Algebra (Linear Equations), In this we are going to learn the solution of linear equations and problems related to it in Algebra. Questions are not directly asked from this topic but it would be very helpful in clearing the concepts.

Let’s try to learn what equation is first and its Components.

Equation: – A statement of equality of which involves variable number(s) is called an equation.

E.g.  4x = 12, 7 – 2z3 = 5 etc.

Linear equations: An equation in which the highest power of the variables involved is one, is called a linear equation.

Eg.   x / 3 = 8, x + y = 10 etc.

• Linear equation in one variable: An equation containing only one variable with the highest power one. (Represent point in the number line.

E.g.  16x = 56, 78z = 81

• Linear equation in two variables:  An equation containing only two variables with the highest power one. A graph of linear equations ax+by=c is a straight line.

E.g.  3x + 2y = 34, 3 = x / (5 – t) etc.

• Linear equation in two variables: A linear equation in which numbers of unknown variables are three, is known as the linear equation in three variables. (Represents a place in 3D)

E.g.  4x + 6y + 7z = 20, x + y + 2z = 5 etc.

We learned about the different types of equations but what are the uses of these equations? How to solve these equations? What is consistency in equations?

Let’s start by studying the consistency of the system.

Consistency of the System of Linear Equations

A set of linear equations is said to be consistent if there exists at least one solution for these equations, otherwise, the equation is inconsistent.

Let us consider a system of two linear equations as shown,

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Consistent System: If the system has at least one solution exists then there would be two possibilities;

• Independent Equation: System has a unique solution or   a1 / a2 ≠ b1 / b2, it represents a pair of intersecting lines.
• Dependent equation: System has infinity solution or a1/a2=b1/b2 =c1/c2, it represents a a pair of parallel lines.

Inconsistent System: If system of equations has no solution then

 a1 / a2 = b1 / b2 ≠ c1 / c2

E.g. Check whether the given system is consistent or not. If yes, then find the solution

x – 2y = 0

3x + 4y – 20 = 0

Here

a1 = 1

a2 = 3

b1 = – 2

b2 = 4

c1 = 0

c2 = -20

a1 / a2 = 1 / 3, b1 / b2 = (-2) / 4 = (-1) / 2, c1 / c2 = 0

Here,

a1 / a2 ≠ b1 / b2

We don’t know how to find solution so let’s try to find out first

Methods of Solving Linear Equations

• Substitution method
• Elimination method
• Cross multiplication method

Substitution method: In this method, first the value of one variable must be represented in the form of another variable and put this value in another equation and solve it. Thus, the value of one variable is obtained and this value is used to find the value of another variable.

2x – y = 3

4x – y = 5

4x – y = 5      (ii)

Putting the value of y in equation (ii)

4x – (2x – 3) = 5

2x + 3 = 5

2x = 2

x = 1

Now putting the value of x in (ii)

⇒ 4(1) – y = 5

⇒4 – y = 5

⇒y = -1

Hence x = 1 and y = -1

Elimination method: In this method, the coefficients of one of the variables of each equation become the same by multiplying a proper multiple. Solve these equations and by which we get the value of another variable and thus with the help of this value, we can find the value of another variable.

E.g. Solve the following equations with elimination method.

2x – y = 3,

4x – y = 5

⇒2x – y = 3                  (i)

⇒4x – y = 5                  (ii)

Subtracting (ii) from (i)

⇒x = 1

By putting the value of x in (i)

⇒2(1) – y = 3

⇒2 – y = 3

⇒y = -1

Yeah, we got the same answer by both methods, let’s try cross-multiplication method.

Cross multiplication method:

 Let

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Are two equations, by cross-multiplication method

x / (b¬¬1c2 – b2c1) = y / (c1a2 – c2a1) = 1 / (a1b2 – a2b1)

Solve following equations with cross-multiplication method.

⇒2x – y – 3 = 0,

⇒4x – y – 5 = 0

By putting the value of a1, b1, c1, a2, b2 and c2 in formula

⇒x / (b1c2 – b2c1) = y / (c1a2 – c2a1) = 1 / (a1b2 – a2b1)

⇒x / (-1 × (-5) – (-1) × (-3)) = y / ((-3) × 4 – (-5) × 2) = 1  / (2 × (-1) – 4 × (-1))

⇒x / (5 – 3) = y / (-12 + 10) = 1 / (-2 + 4)

⇒x / 2 = y / (-2) = 1 / 2 ⇒x = 1, y = -1