# Pair Of Linear Equations In Two Variables

Notes of chapter: Pair Of Linear Equations In Two Variables are presented below. Indepth notes along with worksheets and NCERT Solutions for Class 10.

(1) Two linear equations in the same two variables are called a pair of linear equations in two variables.

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

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

Eg:- Pair of linear equations in two variables are:

2x + 4y = 3 and x + y + 6 = 0

As we know that representation of a linear equation in two variables is a straight line.

Therefore, we will have two straight lines for two linear equations.

Possibilities of the position of two lines in a plane are given below:-

(i) The two lines will intersect at one point.

Eg:- If Ram buys two erasers and 2 pens in Rs 20 and Shyam buys four  erasers and one pen in Rs 30. Represent this situation algebraically and graphically.

Ans-

Let Cost of 1 eraser = Rs x

Cost of 1 pen = Rs y

Algebraic equation for Ram: 2x + 2y = Rs 20

Algebraic equation for Shyam : = 4x + y = Rs 30

Now, find out different value of x and y.

For Ram

 X 0 10 y 10 0

For Shyam

 X 0 30 y 7.5 0

Plot these points in a graph. These straight lines intersect each other at one point.

E(3, 6.5) is the only common point on both the lines, there is one and only one solution for this pair of linear equations in two variables.

Those pairs which have a common solution are known as consistent pair of linear equations.

If lines are

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

(ii) The two lines are parallel.

Eg:- If Ram buys two erasers and two pens in Rs 20 and Shyam buys four  erasers and four pens in Rs 80. Represent this situation algebraically and graphically.

Ans-

Let Cost of 1 eraser = Rs x

Cost of 1 pen = Rs y

Algebraic equation for Ram: 2x + 2y = Rs 20

Algebraic equation for Shyam : = 4x + 4y = Rs 80

Now, find out different value of x and y.

For Ram

 X 0 10 y 10 0

For Shyam

 X 0 20 y 20 0

Plot these points in a graph. These straight lines are parallel to each other at one point.

The lines are parallels and do not cross each other. Therefore, the equations have no common solution.

Those pairs which do not have a common solution are known as inconsistent pair of linear equations.

If lines are

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

(iii)The two lines will be coincident.

Eg:- If Ram buys one erasers and two pens in Rs 30 and Shyam buys two erasers and four pens in Rs 60. Represent this situation algebraically and graphically.

Ans-

Let Cost of 1 eraser = Rs x

Cost of 1 pen = Rs y

Algebraic equation for Ram: x + 2y = Rs 30

Algebraic equation for Shyam : = 2x + 4y = Rs 60

Now, find out different value of x and y.

For Ram

 X 0 15 y 30 0

For Shyam

 X 0 15 y 30 0

Plot these points in a graph. These straight lines are coincide to each other at one point.

Both lines coincide. Therefore, every point on the line is a common solution to both the equations. Hence, they have infinitely many solutions.

Those pairs which have many distinct common solutions are known as dependent pair of linear equations in two variables.

A dependent pair of linear equations is always consistent.

If lines are

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

(2) Algebraic Methods of Solving a Pair of Linear Equations-

(i) Substitution Method – When we substitute value of one variable by expressing it in terms of the other variable to solve the pair of linear equations, the method is known as substitute method.

Eg:- Solve the following pair of equations by substituting method:

2x + y = 8

5x + 4y = 10

Ans-

2x + y = 8…(1)

5x + 4y = 10…(2)

Step 1-  Writing equation (1) one variable in terms of other variable.

2x + y = 8

y = 8 – 2x..(3)

Step 2- Substitute value of y in equation (2)

5x + 4(8 – 2x) = 10

5x + 32 -8x = 10

-3x = 10 – 32

-3x = -22

Verification- To verify put values of variables x and y in any equation.

Now, we are putting values of variables x and y in equation 1.

(ii) Elimination Method- The method in which we eliminate one variable first, to get a linear equation in one variable, is known as elimination Method.

Eg:- The ratio of incomes of two persons is 5:8 and the ratio of their expenditure is 1:2. Each of them manages to save Rs 3000 per month. Find their monthly incomes.

Ans- Let us denote income of two persons by Rs 5x and Rs 8x and the expenditures by Rs 1y and Rs 2y.

Therefore, equations are

5x – y = 2000..(1)

8x – 2y = 2000..(2)

Step 1- Multiply equation (1) by 2

10x – 2y = 4000…(3)

8x – 2y = 2000…(4)

Step 2 – Subtract equation 4 from equation 3

(10x – 2y) – (8x – 2y) = 4000 – 2000

10x – 2y – 8x + 2y = 2000

2x = 2000

x =

Step 3- Put value of x in equation 1

5(1000) – y = 2000

5000– y = 2000

-y = 2000 – 5000

= -3000

y = 3000

Therefore, solutions of the x = 1000 and y = 3000

So, income of first person = Rs 5x

= 5(1000)

=

Income of second person = Rs 8x

= Rs 8(1000)

= Rs 8000

Hence, income of first person is Rs 5000 and income of second person Rs 8000.

Verification:-

Income ratio

Expenditure of first person = 5000 – 2000 = 3000Rs

Expenditure of second person =8000– 2000 = Rs 6000

Expenditure Ratio

Hence, verified.

(iii) Cross – Multiplication Method –

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

We can write the solutions given by above equations in the following form:

Eg:-  Cost of 2 pens and 3 erasers is Rs 20 and cost of 4 pen and 1 eraser is Rs 25. Find the cost of one eraser and one pen.

Ans-

Let cost of one pen = Rs x

Cost of te one eraser = Rs y

2x + 3y = 20 or 2x + 3y – 20 =0…(1)

4x + y = 25 or 4x + y – 25 = 0…(2)

Hence, cost of one eraser = Rs 3

And cost of one pen = Rs = 5.5

Verification – put values of x and y in any of equation (1) and (2)

2x + 3y – 20 = 0

2(5.5) + 3(3) – 20

11 + 9 – 20

=0

Hence, verified.

(3)Equations Reducible to a Pair of Linear Equations in Two Variables-

Eg:- Solve the pair of equation:

Step 3:- Substitute values of p and q in equations 1 and 2

2p + 3q = 13…(3)

5p – 4q = -2….(4)

Step 4:- Solve equation 3 and 4

Multiply equation 3 by 4

(4) (2p)+(4)(3q) = (4) (13)… (5)

8p + 12q =52

Multiply equation 4 by 3

3(5p) – 3(4q) = 3( -2)

15p – 12 q = -6…(6)

Now, add equation 5 and (6)

23p = 46

p = 2

Put value of p in equation 4

5(2) – 4q = -2

10 – 4q = -2

-4q = -2 – 10

-4q = -12

q = 3

Therefore,

5(2) – 4(3) = -2

10 – 12 = -2

-2 = -2

RHS = LHS

Hence, verified.

Helping Topics

NCERT Solutions Class 10

Worksheet Class 10