Solving Simultaneous Equations

Introduction

Key Concepts

Understanding Simultaneous Equations

Simultaneous equations are a set of two or more equations with the same set of variables. The solution to these equations is the set of variable values that satisfy all equations simultaneously. In the context of the Cambridge IGCSE curriculum, students encounter both linear and non-linear simultaneous equations.

Types of Simultaneous Equations

There are primarily two types of simultaneous equations:

• Linear Simultaneous Equations: Equations where the variables are to the first power. For example: $$\begin{align*} 2x + 3y &= 6 \\ x - y &= 2 \end{align*}$$
• Non-linear Simultaneous Equations: Equations where at least one equation is non-linear (e.g., quadratic, exponential). For example: $$\begin{align*} x^2 + y^2 &= 25 \\ x - y &= 1 \end{align*}$$

Methods of Solving Simultaneous Equations

There are several methods to solve simultaneous equations, each with its advantages depending on the nature of the equations involved. The primary methods include:

1. Substitution Method: Involves solving one equation for one variable and substituting this into the other equation(s).
2. Elimination Method: Involves adding or subtracting equations to eliminate one variable, making it easier to solve for the remaining variable.
3. Graphical Method: Involves graphing the equations on the coordinate plane and finding the point(s) of intersection.
4. Matrix Method: Uses matrices and determinants to solve systems of equations, suitable for larger systems.

Substitution Method

The substitution method is particularly useful when one of the equations can be easily solved for one variable. The steps are:

1. Choose one equation and solve for one variable in terms of the others.
2. Substitute this expression into the other equation(s).
3. Solve the resulting equation for the remaining variable.
4. Substitute back to find the value of the other variable.

Example: $$\begin{align*} x + y &= 5 \quad \text{(1)} \\ 2x - y &= 3 \quad \text{(2)} \end{align*}$$ Solving equation (1) for y: $$y = 5 - x$$ Substituting into equation (2): $$2x - (5 - x) = 3 \\ 2x - 5 + x = 3 \\ 3x = 8 \\ x = \frac{8}{3}$$ Substituting back: $$y = 5 - \frac{8}{3} = \frac{7}{3}$$ Thus, the solution is $$x = \frac{8}{3}, \quad y = \frac{7}{3}$$

Elimination Method

The elimination method is effective when the coefficients of one of the variables can be made equal (or opposites) by multiplying the equations by suitable numbers. The steps are:

1. Arrange the equations in standard form: Ax + By = C.
2. Multiply one or both equations by suitable numbers to get the coefficients of one variable to be the same or additive inverses.
3. Add or subtract the equations to eliminate one variable.
4. Solve for the remaining variable.
5. Substitute back to find the other variable.

Example: $$\begin{align*} 3x + 4y &= 10 \quad \text{(1)} \\ 2x - 5y &= -3 \quad \text{(2)} \end{align*}$$ To eliminate y, multiply equation (1) by 5 and equation (2) by 4: $$\begin{align*} 15x + 20y &= 50 \quad \text{(3)} \\ 8x - 20y &= -12 \quad \text{(4)} \end{align*}$$ Adding (3) and (4): $$23x = 38 \\ x = \frac{38}{23} = \frac{38}{23}$$ Substituting back into equation (1): $$3\left(\frac{38}{23}\right) + 4y = 10 \\ \frac{114}{23} + 4y = 10 \\ 4y = 10 - \frac{114}{23} = \frac{230 - 114}{23} = \frac{116}{23} \\ y = \frac{29}{23}$$ Thus, the solution is $$x = \frac{38}{23}, \quad y = \frac{29}{23}$$

Graphical Method

The graphical method involves plotting each equation on the coordinate plane and identifying the point where the lines intersect. This point represents the solution to the system. While this method provides a visual understanding, it is less precise unless using graphing software or tools.

Example: Consider the system: $$\begin{align*} y &= 2x + 1 \quad \text{(1)} \\ y &= -x + 4 \quad \text{(2)} \end{align*}$$ Plotting both equations on the graph, the lines intersect at the point $$\left(1, 3\right)$$. Therefore, the solution is $$x = 1, \quad y = 3$$.

Matrix Method

The matrix method is a powerful technique for solving systems of equations, especially useful for larger systems. It involves representing the system as a matrix and using operations to find the inverse matrix or using determinants (Cramer's Rule). Example: Solve the system: $$\begin{align*} x + 2y &= 5 \\ 3x - y &= 4 \end{align*}$$ Representing as a matrix: $$\begin{bmatrix} 1 & 2 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 5 \\ 4 \end{bmatrix}$$ Using the inverse matrix method: $$\text{Inverse} = \frac{1}{(1)(-1) - (3)(2)} \begin{bmatrix} -1 & -2 \\ -3 & 1 \end{bmatrix} = \frac{1}{-7} \begin{bmatrix} -1 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & -\frac{1}{7} \end{bmatrix}$$ Multiplying by the constants matrix: $$\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & -\frac{1}{7} \end{bmatrix} \begin{bmatrix} 5 \\ 4 \end{bmatrix} = \begin{bmatrix} \frac{1}{7}(5) + \frac{2}{7}(4) \\ \frac{3}{7}(5) - \frac{1}{7}(4) \end{bmatrix} = \begin{bmatrix} \frac{5 + 8}{7} \\ \frac{15 - 4}{7} \end{bmatrix} = \begin{bmatrix} \frac{13}{7} \\ \frac{11}{7} \end{bmatrix}$$ Thus, $$x = \frac{13}{7}, \quad y = \frac{11}{7}$$.

Applications of Simultaneous Equations

Simultaneous equations are widely used to solve real-life problems involving multiple unknowns. Some common applications include:

• Financial Planning: Determining the cost and quantity of products in budgeting.
• Engineering: Solving for stresses and forces in structures.
• Economics: Modeling supply and demand equilibrium.
• Chemistry: Balancing chemical equations.

Example Problem

Problem: A school is planning a trip and needs to rent buses and vans. Each bus can carry 50 students and costs $300 to rent. Each van can carry 12 students and costs $100 to rent. If the school has 312 students and the total cost for renting vehicles is $1,200, how many buses and vans should the school rent? Solution: Let the number of buses be $$b$$ and the number of vans be $$v$$. We have the following system of equations: $$\begin{align*} 50b + 12v &= 312 \quad \text{(1)} \\ 300b + 100v &= 1200 \quad \text{(2)} \end{align*}$$ Simplifying equation (2) by dividing by 100: $$3b + v = 12 \quad \text{(3)}$$ From (3): $$v = 12 - 3b$$ Substituting into equation (1): $$50b + 12(12 - 3b) = 312 \\ 50b + 144 - 36b = 312 \\ 14b = 168 \\ b = 12$$ Substituting back: $$v = 12 - 3(12) = 12 - 36 = -24$$ Since the number of vans cannot be negative, there must be an error in the initial assumptions or the problem must be adjusted for feasibility. This indicates that with the given constraints, it's not possible to accommodate all students within the budget.

Advanced Concepts

Non-linear Simultaneous Equations

While linear simultaneous equations involve variables to the first power, non-linear equations include higher powers or other functions of variables. Solving non-linear systems often requires more sophisticated techniques, such as substitution combined with factoring, using the quadratic formula, or graphical methods.

Solving Quadratic and Linear Systems

Consider a system where one equation is linear and the other is quadratic: $$\begin{align*} y &= 2x + 3 \quad \text{(1)} \\ y &= x^2 + x + 1 \quad \text{(2)} \end{align*}$$ To find the intersection points, set equation (1) equal to equation (2): $$2x + 3 = x^2 + x + 1 \\ x^2 - x - 2 = 0$$ Solving the quadratic equation: $$x = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2} \\ x = 2 \quad \text{or} \quad x = -1$$ Substituting back into equation (1): For $$x = 2$$: $$y = 2(2) + 3 = 7$$ For $$x = -1$$: $$y = 2(-1) + 3 = 1$$ Thus, the solutions are $$(2, 7)$$ and $$(-1, 1)$$.

Systems with Three Variables

Extending simultaneous equations to three variables increases complexity. Methods such as substitution and elimination can still be applied, but they require careful manipulation. Alternatively, matrix methods (like Gaussian elimination) are more efficient for larger systems.

Example: Solve the system: $$\begin{align*} x + y + z &= 6 \quad \text{(1)} \\ 2x - y + 3z &= 14 \quad \text{(2)} \\ -x + 2y - z &= -2 \quad \text{(3)} \end{align*}$$ Using elimination: 1. From (1): $$x = 6 - y - z$$ 2. Substitute into (2): $$2(6 - y - z) - y + 3z = 14 \\ 12 - 2y - 2z - y + 3z = 14 \\ 12 - 3y + z = 14 \\ -3y + z = 2 \quad \text{(4)}$$ 3. Substitute into (3): $$-(6 - y - z) + 2y - z = -2 \\ -6 + y + z + 2y - z = -2 \\ 3y - 6 = -2 \\ 3y = 4 \\ y = \frac{4}{3}$$ 4. Substitute y into (4): $$-3\left(\frac{4}{3}\right) + z = 2 \\ -4 + z = 2 \\ z = 6$$ 5. Substitute y and z into (1): $$x + \frac{4}{3} + 6 = 6 \\ x + \frac{22}{3} = 6 \\ x = 6 - \frac{22}{3} = \frac{-4}{3}$$ Thus, the solution is $$x = \frac{-4}{3}, \quad y = \frac{4}{3}, \quad z = 6$$.

Parametric and Vector Methods

For systems with infinitely many solutions, such as dependent equations, parametric methods introduce parameters to express the solutions. Vector methods represent the system in matrix form and use vector operations to find solutions.

Example: Solve the system: $$\begin{align*} x + y &= 2 \quad \text{(1)} \\ 2x + 2y &= 4 \quad \text{(2)} \end{align*}$$ Observation: Equation (2) is a multiple of equation (1), indicating infinitely many solutions. Expressing in parametric form: Let $$x = t$$, then from (1): $$y = 2 - t$$ Thus, the solution set is $$\{(t, 2 - t) \mid t \in \mathbb{R}\}$$.

Applications in Different Fields

Simultaneous equations extend beyond pure mathematics into various disciplines:

• Physics: Modeling motion where multiple forces act simultaneously.
• Economics: Determining equilibrium prices in supply and demand models.
• Biology: Analyzing population dynamics with interacting species.
• Engineering: Designing electrical circuits involving multiple components.

Understanding the broader applications enhances the relevance and importance of mastering simultaneous equations.

Solving with Technology

Modern technology, including graphing calculators and computer algebra systems, can solve simultaneous equations efficiently. These tools are invaluable for verifying manual calculations and handling complex systems that are impractical to solve by hand.

Example: Using a graphing calculator to solve: $$\begin{align*} 3x + y &= 9 \\ x - 2y &= -4 \end{align*}$$ Entering the equations into the calculator and using the system solver function yields: $$x = 3, \quad y = 0$$.

Understanding Solutions: Unique, No Solution, Infinitely Many Solutions

A system of simultaneous equations can have:

• Unique Solution: One set of values satisfies all equations.
• No Solution: Equations are contradictory, with no common solution.
• Infinitely Many Solutions: Equations are dependent, representing the same line or plane.

Example of No Solution: $$\begin{align*} x + y &= 2 \\ x + y &= 5 \end{align*}$$ These equations represent parallel lines with no intersection.

Example of Infinitely Many Solutions: $$\begin{align*} 2x + 4y &= 8 \\ x + 2y &= 4 \end{align*}$$ The second equation is a multiple of the first, indicating the same line.

Comparison Table

Summary and Key Takeaways