from all denominators.

Distributing errors: When multiplying by the LCD, remember to multiply every term in the equation.

Tips & Tricks

Cross-multiplication shortcut: Only works for equations with a single fraction on each side of the equals sign.

Factor first: Always factor denominators completely before identifying the LCD.

Zero denominator check: Before you start solving, note which values would make denominators zero—these can't be solutions!

Practice Suggestions

  • Start with simple equations like 2/x = 3
  • Progress to equations requiring factoring, like 1/(x²-4) = 1/3
  • Create your own equations and solve them
  • Practice with word problems involving rates and work
  • Always check your solutions—this is half the battle!

Practice problems

6 of the 71, worked through step by step — try them before opening the answer.

1 (x² - 9)/(x - 3) = ?

Hint: Look for a difference of squares pattern in the numerator and consider how it relates to the denominator.

Show the answer

Answer: x + 3

  1. Recognize the numerator** The numerator is x² - 9. Notice that x² - 9 is a difference of squares: x² - 9 = x² - 3² --- **
  2. Apply the difference of squares formula** The formula is: a² - b² = (a - b)(a + b) Here a = x, b = 3. So: x² - 9 = (x - 3)(x + 3) --- **
  3. Rewrite the original expression** (x² - 9)/(x - 3) = [(x - 3)(x + 3)] / (x - 3) --- **
  4. Cancel the common factor** We can cancel (x - 3) from numerator and denominator, provided x ≠ 3 (since division by zero is undefined). So we get: x + 3 --- **
  5. Final answer** Thus, (x² - 9)/(x - 3) = x + 3, for x ≠ 3. --- ANSWER: x + 3

Let's solve step by step. We are given: (x² - 9)/(x - 3) --- **

2 (x + 3)/(x - 2) = 2 = ?

Hint: When solving rational equations, eliminate denominators by multiplying both sides by the denominator, then solve the resulting linear equation.

Show the answer

Answer: 7

  1. Identify the domain restriction. Since we have a denominator x - 2, it cannot be zero. So x ≠ 2.
  2. Eliminate the fraction by multiplying both sides by (x - 2). (x + 3)/(x - 2) = 2 Multiply both sides by (x - 2): x + 3 = 2(x - 2)
  3. Expand the right-hand side. x + 3 = 2x - 4
  4. Rearrange the equation to solve for x. Subtract x from both sides: 3 = x - 4
  5. Add 4 to both sides. 3 + 4 = x 7 = x
  6. Check domain restriction. x = 7 is not equal to 2, so it is valid.
  7. Verify by substituting back into the original equation. (7 + 3)/(7 - 2) = 10/5 = 2, which matches the right-hand side. Final answer: x = 7

We are given the equation: (x + 3)/(x - 2) = 2

3 (x + 3)/(x - 2) = 4 = ?

Hint: To solve rational equations, eliminate denominators by multiplying both sides by the denominator, then solve the resulting linear equation. Remember to check for extraneous solutions that make the original denominator zero.

Show the answer

Answer: 11/3

  1. Identify the domain restriction. The denominator x - 2 cannot be zero, so x ≠ 2.
  2. Eliminate the fraction by multiplying both sides by (x - 2). (x + 3)/(x - 2) = 4 Multiply both sides by (x - 2): x + 3 = 4(x - 2)
  3. Expand the right-hand side. x + 3 = 4x - 8
  4. Rearrange terms to isolate x. Subtract x from both sides: 3 = 3x - 8
  5. Add 8 to both sides. 3 + 8 = 3x 11 = 3x
  6. Divide both sides by 3. x = 11/3
  7. Check the domain. x = 11/3 is not equal to 2, so it is valid.
  8. Check in the original equation (optional but good practice). Left-hand side: (11/3 + 3)/(11/3 - 2) = (11/3 + 9/3)/(11/3 - 6/3) = (20/3)/(5/3) = (20/3) * (3/5) = 20/5 = 4 Matches the right-hand side. Final answer: x = 11/3

We are given the equation: (x + 3)/(x - 2) = 4

4 (x + 3)/(x - 2) = 5 = ?

Hint: To solve rational equations, eliminate denominators by multiplying both sides by the denominator, then solve the resulting linear equation. For example, if you had (y + 1)/(y - 4) = 2, you would multiply both sides by (y - 4) to eliminate the fraction.

Show the answer

Answer: 3.25

  1. Eliminate the fraction by multiplying both sides by (x - 2). We must note that x cannot be 2, because that would make the denominator zero. (x + 3) = 5 * (x - 2)
  2. Expand the right-hand side. x + 3 = 5x - 10
  3. Bring all terms involving x to one side and constants to the other. Subtract x from both sides: 3 = 4x - 10
  4. Add 10 to both sides. 3 + 10 = 4x 13 = 4x
  5. Divide both sides by 4. x = 13/4
  6. Convert to decimal. 13/4 = 3.25 Final answer: x = 3.25

We are given the equation: (x + 3)/(x - 2) = 5

5 (x + 2)/(x - 1) = 3 = ?

Hint: To solve rational equations, eliminate the denominator by multiplying both sides by the expression in the denominator, then solve the resulting linear equation.

Show the answer

Answer: 2.5

  1. Start with the equation (x + 2)/(x - 1) = 3
  2. Multiply both sides by (x - 1) to eliminate the denominator: x + 2 = 3(x - 1)
  3. Distribute the 3 on the right side: x + 2 = 3x - 3
  4. Subtract x from both sides: 2 = 2x - 3
  5. Add 3 to both sides: 5 = 2x
  6. Divide both sides by 2: x = 5/2 = 2.5

The answer is 2.5.

6 (2x + 3)/(x - 1) = 5 = ?

Hint: To solve rational equations, eliminate denominators by multiplying both sides by the denominator, then solve the resulting linear equation. For example, if you had (a)/(b) = c, you would multiply both sides by b to get a = c × b.

Show the answer

Answer: 8/3

  1. Since the equation is (2x + 3)/(x - 1) = 5, we can multiply both sides by (x - 1) to eliminate the denominator. But note: x cannot be 1 because that would make the denominator zero. So, assuming x ≠ 1, multiply both sides by (x - 1): (2x + 3) = 5 * (x - 1)
  2. Expand the right-hand side: 2x + 3 = 5x - 5
  3. Bring like terms together. Subtract 2x from both sides: 3 = 5x - 5 - 2x 3 = 3x - 5
  4. Add 5 to both sides: 3 + 5 = 3x 8 = 3x
  5. Divide both sides by 3: x = 8/3
  6. Check if x ≠ 1 (yes, 8/3 is not 1) and substitute back to verify: (2*(8/3) + 3) / (8/3 - 1) = (16/3 + 9/3) / (8/3 - 3/3) = (25/3) / (5/3) = (25/3) * (3/5) = 25/5 = 5. It works. Final answer: x = 8/3

We are given the equation: (2x + 3)/(x - 1) = 5.

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