Application of Linear Equations – Definition & Examples

Applications of Linear 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. This is done by utilising letters to represent unknowns reframing the 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 linear equations will be discussed in-depth in this article.

What is a Linear Equation?

When two algebraic expressions are equated to each other using an equal \”(=)\” sign, it is called an equation. A linear equation has a degree of one. That is, the maximum power of the variable in the equation is one.

Linear Equation in One Variable

A linear equation in one variable is a set of equations with a degree one and contains only one variable.

For example: \(4 x=12\)
\(19 x+45=68\)

Linear Equation in Two Variables

A linear equation in two variables is a set of equations with degree one and two variables.

For example: \(4 x+5 y=15\)
\(8 x=17-2 y\)

Applications of Linear Equations

There are several examples of linear equations in the real world. These real problems are transformed into mathematical equations, which are subsequently solved through a variety of methods. The link between the data and the unknowns (variables) in the situation should be clearly explained.

Calculations of the following are some of the most common applications of linear equations in real life:

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

Key Words, Translation, and Strategy

The key to successful translation is to read the problem carefully and find essential words and phrases. Below are some keywords and their translation in mathematical form:

Here are some examples of translated key phrases.

When converting a sentence to a mathematical statement, make sure to read it multiple times and find the essential words and phrases.

Example: Translate: Four less than twice some number is \(16\).
Solution: First, pick a variable for the unknown number and write down the essential words and phrases.

Let \(x\) stand for the unknown represented by the phrase “some number.”

Keep in mind that subtraction is not a commutative operation. As a result, be cautious while making differences. \(4-2 x=16\) is an incorrect translation in this case.

Answer: \(2 x-4=16\)

It’s essential first to identify the variable \(-\) let \(x\) stand for it \(-\) and describe the unknown number in words. This step not only improves the readability of your work but also pushes you to consider what you’re searching for. Most of the time, if you know what you’re looking for, you’ll be able to find it.

Example: Translate: When we subtract \(7\) from \(3\) times the sum of a number and \(12\), the result is \(20\).
Solution: Let \(n\) be the unknown number.

Answer: \(3(n+12)-7=20\)

Examine the structures of the following two phrases, as well as their translations, to see why parenthesis are required:

The trick is to concentrate on the words “three times the sum.” This prompts us to put the sum in parenthesis and multiply it by three. Once you’ve converted an application into an algebraic equation, use the techniques you’ve learnt to solve it.

Guidelines for Setting Up and Solving Word Problems

Step 1: Read the problem many times to find essential words and phrases, then organise the data.
Step 2: Assign the unknown quantities a letter or an expression to identify the variables.
Step 3: Translate the problem into an algebraic equation and set it up.
Step 4: Solve the resulting algebraic equation.
Step 5: Finally, answer the question in sentence form and make sense (check it).

Problems Involving Relationships Between Real Numbers

We classify applications involving relationships between real numbers broadly as number problems. These problems can sometimes be solved using some creative arithmetic, guessing, and checking. Begin by working through the basic steps outlined in the general guidelines for solving word problems.

Example: A larger integer is \(2\) less than \(3\) times a smaller integer. The sum of the two integers is \(18\). Find the integers.
Solution: Identify the variables: Start by assigning a variable to the smaller integer.
Let \(x\) represents the smaller integer.
Use the first sentence to identify the larger integer in terms of the variable \(x\) : “A larger integer is \(2\) less than \(3\) times a smaller.”
Let \(3 x-2\) represents the larger integer.

Set up an equation: Add the expressions representing the two integers, and set the resulting expression equal to \(18\) as indicated in the second sentence: “The sum of the two integers is \(18\).”
\(x+(3 x-2)=18\)

Solution: Solve the equation to obtain the value of smaller integer \(x\).
\(x+(3 x-2)=18\)
\(x+3 x-2=18\)
\(4 x-2=18\)
\(4 x-2+2=18+2\)
\(4 x=20\)
\(\frac{4 x}{4}=\frac{20}{4}\)
\(x=5\)

Back substitute: Use the expression \(3x-2\) to find the larger integer—this is called back substitution.
\(3 x-2=3(5)-2=15-2=13\)

Answer the question: Hence, the two integers are \(5\) and \(13\).

Verify: \(5+13=18\). The answer makes sense.

Solved Examples – Applications of Linear Equations

Q.1. A train moving nonstop to its destination may move at a speed of \(72\) miles per hour on average. The train makes several stops on the way back and can only go at a speed of \(48\) miles per hour. What is the travel time if the return journey takes \(2\) hours longer than the original trip to the destination?
Ans: First of all, identify the unknown quantity and organise the data.
Let \(t\) represent the time it takes to arrive at the destination.
Let \(t+2\) represent the time it takes for the return trip.

Distance travelled to the destination: \(D=\) speed \(\times\) time \(=72 \cdot t\)
Distance travelled on the return trip: \(D=\) speed \(\times\) time \(=48 \cdot(t+2)\)
In this case, the distance travelled to the destination and back is the same, and the equation is
\(72 t=48(t+2)\)
Solve for \(t\)
\(72 t=48(t+2)\)
\(72 t=48 t+96\)
\(72 t-48 t=48 t+96-48 t\)
\(24 t=96\)
\(\frac{24 t}{24}=\frac{96}{24}\)
\(t=4\)
The return trip takes \(t+2=4+2=6 \,\text {hours}\)
Hence, it takes \(4 \,\text {hours}\) to arrive at the destination and \(6 \,\text {hours}\) to return.

Q.2. If one side of a triangle is one-third the perimeter, the second side is one-fifth the perimeter, and the third side is \(7\) meters, what is the perimeter of the triangle?
Ans: Let \(p\) the perimeter. Then
First side \(=\frac{p}{3}\)
Second side \(=\frac{p}{5}\)
And third side \(=7 \mathrm{~m}\)
Therefore according to the problem statement
Perimeter \(p=\frac{p}{3}+\frac{p}{5}+7\)
\(\Rightarrow 15 p=5 p+3 p+105\)
\(\Rightarrow 7 p=105\)
\(p=15\)
Hence, the perimeter is \(15 \mathrm{~m}\).

Q.3. Simran is twice as old as Anushka. \(10\) years ago, her age was thrice of Anushka. Find their present ages.
Ans: In this word problem, the ages of Simran and Anushka are unknown quantities.
Let Anushka’s present age \(=x\) years.
Since Simran’s present age is \(2\) times that of Anushka, therefore her present age \(=2x\)
\(10\) years ago, Anushka’s age would have been \(=x-10\)
And Simran’s age would have been \(=2 x-10\)
According to the given problem statement, \(10\) years ago, Simran’s age was thrice of Anushka, i.e. \(2 x-10=3(x-10)\)
\(2 x-10=3 x-30\)
\(\Rightarrow x=20\)
Hence, the present age of Anushka is \(20\) years, and Simran’s age is \(2 x=2 \times 20=40\) years.

Q.4. The difference between two whole numbers is \(66\). The ratio of the two numbers is \(2:5\). What are the two numbers?
Ans: Given, the ratio of the two whole numbers is \(2:5\), let the one number be \(2 x\) and the other be \(5 x\).
The difference between the two numbers \(=(5 x-2 x)\)
It is given that the difference is \(66\).
Therefore, \(5 x-2 x=66\)
\(\Rightarrow 3 x=66\)
\(\Rightarrow x=22\)
Hence, the required numbers are \(2 x=2 \times 22=44\) and \(5 x=5 \times 22=110\)

Q.5. 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?
Ans: 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, \(170 x+250=590\)
\(\Rightarrow 170 x=590-250=340\)
\(\Rightarrow x=\frac{340}{170}=2\)
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 \(=25-8 x=25-(8 \times 2)=25-16=9\)

Summary

We learnt what a linear equation is, what a linear equation in one variable is, and what a linear equation in two variables is in this article. We also understood linear equations, keywords, translation, strategy, guidelines for setting up and solving word problems, problems involving real-number relationships, and solved examples.

Frequently Asked Questions (FAQs)

Frequently asked questions related to linear equations in two variables are listed as follows:

Q.1. What is a linear equation?
Ans: When two algebraic expressions are equated to each other using an equal “\(=\)” sign, it is called an equation. A linear equation has a degree of one. That is, the maximum power of the variable in the equation is one.

Q.2. What are the applications of linear equations in daily life?
Ans: The most common applications of linear equations in daily life:
1. Age problems
2. Speed, time, and distance problems
3. Geometry problems
4. Money and percentage of problems
5. Wages and hourly rate problems
6. Force and pressure problems

Q.3. How to solve a linear equation in two variables?
Ans: Linear equation in two variables can be solved by using
1. Graphical Method
2. Method of Substitution
3. Cross Multiplication Method
4. Method of Elimination
5. Determinant Method

Q.4. How many solutions are there for linear equations in two variables?
Ans: A system of linear equations usually has a single solution, but sometimes it can have no solution (parallel lines) or infinite solutions (same line).

Q.5. Can linear equations have fractions?
Ans: Yes, fractions may be used in linear equations as long as the denominator in the fractional component is constant. In a linear equation, the variables cannot be in the denominator of any fraction.