In this section, we will discuss the Linear Equation. Linear Equation part of our syllabus is comparatively easy and has very few concepts.  

What is an Equation?

An equation is a statement that denotes two different mathematical expressions as equal. It consists of ‘=’ sign between two mathematical expressions.

It is generally used in algebra and is mostly useful in finding the unknown values of variables. They are of different types depending upon the maximum degree of the variables. 

• Linear Equation is an equation of degree 1 i.e. the greatest power on any variable in the equation will be 1. It is an equation of a straight line.

Linear Equation is written as ax + by + c = 0, where a, b and c are real numbers and a and b both are not zero.

Let’s take some examples and identify if the given equations are linear or not.

• E.g. 1: 2x – 3y = 0, It is an equation with maximum power 1 so it’s linear.
• E.g. 2: x(x – 2) = 0, Is it a linear equation? No. because when we expand it, it becomes x² – 2x = 0, so here the maximum power is 2. Hence, it’s not linear.

Linear equations can be of different type depending upon the number of variables, we see only two types of linear equation:

1. Linear equation in one variable – A linear equation with one variable. It represents a point on the line. E.g.: 2x + 7 = 10, 9y + 1 = 4
2. Linear equation in two variables – A linear equation with two variables. It represents a line in a plane. E.g.: 5x + 7y = 8, 13x +  12y = 1

Note: When we move any term from one side to another than its mathematical operation is changed. For ‘+’ it becomes ‘-’ and vice versa and for ‘×’ it becomes ‘/’ and vice versa. 

• How to solve Linear Equation in one variable?

Linear equation in one variable is very easy to solve.

Let’s take some examples to understand it.

E.g. 1: Solve: 2x + 4 = x – 3

Step 1: Try to take the variables on one side and constants on the other side.

2x + 4 = x – 3

⇒ 2x – x = -3 – 4

Step 2: Now solve the basic operations.

⇒ 2x – x = -3 – 4

⇒ x = -7 (Ans).

E.g. 2: If 2x + 3 = 9, then find the value of 3x + 2.

2x + 3 = 9

⇒ 2x = 9 – 3

⇒ 2x = 6

⇒ x = 3

Putting value of x,

∴ Required value = 3x + 2

                             = 3 × 3 + 2

                             = 9 + 2

                             = 11 (Ans).

• Methods to Solve Linear Equations in two variables

Linear equations can be solved in three ways. Let’s understand them one by one.

• Substitution Method:

In this method, first of all, from one equation, we find the value of a variable in terms of another and then use that value to put in the second equation to get the values of the variables.

Let’s understand this method with few examples.

E.g. 1: If 4x + 5y = 14 and x – 5y = 16, then the value of x and y are?

Step 1: Take any of the equation and find the value of one variable in terms of other.

4x + 5y = 14

⇒ 4x = 14 – 5y

⇒ x = (14 – 5y)/4

Step 2: Put this value of x in the second equation.

x – 5y = 16

⇒ (14 – 5y)/4 – 5y = 16

⇒ {14 – 5y – (5y × 4)}/4 = 16          [Taking LCM and doing subtraction of fractions]

⇒ 14 – 5y – 20y = 16 × 4

⇒ -25y = 64 – 14

⇒ y = 50/(-25)

⇒ y = -2

Step 3: Put the value of one variable in any equation.

x – 5y = 16

⇒ x – 5 × (-2) = 16

⇒ x + 10 = 16

⇒ x = 6.

∴ Values of x and y are 6 and -2 respectively (Ans).

E.g. 2: If x + 2y = 27 and x – 2y = -1, find the value of y.

 x + 2y = 27

⇒ x = (27 – 2y)

Putting this value in second equation,

⇒ x – 2y = -1

⇒ 27 – 2y – 2y = -1

⇒ -4y = -1 – 27

⇒ -4y = -28

⇒ y = (-28)/(-4)

⇒ y = 7 (Ans).

• Elimination Method:

In this method, we multiply the coefficients of variables with suitable number so as to make equal coefficient of a variable in both the equations. Now, we have to either add or subtract to get the equation in one variable.

Let’s understand it with an example from  previous year questions.

E.g. 1: If x – 2y = 0 and 3x + 4y = 20, then find the value of x and y.

Step 1: Make the coefficient of one variable equal in both the equations.

Equations are x – 2y = 0 and 3x + 4y = 20

Here, we can multiply in the first equation either by 3 or 2 to make the coefficients equal of at least one variable.

Multiplying by 2 in first equation,

⇒ (x – 2y) × 2 = 0 × 2

⇒ 2x – 4y = 0

Step 2: Now that we have equal coefficient of y, add both the equations.

Adding left hand part and right-hand part separately in the equation,

(2x – 4y) + (3x + 4y) = 0 + 20

⇒ 2x – 4y + 3x + 4y = 20

⇒ 5x = 20                 

⇒ x = 4.

Step 3: Put the known value in any equation to get the value of other variables.

Putting in any equation to get the value of y,

⇒ x – 2y = 0

⇒ x = 2y

⇒ y = x/2

⇒ y = 4/2

⇒ y = 2.

∴ Required value of x and y = 4 and 2 (Ans).

Note: Equation remains the same if we add, subtract, multiply or divide by the same number on both the sides of the equation.

E.g. 2: If 4x + 5y = 14 and x – 5y = 16, then the value of x and y are?

Equations are 4x + 5y = 14 and x – 5y = 16. In this case we have equal coefficients already given.

Add the equations,

(4x + 5y) + (x – 5y) = 14 + 16

⇒ 4x + 5y + x – 5y = 30

⇒ 5x = 30

⇒ x = 6

Putting in second equation,

6 – 5y = 16

⇒ -5y = 16 – 6

⇒ -5y = 10

⇒ y = 10/(-5)

⇒ y = -2

∴ Required values of x and y are 6 and -2 respectively (Ans).

Note: When the coefficients have different sign, add them and when they have same sign either ‘+’ or ‘-’ then subtract them to cancel out one variable. 

• Cross-Multiplication Method:

If we have two equations, a₁x + b₁y + c₁ = 0 and a₁x + b₁y + c₁ = 0, then cross-multiplication can help us to get the direct formula of the variables.

x = (b₁c₂ – b₂c₁)/(a₁b₂ – a₂b₁) and y = (c₁a₂ – c₂a₁)/(a₁b₂ – a₂b₁)

Let’s see an example to understand it in a better way.

E.g.: If 4x + 5y = 14 and x – 5y = 16, then the value of x and y are?

4x + 5y = 14

⇒ 4x + 5y – 14 = 0

And,

x – 5y = 16

⇒ x – 5y – 16 = 0

By direct formula,

x = (b1c2 – b2c1)/(a1b2 – a2b1)

⇒ x = [{5 × (-16)} – {(-5) × (-14)}]/[{4 × (-5)} – {1 × 5}]

⇒ x = {(-80) – 70}/{(-20) – 5}

⇒ x = (-150)/(-25)

⇒ x = 6

Putting this value of x in first equation,

x – 5y = 16

⇒ 6 – 5y = 16

⇒ -5y = 10

⇒ y = -2

∴ Required value of x and y = 6 and -2 respectively (Ans). 

• Consistency for the system of Linear Equations:

A set of linear equations is said to be consistent if it has at least one solution for these equations and it is said to be inconsistent if there is no solution for the equations.

• Consistent System

If we have two equations, a₁x + b₁y + c₁ = 0 and a₁x + b₁y + c₁ = 0, then these equations can be consistent in two cases:

a) Intersecting lines (Unique solution) if a₁/a₂ ≠ b₁/b₂

b) Overlapping or coincident lines (Infinitely many solutions) if a₁/a₂ = b₁/b₂ = c₁/c₂

E.g.: For what value of C₂, the system of equation 6x + 2y = 2 and 3x + y = C₂, will be coincident?

6x + 2y = 2

⇒ 6x + 2y – 2 = 0

And,

3x + y = C2

⇒ 3x + y – C2 = 0

∵ The lines are coincident.

∴ We will get infinitely many solutions.

Applying the rule for infinitely many solutions,

a1/a2 = b1/b2 = c1/c2

⇒ 6/3 = 2/1 = (-2)/(-C2)

Taking a1/a2 = c1/c2,

⇒ 2/1 = (-2)/(-C2)

⇒ -C2 = -1

⇒ C2 = 1 (Ans). 

• Inconsistent System

If we have two equations, a₁x + b₁y + c₁ = 0 and a₁x + b₁y + c₁ = 0, then these equations can be inconsistent if there is no solution, i.e.

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Let’s see an example to understand this concept.

• E.g.: For what value of k, the system of equations kx – 20y – 6 = 0 and 6x – 10y – 14 = 0 has no solution?

Applying the relation for no solution,

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

⇒ k/6 = (-20)/(-10) ≠ (-6)/(-14)

Taking a₁/a₂ = b₁/b₂,

⇒ k/6 = (-20)/(-10)

⇒ k/6 = 20/10

⇒ k/6 = 2

⇒ k = 12 (Ans).