Create Systems

Grade 9 · algebra · 100 practice problems · read aloud

🔊 Listen to this explanation

Creating Systems of Equations

🔍 What is This and Why is it Useful?

A system of equations is a set of two or more equations with the same variables. We use them to find a solution that satisfies all equations at once. This is super useful for solving real-world problems where multiple conditions must be met simultaneously, like figuring out costs or planning events.

📝 How to Create a System from a Word Problem

  1. Identify the Variables: Assign letters (like x and y) to the unknown quantities you need to find.
  2. Translate Words into Equations: Look for key phrases that describe relationships between the variables.
  3. Write the System: Write down the equations, one after the other, using the same variables.

✨ Worked Examples

Example 1: Concert Tickets

Problem: Adult tickets cost $5, child tickets cost $2. You sell 100 tickets and collect $350. Create a system.

Step 1: Let a = number of adult tickets, c = number of child tickets.

Step 2: "100 tickets total" → a + c = 100
"Collect $350" → 5a + 2c = 350

System: a + c = 100
5a + 2c = 350

Example 2: Numbers

Problem: The sum of two numbers is 12 and their difference is 4.

Step 1: Let x = first number, y = second number.

Step 2: "Sum is 12" → x + y = 12
"Difference is 4" → x - y = 4

System: x + y = 12
x - y = 4

⚠️ Common Mistakes to Avoid

  • Inconsistent Variables: Using different letters for the same quantity in different equations.
  • Misreading "Less Than" or "Difference": "5 less than x" is x - 5, not 5 - x.
  • Forgetting the Total: When a problem gives a total number of items and a total cost, you'll often need one equation for the count and one for the cost.

💡 Tips & Tricks

  • Underline Key Phrases: Circle the numbers and underline the relationship words (sum, total, difference, less than) in the problem.
  • Define Variables Clearly: Always write "let x = ..." so you remember what your variables represent.
  • Check Your Logic: After writing the system, ask yourself: "If I solve this, will it actually answer the question?"

🎯 How to Practice

Start simple! Practice by creating systems for situations around you. "If I buy 3 drinks and 2 snacks for $11..." Then, move to textbook problems. The key is getting comfortable turning a story into mathematical sentences. Try solving the systems you create using substitution or elimination to see if your answers make sense!

Practice problems

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

1 2x² - 3x + 1 = 0

Hint: This is a quadratic equation that can be solved using the quadratic formula or factoring. Remember that quadratic equations typically have two solutions.

Show the answer

Answer: x = 1, 0.5

  1. Identify the quadratic equation: 2x² - 3x + 1 = 0
  2. Use the quadratic formula: x = [-b ± √(b² - 4ac)] / (2a), where a = 2, b = -3, c = 1
  3. Calculate the discriminant: b² - 4ac = (-3)² - 4(2)(1) = 9 - 8 = 1
  4. Apply the quadratic formula: x = [3 ± √1] / (4) = [3 ± 1] / 4
  5. Calculate the two solutions: x = (3 + 1)/4 = 4/4 = 1, and x = (3 - 1)/4 = 2/4 = 0.5
  6. The solutions are x = 1 and x = 0.5

2 Solve the system: 2x + 3y = 12 and x - y = 1

Hint: You can use substitution or elimination method. For elimination, try making the coefficients of one variable match by multiplying one equation.

Show the answer

Answer: x = 3, y = 2

  1. Use substitution method. From the second equation x - y = 1, we get x = y + 1
  2. Substitute x = y + 1 into the first equation: 2(y + 1) + 3y = 12
  3. Simplify: 2y + 2 + 3y = 12
  4. Combine like terms: 5y + 2 = 12
  5. Subtract 2 from both sides: 5y = 10
  6. Divide by 5: y = 2
  7. Substitute y = 2 into x = y + 1: x = 2 + 1 = 3
  8. Check: 2(3) + 3(2) = 6 + 6 = 12 ✓ and 3 - 2 = 1 ✓ The solution is x = 3, y = 2.

3 Solve the system: 3x² - 2y = 7 and 2x + y = 4

Hint: Try substituting one equation into the other to eliminate a variable. For example, if you had a² + b = 5 and a + b = 3, you could solve for b in the second equation and substitute into the first.

Show the answer

Answer: x = 1, y = 2

  1. Solve the linear equation for y: 2x + y = 4 → y = 4 - 2x
  2. Substitute into the quadratic equation: 3x² - 2(4 - 2x) = 7
  3. Simplify: 3x² - 8 + 4x = 7
  4. Rearrange: 3x² + 4x - 15 = 0
  5. Solve the quadratic using factoring: (3x - 5)(x + 3) = 0
  6. Find x values: x = 5/3 or x = -3
  7. Find corresponding y values: When x = 5/3: y = 4 - 2(5/3) = 4 - 10/3 = 2/3 When x = -3: y = 4 - 2(-3) = 4 + 6 = 10
  8. Verify both solutions work in both equations
  9. The solutions are (5/3, 2/3) and (-3, 10)

4 Solve the system: 3x² - 2x + y = 7 and 2x + y = 3

Hint: Try substituting one equation into the other to eliminate a variable. For example, if you had a² + b = 5 and a + b = 2, you could solve for b in terms of a from the second equation.

Show the answer

Answer: x = 2, y = -1

  1. From the second equation 2x + y = 3, solve for y: y = 3 - 2x
  2. Substitute y = 3 - 2x into the first equation: 3x² - 2x + (3 - 2x) = 7
  3. Simplify: 3x² - 2x + 3 - 2x = 7 → 3x² - 4x + 3 = 7
  4. Subtract 7 from both sides: 3x² - 4x - 4 = 0
  5. Solve the quadratic equation using factoring: (3x + 2)(x - 2) = 0
  6. Find x values: x = -2/3 or x = 2
  7. Substitute back to find y: When x = -2/3: y = 3 - 2(-2/3) = 3 + 4/3 = 13/3 When x = 2: y = 3 - 2(2) = 3 - 4 = -1
  8. The solution is x = 2, y = -1

5 Isabella bought 2 pens and 7 notebooks for $27. A pen costs $2 more than a notebook. Write a system of equations to represent this situation.

Hint: Define variables for the costs of each item. One equation should represent the total cost, and the other should represent the relationship between the individual costs.

Show the answer

Answer: 2p + 7n = 27, p = n + 2

  1. Let p = cost of one pen, n = cost of one notebook
  2. For total cost: 2 pens + 7 notebooks = $27 → 2p + 7n = 27
  3. For cost relationship: pen costs $2 more than notebook → p = n + 2
  4. The system of equations is: 2p + 7n = 27 and p = n + 2

6 A theater sells tickets for $12 for adults and $8 for children. For a particular show, the theater sold 120 tickets and collected $1240. How many adult tickets were sold?

Hint: Set up a system of equations where one equation represents the total number of tickets and the other represents the total revenue. Use substitution or elimination to solve for the number of adult tickets.

Show the answer

Answer: 70

  1. Solve for one variable from Equation (1). From (1): C = 120 - A
  2. Substitute C into Equation (2). 12A + 8(120 - A) = 1240
  3. Simplify and solve for A. 12A + 960 - 8A = 1240 (12A - 8A) + 960 = 1240 4A + 960 = 1240
  4. Subtract 960 from both sides. 4A = 1240 - 960 4A = 280
  5. Divide both sides by 4. A = 280 / 4 A = 70 So, the number of adult tickets sold is 70.
  6. Check the answer. If A = 70, then C = 120 - 70 = 50. Money collected = 12*70 + 8*50 = 840 + 400 = 1240. This matches the problem statement. Final answer: 70 adult tickets.

Let's define variables for the number of adult tickets and child tickets. Let: A = number of adult tickets C = number of child tickets From the problem: 1. The total number of tickets sold is 120. So: A + C = 120 2. The total money collected is $1240. Adult tickets cost $12 each, child tickets cost $8 each. So: 12A + 8C = 1240 We now have the system of equations: Equation (1): A + C = 120 Equation (2): 12A + 8C = 1240

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