Equivalent Forms

Grade 9 · algebra · 100 practice problems · read aloud

🔊 Listen to this explanation

What Are Equivalent Forms? 🤔

Equivalent forms are different algebraic expressions that represent the same value for all variable inputs. Think of them as different "outfits" for the same mathematical "body"!

Why it's useful: Equivalent forms help us simplify problems, solve equations, and understand relationships between variables more clearly.

How to Create Equivalent Forms

  1. Identify the original expression
  2. Apply algebraic properties (distributive, combining like terms, factoring)
  3. Verify both forms give the same output for test values

Worked Examples

Example 1: Distributive Property

Original: 3(x + 4)

Step 1: Apply distributive property: 3 × x + 3 × 4

Equivalent form: 3x + 12 ✅

Example 2: Combining Like Terms

Original: 2x + 5 + 3x - 2

Step 1: Group like terms: (2x + 3x) + (5 - 2)

Step 2: Combine: 5x + 3 ✅

Example 3: Factoring

Original: x² + 5x + 6

Step 1: Find factors of 6 that add to 5: 2 and 3

Equivalent form: (x + 2)(x + 3) ✅

⚠️ Common Mistakes to Avoid

  • Wrong distribution: 2(x + 3) ≠ 2x + 3 (missing multiplication)
  • Sign errors: -3(x - 2) = -3x + 6 (not -3x - 6)
  • Forgetting to combine all like terms
  • Incorrect factoring: Check your work by expanding back!

💡 Tips & Tricks

  • Test with numbers: Plug in x = 1 to verify both forms give same result
  • FOIL backwards: For factoring quadratics, work in reverse
  • Look for GCF first: Always check for greatest common factors
  • Color code: Use different colors for like terms when combining

Practice Makes Perfect! 📝

Try these practice strategies:

  • Create flashcards with expressions on front, equivalent forms on back
  • Work with a partner - one writes original, other finds equivalent form
  • Use online algebra practice sites for instant feedback
  • Start simple: 2(x + 3) → ? then progress to: x² + 7x + 12 → ?

Practice problems

6 of the 100, worked through step by step — try them before opening the answer.

1 2^(3x-1) = 32

Hint: Express both sides with the same base and then equate the exponents

Show the answer

Answer: 2

  1. Recognize that 32 can be written as a power of 2: 32 = 2^5
  2. Rewrite the equation: 2^(3x-1) = 2^5
  3. Since the bases are equal, set the exponents equal: 3x - 1 = 5
  4. Add 1 to both sides: 3x = 6
  5. Divide both sides by 3: x = 2

The answer is 2.

2 2^(3x-1) = 16

Hint: Express both sides with the same base and then equate the exponents

Show the answer

Answer: 5/3

  1. Write 16 as a power of 2: 16 = 2^4
  2. Substitute into the equation: 2^(3x-1) = 2^4
  3. Since the bases are equal, set the exponents equal: 3x - 1 = 4
  4. Add 1 to both sides: 3x = 5
  5. Divide both sides by 3: x = 5/3

The answer is 5/3.

3 2x² - 8x + 6 = 0

Hint: For quadratic equations of the form ax² + bx + c = 0, you can factor the expression or use the quadratic formula to find the values of the variable that satisfy the equation.

Show the answer

Answer: x = 1, 3

  1. Check for common factors** All coefficients (2, -8, 6) are divisible by 2. Divide the entire equation by 2: (2x²)/2 - (8x)/2 + 6/2 = 0/2 x² - 4x + 3 = 0 --- **
  2. Factor the quadratic** We look for two numbers that multiply to 3 (the constant term) and add to -4 (the coefficient of x). The numbers are -1 and -3 because: (-1) × (-3) = 3 (-1) + (-3) = -4 So we can factor: x² - 4x + 3 = (x - 1)(x - 3) = 0 --- **
  3. Apply the zero-product property** If (x - 1)(x - 3) = 0, then: x - 1 = 0 or x - 3 = 0 --- **
  4. Solve each equation** x - 1 = 0 → x = 1 x - 3 = 0 → x = 3 --- **Final Answer:** x = 1, 3

Let's solve the equation step by step. We have: 2x² - 8x + 6 = 0 --- **

4 √(x² - 6x + 9) = ?

Hint: Look for a perfect square trinomial pattern where the expression under the radical can be rewritten as a squared binomial.

Show the answer

Answer: |x - 3|

  1. Recognize the expression inside the square root** x² - 6x + 9 Notice that this is a perfect square trinomial. --- **
  2. Factor the trinomial** x² - 6x + 9 = (x - 3)² Check: (x - 3)² = x² - 2*x*3 + 3² = x² - 6x + 9. Correct. --- **
  3. Substitute back into the square root** √(x² - 6x + 9) = √( (x - 3)² ) --- **
  4. Recall the property of square roots** For any real number A, √(A²) = |A| This is because the square root function returns only the non-negative root. --- **
  5. Apply the property** Here A = x - 3, so: √( (x - 3)² ) = |x - 3| --- **
  6. Final answer** √(x² - 6x + 9) = |x - 3| --- **Reasoning summary:** We factored the quadratic into a perfect square, then used the rule that the square root of a square is the absolute value, ensuring the result is always non-negative.

Let's solve step by step. We are given: √(x² - 6x + 9) = ? --- **

5 log₂(8) + log₃(9) = ?

Hint: Recall that logarithms represent exponents. For example, log₄(16) asks '4 to what power equals 16?'

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Answer: 5

  1. Evaluate log₂(8). This means '2 to what power equals 8?' Since 2³ = 8, log₂(8) = 3.
  2. Evaluate log₃(9). This means '3 to what power equals 9?' Since 3² = 9, log₃(9) = 2.
  3. Add the results: 3 + 2 = 5.

The answer is 5.

6 2^(x+1) - 2^(x-1) = 24

Hint: Use exponent rules to factor out a common term and simplify the equation

Show the answer

Answer: 4

  1. Factor out 2^(x-1) from both terms: 2^(x-1)(2^2 - 1) = 24
  2. Simplify inside parentheses: 2^(x-1)(4 - 1) = 24
  3. Calculate: 2^(x-1)(3) = 24
  4. Divide both sides by 3: 2^(x-1) = 8
  5. Rewrite 8 as a power of 2: 2^(x-1) = 2^3
  6. Set exponents equal: x - 1 = 3
  7. Solve for x: x = 4

The answer is 4.

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