Systems by Elimination

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Systems by Elimination · Grade 8 · Worksheet 2

1. 2x + 3y = 12; 2x - y = 4 Answer: x = 3, y = 2 Solution: (1) 2x + 3y = 12 (2) 2x - y = 4 Solve one equation for one variable.
   Full step-by-step solution

   We have the system of equations: (1) 2x + 3y = 12 (2) 2x - y = 4 Step 1: Solve one equation for one variable. From equation (2): 2x - y = 4 Add y to both sides: 2x = 4 + y Subtract 4 from both sides: 2x - 4 = y So y = 2x - 4 Step 2: Substitute y = 2x - 4 into equation (1). Equation (1): 2x + 3y = 12 Replace y with (2x - 4): 2x + 3(2x - 4) = 12 Step 3: Simplify and solve for x. 2x + 6x - 12 = 12 8x - 12 = 12 Add 12 to both sides: 8x = 24 Divide both sides by 8: x = 3 Step 4: Substitute x = 3 into y = 2x - 4. y = 2(3) - 4 y = 6 - 4 y = 2 Step 5: Check the solution in both original equations. Equation (1): 2(3) + 3(2) = 6 + 6 = 12 ✓ Equation (2): 2(3) - 2 = 6 - 2 = 4 ✓ Final answer: x = 3, y = 2

2. Emma is planning a science fair project and needs to mix two chemical solutions. Solution A contains 15% salt and Solution B contains 40% salt. She wants to create 500 milliliters of a mixture that is 25% salt. How many milliliters of each solution should Emma use? Answer: (300,200) Solution: Let x be milliliters of Solution A (15% salt) and y be milliliters of Solution B (40% salt).
   Full step-by-step solution

   Step 1: Let x be milliliters of Solution A (15% salt) and y be milliliters of Solution B (40% salt). Step 2: Write the total volume equation: x + y = 500 Step 3: Write the salt content equation: 0.15x + 0.40y = 0.25 × 500 Step 4: Calculate the right side: 0.25 × 500 = 125, so 0.15x + 0.40y = 125 Step 5: Multiply the first equation by -0.15: -0.15x - 0.15y = -75 Step 6: Add this to the salt equation: (0.15x + 0.40y) + (-0.15x - 0.15y) = 125 + (-75) Step 7: Simplify: 0.25y = 50 Step 8: Solve for y: y = 50 ÷ 0.25 = 200 Step 9: Substitute y = 200 into x + y = 500: x + 200 = 500 Step 10: Solve for x: x = 500 - 200 = 300 Emma should use 300 ml of Solution A and 200 ml of Solution B.

3. Liam is planning a school fundraiser and needs to determine how many student tickets and adult tickets were sold. The total number of tickets sold was 120. Student tickets cost $3 each, adult tickets cost $5 each, and the total amount collected was $500. How many student tickets and how many adult tickets were sold? Answer: 50 student tickets and 70 adult tickets Solution: Let s = number of student tickets Let a = number of adult tickets Write the equation for the total number of tickets We know the total tickets sold is 120.
   Full step-by-step solution

   Let's define variables for the number of tickets: Let s = number of student tickets Let a = number of adult tickets --- **Step 1: Write the equation for the total number of tickets** We know the total tickets sold is 120. So: s + a = 120 --- **Step 2: Write the equation for the total money collected** Student tickets cost $3 each → money from students = 3s Adult tickets cost $5 each → money from adults = 5a Total money = $500 So: 3s + 5a = 500 --- **Step 3: Solve the system of equations** From s + a = 120, we get s = 120 - a Substitute into the money equation: 3(120 - a) + 5a = 500 --- **Step 4: Simplify and solve for a** 360 - 3a + 5a = 500 360 + 2a = 500 2a = 500 - 360 2a = 140 a = 70 --- **Step 5: Find s** s = 120 - a = 120 - 70 = 50 --- **Step 6: Check** Number of tickets: 50 + 70 = 120 ✓ Money: 50 × 3 = 150, 70 × 5 = 350, total = 150 + 350 = 500 ✓ --- **Final answer:** 50 student tickets and 70 adult tickets were sold.

4. 8x + 6y = 44; 4x - 6y = -20 Answer: x = 2, y = 14/3 Solution: 8x + 6y = 44 4x - 6y = -20 (8x + 6y) + (4x - 6y) = 44 + (-20) 8x + 4x + 6y - 6y = 24 12x = 24 x = 24 ÷ 12 x = 2 Substitute x = 2 into the first equation: 8(2) + 6y = 44 16 + 6y = 44 6y = 44 - 16 6y = 28 y = 28 ÷ 6 y = 14/3 4(2) - 6(14/3) = 8 - 28 = -20 ✓ The solution is x = 2, y = 14/3.
   Full step-by-step solution

   Step 1: Write the equations: 8x + 6y = 44 4x - 6y = -20 Step 2: Add the equations to eliminate y: (8x + 6y) + (4x - 6y) = 44 + (-20) 8x + 4x + 6y - 6y = 24 12x = 24 Step 3: Solve for x: x = 24 ÷ 12 x = 2 Step 4: Substitute x = 2 into the first equation: 8(2) + 6y = 44 16 + 6y = 44 Step 5: Solve for y: 6y = 44 - 16 6y = 28 y = 28 ÷ 6 y = 14/3 Step 6: Verify with the second equation: 4(2) - 6(14/3) = 8 - 28 = -20 ✓ The solution is x = 2, y = 14/3.

5. A right triangle is drawn on a coordinate plane with vertices at (0,0), (6,0), and (6,8). A second triangle is drawn with vertices at (0,0), (9,0), and (9,12). Are these triangles similar? If so, what is the scale factor from the first triangle to the second? Answer: yes, 1.5 Solution: We have two right triangles. First triangle vertices: A(0,0), B(6,0), C(6,8) Second triangle vertices: A'(0,0), B'(9,0), C'(9,12) From A(0,0) to B(6,0): length AB = 6 From B(6,0) to C(6,8): length BC = 8 From A(0,0) to C(6,8): length AC = sqrt((6-0)^2 + (8-0)^2) = sqrt(36 + 64) = sqrt(100) = 10…
   Full step-by-step solution

   Step 1: Understand the problem We have two right triangles. First triangle vertices: A(0,0), B(6,0), C(6,8) Second triangle vertices: A'(0,0), B'(9,0), C'(9,12) Step 2: Determine side lengths of the first triangle From A(0,0) to B(6,0): length AB = 6 From B(6,0) to C(6,8): length BC = 8 From A(0,0) to C(6,8): length AC = sqrt((6-0)^2 + (8-0)^2) = sqrt(36 + 64) = sqrt(100) = 10 So first triangle sides: 6, 8, 10 Step 3: Determine side lengths of the second triangle From A'(0,0) to B'(9,0): length A'B' = 9 From B'(9,0) to C'(9,12): length B'C' = 12 From A'(0,0) to C'(9,12): length A'C' = sqrt((9-0)^2 + (12-0)^2) = sqrt(81 + 144) = sqrt(225) = 15 So second triangle sides: 9, 12, 15 Step 4: Check if triangles are similar Triangles are similar if their corresponding sides are in proportion. Let's compare each side: First triangle sides: 6, 8, 10 Second triangle sides: 9, 12, 15 Check ratios: 9/6 = 1.5 12/8 = 1.5 15/10 = 1.5 All three ratios are equal to 1.5, so the triangles are similar. Step 5: Determine the scale factor Since all side ratios are 1.5, the scale factor from the first triangle to the second triangle is 1.5. Final answer: yes, 1.5

6. Emma is managing a community garden and needs to buy soil and compost for planting beds. She can purchase bags of soil for $12 each and bags of compost for $18 each. Her budget is $300, and she needs exactly 20 bags total to fill all the beds. How many bags of soil and how many bags of compost should Emma buy to spend exactly her budget and get the right number of bags? Answer: (10, 10) Solution: Let s = number of soil bags, c = number of compost bags s + c = 20 (total bags) 12s + 18c = 300 (total cost) Multiply the first equation by 12 to align coefficients: 12s + 12c = 240 (12s + 18c) - (12s + 12c) = 300 - 240 6c = 60 c = 60 ÷ 6 = 10 Substitute c = 10 into s + c = 20: s + 10 = 20 s =…
   Full step-by-step solution

   Step 1: Let s = number of soil bags, c = number of compost bags Step 2: Write the system of equations: s + c = 20 (total bags) 12s + 18c = 300 (total cost) Step 3: Multiply the first equation by 12 to align coefficients: 12s + 12c = 240 Step 4: Subtract this from the second equation: (12s + 18c) - (12s + 12c) = 300 - 240 6c = 60 Step 5: Solve for c: c = 60 ÷ 6 = 10 Step 6: Substitute c = 10 into s + c = 20: s + 10 = 20 s = 10 Step 7: Check the solution: 10 + 10 = 20 bags ✓ 12(10) + 18(10) = 120 + 180 = $300 ✓ Emma should buy 10 bags of soil and 10 bags of compost.

7. Emma is planning a science club field trip to the planetarium. She needs to arrange transportation using vans and buses. Each van can carry 8 students and requires 1 chaperone. Each bus can carry 40 students and requires 2 chaperones. The trip has 152 students signed up and 10 chaperones available. How many vans and buses should Emma use to transport all students while using exactly the available chaperones? Answer: (4, 3) Solution: Let v = number of vans and b = number of buses.
   Full step-by-step solution

   Step 1: Let v = number of vans and b = number of buses. Step 2: Write the equation for students: 8v + 40b = 152 Step 3: Write the equation for chaperones: 1v + 2b = 10 Step 4: Multiply the chaperone equation by 8: 8v + 16b = 80 Step 5: Subtract this from the student equation: (8v + 40b) - (8v + 16b) = 152 - 80 Step 6: Simplify: 24b = 72 Step 7: Solve for b: b = 72 ÷ 24 = 3 Step 8: Substitute b = 3 into the chaperone equation: v + 2(3) = 10 Step 9: Solve for v: v + 6 = 10, so v = 4 Step 10: Check: 4 vans × 8 students = 32 students, 3 buses × 40 students = 120 students, total = 152 students. 4 vans × 1 chaperone = 4 chaperones, 3 buses × 2 chaperones = 6 chaperones, total = 10 chaperones. The answer is (4, 3).