Linear Systems

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Answer Key & Explanations

Linear Systems · Grade 12 · Worksheet 3

1. Solve the system: 2x + y - z = 7, x - 3y + 2z = -1, 3x + 2y - 4z = 4 Answer: x = 3, y = 2, z = 1 Solution: Solving systems of three equations involves eliminating variables systematically.
   Full step-by-step solution

   Solving systems of three equations involves eliminating variables systematically. You can use substitution (solve one equation for a variable and substitute into others) or elimination (add/subtract equations to cancel variables). The goal is to reduce the three-variable system to a two-variable system, then to a single equation with one variable.

2. Emma, Liam, and Olivia are managing a small business that sells three types of handcrafted candles: lavender, vanilla, and cinnamon. Last month, the total revenue from all candles was $185. The revenue from lavender candles was $15 more than the revenue from vanilla candles, and the revenue from cinnamon candles was $5 less than twice the revenue from vanilla candles. Determine the revenue from each type of candle. Answer: lavender = $55, vanilla = $40, cinnamon = $75 Solution: Define variables. Let v = revenue from vanilla candles. Then lavender revenue = v + 15, and cinnamon revenue = 2v - 5.
   Full step-by-step solution

   Step 1: Define variables. Let v = revenue from vanilla candles. Then lavender revenue = v + 15, and cinnamon revenue = 2v - 5. Step 2: Write the total revenue equation: (v + 15) + v + (2v - 5) = 185. Step 3: Combine like terms: 4v + 10 = 185. Step 4: Subtract 10 from both sides: 4v = 175. Step 5: Divide by 4: v = 43.75. Step 6: Check: lavender = 43.75 + 15 = 58.75, cinnamon = 2(43.75) - 5 = 87.50 - 5 = 82.50. Sum: 58.75 + 43.75 + 82.50 = 185. So the revenues are: lavender = $58.75, vanilla = $43.75, cinnamon = $82.50.

3. 3x + 5y - 7z = 15, 5x - 3y + 9z = 23, 7x + 9y - 5z = 31 Answer: x = 3, y = 5, z = 1 Solution: (1) 3x + 5y - 7z = 15 (2) 5x - 3y + 9z = 23 (3) 7x + 9y - 5z = 31 Eliminate y from (1) and (2).
   Full step-by-step solution

   Step 1: Label the equations: (1) 3x + 5y - 7z = 15 (2) 5x - 3y + 9z = 23 (3) 7x + 9y - 5z = 31 Step 2: Eliminate y from (1) and (2). Multiply (1) by 3 and (2) by 5: 9x + 15y - 21z = 45 25x - 15y + 45z = 115 Add: 34x + 24z = 160 → divide by 2: 17x + 12z = 80 (Equation A) Step 3: Eliminate y from (1) and (3). Multiply (1) by 9 and (3) by 5: 27x + 45y - 63z = 135 35x + 45y - 25z = 155 Subtract: (27x - 35x) + (45y - 45y) + (-63z + 25z) = 135 - 155 -8x - 38z = -20 → multiply by -1: 8x + 38z = 20 → divide by 2: 4x + 19z = 10 (Equation B) Step 4: Solve the system of A and B: 17x + 12z = 80 4x + 19z = 10 Multiply B by 17: 68x + 323z = 170 Multiply A by 4: 68x + 48z = 320 Subtract: (68x - 68x) + (323z - 48z) = 170 - 320 275z = -150 z = -150/275 = -6/11 Step 5: Substitute z = -6/11 into Equation B: 4x + 19(-6/11) = 10 4x - 114/11 = 10 4x = 10 + 114/11 = 110/11 + 114/11 = 224/11 x = 56/11 Step 6: Substitute x = 56/11 and z = -6/11 into equation (1): 3(56/11) + 5y - 7(-6/11) = 15 168/11 + 5y + 42/11 = 15 210/11 + 5y = 15 5y = 15 - 210/11 = 165/11 - 210/11 = -45/11 y = -9/11 Step 7: The solution is x = 56/11, y = -9/11, z = -6/11. Verification in (2): 5(56/11) - 3(-9/11) + 9(-6/11) = 280/11 + 27/11 - 54/11 = 253/11 = 23 ✓ Verification in (3): 7(56/11) + 9(-9/11) - 5(-6/11) = 392/11 - 81/11 + 30/11 = 341/11 = 31 ✓

4. In a three-dimensional coordinate system, a plane is defined by the equation 3x + 4y - 2z = 24. A rectangular box has its vertices at the origin (0,0,0) and at (12, 0, 0), (0, 15, 0), (0, 0, 18), and the corresponding points in the positive x, y, and z directions. The plane cuts through the box, intersecting the x-axis, y-axis, and z-axis at three points, forming a triangular region in the first octant. Find the coordinates of these three intersection points. Answer: (8, 0, 0), (0, 6, 0), (0, 0, -12) Solution: Find the x-intercept. On the x-axis, y = 0 and z = 0. Substitute into 3x + 4(0) - 2(0) = 24, so 3x = 24, x = 8.
   Full step-by-step solution

   Step 1: Find the x-intercept. On the x-axis, y = 0 and z = 0. Substitute into 3x + 4(0) - 2(0) = 24, so 3x = 24, x = 8. Point A: (8, 0, 0). Step 2: Find the y-intercept. On the y-axis, x = 0 and z = 0. Substitute: 3(0) + 4y - 2(0) = 24, so 4y = 24, y = 6. Point B: (0, 6, 0). Step 3: Find the z-intercept. On the z-axis, x = 0 and y = 0. Substitute: 3(0) + 4(0) - 2z = 24, so -2z = 24, z = -12. Point C: (0, 0, -12). The three intersection points are (8, 0, 0), (0, 6, 0), and (0, 0, -12).

5. A particle moves along a path where its position coordinates (x, y, z) at time t satisfy the system: x' = 2x - y + z, y' = x + 3y - 2z, z' = -x + 2y + z. At a critical point where all derivatives are zero, the system becomes: 2x - y + z = 0, x + 3y - 2z = 0, -x + 2y + z = 0. Find the value of x² + y² + z² at this critical point. Answer: 0 Solution: Set up the system of equations from the condition that all derivatives are zero: 2x - y + z = 0 x + 3y - 2z = 0 -x + 2y + z = 0 Solve the system.
   Full step-by-step solution

   Step 1: Set up the system of equations from the condition that all derivatives are zero: 2x - y + z = 0 x + 3y - 2z = 0 -x + 2y + z = 0 Step 2: Solve the system. Add the first and third equations: (2x - y + z) + (-x + 2y + z) = 0 + 0 x + y + 2z = 0 Step 3: Now we have: x + y + 2z = 0 x + 3y - 2z = 0 Step 4: Subtract the first equation from the second: (x + 3y - 2z) - (x + y + 2z) = 0 - 0 2y - 4z = 0 2y = 4z y = 2z Step 5: Substitute y = 2z into x + y + 2z = 0: x + 2z + 2z = 0 x + 4z = 0 x = -4z Step 6: Check if these values satisfy all original equations: First equation: 2(-4z) - (2z) + z = -8z - 2z + z = -9z = 0, so z = 0 Second equation: (-4z) + 3(2z) - 2z = -4z + 6z - 2z = 0 Third equation: -(-4z) + 2(2z) + z = 4z + 4z + z = 9z = 0, so z = 0 Step 7: Since z = 0, then x = -4(0) = 0 and y = 2(0) = 0 Step 8: Calculate x² + y² + z² = 0² + 0² + 0² = 0 The answer is 0.

6. Solve the system: 2x + y - z = 8, x - 3y + 2z = -1, 4x + 2y + 3z = 19 Answer: x = 3, y = 1, z = -1 Solution: Solving systems of three equations involves eliminating one variable at a time to reduce the problem to a simpler system.
   Full step-by-step solution

   Solving systems of three equations involves eliminating one variable at a time to reduce the problem to a simpler system. Methods like substitution or elimination work by systematically removing variables until you can solve for one, then back-substitute to find the others. This process relies on the properties of equality and linear combinations.