Application of Equations: Definition, Types, Examples, Uses

Application of Equations: A mathematical statement in which two expressions on both the left and right sides are equal is an equation. Algebra makes it easier to solve real-world situations by utilising letters to represent unknowns, reframing issues as equations, and providing systematic solutions to those equations. 

To use algebra to solve a problem, first convert the problem’s language into mathematical statements that define the connections between the provided data and the unknowns. Usually, the most challenging part of the procedure is translating the information into mathematical statements. The applications of equations will be discussed in-depth in this article.

What are Equations?

A mathematical statement containing an ‘equal to’ sign between two algebraic expressions with equal values is known as an equation.

For example, \(2 y+4\) is the expression on the left-hand side that is equated with the expression \(24\) on the right-hand side.

Take a look at the examples below. These will help you understand what an equation in math means.

Types of Equations

Depending on the degree of equations, the equations are mainly classified as 

1. Linear Equation
2. Quadratic Equation
3. Cubic Equation

Linear Equation

A linear equation with the variables \(x\) and \(y\) has the following standard form:

\(Ax+By=C\)

Quadratic Equation

The standard form of a quadratic equation with variable \(x\) is:

\(a x^{2}+b x+c=0\), where \(a \neq 0\)

Cubic Equation

The standard form of a cubic equation with variable \(x\) is:

\(a x^{3}+b x^{2}+c x+d=0\), where \(a \neq 0\)

Applications of Linear Equations

There are numerous real-world examples of linear equations. These issues are translated into mathematical equations, which are then solved using several techniques. The relationship between the data and the unknowns (variables) in the situation must be stated clearly.

Some of the most popular applications of linear equations in real life include the following:

1. Age problems
2. Speed, time, and distance problems
3. Geometry problems
4. Money and percentage problems
5. Wages and hourly rate problems
6. Force and pressure problems 

Key Words, Translation, and Strategy

An equation is a mathematical statement in which two expressions on both the left and right sides are equal. Algebra makes it easier to solve problems in the real world. This is done by representing unknowns with letters, reframing problems as equations, and offering systematic answers to those equations. The most challenging aspect of the operation is usually converting the data into mathematical statements. The key to a good translation is to understand the problem and identify relevant words and phrases thoroughly.

Here are some examples of translated key phrases.

Example: Rashi has a total of \(₹590\) as currency notes in the denominations of \(₹50, ₹20\), and \(₹10\). The ratio of the number of \(₹50\) notes and \(₹20\) notes is \(3:5\). If she has a total of \(25\) notes, how many notes of each denomination she has?

Let the number of \(₹50\) notes and \(₹20\) notes are \(3 x\) and \(5 x\), respectively.

But she has \(25\) notes in total.

Therefore, the number of \(₹10\) notes \(=25-(3 x+5 x)=25-8 x\)

The amount she has from \(₹50\) notes \(=3 x \times 50=₹150 x\)

from \(₹20\) notes \(=5 x \times 20=₹100 x\)

from \(₹10\) notes \(=(25-8 x) \times 10=₹(250-80 x)\)

Hence the total money she has \(=₹150 x+₹100 x+₹(250-80 x)=₹(170 x+250)\)

But she has \(₹590\).

Therefore, \(170 x+250=590\)

\(170 x=590-250=340\)

\(x=\frac{340}{170}=2\)

The number of \(₹50\) notes she has \(=3 x=3 \times 2=6\)

The number of \(₹20\) notes she has \(=5 x=5 \times 2=10\)

The number of \(₹10\) notes she has \(=25-8 x=25-(8 \times 2)=25-16=9\)

Applications of Quadratic Equations

In everyday life, we use quadratic formulas to calculate areas, determine the profit of a product, and calculate the speed of an item. Furthermore, a quadratic equation contains at least one squared variable. It refers to a problem that involves multiplying a variable by itself, often known as squares. Furthermore, the size of a square equals its side length multiplied by itself, which is where this terminology comes.

Example: The product of two even consecutive positive integers is \(120\). Find the numbers.

Let the two even integers be \(x, x+2\).

Now, according to the statement, \(x(x+2)=120 \Rightarrow x^{2}+2 x-120=0\)
\(\Rightarrow x=\frac{-2 \pm \sqrt{4+4 \times 120}}{2} \Rightarrow x=\frac{-2 \pm \sqrt{4+480}}{2} \Rightarrow x=\frac{-2 \pm \sqrt{484}}{2} \Rightarrow x=\frac{-2 \pm 22}{2}\)

So, \(x=10\) or, \(x=-12\)
Here, we need to find the positive even integers. So, avoid the negative value of \(x\).
Hence, the two even consecutive integers are \(x=10\) and \(x+2=10+2=12\).

Example: If the speed of a bike is increased by \(10 \mathrm{~km} / \mathrm{hr}\), the time of journey for a distance of \(72 \mathrm{~km}\) is reduced by \(36\) minutes. Find the initial speed of the bike.

Let the initial speed of the bike is \(x \,\mathrm{kmph}\).
Speed of the bike after increasing the speed is \((x+10) \,\mathrm{kmph}\).
Distance is \(72 \mathrm{~km}\) (given).

Time taken to cover the distance is \(\left(\frac{72}{x}\right) \,\text {hours}\)
Time taken after increasing the speed is \(\left(\frac{72}{x+10}\right) \,\text {hours}\)

According to the question, \(\left(\frac{72}{x}\right)-\left(\frac{72}{x+10}\right)=\frac{36}{60}\)

\(\Rightarrow\left(\frac{1}{x}\right)-\left(\frac{1}{x+10}\right)=\frac{36}{60 \times 72} \Rightarrow\left[\frac{x+10-x}{x(x+10)}\right]=\frac{1}{120}\)

\(\Rightarrow\left[\frac{10}{x(x+10)}\right]=\frac{1}{120}\)

\(\Rightarrow\left[\frac{1}{x(x+10)}\right]=\frac{1}{1200}\)

\(\Rightarrow x(x+10)=1200 \Rightarrow x^{2}+10 x-1200=0 \Rightarrow x^{2}+40 x-30 x-1200=0\)

\(\Rightarrow x(x+40)-30(x-40)=0 \Rightarrow(x+40)(x-30)=0 \Rightarrow x=-40\) or, \(x=30\) Since the speed can not
be negative so, the initial speed is \(30 \,\mathrm{kmph}\).

Applications of Cubic Equations 

Some problems involving the cubic equations are

1. Making box 
2. Height of water in a spherical tank 
3. The smallest distance from a parabola 
4. Pumping water out of a tank 
5. Equation of state for real gases 
6. Electrical resistance 
7. Finding interest rate 
8. Break-even points in economics 
9. Trisecting the angle

Solved Examples – Application of Equations

Q.1. There is a hall whose length is five times the width. The area of the floor is \(45 \mathrm{~m}^{2}\). Find the length and width of the hall.
Ans: Let us suppose that \(l\) is the length of the hall and \(w\) is the width of the hall.
Given that the length of the hall is five times the width
So, \(l=5 w\)
Then,
Area of hall \(=\) length \(\times\) width
\(\Rightarrow l \times w=45\)
\(\Rightarrow 5 w \times w=45\)
\(\Rightarrow 5 w^{2}=45\)
\(\Rightarrow w^{2}=9\)
\(\Rightarrow w=\sqrt{9}\)
\(\Rightarrow w=\pm 3\)
The width cannot be negative.
Therefore, the width is \(3\,{\rm{m}},\) and the length is \(5 w=5 \times 3=15 \mathrm{~m}\).

Q.2. The three sides of a right-angled triangle are \(x, x+1\) and \(5\). Find \(x\) and the area if the longest side is \(5\).
Ans: The longest side will be the hypotenuse. Therefore, we can write:
\(x^{2}+(x+1)^{2}=5^{2}\) (Pythagoras’ Theorem)
\(x^{2}+x^{2}+2 x+1=252 x^{2}+2 x+1=25\)
Hence, \(x^{2}+x-12=0\)
\((x-3)(x+4)=0\)
\((x+4)=0\) or \((x-3)=0\)
\(x=-4\) or \(x=3\)
We can only take \(x=3\) here because the length can’t be negative.
Hence, \(x=3\) and therefore,
Area \(=\frac{1}{2} \times 3 \times 4=6\) square units

Q.3. The difference between two whole numbers is \(110\). The ratio of the two numbers is \(2: 7\). What are the two numbers?
Ans: Given, the ratio of the two numbers is \(2:7\), let the one number be \(2 x\) and the other be \(7 x\).
The difference between the two numbers \(=(7 x-2 x)\)
It is given that the difference is \(110\).
Therefore, \(7 x-2 x=110\)
\(5 x=110\)
\(x=22\)
Hence, the required numbers are \(2 x=2 \times 22=44\) and \(7 x=7 \times 22=154\)

Q.4. Kiran is thrice as old as Anu. \(12\) years later, her age will be twice of Anu. Find their present ages.
Ans: In this word problem, the ages of Kiran and Anu are unknown quantities.
Let Anu’s present age \(=x\) years.
Since Kiran’s present age is \(3\) times that of Anu, therefore her present age \(=3 x\)
\(12\) years later, Anu’s age will be \(=x+12\)
And Simran’s age will be \(=3 x+12\)
According to the given problem statement, \(12\) years later, Kiran’s age will be twice of Anu, i.e., \(3 x+12=2(x+12)\)
\(3 x+12=2 x+24\)
\(x=12\)
Hence, the present age of Anu is \(12\) years, and Kiran’s age is \(3 x=3 \times 12=36\) years.

Q.5. If one side of a triangle is half the perimeter, the second side is one-third the perimeter, and the third side is \(10\) meters, what is the perimeter of the triangle?
Ans: Let \(p\) the perimeter. Then
First side \(=\frac{p}{2}\)
Second side \(=\frac{p}{3}\)
And the third side \(=10 \mathrm{~m}\)
Therefore according to the problem statement
Perimeter \(p=\frac{p}{2}+\frac{p}{3}+10\)
\(\Rightarrow 6 p=3 p+2 p+60\)
\(\Rightarrow 6 p-5 p=60\)
\(\Rightarrow p=60\)
Hence, the perimeter is \(60 \mathrm{~m}\).

Summary

We have covered what an equation is in arithmetic and the many forms of equations in this article. Linear equations, keywords, translation, and strategy were also included. We also spoke about how to use linear and quadratic equations and solved examples and commonly asked problems.

Learn About Solution of an Equation

Frequently Asked Questions – Application of Equations

Q.1. What is the purpose of an equation in math?
Ans: An equation is a mathematical statement stating the equivalence of two expressions linked by the equality symbol “\(=\)”.

Q.2. What is the application of the quadratic equation?
Ans: In everyday life, we use quadratic formulas to calculate areas, determine the profit of a product, and calculate the speed of an item.

Q.3. What are the ways to solve systems of equations?
Ans: Substitution, elimination, and graphing are the three methods for solving systems of linear equations.

Q.4. What is the application of equations?
Ans: Develop problem-solving methods through identifying essential words and phrases, converting sentences to mathematical equations, and identifying essential words and phrases. Identify and solve word issues using numerical connections. Work out perimeter-related geometry issues.

Q.5. What makes a problem quadratic?
Ans: It refers to an issue that involves multiplying a variable by itself, often known as squares. Furthermore, the size of a square equals its side length multiplied by itself, which is where this terminology comes from.