Create Equations: Two More Variables

Grade 9 · algebra · 100 practice problems · read aloud

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Creating Equations in Algebra

What is Creating Equations?

Creating an equation means translating a real-world situation or a word problem into a mathematical sentence using variables, numbers, and operations. This is the crucial first step in solving problems. 🧩

Why it's useful: It turns complex situations into solvable math problems. You use this skill in budgeting, planning trips, and understanding science concepts.

How to Create Equations: A Step-by-Step Guide

  1. Read Carefully: Understand the problem. Identify what you know and what you need to find.
  2. Define a Variable: Choose a letter (like x) to represent the unknown quantity you're solving for.
  3. Look for Key Words: Words like "total," "sum," "difference," "product," or "is" signal operations and the equals sign.
  4. Write the Equation: Translate the words into a mathematical expression using your variable.
  5. Check Your Work: Read the problem again with your equation. Does it make sense?

Worked Examples

Example 1: The Concert

Problem: Tickets for a concert cost $15 each. You have $60 total. How many tickets can you buy?

  1. Variable: Let t = number of tickets.
  2. Key Info: Cost per ticket ($15) times number of tickets equals total cost ($60).
  3. Equation: 15t = 60

Example 2: Consecutive Integers

Problem: The sum of three consecutive integers is 48. Find the integers.

  1. Variable: Let n = the first integer. Then n+1 is the second, and n+2 is the third.
  2. Key Info: "Sum" means addition. "Is" means equals.
  3. Equation: n + (n+1) + (n+2) = 48

Common Mistakes to Avoid

Misplacing the Equals Sign: The equals sign means "is the same as." Don't just put it at the end. Make sure both sides balance.

Incorrect Variable Definition: Be precise. "Let x = Sarah's age" is better than "Let x = Sarah."

Ignoring Order of Operations: Phrases like "5 more than twice a number" translate to 2x + 5, not 2(x+5).

Tips & Tricks

Keyword Translation Guide:

  • "more than" or "sum" → +
  • "less than" or "difference" → - (watch the order!)
  • "product" or "times" → ×
  • "quotient" or "per" → ÷
  • "is" or "total" → =

Check with a Simple Number: Replace your variable with a small number to see if your equation logic holds.

How to Practice

  • Start with simple one-step problems and gradually move to multi-step ones.
  • Rewrite word problems from your textbook as equations, even if you don't solve them.
  • Create your own word problems for a given equation, like 2x - 7 = 15.
  • Practice consistently! This skill builds with repetition.

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 specific formula that involves the coefficients and the square root operation.

Show the answer

Answer: x = 1, x = 1.5

  1. Identify coefficients** The equation is in the form: a x² + b x + 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. The numbers -2 and -3 multiply to 6 but add to -5? No: (-2) + (-3) = -5, yes. So we split the middle term -5x into -2x - 3x. --- **
  3. Factor by grouping** 2x² - 5x + 3 = 2x² - 2x - 3x + 3 Group terms: (2x² - 2x) + (-3x + 3) Factor each group: 2x(x - 1) - 3(x - 1) Now factor out (x - 1): (x - 1)(2x - 3) = 0 --- **
  4. Solve for x** Set each factor equal to 0: x - 1 = 0 → x = 1 2x - 3 = 0 → 2x = 3 → x = 3/2 = 1.5 --- **Final Answer:** x = 1, x = 1.5

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

2 2x² - 5x - 3 = 0

Hint: This is a quadratic equation that can be solved by factoring, completing the square, or using the quadratic formula. Look for two numbers whose product equals the constant term times the leading coefficient and whose sum equals the middle coefficient.

Show the answer

Answer: x = 3, x = -0.5

  1. Identify coefficients** This is a quadratic equation in the form ax² + bx + c = 0. Here: a = 2, b = -5, c = -3. --- **
  2. Factor the quadratic** We look for two numbers that multiply to a * c = 2 * (-3) = -6, and add to b = -5. The numbers -6 and +1 multiply to -6 and add to -5. --- **
  3. Rewrite the middle term** 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 binomial** (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** x - 3 = 0 → x = 3 2x + 1 = 0 → 2x = -1 → x = -1/2 --- **Final answer:** x = 3, x = -1/2 (which is -0.5)

Let's solve the equation step by step. We have: 2x² - 5x - 3 = 0 --- **

3 (2x² - 5x - 3) ÷ (x - 3) = ?

Hint: Use polynomial long division to divide the quadratic expression by the linear binomial. Focus on matching the leading terms at each step.

Show the answer

Answer: 2x + 1

  1. Set up polynomial long division with (2x² - 5x - 3) ÷ (x - 3)
  2. Divide the first term: 2x² ÷ x = 2x
  3. Multiply: 2x × (x - 3) = 2x² - 6x
  4. Subtract: (2x² - 5x - 3) - (2x² - 6x) = x - 3
  5. Divide the next term: x ÷ x = 1
  6. Multiply: 1 × (x - 3) = x - 3
  7. Subtract: (x - 3) - (x - 3) = 0
  8. The quotient is 2x + 1 with remainder 0

The answer is 2x + 1.

4 (3x² - 7x - 6) ÷ (x - 3) = ?

Hint: Use polynomial long division or synthetic division to divide the quadratic expression by the linear binomial. Consider what polynomial multiplied by (x - 3) gives you the original quadratic.

Show the answer

Answer: 3x + 2

  1. Set up polynomial long division: (3x² - 7x - 6) ÷ (x - 3)
  2. Divide the first term: 3x² ÷ x = 3x
  3. Multiply: 3x × (x - 3) = 3x² - 9x
  4. Subtract: (3x² - 7x - 6) - (3x² - 9x) = 2x - 6
  5. Divide the next term: 2x ÷ x = 2
  6. Multiply: 2 × (x - 3) = 2x - 6
  7. Subtract: (2x - 6) - (2x - 6) = 0
  8. The quotient is 3x + 2 with remainder 0
  9. Therefore, (3x² - 7x - 6) ÷ (x - 3) = 3x + 2

5 Aroha is designing a triangular banner where the base is 7 cm longer than the height. Write an equation for the area A in terms of the height h.

Hint: Remember that the area of a triangle is half the product of base and height. Consider how the base relates to the height in this scenario.

Show the answer

Answer: A = (1/2)h(h + 7)

When creating equations from geometric contexts, identify the given relationships between variables and substitute them into standard formulas. For a triangle, the area formula is A = (1/2) × base × height. If the problem states a relationship between base and height, substitute that relationship into the formula to create an equation with multiple variables.

6 The sum of two numbers is 24. When the larger number is squared and then decreased by the square of the smaller number, the result is 96. What is the larger number?

Hint: Define variables for the two numbers and set up a system of equations based on the given relationships. Consider using the identity for the difference of squares to simplify the system.

Show the answer

Answer: 14

  1. Write the equations from the problem. From "The sum of two numbers is 24": L + S = 24 From "When the larger number is squared and then decreased by the square of the smaller number, the result is 96": L^2 - S^2 = 96
  2. Notice that L^2 - S^2 is a difference of squares. We can factor it: L^2 - S^2 = (L - S)(L + S) = 96
  3. Substitute L + S = 24 into the factored equation. (L - S)(24) = 96
  4. Solve for (L - S). Divide both sides by 24: L - S = 96 / 24 L - S = 4
  5. Now we have a system of two equations: Equation 1: L + S = 24 Equation 2: L - S = 4
  6. Add the two equations to eliminate S. (L + S) + (L - S) = 24 + 4 L + S + L - S = 28 2L = 28
  7. Solve for L. L = 28 / 2 L = 14
  8. Find S for completeness (optional check). From L + S = 24, 14 + S = 24, so S = 10.
  9. Check the conditions. Sum: 14 + 10 = 24 (correct) L^2 - S^2 = 196 - 100 = 96 (correct) Therefore, the larger number is 14.

Let's call the larger number L and the smaller number S.

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