3.1 Linear Simultaneous Equations
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Learning Objectives
- Understand what simultaneous equations are and when they arise
- Solve linear simultaneous equations using the elimination method
- Solve linear simultaneous equations using the substitution method
- Solve linear simultaneous equations using matrices
- Apply simultaneous equations to solve real-world problems
Key Concepts
Introduction to Simultaneous Equations
Simultaneous equations are a set of equations that must be satisfied at the same time.
Linear simultaneous equations involve two or more linear equations with two or more unknowns.
A solution to a system of simultaneous equations is a set of values for the unknowns that satisfies all equations in the system.
The Elimination Method
The elimination method involves adding or subtracting equations to eliminate one of the variables.
To use this method, we first need to ensure that the coefficients of one of the variables are equal in magnitude but opposite in sign.
After eliminating one variable, we can solve for the remaining variable and then substitute back to find the value of the eliminated variable.
Example 1
Solve the following system of equations using the elimination method: 2x + 3y = 7 4x - 5y = 3
Solution:
Step 1: Multiply the first equation by 2 to make the coefficients of x match in magnitude. 2(2x + 3y = 7) → 4x + 6y = 14 Step 2: Add this equation to the second equation to eliminate x. 4x + 6y = 14 4x - 5y = 3 -------------- 11y = 17 Step 3: Solve for y. 11y = 17 y = 17/11 Step 4: Substitute this value of y back into the first equation to find x. 2x + 3y = 7 2x + 3(17/11) = 7 2x + 51/11 = 7 2x = 7 - 51/11 2x = (77 - 51)/11 2x = 26/11 x = 13/11 So the solution is x = 13/11, y = 17/11.
Example 2
Solve the following system of equations using the elimination method: 3x + 2y = 1 5x - 2y = 17
Solution:
Step 1: Add the two equations to eliminate y. 3x + 2y = 1 5x - 2y = 17 -------------- 8x = 18 Step 2: Solve for x. 8x = 18 x = 18/8 = 9/4 Step 3: Substitute this value of x back into the first equation to find y. 3x + 2y = 1 3(9/4) + 2y = 1 27/4 + 2y = 1 2y = 1 - 27/4 2y = (4 - 27)/4 2y = -23/4 y = -23/8 So the solution is x = 9/4, y = -23/8.
The Substitution Method
The substitution method involves expressing one variable in terms of the other from one equation, and then substituting this expression into the other equation.
This method is particularly useful when one of the equations already has a variable isolated, or when the coefficient of one of the variables is 1.
After substituting, we solve for one variable and then back-substitute to find the other variable.
Example 1
Solve the following system of equations using the substitution method: x + y = 5 2x - 3y = 4
Solution:
Step 1: Express y in terms of x from the first equation. x + y = 5 y = 5 - x Step 2: Substitute this expression for y into the second equation. 2x - 3y = 4 2x - 3(5 - x) = 4 2x - 15 + 3x = 4 5x - 15 = 4 5x = 19 x = 19/5 Step 3: Substitute this value of x back to find y. y = 5 - x y = 5 - 19/5 y = (25 - 19)/5 y = 6/5 So the solution is x = 19/5, y = 6/5.
Example 2
Solve the following system of equations using the substitution method: 2x - y = 3 3x + 2y = 4
Solution:
Step 1: Express y in terms of x from the first equation. 2x - y = 3 y = 2x - 3 Step 2: Substitute this expression for y into the second equation. 3x + 2y = 4 3x + 2(2x - 3) = 4 3x + 4x - 6 = 4 7x - 6 = 4 7x = 10 x = 10/7 Step 3: Substitute this value of x back to find y. y = 2x - 3 y = 2(10/7) - 3 y = 20/7 - 3 y = 20/7 - 21/7 y = -1/7 So the solution is x = 10/7, y = -1/7.
Using Matrices to Solve Simultaneous Equations
Linear simultaneous equations can be represented and solved using matrices.
For a system of equations AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix, the solution is X = A^(-1)B, where A^(-1) is the inverse of A.
This method is particularly useful for solving systems with more than two equations and two unknowns.
Example 1
Solve the following system of equations using matrices: 2x + 3y = 7 4x - 5y = 3
Solution:
Step 1: Write the system in matrix form. [2 3] [x] = [7] [4 -5] [y] [3] Step 2: Find the determinant of the coefficient matrix. det([2 3]) = 2 × (-5) - 3 × 4 = -10 - 12 = -22 [4 -5] Step 3: Find the inverse of the coefficient matrix. A^(-1) = (1/det(A)) × [[-5 -3], [-4 2]] = (1/-22) × [[-5 -3], [-4 2]] = (1/22) × [[5 3], [4 -2]] Step 4: Multiply the inverse by the constant matrix to find the solution. [x] = (1/22) × [[5 3], [4 -2]] × [7] [y] [3] = (1/22) × [[5×7 + 3×3], [4×7 + (-2)×3]] = (1/22) × [[35 + 9], [28 - 6]] = (1/22) × [[44], [22]] = [[44/22], [22/22]] = [[2], [1]] So the solution is x = 2, y = 1.
Example 2
Solve the following system of equations using matrices: x + 2y + 3z = 14 2x - y + z = 3 3x + y - 2z = -1
Solution:
Step 1: Write the system in matrix form. [1 2 3] [x] = [14] [2 -1 1] [y] = [3] [3 1 -2] [z] = [-1] Step 2: Find the determinant of the coefficient matrix. det(A) = 1 × ((-1) × (-2) - 1 × 1) - 2 × (2 × (-2) - 3 × 1) + 3 × (2 × 1 - 3 × (-1)) = 1 × (2 - 1) - 2 × (-4 - 3) + 3 × (2 + 3) = 1 × 1 - 2 × (-7) + 3 × 5 = 1 + 14 + 15 = 30 Step 3: Find the inverse of the coefficient matrix (this is a complex calculation, so we'll skip the details). Step 4: Multiply the inverse by the constant matrix to find the solution. [x] = A^(-1) × [14] [y] [3] [z] [-1] Performing this calculation gives x = 1, y = 2, z = 3.
Applications of Simultaneous Equations
Simultaneous equations are used to solve many real-world problems involving multiple unknowns and constraints.
Common applications include mixture problems, rate problems, and problems involving motion or geometry.
The key to solving these problems is to identify the unknowns and set up the appropriate equations based on the given information.
Example 1
A coffee shop sells regular coffee for £2.50 and specialty coffee for £3.50. On a particular day, they sold 120 cups in total and took in £350. How many of each type of coffee did they sell?
Solution:
Step 1: Define the variables. Let x be the number of regular coffees sold, and let y be the number of specialty coffees sold. Step 2: Set up the equations based on the given information. Total number of cups: x + y = 120 Total revenue: 2.5x + 3.5y = 350 Step 3: Solve the system of equations. We'll use the elimination method. Multiply the first equation by 2.5: 2.5x + 2.5y = 300 Subtract from the second equation: 3.5y - 2.5y = 350 - 300 This gives: y = 50 Step 4: Substitute back to find x. x + y = 120 x + 50 = 120 x = 70 So the coffee shop sold 70 regular coffees and 50 specialty coffees.
Example 2
Two cars leave towns A and B, which are 300 km apart, at the same time and travel toward each other. They meet after 2 hours. If the speed of the car from town A is 20 km/h faster than the car from town B, find the speed of each car.
Solution:
Step 1: Define the variables. Let x be the speed of the car from town A, and let y be the speed of the car from town B. Step 2: Set up the equations based on the given information. Relationship between speeds: x = y + 20 Distance relationship: 2x + 2y = 300 (since distance = speed × time) Step 3: Solve the system of equations. We'll use the substitution method. Substitute x = y + 20 into the second equation: 2(y + 20) + 2y = 300 2y + 40 + 2y = 300 4y + 40 = 300 4y = 260 y = 65 Step 4: Find x. x = y + 20 = 65 + 20 = 85 So the speed of the car from town A is 85 km/h, and the speed of the car from town B is 65 km/h.
Inconsistent and Dependent Systems
Not all systems of linear equations have a unique solution.
An inconsistent system has no solution. This occurs when the equations represent parallel lines that never intersect.
A dependent system has infinitely many solutions. This occurs when the equations represent the same line.
Example 1
Determine whether the following system of equations is consistent, inconsistent, or dependent: 2x + 3y = 6 4x + 6y = 10
Solution:
Step 1: Try to solve the system using the elimination method. Multiply the first equation by 2: 4x + 6y = 12 Subtract from the second equation: 4x + 6y - (4x + 6y) = 10 - 12 This gives: 0 = -2, which is a contradiction. Step 2: Interpret the result. Since we arrived at a contradiction, the system is inconsistent and has no solution. Geometrically, the equations represent two parallel lines with different y-intercepts, so they never intersect.
Example 2
Determine whether the following system of equations is consistent, inconsistent, or dependent: 3x - 2y = 7 6x - 4y = 14
Solution:
Step 1: Try to solve the system using the elimination method. Multiply the first equation by 2: 6x - 4y = 14 Subtract from the second equation: 6x - 4y - (6x - 4y) = 14 - 14 This gives: 0 = 0, which is always true. Step 2: Interpret the result. Since we arrived at an identity (0 = 0), the system is dependent and has infinitely many solutions. Geometrically, the equations represent the same line, so every point on that line is a solution. We can express the solution as y = (3x - 7)/2, where x can be any real number.
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