Rearrange Formulas

Grade 9 · algebra · 100 practice problems · read aloud

🔊 Listen to this explanation

Rearranging Formulas

🧠 What is it and Why is it Useful?

Rearranging formulas (or "changing the subject") is the process of isolating a specific variable in an equation. This is incredibly useful because it allows you to solve for any quantity you want, depending on the information you have. For example, if you know the area of a rectangle, you can rearrange the area formula to solve for its length or width.

📝 Step-by-Step Guide

  1. Identify the variable you want to solve for (the "subject").
  2. Use inverse operations to "undo" what's being done to your variable. Perform the same operation to both sides of the equation to keep it balanced.
  3. Simplify both sides of the equation after each step.
  4. Check your work by substituting numbers back into the original formula.

🔍 Worked Examples

Example 1: Solve for l in the area of a rectangle formula.

Formula: A = l × w

Step 1: The variable l is being multiplied by w. To isolate l, do the inverse: divide both sides by w.

Step 2: A / w = (l × w) / w

Step 3: A / w = l

Answer: l = A / w

Example 2: Solve for h in the volume of a cylinder formula.

Formula: V = πr²h

Step 1: h is being multiplied by π and r². Divide both sides by πr².

Step 2: V / (πr²) = (πr²h) / (πr²)

Step 3: V / (πr²) = h

Answer: h = V / (πr²)

⚠️ Common Mistakes

  • Forgetting to perform operations on both sides: Whatever you do to one side, you MUST do to the other.
  • Incorrect order of operations: Undo addition/subtraction before multiplication/division. Think of "undoing" the operations in reverse PEMDAS order.
  • Not distributing completely: When a term outside a bracket is multiplied or divided, it must be applied to every term inside.

💡 Tips & Tricks

  • Think of the formula as a balance scale—it must stay level!
  • Before you start, identify all the operations being performed on your target variable.
  • If your variable appears in more than one term, you'll need to factor it out first.

🎯 Practice Suggestions

  • Start with simple geometric formulas (like A = lw or P = 2l + 2w).
  • Create flashcards with a formula on one side and the rearranged version on the other.
  • Use online practice tools that generate endless problems for you to solve.
  • Challenge yourself by trying to rearrange the same formula for every variable in it.

Practice problems

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

1 2x² + 5x - 3 = 0

Hint: For equations of the form ax² + bx + c = 0, you can use a special formula that involves the coefficients a, b, and c along with square roots.

Show the answer

Answer: x = 0.5, x = -3

  1. Identify coefficients The equation is in the form ax² + bx + c = 0. Here: a = 2, b = 5, c = -3.
  2. Check if factoring is possible We look for two numbers that multiply to a * c = 2 * (-3) = -6, and add to b = 5. Possible pairs for -6: (1, -6) sum = -5, (-1, 6) sum = 5. Yes, -1 and 6 work.
  3. Rewrite the middle term using these numbers 2x² + 6x - 1x - 3 = 0.
  4. Factor by grouping Group terms: (2x² + 6x) + (-1x - 3) = 0. Factor each group: 2x(x + 3) - 1(x + 3) = 0.
  5. Factor out the common factor (x + 3)(2x - 1) = 0.
  6. Apply the zero product property If (x + 3)(2x - 1) = 0, then: x + 3 = 0 or 2x - 1 = 0.
  7. Solve each equation From x + 3 = 0: x = -3. From 2x - 1 = 0: 2x = 1, x = 1/2 = 0.5. Final answer: x = 0.5, x = -3.

We are solving the quadratic equation: 2x² + 5x - 3 = 0.

2 √(x² - 8x + 16) = 7

Hint: Remember that the square root of a squared expression equals the absolute value of that expression. Consider both positive and negative cases when solving.

Show the answer

Answer: 11

  1. Square both sides to eliminate the square root: (√(x² - 8x + 16))² = 7²
  2. This gives us: x² - 8x + 16 = 49
  3. Recognize that x² - 8x + 16 is a perfect square: (x - 4)² = 49
  4. Take square root of both sides: |x - 4| = 7
  5. Solve the absolute value equation: x - 4 = 7 or x - 4 = -7
  6. Solve first case: x - 4 = 7 → x = 11
  7. Solve second case: x - 4 = -7 → x = -3
  8. Check both solutions in the original equation: For x = 11: √(121 - 88 + 16) = √49 = 7 ✓ For x = -3: √(9 + 24 + 16) = √49 = 7 ✓ Both solutions are valid.

The answer is 11.

3 √(x² - 10x + 25) = 7

Hint: Notice that the expression under the square root is a perfect square trinomial. Consider what value squared equals the number on the right side.

Show the answer

Answer: 12

  1. Recognize that x² - 10x + 25 is a perfect square trinomial
  2. Factor the expression: x² - 10x + 25 = (x - 5)²
  3. Rewrite the equation: √((x - 5)²) = 7
  4. Simplify the square root: |x - 5| = 7
  5. Solve the absolute value equation: x - 5 = 7 or x - 5 = -7
  6. Solve the first case: x - 5 = 7 → x = 12
  7. Solve the second case: x - 5 = -7 → x = -2
  8. Check both solutions in the original equation
  9. For x = 12: √(144 - 120 + 25) = √49 = 7 ✓
  10. For x = -2: √(4 + 20 + 25) = √49 = 7 ✓ Both solutions are valid, but the problem asks for a single answer. The positive solution is 12.

4 Solve for x: 2^(x+1) = 16

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

Show the answer

Answer: 3

  1. Write 16 as a power of 2. 16 = 2^4.
  2. Substitute this into the equation: 2^(x+1) = 2^4.
  3. Since the bases are the same, the exponents must be equal: x + 1 = 4.
  4. Subtract 1 from both sides to solve for x: x = 4 - 1.
  5. x = 3.

The answer is 3.

5 Solve for x: 2^(x+1) = 32

Hint: Express both sides with the same base to compare exponents

Show the answer

Answer: 4

  1. Write 32 as a power of 2: 32 = 2^5
  2. Substitute into the equation: 2^(x+1) = 2^5
  3. Since the bases are equal, set the exponents equal: x + 1 = 5
  4. Subtract 1 from both sides: x = 5 - 1
  5. x = 4

The answer is 4.

6 Solve for h: A = 2πr² + 2πrh

Hint: Isolate the variable by moving all other terms to the opposite side of the equation, then divide by the coefficient of the variable.

Show the answer

Answer: h = (A - 2πr²)/(2πr)

  1. Our goal is to solve for h. First, isolate the term containing h. Subtract 2πr² from both sides: A - 2πr² = 2πrh
  2. Now, h is multiplied by 2πr. To solve for h, divide both sides by 2πr: h = (A - 2πr²) / (2πr)
  3. This is the final expression for h in terms of A and r. We can check: if we substitute this h back into the original equation, we get: 2πr² + 2πr * [ (A - 2πr²) / (2πr) ] = 2πr² + (A - 2πr²) = A So the solution is correct. Final answer: h = (A - 2πr²)/(2πr)

We start with the equation: A = 2πr² + 2πrh

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