Equations in Maths

Ever tried to determine how much you have to spend to purchase two notebooks and a pen if you only know the sum of the costs? Or to determine how far away your friend is if your friend is running at some speed? In all of these cases, you are unconsciously applying equations! Equations are the foundation of mathematics and are used extensively in everyday situations. Here, we’ll explore everything from solutions to real-life examples of equations. So, let's begin!

1.0What is an Equation? 

An equation is a mathematical statement indicating that two things are equal to each other. It always contains an equal sign (=) between two sides. The two sides are:

• Left-Hand Side (LHS)
• Right-Hand Side (RHS)

When both sides have the same value, then it is said to be a valid equation. For example, take an equation, 6 + 4 = 10

In the equation, the left-hand side (6 + 4) equals the right-hand side (10), so it is a valid equation. 

2.0Algebraic Equations: A Brief

Equations become more interesting when they have single or multiple variables (such as x, y, or z), representing unknown figures. These are called algebraic equations. Algebraic equations, therefore, refer to equations with variables, constants, and arithmetic operations such as addition, subtraction, multiplication, and division. For example, take an equation: x + 3 = 7 

The example signifies the value that needs to be added to 3, such that it gives 7 as a result. 

3.0Different Types of Algebraic Equations

Algebraic equations can be formed in many ways depending on the variables, constants, powers, etc. Understanding different types of algebraic equations helps us understand the perfect solving technique needed for that equation. Let’s take a closer look at the three most common types of these equations: 

A. Linear Equations: 

A linear equation is an equation in which the highest power of the variables doesn’t exceed 1. It is named so due to its nature as a graph during its visual representation. It forms a straight line on a graph. The linear equations can have single or multiple variables, for example: 

x + 2y = 5

3x + 4 = 19

B. Quadratic Equations: 

Quadratic equations are identified by the highest power of the variable in the given equation being 2. It forms a U-shaped curve or parabola when graphed on the graph. Generally, quadratic equations contain a single variable with a standard form written as: 

$a{x}^2+bx+c=0$

Here, 

• a, b, and c are constants, where $a\ne 0$. 
• x is the single variable used in the equation. 

C. System of Equations: 

A system of equations is not a separate type of equation. However, it is a special case of the equations discussed above. When two or more equations (mostly linear) are solved together, it is known as a system of equations. The main aim of this is to find the unknown values of all the variables of all the equations involved in the system. It generally includes multiple variables in the equations. For example: 

x + y = 10

x – y = 2

Here, both equations are related to each other in some way. This relation helps solve these equations using different methods. 

4.0Solving Equations

Now that we understand the various forms of equations, let's study how to solve them. Every form of algebraic equation has its own way of being solved based on the level of complexity and the number of variables involved. Let's divide it by form:

A. Solving Linear Equations: 

• Simplify both sides (eliminate brackets, consolidate like terms)
• Shift constants to one side and variables to the other side of the equation.
• Isolate the variable with simple arithmetic operations.
• Test your solution by putting it back

B. Solving Quadratic Equations: 

Solving quadratic equations can be tricky. Therefore, different methods are used depending on the constants and numbers involved in the equation. These methods are: 

1. Factorisation: This method involves splitting the equation into two brackets, then taking the common terms out of these brackets and finally solving for x. 
2. Completing the Square: When the method of factorisation can not be performed due to the inability to form a perfect square equation. Hence, the main focus of this method involves converting the given equation into a perfect square form. 
3. Quadratic Formula: It is the direct formula to find the values of the unknown variable x. The formula for the equation $a{x}^2+bx+c=0$ is: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

5.0Solving a System of Equations

The system of equations is solved using the following methods: 

• Substitution Method: Solve one equation for one variable and then substitute it for another.
• Elimination Method: Add or subtract the equations to eliminate one variable.
• Graphing Technique: Graph both equations on a graph. Where they cross is the solution.

6.0Real-Life Examples of Equations: Solved Problems

Problem 1: Five years ago, A was three times as old as B. Ten years from now, A will be twice as old as B. What are their present ages?

Solution: Let the present age of A = x 

Let the present age of B = y 

According to the question,

Five years ago, 

A – 5 = 3(B – 5) 

A – 5 = 3B – 15 

A – 3B = – 10 …..(1)

In ten years from now, 

A + 10 = 2(B + 10) 

A + 10 = 2B + 20

A – 2B = 10 ….(2) 

Using the elimination method in equations 1 and 2. 

A – 3B = – 10

–(A – 2B = 10)

_____________

– B = –20

B = 20

Putting the value of B in either equation 1 or 2: 

A – 2(20) = 10

A = 10 + 40 = 50years 

Hence, the ages of A and B are 50 years and 20 years, respectively. 

Problem 2: A water tank is being filled by a pipe that adds 1/4th of the tank’s capacity every hour. However, due to a leakage, 1/6th of the tank’s capacity drains out every hour. If the tank was already 1/3rd full, how long will it take to fill the tank? 

Solution: Let the number of hours needed to fill the tank = x

According to the question, 

Water added per hour = ¼

Water lost per hour = ⅙ 

The net water gain per hour: $\frac{1}{4}-\frac{1}{6}=\frac{3-2}{12}=\frac{1}{12}$

As the tank was already ⅓ filled, hence, 

Water needed to be filled = 1 – ⅓ = ⅔ 

If 1/12 of the tank is filled every hour, and you need to fill ⅔, then:

$\frac{1}{12}\times x=\frac{2}{3}\Rightarrow x=\frac{12\times 2}{3}=8hours$

Problem 3: The length of a rectangular garden is 4 meters more than its width. If the area of the garden is 96 m², find its dimensions. 

Solution: Let the width of the garden = x 

Then the length of the garden = x + 4

Area of the garden = 96 m² 

Length × Width = 96

x × (x+4) = 96

x² + 4x – 96 = 0 

x² + 12x – 8x – 96 = 0 

x(x+12) – 8(x+12) = 0 

(x – 8) (x + 12) = 0

x = 8, –12 

Hence, the width of the garden = 8 m 

And the length of the garden = x + 4 = 12m