CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variable

Class Notes of Ch 3 : Pair of Linear Equations in Two Variables

Class 10th Maths

Pair of Linear Equations in Two Variables

Topics:

Pair Linear Equations in Two Variable
Geometrical Representation of Linear equation
Solution of Linear Equation in Two Variables
Algebraic Method
Substitution Method
Elimination Method
Cross Multiplication Method
Equations Reducible to Pair of Linear Equations

Pair Linear Equations in Two Variable

Equation which can be put in 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 x and y. E.g. 3x + 10y =8

If it is given that sum of age of Students Ram & Sita is 9 & the difference in their age is 1, how can we find their age?

Here we can assume age of Ram to be x & that of Sita to be y & then create linear equations in 2 variable.

Sum of their age is 9, implies x+y=9

Difference of their age is 1, implies x-y=1

Now we can solve this pair of equation to find value of x & y, that is age of Ram & Sita.

These two linear equations are in the same two variables x and y. Equations like these are 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

a1x + b1y + c1 = 0 and a2x + b2 y + c2 = 0,

Where a1, b1, c1, a2, b2, c2 are all real numbers and a12 + b12 ≠ 0, a22 + b22 ≠ 0.

Geometrical Representation of Linear Equation

Geometrically, each solution (x, y) of a linear equation in two variables, ax + by + c = 0, corresponds to a point on the line representing the equation, and vice versa. E.g. Linear equation in two variable: x + y=3 is plotted in graph below. It has many solutions (3,0), (2,1), (1,2), (0,3), (-1,4) etc.

Solution of Linear Equation in Two Variables

There are two approaches for solve linear equation in 2 variables

Graphical Approach
Algebraic Approach

Graphical Approach: Draw the graph of both equations

If lines intersect at a single point, the it has a unique solution
If lines are parallel, then equations have no solution
If lines are coincident, then equations have infinitely many solutions.

A pair of linear equations which has no solution, is called an inconsistent pair of linear equations.

A pair of linear equations in two variables, which has a solution, is called a consistent pair of linear equations.

A pair of linear equations which are equivalent has infinitely many distinct common solutions. Such a pair is called a dependent pair of linear equations in two variables. Note that a dependent pair of linear equations is always consistent.

Algebraic Method

There are 3 Algebraic methods

Substitution Method
Elimination Method
Cross - Multiplication Method

Substitution Method

Step 1: Find the value of one variable, say y in terms of the other variable, i.e., x .

Step 2: Substitute this value of y in the other equation, and reduce it to an equation in one variable, i.e., in terms of x, which can be solved

Step 3: If in Step 2, we obtain a true statement involving no variable, then the original pair of equations has infinitely many solutions.

Step 4: If in Step 2, we obtain a false statement involving no variable, then the original pair of equations has no solution, i.e., it is inconsistent

Step 5: Substitute the value of x (or y) obtained in Step 2 in the equation used in Step 1 to obtain the value of the other variable.

Numerical 1: solve x + y = 14 -- (i) ; x – y = 4 --(ii) using substitution method.

Solution:

x = 4+y using equation (ii) the value of one variable, say x in terms of the other variable, i.e., y .

(4 +y) + y = 14 Substituting value of y in equation (i) to reduce it to equation in one variable

2y = 14-4 or 2y = 10 or y= 5 (Obtain the solution for y)

X = 4 + y = 4 + 5 =9 (Substitute the value of y obtained in Step 2 to get value of x)

Thus solution is x,y = 9,5

Numerical 2: Solve x + y = 3 -- (i) ; 2x + 2y = 6 -- (ii) using substitution method.

Solution:

x = 3-y using equation (i) the value of one variable, say x in terms of the other variable, i.e., y.

2(3-y) + 2y = 6 Substituting value of y in equation (i) to reduce it to equation in one variable

6 – 2y + 2y = 6 or 6 =6 (Try to Obtain the solution for y)

Since we obtain a true statement involving no variable, then the original pair of equations has infinitely many solutions.

Numerical 3: Solve x + y = 3 -- (i) ; 2x + 2y = 8 -- (ii) using substitution method.

Solution:

x = 3-y using equation (i) the value of one variable, say x in terms of the other variable, i.e., y.

2(3-y) + 2y = 8 Substituting value of y in equation (i) to reduce it to equation in one variable

6 – 2y + 2y =8 or 6 =8 (Try to Obtain the solution for y)

Since we obtain a false statement involving no variable, then the original pair of equations has no solution, i.e., it is inconsistent.

Elimination Method

Step 1 : First multiply both the equations by some suitable non-zero constants to make the coefficients of one variable (either x or y) numerically equal.

Step 2 : Then add or subtract one equation from the other so that one variable gets eliminated. If you get an equation in one variable, go to Step 3.

If in Step 2, we obtain a true statement involving no variable, then the original pair of equations has infinitely many solutions.

If in Step 2, we obtain a false statement involving no variable, then the original pair of equations has no solution, i.e., it is inconsistent.

Step 3 : Solve the equation in one variable (x or y) so obtained to get its value.

Step 4 : Substitute this value of x (or y) in either of the original equations to get the value of the other variable.

Numerical 4: solve x +3y = 8 -- (i) ; 2x – 2y = 8 --(ii) using elimination method.

Solution:

Multiple equation (i) by 2 to match coefficients of x in both the equation

2x + 6y =16 --(i)

2x – 2y = 8 --(ii) Subtracting (ii) from (i) we get

----------------------

8y = 8 or y =1

Putting the value of y in equation 1, we get x =5

Numerical 5: solve x +y =4 -- (i) ; 2x + 2y = 16 --(ii) using elimination method.

Solution:

Multiple equation (i) by 2 to match coefficients of x in both the equation

2x + 2y =16 --(i)

2x – 2y = 16 --(ii) Subtracting (ii) from (i) we get

----------------------

0 = 0

Since, we obtain a true statement involving no variable, original pair of equations has infinitely many solutions

Numerical 6: solve x +y =4 -- (i) ; 2x + 2y = 14 --(ii) using elimination method.

Solution:

Multiple equation (i) by 2 to match coefficients of x in both the equation

2x + 2y =16 --(i)

2x – 2y = 14 --(ii) Subtracting (ii) from (i) we get

----------------------

0 = -2

Since, we obtain a false statement involving no variable, the original pair of equations has no solution, i.e., it is inconsistent.

Cross Multiplication Method

Let’s assume that we have to find solution for a1x + b1y + c1 = 0 (i) and a2x + b2y + c2 = 0 (ii)

Step 1: Multiply Equation (i) by b2 and Equation (ii) by b1, to get

b2a1x + b2b1y + b2c1 = 0 ---(iii)

b1a2x + b1b2 y + b1c2 = 0 ----(iv)

Step 2: Subtracting Equation (4) from (3), we get:

(b2a1 – b1a2) x + (b2b1 – b1b2) y + (b2c1– b1c2) = 0

Or (b2a1 – b1a2) x + (b2c1– b1c2) = 0

Or x = (b1c2– b2c1)/ (a1b2 – a2b1)

Step 3: Substituting this value of x in (i) or (ii), we get

y = (c1a2 – c2a1)/ (a1b2 – a2b1)

Step 4: Calculate value of a1b2 – a2b1

Step 5: if a1b2 – a2b1 ≠ 0 or a1/a2 ≠ b1/b2, then equation has definite solution.

Step 6: if a1b2 –a2b1 =0 or a1/a2 = b1/b2, then there are two possibilities

When a1/a2 = b1/b2 = c1/c2 , then there are infinitely many solutions.
When a1/a2 = b1/b2 ≠ c1/c2, then there is no solution.

Equations Reducible to Pair of Linear Equations in Two Variables

Sometimes we come across pair of equation which are not linear but can be reduced to linear form by making some suitable substitutions.

E.g. 2/x + 3/y = 13 & 5/x + 4/y = 2

This equation is not a linear equation, but if we substitute 1/x with a & 1/y with b, then the equation becomes

2a+ 3b =13 & 5a + 4b = 2. This is now a pair of linear equation in 2 variables.

After finding the values of variables a & b, we can easily find x as 1/a & y as 1/b.

Also Read :

MATHS Revision Notes

Chapter:01 Real Numbers System
Chapter:02 Polynomials
Chapter:03 Pair of Linear Equations in Two Variables
Chapter:04 Quadratic Equation
Chapter:05 Arithemetic Progressions
Chapter:06 Triangles
Chapter:07 Coordinate Geometry
Chapter:08 Introduction to Trignometry
Chapter:09 Some Application Of Trignometry
Chapter:10 Circles
Chapter:11 Constructions
Chapter:12 Area Related to Cirles
Chapter:13 Surface Area Volume
Chapter:14 Stastistics
Chapter:15 Probability

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