Systems of Equations

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Systems of Equations · Grade 8 · Worksheet 3

1. Hana is helping to organize the school's sports equipment. She needs to order basketballs and soccer balls. Each basketball costs $24 and each soccer ball costs $18. Hana has a total budget of $360 for balls. She also needs to order exactly 6 more soccer balls than basketballs so that there are enough for the teams. How many basketballs and how many soccer balls should Hana order to spend exactly her budget and meet the requirement? Answer: 6 basketballs and 12 soccer balls Solution: Let b = number of basketballs and s = number of soccer balls. Write the cost equation: 24b + 18s = 360. Write the relationship equation: s = b + 6.
   Full step-by-step solution

   Step 1: Let b = number of basketballs and s = number of soccer balls. Step 2: Write the cost equation: 24b + 18s = 360. Step 3: Write the relationship equation: s = b + 6. Step 4: Substitute s = b + 6 into the cost equation: 24b + 18(b + 6) = 360. Step 5: Distribute: 24b + 18b + 108 = 360. Step 6: Combine like terms: 42b + 108 = 360. Step 7: Subtract 108 from both sides: 42b = 252. Step 8: Divide both sides by 42: b = 6. Step 9: Substitute b = 6 into s = b + 6: s = 6 + 6 = 12. Step 10: Check: 24(6) + 18(12) = 144 + 216 = 360. Correct. Hana should order 6 basketballs and 12 soccer balls.

2. 2x + 3y = 12; x - y = 1; x = ? Answer: 3 Solution: 1) 2x + 3y = 12 2) x - y = 1 We want to find x. Solve one of the equations for one variable. From equation (2): x - y = 1 So x = y + 1.
   Full step-by-step solution

   We are given the system of equations: 1) 2x + 3y = 12 2) x - y = 1 We want to find x. Step 1: Solve one of the equations for one variable. From equation (2): x - y = 1 So x = y + 1. Step 2: Substitute x = y + 1 into equation (1). Equation (1): 2x + 3y = 12 Replace x with (y + 1): 2(y + 1) + 3y = 12 Step 3: Simplify and solve for y. 2y + 2 + 3y = 12 5y + 2 = 12 5y = 12 - 2 5y = 10 y = 10/5 y = 2 Step 4: Substitute y = 2 back into x = y + 1. x = 2 + 1 x = 3 Step 5: Check the solution. Equation (1): 2(3) + 3(2) = 6 + 6 = 12 ✓ Equation (2): 3 - 2 = 1 ✓ Thus, x = 3.

3. A triangular garden is drawn on a coordinate plane with vertices at (1, 2), (5, 6), and (9, 2). A straight path is planned that connects the midpoint of the left side (from (1,2) to (5,6)) to the midpoint of the base (from (1,2) to (9,2)). Find the coordinates of the point where this path intersects with a vertical line drawn through x = 4. Answer: (4, 3) Solution: Find the midpoint of the left side from (1,2) to (5,6) Midpoint = ((1+5)/2, (2+6)/2) = (6/2, 8/2) = (3, 4) Find the midpoint of the base from (1,2) to (9,2) Midpoint = ((1+9)/2, (2+2)/2) = (10/2, 4/2) = (5, 2) Find the equation of the line through (3,4) and (5,2) Slope = (2-4)/(5-3) = (-2)/2 =…
   Full step-by-step solution

   Step 1: Find the midpoint of the left side from (1,2) to (5,6) Midpoint = ((1+5)/2, (2+6)/2) = (6/2, 8/2) = (3, 4) Step 2: Find the midpoint of the base from (1,2) to (9,2) Midpoint = ((1+9)/2, (2+2)/2) = (10/2, 4/2) = (5, 2) Step 3: Find the equation of the line through (3,4) and (5,2) Slope = (2-4)/(5-3) = (-2)/2 = -1 Using point-slope form with point (3,4): y - 4 = -1(x - 3) Simplify: y - 4 = -x + 3 Final equation: y = -x + 7 Step 4: Find where this line intersects the vertical line x = 4 Substitute x = 4 into y = -x + 7 y = -4 + 7 = 3 Step 5: The intersection point is (4, 3) The answer is (4, 3).

4. Mere is helping to organize a school cultural festival. She needs to arrange seating using small tables and large tables. Each small table seats 4 people and each large table seats 6 people. There are 80 people attending the festival, and Mere has a total of 16 tables available. How many small tables and how many large tables should she use to seat everyone exactly? Answer: 8 small tables and 8 large tables Solution: Let x = number of small tables and y = number of large tables.
   Full step-by-step solution

   Step 1: Let x = number of small tables and y = number of large tables. Step 2: Equation for total tables: x + y = 16 Step 3: Equation for total seats: 4x + 6y = 80 Step 4: Solve the first equation for x: x = 16 - y Step 5: Substitute into the second equation: 4(16 - y) + 6y = 80 Step 6: Distribute: 64 - 4y + 6y = 80 Step 7: Combine like terms: 64 + 2y = 80 Step 8: Subtract 64 from both sides: 2y = 16 Step 9: Divide both sides by 2: y = 8 Step 10: Substitute y = 8 into x = 16 - y: x = 16 - 8 = 8 So Mere needs 8 small tables and 8 large tables. The answer is 8 small tables and 8 large tables.

5. Liam is planning a school fundraiser and needs to determine how many small and large candles to sell. Small candles cost $2 each to make and sell for $5, while large candles cost $4 each to make and sell for $8. Liam has a budget of $60 for materials and wants to make exactly 20 candles. How many of each size should he make to maximize his profit? Answer: 10 small candles and 10 large candles Solution: x = number of small candles y = number of large candles Small candles cost $2 each to make, large cost $4 each. Budget = $60. So: 2x + 4y ≤ 60 x + y = 20 x ≥ 0, y ≥ 0 From x + y = 20, we get y = 20 - x.
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Define the variables** Let x = number of small candles y = number of large candles --- **Step 2: Write the constraints** From the problem: - Material cost constraint: Small candles cost $2 each to make, large cost $4 each. Budget = $60. So: 2x + 4y ≤ 60 - Total number of candles: x + y = 20 - Non-negativity: x ≥ 0, y ≥ 0 --- **Step 3: Simplify the constraints** From x + y = 20, we get y = 20 - x. Substitute into the cost constraint: 2x + 4(20 - x) ≤ 60 2x + 80 - 4x ≤ 60 -2x + 80 ≤ 60 -2x ≤ -20 x ≥ 10 Also, since y = 20 - x ≥ 0, we get x ≤ 20. So x must be between 10 and 20 inclusive: 10 ≤ x ≤ 20. --- **Step 4: Write the profit function** Profit per small candle = selling price - cost = 5 - 2 = $3 Profit per large candle = 8 - 4 = $4 Total profit: P = 3x + 4y Substitute y = 20 - x: P = 3x + 4(20 - x) P = 3x + 80 - 4x P = 80 - x --- **Step 5: Maximize profit** P = 80 - x is a decreasing function of x. To maximize P, we want the smallest possible x. From Step 3, smallest x allowed is 10. So x = 10, then y = 20 - 10 = 10. --- **Step 6: Check cost constraint for (10, 10)** Cost = 2*10 + 4*10 = 20 + 40 = 60 (exactly budget) Profit = 3*10 + 4*10 = 30 + 40 = $70 --- **Step 7: Verify other corner points** If x = 20, y = 0: Cost = 2*20 + 4*0 = 40 ≤ 60 (ok) Profit = 3*20 + 4*0 = 60 (less than 70) If x = 10, y = 10: profit = 70 (maximum) If x = 0, y = 20: Cost = 0 + 80 = 80 > 60 (not allowed) So indeed, the only feasible range was 10 ≤ x ≤ 20, and profit decreases as x increases, so x = 10 gives maximum profit. --- **Final answer:** Liam should make 10 small candles and 10 large candles.

6. 3x + 2y = 19; 2x - 3y = 4 Answer: x = 5, y = 2 Solution: Multiply the first equation by 3: 3(3x + 2y) = 3(19) → 9x + 6y = 57 Multiply the second equation by 2: 2(2x - 3y) = 2(4) → 4x - 6y = 8 Add the two equations: (9x + 6y) + (4x - 6y) = 57 + 8 → 13x = 65 Solve for x: x = 65 ÷ 13 = 5 Substitute x = 5 into the first equation: 3(5) + 2y = 19 → 15 + 2y…
   Full step-by-step solution

   Step 1: Multiply the first equation by 3: 3(3x + 2y) = 3(19) → 9x + 6y = 57 Step 2: Multiply the second equation by 2: 2(2x - 3y) = 2(4) → 4x - 6y = 8 Step 3: Add the two equations: (9x + 6y) + (4x - 6y) = 57 + 8 → 13x = 65 Step 4: Solve for x: x = 65 ÷ 13 = 5 Step 5: Substitute x = 5 into the first equation: 3(5) + 2y = 19 → 15 + 2y = 19 Step 6: Solve for y: 2y = 19 - 15 → 2y = 4 → y = 2 The solution is x = 5, y = 2.

7. 5x + 2y = 20; 3x - 4y = -14. Solve for x and y. Answer: x = 2, y = 5 Solution: Write the system: 5x + 2y = 20 and 3x - 4y = -14. Multiply the first equation by 2 to match the y coefficients: 2(5x + 2y) = 2(20) → 10x + 4y = 40.
   Full step-by-step solution

   Step 1: Write the system: 5x + 2y = 20 and 3x - 4y = -14. Step 2: Multiply the first equation by 2 to match the y coefficients: 2(5x + 2y) = 2(20) → 10x + 4y = 40. Step 3: Now add this to the second equation: (10x + 4y) + (3x - 4y) = 40 + (-14) → 13x = 26. Step 4: Solve for x: x = 26 ÷ 13 = 2. Step 5: Substitute x = 2 into the first original equation: 5(2) + 2y = 20 → 10 + 2y = 20. Step 6: Solve for y: 2y = 20 - 10 → 2y = 10 → y = 5. Step 7: Check in the second original equation: 3(2) - 4(5) = 6 - 20 = -14 ✓. The solution is x = 2, y = 5.