Linear Systems

lessonbunny.com

Answer Key & Explanations

Linear Systems · Grade 12 · Worksheet 2

1. A civil engineering firm is designing a suspension bridge where the main cables follow a parabolic path. The cable's height above the roadway is modeled by the equation h(x) = ax² + bx + c, where x is the horizontal distance from the left tower in meters. Measurements show that at x = 0 (left tower), the cable is 50 meters high; at x = 100 meters (center), the cable is 10 meters high; and at x = 200 meters (right tower), the cable is 50 meters high. Determine the coefficients a, b, and c that define the cable's parabolic path. Answer: a = 0.004, b = -0.8, c = 50 Solution: For a parabola in standard form, substituting the coordinates of three distinct points gives us three equations that can be solved simultaneously to find the coefficients that define the curve's specific shape and position.
   Full step-by-step solution

   When modeling real-world phenomena with quadratic functions, we can use known points to create a system of equations. For a parabola in standard form, substituting the coordinates of three distinct points gives us three equations that can be solved simultaneously to find the coefficients that define the curve's specific shape and position.

2. A chemical engineering company is designing a reactor system with three interconnecting tanks. Tank A receives a solution containing x grams of compound P, y grams of compound Q, and z grams of compound R. Tank B receives a mixture where the amount of compound P is twice that of compound Q, and the amount of compound R is 5 grams more than compound P. Tank C receives a mixture where the total mass is 45 grams, with compound P having 3 grams less than compound Q, and compound R having twice the mass of compound P. If the total mass across all three tanks is 120 grams, and the total amount of compound P is equal to the total amount of compound Q, determine the mass of each compound in Tank A. Answer: x = 8, y = 10, z = 7 Solution: In chemical mixture problems with multiple constraints, we create a system of equations where each condition becomes an equation. The key is to carefully translate the word descriptions into mathematical relationships.
   Full step-by-step solution

   In chemical mixture problems with multiple constraints, we create a system of equations where each condition becomes an equation. The key is to carefully translate the word descriptions into mathematical relationships. For example, 'twice as much' becomes multiplication by 2, and 'more than' indicates addition. Once you have all equations, methods like substitution or elimination can help find the values of the unknown quantities. These types of problems demonstrate how algebra models real-world scenarios in fields like chemistry and engineering.

3. Isabella, Mason, and Charlotte are partners in a landscaping business. They invested different amounts of money and agreed to share the annual profit proportionally to their investments. Their total investment was $84,000. Mason invested $9,000 more than Isabella, and Charlotte invested $12,000 less than twice Mason's investment. Determine how much each person invested. Answer: Isabella: $15,000, Mason: $24,000, Charlotte: $45,000 Solution: Let x = Isabella's investment, y = Mason's investment, z = Charlotte's investment. Write the equations.
   Full step-by-step solution

   Let x = Isabella's investment, y = Mason's investment, z = Charlotte's investment. Step 1: Write the equations. Total: x + y + z = 84000 Mason compared to Isabella: y = x + 9000 Charlotte compared to Mason: z = 2y - 12000 Step 2: Substitute y and z in terms of x into the total equation. From y = x + 9000, and z = 2(x + 9000) - 12000 = 2x + 18000 - 12000 = 2x + 6000. Step 3: Substitute into x + y + z = 84000. x + (x + 9000) + (2x + 6000) = 84000 4x + 15000 = 84000 4x = 69000 x = 17250 Step 4: Find y and z. y = 17250 + 9000 = 26250 z = 2(26250) - 12000 = 52500 - 12000 = 40500 Step 5: Check the total: 17250 + 26250 + 40500 = 84000. Correct. Final answer: Isabella invested $17,250, Mason invested $26,250, and Charlotte invested $40,500.

4. Solve the system: 2x + 3y - z = 7, x - y + 2z = -1, 3x + 2y + z = 12 Answer: x = 2, y = 1, z = 0 Solution: Systems of three equations can be solved by eliminating one variable at a time. Once you find x and y, substitute back to find z.
   Full step-by-step solution

   Systems of three equations can be solved by eliminating one variable at a time. For example, with equations a, b, and c, you can combine a and b to eliminate z, then combine b and c to also eliminate z, giving you two equations in x and y. Once you find x and y, substitute back to find z.

5. A chemical engineering company is designing a reactor system that requires precise temperature control. The system uses three heating elements whose temperatures (x, y, z in °C) must satisfy the following conditions: The sum of all three temperatures is 120°C, the difference between the first and second element is twice the third element's temperature, and the first element's temperature equals the sum of the other two. Determine the exact temperature setting for each heating element. Answer: (60, 40, 20) Solution: Systems of equations with three variables appear frequently in engineering and scientific applications where multiple constraints must be satisfied simultaneously.
   Full step-by-step solution

   Systems of equations with three variables appear frequently in engineering and scientific applications where multiple constraints must be satisfied simultaneously. The key is to identify the relationships described in the problem and represent them mathematically. Once you have your equations, various methods like substitution or elimination can help you find values that satisfy all conditions. This type of problem demonstrates how mathematics helps solve real-world design challenges where multiple factors interact.

6. 2x + 5y - z = 15, x - 3y + 2z = 10, 4x + y + 3z = 25. Solve for x, y, z. Answer: x = 5, y = 3, z = 10 Solution: (1) 2x + 5y - z = 15 (2) x - 3y + 2z = 10 (3) 4x + y + 3z = 25 Eliminate z from (1) and (2). Multiply (1) by 2: 4x + 10y - 2z = 30. Add to (2): (4x + 10y - 2z) + (x - 3y + 2z) = 30 + 10 → 5x + 7y = 40.
   Full step-by-step solution

   Step 1: Label the equations: (1) 2x + 5y - z = 15 (2) x - 3y + 2z = 10 (3) 4x + y + 3z = 25 Step 2: Eliminate z from (1) and (2). Multiply (1) by 2: 4x + 10y - 2z = 30. Add to (2): (4x + 10y - 2z) + (x - 3y + 2z) = 30 + 10 → 5x + 7y = 40. (Equation A) Step 3: Eliminate z from (1) and (3). Multiply (1) by 3: 6x + 15y - 3z = 45. Add to (3): (6x + 15y - 3z) + (4x + y + 3z) = 45 + 25 → 10x + 16y = 70. Divide by 2: 5x + 8y = 35. (Equation B) Step 4: Solve the system of A and B: A: 5x + 7y = 40 B: 5x + 8y = 35 Subtract A from B: (5x + 8y) - (5x + 7y) = 35 - 40 → y = -5. Step 5: Substitute y = -5 into A: 5x + 7(-5) = 40 → 5x - 35 = 40 → 5x = 75 → x = 15. Step 6: Substitute x = 15 and y = -5 into (1): 2(15) + 5(-5) - z = 15 → 30 - 25 - z = 15 → 5 - z = 15 → z = -10. Step 7: Verify with (2): 15 - 3(-5) + 2(-10) = 15 + 15 - 20 = 10 ✓ Verify with (3): 4(15) + (-5) + 3(-10) = 60 - 5 - 30 = 25 ✓ The solution is x = 15, y = -5, z = -10.