Systems Word Problems

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Answer Key & Explanations

Systems Word Problems · Grade 8 · Worksheet 3

1. Olivia is organizing the school's book fair. She has two types of bookmarks to sell: small bookmarks cost $2 each and large bookmarks cost $5 each. Olivia sells a total of 60 bookmarks and collects $210 in sales. How many small bookmarks and how many large bookmarks did Olivia sell? Answer: 30 small bookmarks and 30 large bookmarks Solution: Let x = number of small bookmarks sold and y = number of large bookmarks sold.
   Full step-by-step solution

   Step 1: Let x = number of small bookmarks sold and y = number of large bookmarks sold. Step 2: Write the equation for total bookmarks: x + y = 60 Step 3: Write the equation for total money: 2x + 5y = 210 Step 4: Solve the first equation for x: x = 60 - y Step 5: Substitute into the second equation: 2(60 - y) + 5y = 210 Step 6: Simplify: 120 - 2y + 5y = 210 Step 7: Combine like terms: 120 + 3y = 210 Step 8: Subtract 120 from both sides: 3y = 90 Step 9: Divide by 3: y = 30 Step 10: Substitute back to find x: x = 60 - 30 = 30 Olivia sold 30 small bookmarks and 30 large bookmarks.

2. 2x + 3y = 19 and 5x - 2y = 14, find x and y Answer: x = 4, y = 3 Solution: Step 1: Multiply the first equation by 2: 2(2x + 3y) = 2(19) → 4x + 6y = 38 Step 2: Multiply the second equation by 3: 3(5x - 2y) = 3(14) → 15x - 6y = 42 Step 3: Add the two new equations: (4x + 6y) + (15x - 6y) = 38 + 42 → 19x = 80 Step 4: Solve for x: x = 80/19 = 4 Step 5: Substitute x = 4…
   Full step-by-step solution

   Step 1: Multiply the first equation by 2: 2(2x + 3y) = 2(19) → 4x + 6y = 38 Step 2: Multiply the second equation by 3: 3(5x - 2y) = 3(14) → 15x - 6y = 42 Step 3: Add the two new equations: (4x + 6y) + (15x - 6y) = 38 + 42 → 19x = 80 Step 4: Solve for x: x = 80/19 = 4 Step 5: Substitute x = 4 into the first equation: 2(4) + 3y = 19 → 8 + 3y = 19 → 3y = 11 → y = 11/3 = 3 Step 6: Verify with second equation: 5(4) - 2(3) = 20 - 6 = 14 ✓ The solution is x = 4, y = 3.

3. Emma is organizing a school bake sale. She sells cupcakes for $2 each and brownies for $3 each. She sold a total of 70 items and raised $180. How many cupcakes and how many brownies did Emma sell? Answer: 30 cupcakes and 40 brownies Solution: Let c = number of cupcakes and b = number of brownies.
   Full step-by-step solution

   Let c = number of cupcakes and b = number of brownies. Equation 1 (total items): c + b = 70 Equation 2 (total money): 2c + 3b = 180 Solve the first equation for c: c = 70 - b Substitute into the second equation: 2(70 - b) + 3b = 180 Simplify: 140 - 2b + 3b = 180 Combine like terms: 140 + b = 180 Subtract 140 from both sides: b = 40 Substitute back to find c: c = 70 - 40 = 30 Emma sold 30 cupcakes and 40 brownies.

4. Matiu is helping organize the school sports day. He needs to buy two types of balls: soccer balls cost $24 each and basketballs cost $18 each. The school has a budget of $540 for balls, and Matiu needs exactly 26 balls in total for all the teams. How many soccer balls and how many basketballs should Matiu buy? Answer: 12 soccer balls, 14 basketballs Solution: Let s = number of soccer balls and b = number of basketballs.
   Full step-by-step solution

   Let s = number of soccer balls and b = number of basketballs. Equation 1 (total balls): s + b = 26 Equation 2 (total cost): 24s + 18b = 540 Solve equation 1 for s: s = 26 - b Substitute into equation 2: 24(26 - b) + 18b = 540 Simplify: 624 - 24b + 18b = 540 Combine like terms: 624 - 6b = 540 Subtract 624 from both sides: -6b = -84 Divide by -6: b = 14 Substitute back: s = 26 - 14 = 12 Matiu should buy 12 soccer balls and 14 basketballs.

5. Emma is planning a science fair project and needs to create two different chemical solutions. Solution A requires 3 grams of salt per liter, and Solution B requires 5 grams of salt per liter. She has exactly 40 grams of salt available and wants to make a total of 10 liters of combined solution. How many liters of each solution should Emma make to use all her salt and create exactly 10 liters? Answer: 5 Solution: Let x = liters of Solution A, y = liters of Solution B.
   Full step-by-step solution

   Let x = liters of Solution A, y = liters of Solution B. Equation 1 (total volume): x + y = 10 Equation 2 (total salt): 3x + 5y = 40 From Equation 1: y = 10 - x Substitute into Equation 2: 3x + 5(10 - x) = 40 3x + 50 - 5x = 40 -2x + 50 = 40 -2x = 40 - 50 -2x = -10 x = 5 Then y = 10 - 5 = 5 Emma should make 5 liters of Solution A and 5 liters of Solution B.

6. Mere is organizing the school's annual fun run. She needs to order two types of medals: gold medals for the winners and silver medals for all participants. Gold medals cost $4 each and silver medals cost $2 each. She has a budget of $200 for medals. She also knows that she needs exactly 10 more silver medals than gold medals to have enough for all the runners. How many gold medals and how many silver medals should Mere order to use her entire budget and meet the requirement for the number of medals? Answer: 30 gold medals and 40 silver medals Solution: Let g = number of gold medals and s = number of silver medals. Write the equation for the total cost. Gold medals cost $4 each, silver medals cost $2 each, and the total budget is $200.
   Full step-by-step solution

   Let g = number of gold medals and s = number of silver medals. Step 1: Write the equation for the total cost. Gold medals cost $4 each, silver medals cost $2 each, and the total budget is $200. 4g + 2s = 200 Step 2: Write the equation for the relationship between the medals. There are 10 more silver medals than gold medals. s = g + 10 Step 3: Substitute the second equation into the first equation. 4g + 2(g + 10) = 200 Step 4: Simplify and solve for g. 4g + 2g + 20 = 200 6g + 20 = 200 6g = 180 g = 30 Step 5: Substitute g = 30 back into the equation s = g + 10 to find s. s = 30 + 10 = 40 Mere should order 30 gold medals and 40 silver medals.

7. 2x + 3y = 12 and 4x - y = 10, find x and y Answer: x = 3, y = 2 Solution: (1) 2x + 3y = 12 (2) 4x - y = 10 Solve one equation for one variable. From equation (2): 4x - y = 10 Add y to both sides: 4x = 10 + y Subtract 10 from both sides: y = 4x - 10 Substitute y = 4x - 10 into equation (1).
   Full step-by-step solution

   We are given the system of equations: (1) 2x + 3y = 12 (2) 4x - y = 10 Step 1: Solve one equation for one variable. From equation (2): 4x - y = 10 Add y to both sides: 4x = 10 + y Subtract 10 from both sides: y = 4x - 10 Step 2: Substitute y = 4x - 10 into equation (1). Equation (1): 2x + 3y = 12 Replace y with (4x - 10): 2x + 3(4x - 10) = 12 Step 3: Simplify and solve for x. 2x + 12x - 30 = 12 14x - 30 = 12 Add 30 to both sides: 14x = 42 Divide both sides by 14: x = 42/14 x = 3 Step 4: Substitute x = 3 into y = 4x - 10. y = 4(3) - 10 y = 12 - 10 y = 2 Step 5: Check the solution in both original equations. In (1): 2(3) + 3(2) = 6 + 6 = 12 ✓ In (2): 4(3) - 2 = 12 - 2 = 10 ✓ Final answer: x = 3, y = 2