Constraint Systems

Grade 9 · algebra · 100 practice problems · read aloud

🔊 Listen to this explanation

Constraint Systems: Solving with Multiple Rules

🔍 What is a Constraint System?

A constraint system is a set of two or more equations (or inequalities) that must be true at the same time. They are also called systems of equations. We use them to find values that satisfy all given conditions, like finding where two lines cross on a graph.

📝 How to Solve: The Substitution Method

  1. Solve one equation for one variable.
  2. Substitute that expression into the other equation.
  3. Solve the new equation for the one remaining variable.
  4. Substitute back to find the other variable.
  5. Check your solution in both original equations!

✨ Worked Examples

Example 1: Solve the system.

y = 2x + 1
x + y = 10

  1. The first equation is already solved for y: y = 2x + 1
  2. Substitute into the second equation: x + (2x + 1) = 10
  3. Solve for x: 3x + 1 = 10 → 3x = 9 → x = 3
  4. Substitute back: y = 2(3) + 1 → y = 7
  5. Check: 7 = 2(3)+1 ✔️ and 3+7=10 ✔️

Solution: (3, 7)

Example 2: Solve the system.

2x + y = 11
3x - y = 9

  1. Solve the first equation for y: y = 11 - 2x
  2. Substitute into the second: 3x - (11 - 2x) = 9
  3. Solve for x: 3x - 11 + 2x = 9 → 5x = 20 → x = 4
  4. Substitute back: y = 11 - 2(4) → y = 3

Solution: (4, 3)

⚠️ Common Mistakes to Avoid

  • Sign errors: Be careful with negative signs when substituting! Example: -(2x+1) is -2x-1, not -2x+1.
  • Forgetting to check: Always plug your solution back into both original equations to verify.
  • Mixing variables: Keep your x's and y's straight. Label your work clearly.

💡 Tips & Tricks

  • Look for the easy solve: Choose to solve the equation that has a variable with a coefficient of 1 (like just "x" or "y").
  • Graphical check: Sketch the lines. The solution is where they intersect!
  • Stay organized: Use parentheses when substituting to avoid sign mistakes.

🎯 How to Practice

Start with systems where one equation is already solved for a variable (like Example 1). Then move to systems where you need to do the first solving step (like Example 2). Create your own problems and solve them, or use online practice sites. The key is repetition to build confidence!

Practice problems

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

1 Solve for x: 3^(2x - 1) = 27

Hint: Express both sides with the same base, then set the exponents equal to each other

Show the answer

Answer: 2

  1. Recognize that 27 can be written as 3^3, so the equation becomes 3^(2x - 1) = 3^3
  2. Since the bases are equal, set the exponents equal to each other: 2x - 1 = 3
  3. Add 1 to both sides: 2x = 4
  4. Divide both sides by 2: x = 2

The answer is 2.

2 √(x² - 4x + 4) = 5, solve for x

Hint: Remember that the square root of a squared expression equals the absolute value of that expression

Show the answer

Answer: 7

  1. Recognize that x² - 4x + 4 is a perfect square trinomial
  2. Factor the expression: x² - 4x + 4 = (x - 2)²
  3. The equation becomes √((x - 2)²) = 5
  4. The square root of a square gives the absolute value: |x - 2| = 5
  5. Solve the absolute value equation: x - 2 = 5 or x - 2 = -5
  6. For x - 2 = 5: x = 7
  7. For x - 2 = -5: x = -3
  8. Check both solutions in the original equation
  9. For x = 7: √(49 - 28 + 4) = √25 = 5 ✓
  10. For x = -3: √(9 + 12 + 4) = √25 = 5 ✓ Both solutions are valid, but the problem asks for one solution, so we can use x = 7.

3 A right triangle is drawn on a coordinate plane with vertices at A(0,0), B(4,0), and C(4,3). A circle is inscribed inside this triangle, tangent to all three sides. What is the radius of this inscribed circle?

Hint: Consider the geometric relationship between a right triangle's side lengths and its inscribed circle. The radius can be found using a formula involving the triangle's perimeter and area, but with different measurements.

Show the answer

Answer: 1

  1. Understand the problem. We have a right triangle with vertices A(0,0), B(4,0), and C(4,3). Side AB is along the x-axis from (0,0) to (4,0), so length AB = 4. Side BC is vertical from (4,0) to (4,3), so length BC = 3. Side AC is the hypotenuse from (0,0) to (4,3). We can find AC using the Pythagorean theorem: AC = sqrt((4-0)^2 + (3-0)^2) = sqrt(16 + 9) = sqrt(25) = 5. So the triangle has sides: 3, 4, 5.
  2. Recall the formula for the inradius (radius of inscribed circle) of a right triangle. For a right triangle with legs a and b and hypotenuse c, the inradius r is given by: r = (a + b - c)/2.
  3. Apply the formula. Here, a = 3, b = 4, c = 5. r = (3 + 4 - 5)/2 = (7 - 5)/2 = 2/2 = 1.
  4. Conclusion. The radius of the inscribed circle is 1. Final answer: 1

4 A right triangle is drawn on a coordinate plane with vertices at A(0,0), B(6,0), and C(0,8). A circle is inscribed inside this triangle such that it touches all three sides. What is the radius of this inscribed circle?

Hint: Consider the formula for the inradius of a right triangle in terms of its side lengths. The inradius can be expressed using the triangle's perimeter and area, or through a simpler relationship specific to right triangles.

Show the answer

Answer: 2

  1. Understand the problem We have a right triangle with vertices A(0,0), B(6,0), and C(0,8). The circle is inscribed inside the triangle, meaning it touches all three sides. This is called the incircle. We need to find the radius of the incircle.
  2. Recall the formula for the inradius of a right triangle For a right triangle with legs a and b and hypotenuse c, the inradius r is given by: r = (a + b - c) / 2 where c = sqrt(a^2 + b^2).
  3. Identify the side lengths From the coordinates: Side AB is from (0,0) to (6,0), length a = 6. Side AC is from (0,0) to (0,8), length b = 8. Side BC is from (6,0) to (0,8), length c = sqrt((6-0)^2 + (0-8)^2) = sqrt(36 + 64) = sqrt(100) = 10. So: a = 6, b = 8, c = 10.
  4. Apply the formula r = (a + b - c) / 2 r = (6 + 8 - 10) / 2 r = (14 - 10) / 2 r = 4 / 2 r = 2.
  5. Conclusion The radius of the inscribed circle is 2. We can also check: Area of triangle = (1/2) * base * height = (1/2) * 6 * 8 = 24. Semi-perimeter s = (6 + 8 + 10) / 2 = 24 / 2 = 12. Inradius from area formula: r = Area / s = 24 / 12 = 2. Both methods match.

5 Matiu is buying art supplies with a budget of $120. He needs at least 8 sketchbooks and 6 paint sets. Sketchbooks cost $4 each and paint sets cost $8 each. Write the system of inequalities that represents this situation.

Hint: Identify the variables for each item, then create inequalities for the budget constraint and minimum quantity requirements.

Show the answer

Answer: 4x + 8y ≤ 120, x ≥ 8, y ≥ 6

  1. Let x = number of sketchbooks, y = number of paint sets
  2. Budget constraint: 4x + 8y ≤ 120
  3. Minimum sketchbooks: x ≥ 8
  4. Minimum paint sets: y ≥ 6
  5. The system of inequalities is: 4x + 8y ≤ 120, x ≥ 8, y ≥ 6

6 Isabella is organizing a school fundraiser with a budget of $272. She needs to buy at least 12 items total. T-shirts cost $17 each and hats cost $12 each. Write the system of inequalities that represents these constraints.

Hint: Identify the cost constraint and the minimum quantity constraint. Use x for one item type and y for the other, with both being non-negative.

Show the answer

Answer: 17x + 12y ≤ 272, x + y ≥ 12, x ≥ 0, y ≥ 0

  1. Let x = number of T-shirts and y = number of hats
  2. Cost constraint: 17x + 12y ≤ 272 (total cost cannot exceed $272)
  3. Quantity constraint: x + y ≥ 12 (must buy at least 12 items total)
  4. Non-negativity constraints: x ≥ 0, y ≥ 0 (cannot buy negative items)
  5. The complete system is: 17x + 12y ≤ 272, x + y ≥ 12, x ≥ 0, y ≥ 0
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