Linear Systems 3x3

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Linear Systems 3x3 · Grade 12 · Worksheet 1

1. 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.

2. 2x + y - z = 5, x - 2y + z = -5, 3x + y + 2z = 10 Answer: x = 2, y = 1, z = 0 Solution: Step 1: Label the equations: (1) 2x + y - z = 5 (2) x - 2y + z = -5 (3) 3x + y + 2z = 10 Step 2: Add equations (1) and (2) to eliminate z: (1) + (2): (2x + x) + (y - 2y) + (-z + z) = 5 + (-5) 3x - y = 0 So y = 3x Step 3: Substitute y = 3x into equation (1): 2x + (3x) - z = 5 5x - z = 5 So z = 5x…
   Full step-by-step solution

   Step 1: Label the equations: (1) 2x + y - z = 5 (2) x - 2y + z = -5 (3) 3x + y + 2z = 10 Step 2: Add equations (1) and (2) to eliminate z: (1) + (2): (2x + x) + (y - 2y) + (-z + z) = 5 + (-5) 3x - y = 0 So y = 3x Step 3: Substitute y = 3x into equation (1): 2x + (3x) - z = 5 5x - z = 5 So z = 5x - 5 Step 4: Substitute y = 3x and z = 5x - 5 into equation (3): 3x + (3x) + 2(5x - 5) = 10 3x + 3x + 10x - 10 = 10 16x - 10 = 10 16x = 20 x = 20/16 = 5/4 Step 5: Find y and z: y = 3x = 3(5/4) = 15/4 z = 5x - 5 = 5(5/4) - 5 = 25/4 - 20/4 = 5/4 Step 6: Verify with equation (3): 3(5/4) + 15/4 + 2(5/4) = 15/4 + 15/4 + 10/4 = 40/4 = 10 ✓ The solution is x = 5/4, y = 15/4, z = 5/4.

3. Solve: 3x + y - z = 5, x - 3y + 2z = -7, 2x + y + z = 7 Answer: x = 2, y = 1, z = 2 Solution: Step 1: Label the equations: (1) 3x + y - z = 5 (2) x - 3y + 2z = -7 (3) 2x + y + z = 7 Step 2: Add equations (1) and (3) to eliminate z: (1) + (3): (3x + y - z) + (2x + y + z) = 5 + 7 5x + 2y = 12 (equation 4) Step 3: Multiply equation (1) by 2 and add to equation (2): 2*(1): 6x + 2y - 2z = 10…
   Full step-by-step solution

   Step 1: Label the equations: (1) 3x + y - z = 5 (2) x - 3y + 2z = -7 (3) 2x + y + z = 7 Step 2: Add equations (1) and (3) to eliminate z: (1) + (3): (3x + y - z) + (2x + y + z) = 5 + 7 5x + 2y = 12 (equation 4) Step 3: Multiply equation (1) by 2 and add to equation (2): 2*(1): 6x + 2y - 2z = 10 Add to (2): (6x + 2y - 2z) + (x - 3y + 2z) = 10 + (-7) 7x - y = 3 (equation 5) Step 4: Solve the system of equations (4) and (5): (4) 5x + 2y = 12 (5) 7x - y = 3 Multiply equation (5) by 2: 14x - 2y = 6 Add to equation (4): (5x + 2y) + (14x - 2y) = 12 + 6 19x = 18 x = 18/19 Step 5: Substitute x = 18/19 into equation (5): 7(18/19) - y = 3 126/19 - y = 3 y = 126/19 - 57/19 y = 69/19 Step 6: Substitute x = 18/19 and y = 69/19 into equation (1): 3(18/19) + 69/19 - z = 5 54/19 + 69/19 - z = 5 123/19 - z = 95/19 z = 123/19 - 95/19 z = 28/19 Step 7: Verify with equation (2): (18/19) - 3(69/19) + 2(28/19) = 18/19 - 207/19 + 56/19 = -133/19 = -7 ✓ Step 8: Verify with equation (3): 2(18/19) + 69/19 + 28/19 = 36/19 + 69/19 + 28/19 = 133/19 = 7 ✓ The solution is x = 18/19, y = 69/19, z = 28/19.

4. Charlotte is managing a small bakery that produces three types of cookies: chocolate chip, oatmeal raisin, and sugar cookies. On Monday, she sold 2 boxes of chocolate chip, 7 boxes of oatmeal raisin, and 3 boxes of sugar cookies for a total of $142. On Tuesday, she sold 4 boxes of chocolate chip, 2 boxes of oatmeal raisin, and 7 boxes of sugar cookies for a total of $172. On Wednesday, she sold 7 boxes of chocolate chip, 3 boxes of oatmeal raisin, and 2 boxes of sugar cookies for a total of $147. What is the price per box of oatmeal raisin cookies? Answer: 12 Solution: Let x = price of chocolate chip cookies, y = price of oatmeal raisin cookies, z = price of sugar cookies.
   Full step-by-step solution

   Step 1: Let x = price of chocolate chip cookies, y = price of oatmeal raisin cookies, z = price of sugar cookies. Step 2: Write the system of equations: Monday: 2x + 7y + 3z = 142 Tuesday: 4x + 2y + 7z = 172 Wednesday: 7x + 3y + 2z = 147 Step 3: Eliminate x by multiplying Monday's equation by 2 and subtracting from Tuesday's: (4x + 2y + 7z) - 2*(2x + 7y + 3z) = 172 - 2*142 4x + 2y + 7z - 4x - 14y - 6z = 172 - 284 -12y + z = -112 Step 4: Eliminate x using Monday and Wednesday equations: Multiply Monday by 7 and Wednesday by 2: 14x + 49y + 21z = 994 14x + 6y + 4z = 294 Subtract: (14x + 49y + 21z) - (14x + 6y + 4z) = 994 - 294 43y + 17z = 700 Step 5: Solve the system of two equations: -12y + z = -112 => z = -112 + 12y Substitute into 43y + 17z = 700: 43y + 17(-112 + 12y) = 700 43y - 1904 + 204y = 700 247y = 2604 y = 2604/247 = 12 The price per box of oatmeal raisin cookies is $12.

5. Solve: 2x + y - z = 7, x - 3y + 2z = -2, 3x + 2y + z = 12 Answer: x = 3, y = 1, z = 0 Solution: Step 1: Label the equations: (1) 2x + y - z = 7 (2) x - 3y + 2z = -2 (3) 3x + 2y + z = 12 Step 2: Add equations (1) and (3) to eliminate z: (1) + (3): (2x + y - z) + (3x + 2y + z) = 7 + 12 5x + 3y = 19 (equation 4) Step 3: Multiply equation (1) by 2 and add to equation (2): 2*(1): 4x + 2y - 2z =…
   Full step-by-step solution

   Step 1: Label the equations: (1) 2x + y - z = 7 (2) x - 3y + 2z = -2 (3) 3x + 2y + z = 12 Step 2: Add equations (1) and (3) to eliminate z: (1) + (3): (2x + y - z) + (3x + 2y + z) = 7 + 12 5x + 3y = 19 (equation 4) Step 3: Multiply equation (1) by 2 and add to equation (2): 2*(1): 4x + 2y - 2z = 14 Add to (2): (4x + 2y - 2z) + (x - 3y + 2z) = 14 + (-2) 5x - y = 12 (equation 5) Step 4: Solve the system of equations (4) and (5): (4) 5x + 3y = 19 (5) 5x - y = 12 Subtract (5) from (4): (5x + 3y) - (5x - y) = 19 - 12 4y = 7 y = 7/4 Step 5: Substitute y = 7/4 into equation (5): 5x - 7/4 = 12 5x = 12 + 7/4 = 48/4 + 7/4 = 55/4 x = 11/4 Step 6: Substitute x = 11/4 and y = 7/4 into equation (1): 2(11/4) + 7/4 - z = 7 22/4 + 7/4 - z = 7 29/4 - z = 7 z = 29/4 - 28/4 = 1/4 Step 7: Verify with equation (2): (11/4) - 3(7/4) + 2(1/4) = 11/4 - 21/4 + 2/4 = -8/4 = -2 ✓ Step 8: Verify with equation (3): 3(11/4) + 2(7/4) + 1/4 = 33/4 + 14/4 + 1/4 = 48/4 = 12 ✓ The solution is x = 11/4, y = 7/4, z = 1/4.

6. Solve: 4x + 3y - 2z = 18, 2x - y + 5z = 7, x + 2y + 3z = 15 Answer: x = 2, y = 4, z = 1 Solution: Step 1: Write the system: (1) 4x + 3y - 2z = 18 (2) 2x - y + 5z = 7 (3) x + 2y + 3z = 15 Step 2: Multiply equation (2) by 2: 4x - 2y + 10z = 14 Subtract equation (1) from this: (4x - 2y + 10z) - (4x + 3y - 2z) = 14 - 18 -5y + 12z = -4 (equation 4) Step 3: Multiply equation (3) by 4: 4x + 8y +…
   Full step-by-step solution

   Step 1: Write the system: (1) 4x + 3y - 2z = 18 (2) 2x - y + 5z = 7 (3) x + 2y + 3z = 15 Step 2: Multiply equation (2) by 2: 4x - 2y + 10z = 14 Subtract equation (1) from this: (4x - 2y + 10z) - (4x + 3y - 2z) = 14 - 18 -5y + 12z = -4 (equation 4) Step 3: Multiply equation (3) by 4: 4x + 8y + 12z = 60 Subtract equation (1) from this: (4x + 8y + 12z) - (4x + 3y - 2z) = 60 - 18 5y + 14z = 42 (equation 5) Step 4: Add equations (4) and (5): (-5y + 12z) + (5y + 14z) = -4 + 42 26z = 38 z = 38/26 = 19/13 Step 5: Substitute z = 19/13 into equation (5): 5y + 14(19/13) = 42 5y + 266/13 = 42 5y = 42 - 266/13 = 546/13 - 266/13 = 280/13 y = (280/13) ÷ 5 = 280/65 = 56/13 Step 6: Substitute y = 56/13 and z = 19/13 into equation (3): x + 2(56/13) + 3(19/13) = 15 x + 112/13 + 57/13 = 15 x + 169/13 = 15 x = 15 - 169/13 = 195/13 - 169/13 = 26/13 = 2 Step 7: Verify with equation (1): 4(2) + 3(56/13) - 2(19/13) = 8 + 168/13 - 38/13 = 8 + 130/13 = 8 + 10 = 18 ✓ The solution is x = 2, y = 56/13, z = 19/13.

7. Isabella invests in three stocks. Stock A: 4x + 2y - 3z = 15, Stock B: 3x - 4y + 2z = -8, Stock C: 2x + 3y - z = 12. Solve for x, y, and z. Answer: x = 2, y = 3, z = 1 Solution: 4x + 2y - 3z = 15 (1) 3x - 4y + 2z = -8 (2) 2x + 3y - z = 12 (3) Multiply equation (3) by 3: 6x + 9y - 3z = 36 (4) Subtract equation (1) from (4): (6x - 4x) + (9y - 2y) + (-3z - (-3z)) = 36 - 15 2x + 7y = 21 (5) Multiply equation (3) by 2: 4x + 6y - 2z = 24 (6) Subtract equation (2) from…
   Full step-by-step solution

   Step 1: Write the system: 4x + 2y - 3z = 15 (1) 3x - 4y + 2z = -8 (2) 2x + 3y - z = 12 (3) Step 2: Multiply equation (3) by 3: 6x + 9y - 3z = 36 (4) Subtract equation (1) from (4): (6x - 4x) + (9y - 2y) + (-3z - (-3z)) = 36 - 15 2x + 7y = 21 (5) Step 3: Multiply equation (3) by 2: 4x + 6y - 2z = 24 (6) Subtract equation (2) from (6): (4x - 3x) + (6y - (-4y)) + (-2z - 2z) = 24 - (-8) x + 10y - 4z = 32 (7) Step 4: From equation (3): -z = 12 - 2x - 3y, so z = 2x + 3y - 12 Substitute into equation (7): x + 10y - 4(2x + 3y - 12) = 32 x + 10y - 8x - 12y + 48 = 32 -7x - 2y = -16 (8) Step 5: Solve system of equations (5) and (8): From (5): 2x + 7y = 21 From (8): -7x - 2y = -16 Multiply (5) by 7: 14x + 49y = 147 (9) Multiply (8) by 2: -14x - 4y = -32 (10) Add (9) and (10): 45y = 115, y = 115/45 = 23/9 Step 6: Substitute y = 23/9 into (5): 2x + 7(23/9) = 21 2x + 161/9 = 21 2x = 21 - 161/9 = 189/9 - 161/9 = 28/9 x = 14/9 Step 7: Substitute x = 14/9, y = 23/9 into (3): 2(14/9) + 3(23/9) - z = 12 28/9 + 69/9 - z = 12 97/9 - z = 12 z = 97/9 - 108/9 = -11/9 The solution is x = 14/9, y = 23/9, z = -11/9.

8. Olivia is analyzing the nutritional content of a custom trail mix containing almonds, cashews, and peanuts. In a 45-gram sample, the almond content is 5 grams more than the cashew content. The peanut content is 10 grams less than twice the cashew content. The total mass of almonds and peanuts together is 35 grams. How many grams of cashews are in the sample? Answer: 15 Solution: Define variables: Let A = grams of almonds, C = grams of cashews, P = grams of peanuts.
   Full step-by-step solution

   Step 1: Define variables: Let A = grams of almonds, C = grams of cashews, P = grams of peanuts. Step 2: Write equations from the problem: Total mass: A + C + P = 45 Almonds vs cashews: A = C + 5 Peanuts vs cashews: P = 2C - 10 Almonds and peanuts total: A + P = 35 Step 3: Substitute A = C + 5 and P = 2C - 10 into A + P = 35: (C + 5) + (2C - 10) = 35 3C - 5 = 35 3C = 40 C = 40/3 Step 4: Check consistency with total mass equation: A = C + 5 = 40/3 + 15/3 = 55/3 P = 2C - 10 = 80/3 - 30/3 = 50/3 A + C + P = 55/3 + 40/3 + 50/3 = 145/3 ≈ 48.33, which contradicts the given total of 45 grams. Step 5: Realize the fourth equation (A + P = 35) is redundant with the first three. Use only the first three equations: A + C + P = 45 A = C + 5 P = 2C - 10 Step 6: Substitute into the total mass equation: (C + 5) + C + (2C - 10) = 45 4C - 5 = 45 4C = 50 C = 12.5 Step 7: Verify: A = 12.5 + 5 = 17.5 P = 2(12.5) - 10 = 25 - 10 = 15 Total: 17.5 + 12.5 + 15 = 45 ✓ The cashew content is 12.5 grams.