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Basic Mathematics is a part of one's daily life, and because of that, the introduction of Mathematics also takes place with real-life examples. Everyone has come across the age-old question "if you have 6 candies and give your friend 2, how many candies are left with you?". While with the advancement of classes, Maths becomes more about theorems and formulas, a part of it is always derived from practical experiences. The section, algebra deals with such real-world problems and algebra linear equations applications is one of the essential topics under it.

Linear equations real-world problems let a student formulate their equations to solve them. Every sum will require a new approach, though it is very possible to categorise certain situations under one type of equation. To understand this type of equation, one first needs to know different aspects of it.

A linear equation, according to Mathematics, is an equation of first-order that is defined for lines in a coordinate system.

The most common form of a linear equation is,

y= ax + b

Where a is the slope of the line and b is the y-intercept.

This can also help one to understand the relation between two variables that is y and x. Different forms of this equation can be used to solve different types of linear algebra real-world problems. Let's solve a real-world problem to understand how one can formulate a linear equation and solve it.

Q1: Raj went to buy flowers pots in his local nursery. The shopkeeper said that there are two types of pots, terracotta and plastic. He also said that the price of a terracotta pot is three times the price of a plastic pot. If Raj can buy one of each at 100 rupees, what is the price of each pot?

Ans: First, understand the relation between the two costs.

Now, let's assume that the cost of a plastic pot is x, then the cost of a terracotta pot will be three times x that id 3x.

It is stated that one of each pot can be bought using 100 rupees, that means

x + 3x = 100, this is the linear equation for this given problem.

Now solve it for x, that is

4x = 100

x = 100/4

x = 20

So, the price of a plastic pot is 25 rupees. Since a terracotta pot costs three times, its price will be 3 × 25 = 75 rupees.

Under such circumstances where one only needs to find one variable, solving word problems with linear equations are usually like the example above. But if one needs to find the value of more than one variable, a system of linear equations is used.

Linear system or system of linear equations is a set of one or more linear equations that involve a common set of variables. For example,

3x + 2y = 3

-2x + y = 4

These two equations are together a system of linear equations. Algebra linear equations applications are not much different from that of solving for a single variable, one has to formulate their own equations here as well. This can be better understood by practising real-life linear equations word problems. Let's solve one such problem to see the course of equation formulation.

Q2: The cost of passes for a theme park is 162 rupees for 12 children and 3 adults. For the same theme park, the cost of passes for 8 children and 3 adults was 122 rupees. What is the cost of one pass for each child and adult?

Ans: From the question, it is clear that these two situations will translate into two equations. Now, put your understanding into these two situations and form equations just the way it was done above. System of equations real-world problems can have multiple variables; in this case, there are two children and adults.

Let the cost of a child's pass be x, and the cost of an adult's pass be y. So, according to the first situation,

12x + 3y = 162

And according to the second situation

8x + 3y = 122

Now, subtract the first equation from the second one, like

12x + 3y = 162

- 8x + 3y =122

4x = 40

x = 40/4

x = 10

Put the value of x in any of the equations to y’s value, like

8(10) + 3y = 122

80 + 3y = 122

3y = 42

y = 14

So, the cost of a child’s pass is 10 rupees and 14 rupees for an adult’s pass.

Algebra linear equations applications simplify relations between two or some times more variables. Yet, there are several aspects that a student should understand before solving its questions. Vedantu provides live coaching classes on a variety of mathematical topics including linear equations and its different applications. Check their sample sheets to practice more real-world problems and master the skill of equation formulation.

FAQ (Frequently Asked Questions)

1. What is Homogeneous System Linear Algebra?

Ans. Linear algebra or a system of linear equations can be either homogeneous or non-homogeneous. It is said to be homogeneous when all of its constant terms are zero. It is equivalent to a matrix equation -

Ax = 0

Here, A is a k × n matrix; x is a column vector that has n entries, and 0 is its zero vector that has m entries.

2. What are Systems of Linear Equations?

Ans. A set of linear equations with the same variables is called systems of linear equations. To better understand let’s consider two equations kx + jy = l and ax + sy = d. As these equations represent two lines on a graph in the xy-plane, when they are simultaneously solved, its solution is the interaction between the two lines. In the system of equations real-world problems, there are three possibilities, i.e. these two can be two parallel non-identical lines, non-parallel line and identical lines.

3. What is the Substitution Method?

Ans. The substitution method is a way of solving linear equations in real-world problems. It is the process where you substitute the one y-value with the other. For example, in the two equations -

2x – y = -4

4x + y = 9

The first equation can be easily written as

y = 2x + 4

Now when y's value is put in the second equation

4x + 2x + 4 = 9

5x = 9 – 4

5x = 5

x = 5/5

x = 1

Like this system of linear equations can be further simplified.