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Linear equations are the simplest form of equations in mathematics. These equations are called linear equations because on graphing they give straight lines. A pair of linear equations are used to solve many real-life problems based on money, geometry, speed, and distance, etc.

Let’s understand what are pair of linear equations in two variables and what are its different types and its solutions with examples.

## Pair of Linear Equations in Two Variables

Consider the statement “Cost of $20$ pens and $15$ pencils is ₹$450$”. Based on this statement we want to find the costs of a $1$ pen and $1$ pencil.

To find the cost of a pen and a pencil, we start by assuming the costs of $1$ pen and $1$ pencil as $x$ and $y$ (in rupees) respectively.

The statement in mathematical form can be written as $20x + 15y = 450 =>20x + 15y – 450 = 0$.

You also know that an equation that can be put in the form $ax + by + c = 0$, where $a$, $b$, and $c$ are real numbers, and $a$ and $b$ are not both zero, is called a linear equation in two variables.

Can we find the cost of $1$ piece of each of the above items?

To do so, we need one more such statement say “Cost of $25$ pens and $30$ pencils is ₹$675$”

This can be written as $25x + 30y = 675 => 25x + 30y – 675 = 0$.

It can be observed from the above example, to solve such types of real-world problems, we need two equations.

Such a combination of two linear equations in two variables is called a pair of linear equations in two variables.

The general form for a pair of linear equations in two variables $x$ and $y$ is

$a_{1}x + b_{1}y + c_{1} = 0$

and $a_{2}x + b_{2}y + c_{2} = 0$

where $a_{1}$, $b_{1}$, $c_{1}$, $a_{2}$, $b_{2}$, and $c_{2}$ are all real numbers and $a_{1}$, $b_{1}$ and $a_{2}$, $b_{2}$ cannot be simultaneously $0$(zero).

## Framing a Pair of Linear Equations in Two Variables

As observed above, we can convert many real-world problems into a pair of linear equations in two variables. Let’s understand the process of framing a pair of linear equations in two variables from real-world problems.

**Ex 1:** There are some students in the two examination halls A and B. To make the number of students equal in each hall, $10$ students were sent from A to B. But if $20$ students are sent from B to A, the number of students in A becomes double the number of students in B. Find the number of students in the two halls.

Let the number of students in hall A = $x$

And let the number of students in hall B = $y$

When $10$ students are sent from A to B, then

Number of students left in hall A = $x – 10$

And the number of students in hall B = $y + 10$

Now, the number of students is equal in both halls, therefore, $x – 10 = y + 10 => x – y – 10 – 10 = 0 => x – y – 20 = 0$ —————- (1)

When $20$ students are sent from B to A, then

Number of students left in hall B = $x – 20$

And the number of students in hall A = $y + 20$

Now, the number of students in A is double the number of students in B, therefore,

$y + 20 = 2\left(x – 20 \right) => y + 20 = 2x – 40 => -2x + y + 20 + 40 = 0 $

$=> -2x + y + 60 = 0$ ——————– (2)

**Ex 2:** A shopkeeper gives books on rent for reading. She takes a fixed charge for the first two days, and an additional charge for each day thereafter. Latika paid ₹$22$ for a book kept for six days, while Arvind paid ₹$16$ for a book kept for four days. Find the fixed charges and the charge for each extra day.

Let the fixed amount charged by the shopkeeper for a book = ₹$x$

And the per day amount charged by the shopkeeper for a book = ₹$y$

Latika kept the book for $6$ days

Therefore, the per day charge is for $4 \left(6 – 2 \right)$ (Fixed charge includes first two days).

Amount paid by Latika = $x + 4y$

Therefore, $x + 4y = 22$ ————————— (1)

Arvind kept the book for $4$ days

Therefore, the per day charge is for $2 \left(4 – 2 \right)$

Amount paid by Arvind = $x + 2y$

Therefore, $x + 2y = 16$ ————————— (2)

**Ex 3:** It can take $12$ hours to fill a swimming pool using two pipes. If the pipe of a larger diameter is used for $4$ hours and the pipe of a smaller diameter for $9$ hours, only half the pool can be filled. How long would it take for each pipe to fill the pool separately?

Let the time taken by a pipe of a larger diameter to fill a tank = $x$ hours.

And the time taken by a pipe of a smaller diameter to fill a tank = $y$ hours.

Therefore, part of a tank filled by a pipe of a larger diameter in $1$ hour = $\frac{1}{x}$.

And part of a tank filled by a pipe of a smaller diameter in $1$ hour = $\frac{1}{y}$.

When both pipes are opened for $12$ hours, then part of tank filled = $12 \times \frac{1}{x} + 12 \times \frac{1}{y} = \frac{12}{x} + \frac{12}{y}$

Since, in $12$ hours the tank is completely filled, therefore, $\frac{12}{x} + \frac{12}{y} = 1$ ————————– (1)

Also, part of a tank filled by a pipe of a larger diameter in $4$ hours = $4 \times \frac{1}{x} = \frac{4}{x}$

Similarly, part of a tank filled by a pipe of a smaller diameter in $9$ hours = $9 \times \frac{1}{y} = \frac{9}{x}$

Therefore, part of a tank filled by a pipe of a larger diameter in $4$ hours and a pipe of a smaller diameter in $9$ hours = $\frac {4}{x} + \frac{9}{y}$.

Therefore, $\frac {4}{x} + \frac{9}{y} = \frac{1}{2}$ ——————————— (2)

**Ex 4:** Vijay had some bananas, and he divided them into two lots A and B. He sold the first lot at the rate of ₹$2$ for $3$ bananas and the second lot at the rate of ₹$1$ per banana and got a total of ₹$400$. If he had sold the first lot at the rate of ₹$1$ per banana, and the second lot at the rate of ₹$4$ for $5$ bananas, his total collection would have been ₹$460$. Find the total number of bananas he had.

Let the number of bananas in lot A = $x$

And the number of bananas in lot B = $y$

Case 1:

Selling rate of lot A bananas = ₹$2$ for $3$ banana => $1$ banana for ₹$\frac{2}{3}$

Selling rate of lot B bananas = ₹$1$ for $1$ banana=> $1$ banana for ₹$1$

Selling price of all bananas = ₹$\frac{2}{3}x + y$

Therefore, $\frac{2}{3}x + y = 400 => 2x + 3y = 1200$ ——————- (1)

Case 2:

Selling rate of lot A bananas = ₹$1$ for $1$ banana => $1$ banana for ₹$1$

Selling rate of lot B bananas = ₹$4$ for $5$ banana=> $1$ banana for ₹$\frac{4}{5}$

Selling price of all bananas = ₹$x + \frac{4}{5}y$

Therefore, $x + \frac{4}{5}y = 460 => 5x + 4y = 2300$ ——————- (1)

**Methods of Solving Pair of Linear Equations**

There are four methods to solve a system of linear equations in two variables. Those methods are used to solve a pair of linear equations in two variables are:

- Graphical Method
- Substitution Method
- Cross Multiplication Method
- Elimination Method

### Graphical Method of Solving a Pair of Linear Equations

The steps to solve linear equations in two variables graphically are given below:

**Step 1:** Graph each equation

**Step 2:** To graph an equation manually, first convert it to the form $y = mx + b$ by solving the equation for $y$

**Step 3:** Create a table of values of $x$ and $y$ (Two pairs of values are enough to plot a line)

**Step 4:** Identify the point where both lines meet

**Step 5:** The point of intersection is the solution of the given system

### Examples

**Ex 1:** Solve the following system of equations graphically: $x + y = 4$ and $x – y = 2$

From first equation $x + y = 4 => y = 4 – x$

From second equation $x – y = 2 => y = x – 2$

Plot the equation $x + y = 4$

Plot the second equation $x – y = 2$

The point of intersection is $\left(3, 1\right)$

Therefore, the solution of the given pair of equations is $x = 3$ and $y = 1$.

### Substitution Method of Solving a Pair of Linear Equations

The steps to solve linear equations in two variables using the substitution method are given below:

**Step 1:** Solve one of the equations for one variable.

**Step 2:** Substitute this in the other equation to get an equation in terms of a single variable.

**Step 3:** Solve it for the variable.

**Step 4:** Substitute it in any of the equations to get the value of another variable.

### Examples

**Ex 1:** Solve the given pair of linear equations using the substitution method: $x + 2y – 7 = 0$ and $3x – 5y + 12 = 0$

Let $x + 2y – 7 = 0$ ————————- (1)

and $3x – 5y + 12 = 0$ ————————- (2)

From (1)

$x = -2y + 7$ ———————————- (3)

Substitute $x = -2y + 7$ from (3) in (2)

$3\left(-2y + 7 \right) – 5y + 12 = 0 => -6y + 21 – 5y + 12 = 0 => -6y – 5y + 21 + 12 = 0$

$ => -11y + 33 = 0 => -11y = -33 => y = \frac{-33}{-11} => y = 3$

Substitute $y = 3$ in (3)

$x = -2 \times 3 + 7 = -6 + 7 = 1$

Therefore, solution is $x = 1$ and $y = 3$.

### Cross Multiplication Method of Solving a Pair of Linear Equations

Consider a system of linear equations: $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$.

To solve this using the cross multiplication method, we first write the coefficients of each of $x$ and $y$ and constants as follows:

The arrows indicate that those coefficients have to be multiplied. Now we write the following equation by cross-multiplying and subtracting the products.

$\frac{x}{b_{1}c{2} – b_{2}c_{1}} = \frac{y}{c_{1}a{2} – c_{2}a_{1}} = \frac{1}{a_{1}b{2} – a_{2}b_{1}}$

From this, we get two equations:

$\frac{x}{b_{1}c{2} – b_{2}c_{1}} = \frac{1}{a_{1}b{2} – a_{2}b_{1}}$

And $\frac{y}{c_{1}a{2} – c_{2}a_{1}} = \frac{1}{a_{1}b{2} – a_{2}b_{1}}$

Solving these two we get

$x = \frac{b_{1}c{2} – b_{2}c_{1}}{a_{1}b{2} – a_{2}b_{1}}$

$y = \frac{c_{1}a{2} – c_{2}a_{1}}{a_{1}b{2} – a_{2}b_{1}}$

### Examples

**Ex 1:** Solve the given pair of linear equations using the substitution method: $x + 2y – 7 = 0$ and $3x – 5y + 12 = 0$

Let $x + 2y – 7 = 0$ ————————- (1)

and $3x – 5y + 12 = 0$ ————————- (2)

The coefficients from the above two equations are

$a_{1} = 1$, $b_{1} = 2$, $c_{1} = -7$

And $a_{2} = 3$, $b_{2} = -5$, $c_{2} = 12$

Now, using the expressions for finding $x$ and $y$, we get

$x = \frac{b_{1}c{2} – b_{2}c_{1}}{a_{1}b{2} – a_{2}b_{1}} = \frac{2 \times 12 – \left(-5 \right) \times \left(-7 \right)}{1 \times \left(-5 \right) – 3 \times 2} = \frac{24 – 35}{-5 – 6} = \frac{-11}{-11} = 1$

$y = \frac{c_{1}a{2} – c_{2}a_{1}}{a_{1}b{2} – a_{2}b_{1}} = \frac{-7 \times 3 – 12 \times 1}{1 \times \left(-5 \right) – 3 \times 2} = \frac{-21 – 12}{-5 – 6} = \frac{-33}{-11} = 3$

Therefore, solution is $x = 1$ and $y = 3$.

### Elimination Method of Solving a Pair of Linear Equations

The steps to solve linear equations in two variables using the elimination method are given below:

**Step 1:** Arrange the equations in the standard form: $ax + by + c = 0$ or $ax + by = c$

**Step 2:** Check if adding or subtracting the equations would result in the cancellation of a variable.

**Step 3:** If not, multiply one or both equations by either the coefficient of $x$ or $y$ such that their addition or subtraction would result in the cancellation of any one of the variables.

**Step 4:** Solve the resulting single variable equation.

**Step 5:** Substitute it in any of the equations to get the value of another variable.

### Examples

**Ex 1:** Solve the following system of equations using the elimination method: $2x + 3y – 11 = 0$ and $3x + 2y – 9 = 0$

Adding or subtracting these two equations would not result in the cancellation of any variable. Let us aim at the cancellation of $x$.

The coefficients of $x$ in both equations are $2$ and $3$. Their LCM is $6$. We will make the coefficients of $x$ in both equations $6$ and $-6$ such that the $x$ terms get canceled when we add the equations.

$3 \times \left(2x + 3y – 11 = 0 \right)$

$=>6x + 9y – 33 = 0$

$-2 \times \left(3x + 2y – 9 = 0 \right)$

$=> -6x – 4y + 18 = 0$

Now we will add these two equations:

$6x + 9y – 33 = 0$

$-6x -4y + 18 = 0$

On adding both the above equations we get,

$0 + 5y – 15 = 0$

$ => 5y – 15 = 0$

$ => 5y = 15$

$ => y = \frac{15}{5}$

$ => y = 3$

Substitute this in one of the given two equations and solve the resultant variable for $x$.

$2x + 3y – 11 = 0$

$=> 2x +3 \times 3 – 11 = 0$

$=> 2x + 9 – 11 = 0$

$=> 2x = 2$

$=> x = \frac{2}{2}$

$=> x = 1$

Therefore, the solution of the given system of equations is $x = 1$ and $y = 3$.

## Types of Solutions for a Pair of Linear Equations

The graph of the two linear equations may not intersect always. Sometimes they may be parallel. In that case, the system of linear equations in two variables has no solution. In some other cases, both lines coincide with each other. In that case, each point on that line is a solution of the given system and hence the given system has an infinite number of solutions.

**Consistent and Inconsistent System of Linear Equations:**- If the system has a solution, then it is said to be consistent;
- If the system has no solution,, it is said to be inconsistent.

**Independent and Dependent System of Linear Equations:**- If the system has a unique solution, then it is independent.
- If it has an infinite number of solutions, then it is dependent. It means that one variable depends on the other.

## Practice Problems

- Solve the following pairs of linear equations using the graphical method
- $2y = 4x – 6$ and $2x = y + 3$
- $x + 3y = 6$ and $2x – 3y = 12$

- Solve the following pairs of linear equations using the elimination method
- $141x + 93y = 189$ and $93x + 141y = 45$
- $3x = y + 5$ and $5x – y = 11$
- $27x + 31y = 85$ and $31x + 27y = 89$

- Solve the following pairs of linear equations using the substitution method
- $3x – y – 7 = 0$ and $2x + 5y + 1 = 0$
- Find the two numbers whose sum is $75$ and the difference is $15$

- Solve the following pairs of linear equations using the cross-multiplication method
- $x + 2y – 2 = 0$ and $x – 3y – 7 = 0$
- The sum of the digits of a two-digit number is $8$ and the difference between the number and that formed by reversing the digits is $18$. Find the number.

## FAQs

### What is meant by linear equations in two variables?

A linear equation is an equation of degree one. A linear equation in two variables is a type of linear equation in which there are two variables present.

For example, $2x – y = 9$ and $x + 3y =13$

### How do you identify linear equations in two variables?

We can identify a linear equation in two variables if it is expressed in the form $ax + by + c = 0$, consisting of two variables $x$ and $y$, and the highest degree of the given equation is $1$.

### What are the different methods of solving a pair of linear equations in two variables?

There are four methods to solve a system of linear equations in two variables. Those methods are used to solve a pair of linear equations in two variables are:

a) Graphical Method

b) Substitution Method

c) Cross Multiplication Method

d) Elimination Method

## Conclusion

A pair of linear equations in two variables are the equations in which each of the two variables is of the highest exponent order of $1$ and has one, none, or infinitely many solutions. There are four widely used methods of solving these equations – graphical method, substitution method, cross multiplication method, and elimination method.

## Recommended Reading

- Linear Equations in Two Variables – Definition, Types, and Graphs
- Linear Equations in One Variable – Graph & Method of Solving
- What are Algebraic Identities(With Definition, Types & Derivations)
- What is the Meaning of Equation – Definition, Types & Examples
- Division of Algebraic Expressions(With Methods & Examples)
- Multiplication of Algebraic Expressions(With Methods & Examples)
- Subtraction of Algebraic Expressions(With Methods & Examples)
- Addition of Algebraic Expressions(With Methods & Examples)
- What is Algebraic Expression(Definition, Formulas & Examples)
- What is Algebra – Definition, Basics & Examples
- What is Pattern in Math (Definition, Types & Examples)