Solving Simultaneous Equations in Two Unknowns Using Elimination or Substitution

Introduction

Key Concepts

Understanding Simultaneous Equations

Simultaneous equations are sets of two or more equations containing multiple variables. Solving these equations involves finding the values of the variables that satisfy all equations simultaneously. In the case of two unknowns, the system typically consists of two linear equations. These can be represented as:

$$ \begin{align} a_1x + b_1y &= c_1 \\ a_2x + b_2y &= c_2 \end{align} $$

Where \(a_1, a_2, b_1, b_2\), and \(c_1, c_2\) are constants.

Graphical Interpretation

Graphically, each equation represents a straight line in the Cartesian plane. The solution to the system is the point of intersection of these two lines. If the lines intersect at exactly one point, there is a unique solution. If the lines are parallel, there is no solution, and if they coincide, there are infinitely many solutions.

Elimination Method

The elimination method involves adding or subtracting the equations to eliminate one of the variables, making it easier to solve for the remaining variable. The steps are as follows:

1. Align the equations: Ensure that the equations are written in standard form, with like terms aligned.
2. Multiply if necessary: If the coefficients of one variable are not the same or additive inverses, multiply one or both equations by suitable numbers to make them so.
3. Add or subtract the equations: This will eliminate one variable, resulting in a single equation with one unknown.
4. Solve for the remaining variable: Once one variable is found, substitute its value back into one of the original equations to find the other variable.

Example:

Solve the following system using elimination:

$$ \begin{align} 2x + 3y &= 8 \\ 4x - y &= 2 \end{align} $$

Solution:

Multiply the second equation by 3 to align the coefficients of \(y\):

$$ 4x - y = 2 \quad \Rightarrow \quad 12x - 3y = 6 $$

Add this to the first equation:

$$ 2x + 3y = 8 \\ 12x - 3y = 6 \\ \hline 14x = 14 \quad \Rightarrow \quad x = 1 $$>

Substitute \(x = 1\) into the first equation:

$$ 2(1) + 3y = 8 \quad \Rightarrow \quad 3y = 6 \quad \Rightarrow \quad y = 2 $$>

The solution is \(x = 1\) and \(y = 2\).

Substitution Method

The substitution method involves solving one of the equations for one variable and then substituting that expression into the other equation. The steps are as follows:

1. Solve one equation for one variable: Choose the equation that is easiest to solve and isolate one variable.
2. Substitute the expression into the other equation: Replace the isolated variable in the second equation with the expression obtained.
3. Solve for the remaining variable: This will yield the value of one variable.
4. Substitute back to find the other variable: Use the value obtained to find the value of the second variable.

Example:

Solve the following system using substitution:

$$ \begin{align} x + y &= 5 \\ 2x - y &= 1 \end{align} $$>

Solution:

From the first equation:

$$ x = 5 - y $$>

Substitute \(x = 5 - y\) into the second equation:

$$ 2(5 - y) - y = 1 \\ 10 - 2y - y = 1 \\ 10 - 3y = 1 \\ -3y = -9 \\ y = 3 $$>

Substitute \(y = 3\) back into \(x = 5 - y\):

$$ x = 5 - 3 = 2 $$>

The solution is \(x = 2\) and \(y = 3\).

Types of Solutions

Depending on the coefficients and constants in the equations, there are three possible types of solutions:

• Unique Solution: The system has exactly one solution where the two lines intersect.
• No Solution: The system is inconsistent; the lines are parallel and never intersect.
• Infinite Solutions: The system is dependent; the two equations represent the same line.

Determining the Type of Solution

To determine the type of solution without graphing, compare the ratios of the coefficients:

• If \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), the system has a unique solution.
• If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}\), the system has no solution.
• If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\), the system has infinitely many solutions.

Applications of Simultaneous Equations

Simultaneous equations are widely used in various fields such as economics, engineering, physics, and computer science. They help in solving real-world problems like determining the break-even point in business, analyzing electrical circuits, and modeling population growth.

Word Problem Example

Problem: A teacher has a total of 30 students in her class. If the number of boys is twice the number of girls, how many boys and girls are there?

Solution:

Let \(b\) represent the number of boys and \(g\) the number of girls.

We have:

$$ \begin{align} b + g &= 30 \\ b &= 2g \end{align} $$>

Substitute \(b = 2g\) into the first equation:

$$ 2g + g = 30 \\ 3g = 30 \\ g = 10 $$>

Then, \(b = 2(10) = 20\).

The class consists of 20 boys and 10 girls.

System of Equations with Fractions

Systems of equations may sometimes involve fractions, making the elimination or substitution process a bit more intricate. To simplify, it's often useful to eliminate denominators by multiplying all terms by the least common denominator (LCD).

Example:

Solve the following system:

$$ \begin{align} \frac{1}{2}x + \frac{1}{3}y &= 4 \\ \frac{2}{3}x - \frac{1}{4}y &= 1 \end{align} $$>

Solution:

First, eliminate the fractions by finding the LCD for each equation.

For the first equation, LCD of 2 and 3 is 6:

$$ 6 \left(\frac{1}{2}x + \frac{1}{3}y\right) = 6 \times 4 \\ 3x + 2y = 24 $$>

For the second equation, LCD of 3 and 4 is 12:

$$ 12 \left(\frac{2}{3}x - \frac{1}{4}y\right) = 12 \times 1 \\ 8x - 3y = 12 $$>

Now, solve the simplified system:

$$ \begin{align} 3x + 2y &= 24 \\ 8x - 3y &= 12 \end{align} $$>

Using the elimination method, multiply the first equation by 3 and the second by 2:

$$ \begin{align} 9x + 6y &= 72 \\ 16x - 6y &= 24 \end{align} $$>

Add the equations:

$$ 25x = 96 \quad \Rightarrow \quad x = \frac{96}{25} = 3.84 $$>

Substitute \(x = 3.84\) into the first simplified equation:

$$ 3(3.84) + 2y = 24 \\ 11.52 + 2y = 24 \\ 2y = 12.48 \quad \Rightarrow \quad y = 6.24 $$>

The solution is \(x = 3.84\) and \(y = 6.24\).

System of Equations with Negative Coefficients

Negative coefficients in systems of equations can be handled seamlessly using elimination or substitution methods. The key is to maintain consistency in signs while performing operations.

Example:

Solve the following system:

$$ \begin{align} -2x + 5y &= 3 \\ 3x + 2y &= 14 \end{align} $$>

Solution:

Using the elimination method, multiply the first equation by 3 and the second by 2 to align the coefficients of \(x\):

$$ \begin{align} -6x + 15y &= 9 \\ 6x + 4y &= 28 \end{align} $$>

Add the equations:

$$ 19y = 37 \quad \Rightarrow \quad y = \frac{37}{19} = 1.947 $$>

Substitute \(y = 1.947\) into the second original equation:

$$ 3x + 2(1.947) = 14 \\ 3x + 3.894 = 14 \\ 3x = 10.106 \quad \Rightarrow \quad x = \frac{10.106}{3} \approx 3.368 $$>

The solution is \(x \approx 3.368\) and \(y \approx 1.947\).

Advanced Concepts

Matrix Representation and Solutions

Simultaneous equations can be represented and solved using matrices, which streamlines the process, especially for larger systems. A system of two equations can be expressed in matrix form as:

$$ \begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} c_1 \\ c_2 \end{pmatrix} $$>

The solution can be found using the inverse of the coefficient matrix, provided it exists:

$$ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \end{pmatrix}^{-1} \begin{pmatrix} c_1 \\ c_2 \end{pmatrix} $$>

The inverse of a 2x2 matrix is given by:

$$ \begin{pmatrix} a & b \\ c & d \end{pmatrix}^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} $$>

Example:

Find the solution using matrices:

$$ \begin{align} x + 2y &= 5 \\ 3x - y &= 4 \end{align} $$>

Solution:

Matrix form:

$$ A = \begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 5 \\ 4 \end{pmatrix} $$>

Find the inverse of \(A\):

$$ \text{det}(A) = (1)(-1) - (3)(2) = -1 -6 = -7 \\ A^{-1} = \frac{1}{-7} \begin{pmatrix} -1 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & -\frac{1}{7} \end{pmatrix} $$>

Multiply \(A^{-1}\) by \(\mathbf{b}\):

$$ \begin{pmatrix} \frac{1}{7} & \frac{2}{7} \\ \frac{3}{7} & -\frac{1}{7} \end{pmatrix} \begin{pmatrix} 5 \\ 4 \end{pmatrix} = \begin{pmatrix} \frac{1 \times 5 + 2 \times 4}{7} \\ \frac{3 \times 5 - 1 \times 4}{7} \end{pmatrix} = \begin{pmatrix} \frac{13}{7} \\ \frac{11}{7} \end{pmatrix} $$>

The solution is \(x = \frac{13}{7} \approx 1.857\) and \(y = \frac{11}{7} \approx 1.571\).

Cramer's Rule

Cramer's Rule provides an explicit formula for the solution of a system of linear equations with as many equations as unknowns, using determinants. For a system of two equations:

$$ \begin{align} a_1x + b_1y &= c_1 \\ a_2x + b_2y &= c_2 \end{align} $$>

The solutions are given by:

$$ x = \frac{D_x}{D}, \quad y = \frac{D_y}{D} $$>

Where:

$$ D = \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = a_1b_2 - a_2b_1 $$> $$ D_x = \begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix} = c_1b_2 - c_2b_1 $$> $$ D_y = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} = a_1c_2 - a_2c_1 $$>

Example:

Find the solution using Cramer's Rule:

$$ \begin{align} 2x + 3y &= 8 \\ 4x - y &= 2 \end{align} $$>

Solution:

Calculate determinants:

$$ D = \begin{vmatrix} 2 & 3 \\ 4 & -1 \end{vmatrix} = (2)(-1) - (4)(3) = -2 -12 = -14 $$> $$ D_x = \begin{vmatrix} 8 & 3 \\ 2 & -1 \end{vmatrix} = (8)(-1) - (2)(3) = -8 -6 = -14 $$> $$ D_y = \begin{vmatrix} 2 & 8 \\ 4 & 2 \end{vmatrix} = (2)(2) - (4)(8) = 4 -32 = -28 $$>

Thus:

$$ x = \frac{-14}{-14} = 1, \quad y = \frac{-28}{-14} = 2 $$>

The solution is \(x = 1\) and \(y = 2\).

Nonlinear Simultaneous Equations

While the focus is primarily on linear equations, systems involving nonlinear equations, such as quadratic or exponential equations, also exist. Solving these systems may require combining elimination or substitution with methods like factoring, completing the square, or numerical approximation.

Example:

Solve the following system:

$$ \begin{align} x + y &= 10 \\ x^2 + y^2 &= 58 \end{align} $$>

Solution:

From the first equation:

$$ y = 10 - x $$>

Substitute into the second equation:

$$ x^2 + (10 - x)^2 = 58 \\ x^2 + 100 - 20x + x^2 = 58 \\ 2x^2 - 20x + 100 -58 = 0 \\ 2x^2 -20x +42 = 0 \\ x^2 -10x +21 = 0 $$>

Solve the quadratic equation:

$$ x = \frac{10 \pm \sqrt{100 -84}}{2} = \frac{10 \pm \sqrt{16}}{2} = \frac{10 \pm 4}{2} $$>

Thus, \(x = 7\) or \(x = 3\).

If \(x = 7\), then \(y = 3\). If \(x = 3\), then \(y = 7\).

The solutions are \((7, 3)\) and \((3, 7)\).

Parametric Representation

In some cases, systems are represented using parameters, leading to parametric equations. Solving such systems often involves expressing variables in terms of the parameter and interpreting the solutions in a geometric context.

Example:

Find the solutions for the system in terms of parameter \(t\):

$$ \begin{align} x + 2y &= 5 \\ 3x - y &= 4 \end{align} $$>

Solution:

Let \(y = t\). Then from the first equation:

$$ x = 5 - 2t $$>

Substitute into the second equation:

$$ 3(5 - 2t) - t = 4 \\ 15 -6t - t = 4 \\ -7t = -11 \\ t = \frac{11}{7} $$>

Thus, \(y = \frac{11}{7}\) and \(x = 5 - 2\left(\frac{11}{7}\right) = \frac{35}{7} - \frac{22}{7} = \frac{13}{7}\).

The solution is \(x = \frac{13}{7}\) and \(y = \frac{11}{7}\).

Applications in Real Life

Advanced applications of simultaneous equations include optimization problems, such as minimizing costs or maximizing profits, and in determining the equilibrium in supply and demand models. They are also essential in computer algorithms that underpin various technologies like cryptography and graphics rendering.

Interdisciplinary Connections

Simultaneous equations intersect with disciplines like physics for solving motion problems, economics for market equilibrium analysis, and engineering for circuit design and structural analysis. Understanding these connections enhances the applicability and relevance of mathematical concepts in various fields.

Iterative Methods

For larger systems or more complex equations, iterative methods like the Gauss-Seidel or Jacobi method are employed. These techniques approximate solutions through successive iterations and are fundamental in computational mathematics and numerical analysis.

Determinants and Their Properties

Beyond Cramer's Rule, determinants play a significant role in understanding the properties of matrix systems, such as invertibility and the nature of solutions. They are pivotal in linear algebra and have applications in areas like differential equations and quantum mechanics.

Homogeneous Systems

A homogeneous system is one in which all constant terms are zero. Such systems always have at least one solution, known as the trivial solution. Exploring non-trivial solutions involves deeper insights into linear dependence and vector spaces.

Example:

Consider the homogeneous system:

$$ \begin{align} 2x + 3y &= 0 \\ 4x - y &= 0 \end{align} $$>

Solution:

Using elimination:

Multiply the second equation by 3:

$$ 4x - y = 0 \quad \Rightarrow \quad 12x - 3y = 0 $$>

Add to the first equation:

$$ 2x + 3y = 0 \\ 12x - 3y = 0 \\ \hline 14x = 0 \quad \Rightarrow \quad x = 0 $$>

Substitute \(x = 0\) into the second equation:

$$ 4(0) - y = 0 \quad \Rightarrow \quad y = 0 $$>

The only solution is the trivial one: \(x = 0\) and \(y = 0\).

Comparison Table

Method	Description	Advantages	Disadvantages
Elimination	Eliminates one variable by adding or subtracting equations.	Efficient for systems where variables can be easily eliminated.	Can become cumbersome with fractions or complex coefficients.
Substitution	Solves one equation for one variable and substitutes into the other.	Simple and intuitive, especially when one equation is already solved.	Less efficient for larger systems, can involve complex algebra.
Matrix Methods	Uses matrices and determinants to find solutions.	Systematic and easily extendable to larger systems.	Requires understanding of matrix operations and determinants.
Cramer's Rule	Uses determinants to solve for each variable.	Provides explicit solutions without iteration.	Not practical for systems with more than three variables.

Summary and Key Takeaways