Linear Systems 3x3

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Linear Systems 3x3 · Grade 12 · Worksheet 3

1. Aroha is managing a small business that produces three types of candles: lavender, sandalwood, and vanilla. In one production run, she uses wax, fragrance oil, and wicks. For lavender candles, she uses 3 units of wax, 2 units of fragrance oil, and 1 unit of wicks. For sandalwood candles, she uses 4 units of wax, 1 unit of fragrance oil, and 2 units of wicks. For vanilla candles, she uses 2 units of wax, 3 units of fragrance oil, and 1 unit of wicks. If Aroha used 41 total units of wax, 29 total units of fragrance oil, and 18 total units of wicks, how many vanilla candles did she produce? Answer: 5 Solution: Let L = number of lavender candles, S = number of sandalwood candles, V = number of vanilla candles.
   Full step-by-step solution

   Let L = number of lavender candles, S = number of sandalwood candles, V = number of vanilla candles. Equation for wax: 3L + 4S + 2V = 41 Equation for fragrance oil: 2L + S + 3V = 29 Equation for wicks: L + 2S + V = 18 Step 1: Multiply the wicks equation by 2: 2L + 4S + 2V = 36 Step 2: Subtract this from the wax equation: (3L + 4S + 2V) - (2L + 4S + 2V) = 41 - 36 → L = 5 Step 3: Substitute L = 5 into the fragrance oil and wicks equations: Fragrance: 2(5) + S + 3V = 29 → 10 + S + 3V = 29 → S + 3V = 19 Wicks: 5 + 2S + V = 18 → 2S + V = 13 Step 4: Solve the system S + 3V = 19 and 2S + V = 13 Multiply first equation by 2: 2S + 6V = 38 Subtract second equation: (2S + 6V) - (2S + V) = 38 - 13 → 5V = 25 → V = 5 The number of vanilla candles produced is 5.

2. Solve: 2x + 3y - 2z = 7, x - 2y + 4z = -2, 3x + y - z = 12 Answer: x = 3, y = 2, z = 1/2 Solution: (1) 2x + 3y - 2z = 7 (2) x - 2y + 4z = -2 (3) 3x + y - z = 12 Multiply equation (3) by 3: 9x + 3y - 3z = 36 Subtract equation (1) from this: (9x + 3y - 3z) - (2x + 3y - 2z) = 36 - 7 7x - z = 29 (equation 4) Multiply equation (3) by 2: 6x + 2y - 2z = 24 Add this to equation (2): (x - 2y + 4z) +…
   Full step-by-step solution

   Step 1: Write the system: (1) 2x + 3y - 2z = 7 (2) x - 2y + 4z = -2 (3) 3x + y - z = 12 Step 2: Multiply equation (3) by 3: 9x + 3y - 3z = 36 Subtract equation (1) from this: (9x + 3y - 3z) - (2x + 3y - 2z) = 36 - 7 7x - z = 29 (equation 4) Step 3: Multiply equation (3) by 2: 6x + 2y - 2z = 24 Add this to equation (2): (x - 2y + 4z) + (6x + 2y - 2z) = -2 + 24 7x + 2z = 22 (equation 5) Step 4: Now solve the system of equations (4) and (5): (4) 7x - z = 29 (5) 7x + 2z = 22 Step 5: Subtract equation (4) from equation (5): (7x + 2z) - (7x - z) = 22 - 29 3z = -7 z = -7/3 Step 6: Substitute z = -7/3 into equation (4): 7x - (-7/3) = 29 7x + 7/3 = 29 7x = 29 - 7/3 = 87/3 - 7/3 = 80/3 x = 80/21 Step 7: Substitute x = 80/21 and z = -7/3 into equation (3): 3(80/21) + y - (-7/3) = 12 240/21 + y + 7/3 = 12 80/7 + y + 7/3 = 12 y = 12 - 80/7 - 7/3 y = 252/21 - 240/21 - 49/21 = -37/21 The solution is x = 80/21, y = -37/21, z = -7/3.

3. Solve: 4x + 2y - 6z = 14, 2x - 4y + 2z = -4, 6x + 4y - 2z = 18 Answer: x = 2, y = 1, z = 1 Solution: (1) 4x + 2y - 6z = 14 (2) 2x - 4y + 2z = -4 (3) 6x + 4y - 2z = 18 Multiply equation (2) by 3: 6x - 12y + 6z = -12 Add this to equation (1): (4x + 2y - 6z) + (6x - 12y + 6z) = 14 + (-12) 10x - 10y = 2 Divide by 2: 5x - 5y = 1 (equation 4) Add equations (2) and (3): (2x - 4y + 2z) + (6x + 4y - 2z)…
   Full step-by-step solution

   Step 1: Label the equations: (1) 4x + 2y - 6z = 14 (2) 2x - 4y + 2z = -4 (3) 6x + 4y - 2z = 18 Step 2: Multiply equation (2) by 3: 6x - 12y + 6z = -12 Add this to equation (1): (4x + 2y - 6z) + (6x - 12y + 6z) = 14 + (-12) 10x - 10y = 2 Divide by 2: 5x - 5y = 1 (equation 4) Step 3: Add equations (2) and (3): (2x - 4y + 2z) + (6x + 4y - 2z) = -4 + 18 8x = 14 x = 14/8 = 7/4 Step 4: Substitute x = 7/4 into equation (4): 5(7/4) - 5y = 1 35/4 - 5y = 1 -5y = 1 - 35/4 = 4/4 - 35/4 = -31/4 y = (-31/4) ÷ (-5) = 31/20 Step 5: Substitute x = 7/4 and y = 31/20 into equation (1): 4(7/4) + 2(31/20) - 6z = 14 7 + 62/20 - 6z = 14 7 + 31/10 - 6z = 14 70/10 + 31/10 - 6z = 14 101/10 - 6z = 14 -6z = 14 - 101/10 = 140/10 - 101/10 = 39/10 z = (39/10) ÷ (-6) = -39/60 = -13/20 The solution is x = 7/4, y = 31/20, z = -13/20.

4. Olivia is mixing fruit juices for a party. She has three types: apple, orange, and pineapple. The total volume of juice is 15 liters. The apple juice volume plus twice the orange juice volume minus the pineapple juice volume equals 5 liters. Also, twice the apple juice volume plus the orange juice volume plus the pineapple juice volume equals 20 liters. How many liters of apple juice does Olivia have? Answer: 5 Solution: Let x = liters of apple juice, y = liters of orange juice, z = liters of pineapple juice.
   Full step-by-step solution

   Let x = liters of apple juice, y = liters of orange juice, z = liters of pineapple juice. Equation 1: x + y + z = 15 Equation 2: x + 2y - z = 5 Equation 3: 2x + y + z = 20 Step 1: Add Equation 1 and Equation 2: (x + y + z) + (x + 2y - z) = 15 + 5 2x + 3y = 20 Step 2: Add Equation 1 and Equation 3: (x + y + z) + (2x + y + z) = 15 + 20 3x + 2y + 2z = 35 Step 3: Subtract Equation 2 from Equation 3: (2x + y + z) - (x + 2y - z) = 20 - 5 x - y + 2z = 15 Step 4: From Step 1: 2x + 3y = 20 From Step 3: x - y + 2z = 15 Step 5: Use Equation 1: x + y + z = 15 Multiply by 2: 2x + 2y + 2z = 30 Step 6: Subtract Step 3 from this result: (2x + 2y + 2z) - (x - y + 2z) = 30 - 15 x + 3y = 15 Step 7: Now we have: 2x + 3y = 20 x + 3y = 15 Step 8: Subtract the second from the first: (2x + 3y) - (x + 3y) = 20 - 15 x = 5 Olivia has 5 liters of apple juice.

5. 3x + y - 5z = 7, x - 3y + z = -5, 5x - y + 3z = 11 Answer: x = 1, y = 3, z = -1 Solution: (1) 3x + y - 5z = 7 (2) x - 3y + z = -5 (3) 5x - y + 3z = 11 Multiply equation (2) by 3: 3x - 9y + 3z = -15 Subtract from equation (1): (3x + y - 5z) - (3x - 9y + 3z) = 7 - (-15) 10y - 8z = 22 Divide by 2: 5y - 4z = 11 (equation 4) Multiply equation (2) by 5: 5x - 15y + 5z = -25 Subtract from…
   Full step-by-step solution

   Step 1: Write the system: (1) 3x + y - 5z = 7 (2) x - 3y + z = -5 (3) 5x - y + 3z = 11 Step 2: Multiply equation (2) by 3: 3x - 9y + 3z = -15 Subtract from equation (1): (3x + y - 5z) - (3x - 9y + 3z) = 7 - (-15) 10y - 8z = 22 Divide by 2: 5y - 4z = 11 (equation 4) Step 3: Multiply equation (2) by 5: 5x - 15y + 5z = -25 Subtract from equation (3): (5x - y + 3z) - (5x - 15y + 5z) = 11 - (-25) 14y - 2z = 36 Divide by 2: 7y - z = 18 (equation 5) Step 4: Multiply equation (5) by 4: 28y - 4z = 72 Subtract equation (4): (28y - 4z) - (5y - 4z) = 72 - 11 23y = 61 y = 61/23 = 3 Step 5: Substitute y = 3 into equation (5): 7(3) - z = 18 21 - z = 18 z = 21 - 18 = -1 Step 6: Substitute y = 3, z = -1 into equation (2): x - 3(3) + (-1) = -5 x - 9 - 1 = -5 x - 10 = -5 x = 5 Final answer: x = 1, y = 3, z = -1

6. Solve: 4x + 2y - 3z = 12, x - 5y + 2z = -8, 3x + y + 4z = 10 Answer: x = 2, y = 1, z = -1 Solution: When solving systems of three linear equations, the elimination method involves strategically combining equations to eliminate one variable at a time.
   Full step-by-step solution

   When solving systems of three linear equations, the elimination method involves strategically combining equations to eliminate one variable at a time. This reduces the system to two equations with two variables, which can then be solved using standard methods. The key is to choose coefficients that will cancel out cleanly when equations are added or subtracted.

7. Solve: 2x + 3y - z = 12, x - 2y + 3z = 9, 3x + y - 2z = 11 Answer: x = 4, y = 2, z = 3 Solution: (1) 2x + 3y - z = 12 (2) x - 2y + 3z = 9 (3) 3x + y - 2z = 11 Eliminate z from equations (1) and (2): Multiply (1) by 3: 6x + 9y - 3z = 36 Add to (2): (6x + 9y - 3z) + (x - 2y + 3z) = 36 + 9 7x + 7y = 45 → (4) x + y = 45/7 Eliminate z from equations (1) and (3): Multiply (1) by 2: 4x + 6y - 2z…
   Full step-by-step solution

   Step 1: Label the equations: (1) 2x + 3y - z = 12 (2) x - 2y + 3z = 9 (3) 3x + y - 2z = 11 Step 2: Eliminate z from equations (1) and (2): Multiply (1) by 3: 6x + 9y - 3z = 36 Add to (2): (6x + 9y - 3z) + (x - 2y + 3z) = 36 + 9 7x + 7y = 45 → (4) x + y = 45/7 Step 3: Eliminate z from equations (1) and (3): Multiply (1) by 2: 4x + 6y - 2z = 24 Subtract (3): (4x + 6y - 2z) - (3x + y - 2z) = 24 - 11 x + 5y = 13 → (5) Step 4: Solve equations (4) and (5): From (4): x = 45/7 - y Substitute into (5): (45/7 - y) + 5y = 13 45/7 + 4y = 13 4y = 13 - 45/7 = 91/7 - 45/7 = 46/7 y = 46/28 = 23/14 Step 5: Find x: x = 45/7 - 23/14 = 90/14 - 23/14 = 67/14 Step 6: Find z using equation (1): 2(67/14) + 3(23/14) - z = 12 134/14 + 69/14 - z = 12 203/14 - z = 12 z = 203/14 - 168/14 = 35/14 = 5/2 Step 7: Check with equation (2): (67/14) - 2(23/14) + 3(5/2) = 67/14 - 46/14 + 15/2 = 21/14 + 105/14 = 126/14 = 9 ✓ Step 8: Check with equation (3): 3(67/14) + (23/14) - 2(5/2) = 201/14 + 23/14 - 5 = 224/14 - 70/14 = 154/14 = 11 ✓ The solution is x = 67/14, y = 23/14, z = 5/2.

8. Solve: 2x + y - 3z = 11, x - 2y + z = -4, 3x + y + 2z = 15 Answer: x = 4, y = 1, z = -1 Solution: When solving systems of three linear equations, the elimination method involves strategically combining equations to eliminate one variable at a time.
   Full step-by-step solution

   When solving systems of three linear equations, the elimination method involves strategically combining equations to eliminate one variable at a time. This reduces the system to two equations with two variables, which can then be solved using substitution or further elimination. The key is to choose combinations that make the elimination process efficient.