03. LINEAR EQUATIONS IN TWO VARIABLES

Linear equations in two variables

Linear equations in two variables is an equation of the form ax + by + c = 0, where x and y are the two variables and a, b, and c are real numbers and a and b are non-zero.

Pair of Lineaar Equations in two variables

The two linear equations with similar two variables are considered as the pair of linear equations in two variables. There are two methods to solve a pair of linear equations in two variables.

1. Algebraic Method

If the pair of linear equations are given in the form of a₁x + b₁y + c₁ = 0 and a₂x + b₂ y + c₂ = 0, then the following situation takes place.

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

In this condition a₁/a₂ ≠ b₁/b₂

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

 In this condition  a₁/a₂ = b₁/b₂ = c₁/c₂

3. Dependent pair of linear equation - A pair of linear equations which are equivalent has infinitely many solutions. Such a pair is called a dependent pair of linear equation. Dependent pair of linear equations is always consistent.

In this condition  a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Graphical Method of Solution of a Pair of Linear Equations

The graphical representation of linear equations in two variables always forms a straight line.  There are three possibilities when two lines are drawn in a plane.   

1. Two lines intersect with each other at one point.

2. Two lines which are drawn are parallel to each other.

3. The two lines will coincide with each other.     

Example 1 :  Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel and play Hoopla (a game in which you throw a ring on the items kept in the stall, and if the ring covers any object completely you get it). The number of times she played Hoopla half the number of rides she had on the Giant Wheel. Each ride costs Rs.3, and a game Hoopla costs Rs.4. If she spent Rs.20 in the fair,Represent this situation algebraically and graphically (geometrically).

Let the number of giant wheel ride be x and the number of times she played Hoopla be y.

⇒ y = ½ x 

⇒ x - 2y   =   0........(1)

⇒ 3x + 4y = 20 ..... (2)

Hence algebraic equations are x = 2y and 3x + 4y = 20

Graphically representation

x		0	4	6
y =	x	0	2	3
	2

x		0	4	8
y =	20 - 3x	2	5	-1
	4

Example 2 :  Romila went to a stationery shop and purchased 2 pencils and 3 erasers

for Rs  9. Her friend Sonali saw the new variety of pencils and erasers with Romila, and he also bought 4 pencils and 6 erasers of the same kind for Rs 18. Represent this

situation algebraically and graphically. 

Let  us denote the cost of 1 pencil by  x and one eraser by  y. Then the algebraic representation is given by the following  equations:

⇒ 2x + 3y = 9 (1)

⇒ 4x + 6y = 18 (2)

Graphically representation

x		-3	0	3
y =	9 - 2x	5	3	1
	3

x		-3	0	3
y =	18 - 4x	5	3	1
	6

Example 3:  Two rails are represented by the equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Represent this situation geometrically.

⇒ x + 2y – 4 =0...........................(1) 

⇒ 2x + 4y – 12 =0.......................(2)

x		-2	0	4
y =	4  - x	3	2	0
	2

x		-2	0	6
y =	12 - 2x	4	3	0
	4

EXERCISE 3.1

1. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

Let the current age of Aftab be ‘x’years and, the current age of his daughter be ‘y’ years
Seven years ago father's age = (x-7) years
Seven years ago daughter's age = (y-7) years
According to the Question
⇒ (x-7)=7(y-7)
⇒ x-7 = 7y - 49
⇒ x - 7y = -42 …………(1)
After 3 years father's age = (x + 3) years
After 3 years daughter's age = (y + 3) years
According to the condition given in the question
⇒ x + 3 = 3(y + 3)
⇒ x - 3y = 6 …………..(2)

x		-7	0	7
y =	x  - 42	5	6	7
	7

x		-3	0	6
y =	x - 6	-3	-2	0
	3

2. The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 3 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.

Let us denote cost of one bat by Rs ‘x’ and one ball Rs ‘y’Then the algebraic representation is given by the following equations:
⇒ 3x + 6y = 3900
⇒ x + 3y = 1300

x		0	600	1300
y =	3900  - 3x	650	350	0
	6

x		100	400	1300
y =	1300 - x	400	300	0
	3

 

3. The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.

Let us denote cost of one Kg of apples by Rs ‘x’ and one Kg of grapes Rs ‘y’Then the algebraic representation is given by the following equations:

⇒ 2x + y = 160

⇒ 4x + 2y = 300

x		40	70
y =	160  - 2x	80	20
	1

x		50	60
y =	300 - 4x	50	30
	2

Example 4 :  Check graphically whether the pair of equations x + 3y = 6    and 2x – 3y = 12 is consistent. If so, solve them graphically.
⇒ x + 3y = 6 ……………….…(1)
⇒ 2x – 3y = 12 …………..(2)

x		0	6
y =	60 -  x	2	0
	3

x		0	3
y =	2x -12	-3	-2
	3

Plot the points A(0, 2), B(6, 0), P(0, – 4) and Q(3, – 2) on graph paper, and join the points to form the lines AB and PQ 

We observe that there is a point B (6, 0) common to both the lines AB and PQ. So, the solution of the pair of linear equations is x = 6 and y = 0, i.e., the given pair of equations is consistent.

Example 5 : Graphically, find whether the following pair of equations has no solution,

unique solution or infinitely many solutions:

⇒ 5x – 8y + 1 = 0 ...........................(i)  

⇒ 3x –24/5y + 3/5 = 0.................(ii)

Multiplying Equation (ii) by 5/3 we get

⇒ 5x – 8y + 1 = 0

this is the same as Equation (i). Hence the lines represented by Equations (i) and (ii) are coincident. Therefore, Equations (i) and (ii) have infinitely many solutions.

Example 6 : Champa went to a ‘Sale’ to purchase some pants and skirts. When  her friends asked her how many of each she had bought, she answered, “The number of skirts is two less than twice the number of pants purchased. Also, the number of skirts is four less than four times the number of pants purchased”. Help her friends to find how many pants and skirts Champa bought.

Let us denote the number of pants by x and the number of skirts by y. Then

the equations formed are :

      ⇒  y = 2x – 2 

      ⇒ 2x -y  = 2   ..........................(1)

       ⇒ y = 4x – 4 

      ⇒ 4x  -  y = 4 ..........................(2)

x	2	0
y = 2x  -  2	2	-2

x	0	1
y = 4x  -  4	-4	0

Plot the points and draw the lines passing through them to represent the equations, The two lines intersect at the point (1, 0). So, x = 1, y = 0 is the required solution of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did not buy any skirt.

EXERCISE 3.2

Form the pair of linear equations in the following problems, and find their solutions

graphically.

(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together

cost Rs 46. Find the cost of one pencil and that of one pen.

(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4

more than the number of boys, find the number of boys and girls who took part in

the quiz.

Let  number of girls who took part in the quiz be x and number of boys who took part in the quiz be y.

x + y = 10  ……………(1)

and y = x + 4

     -x + y =4………………(2)

x	2	5
y = 10 –x	8	5

x	2	4
y = x +4	6	8

 

  

(ii)  Let the cost of one pencil be Rs x and the cost of one pen be Rs y.

According to the question

 

5x + 7y = 50

7x + 5y = 46

 

x		3	10	-4
y =	50 -  5x	5	0	10
	7

x		8	3	-2
y =	46 – 7x	-5	5	5
	5

 

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes  if we only add 1 to the denominator. What is the fraction?

Solution

Let the numerator be x and denominator be y

According to question

  =1

  x + 1 = y - 1

  x  -y = - 2………..(1)

    =

   2x = y +1

  2x –y =1…………..(2)

Subtracting Equation (2) from (1)

 x -  y = - 2

2x – y =   1

-    +        -  

-x        = -3

 x = 3

Putting this value of x in Equation (1)

3 -y = -  2

-  y = - 2 - 3

 - y = - 5

    y = 5

Therefore, fraction is  =

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Solution

Let the present age of Nuri be  x years and the present age of Sonu be y years

5 years ago, age of Nuri = (x – 5) years

5 years ago, age of Sonu = (y – 5) years

According to question

    (x − 5) = 3 (y − 5)

⇒ x – 5 = 3y – 15

⇒ x − 3y = −10……… (1)

10 years later from present, age of Nuri = (x + 10) years

10 years later from present, age of Sonu = (y + 10) years

According to question

    (x + 10) = 2 (y + 10)

⇒ x + 10 = 2y + 20

⇒ x − 2y = 10 ……… (2)

Subtracting equation (1) from (2), we get

x − 2y = 10

x − 3y = −10

-    +       +  

       y = 20

Putting the value of y in equation (1)

    x – 3 (20) = −10

⇒ x – 60 = −10

⇒ x = 50

Therefore, present age of Nuri = 50 years and present age of Sonu = 20 years

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Solution

Let the digit at ten’s place be x and the digit at one’s place be y

So, two-digit number = 10x +y

by reversing digit the two-digit number = 10y +x

According to question

         x + y = 9

   ⇒ x + y - 9 =0….. (1)

And  9 (10x + y) = 2 (10y + x)

  ⇒ 90x + 9y = 20y + 2x

  ⇒ 88x = 11y

  ⇒ 8x = y

 ⇒ 8x – y = 0 ….. (2)

Adding equation  (1) and (2)

      x   + y - 9 =0

     8x – y = 0

     9x       = 9

⇒ x = 1

Putting the value of x in equation (1)

     1 + y = 9

⇒ y = 9 – 1 = 8

Therefore, number = 10x + y

                               = 10 (1) + 8

                               = 18

(iv) Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.
Solution
Let the number of Rs 100 notes be x and the number of Rs 50 notes be y
According to given conditions,
         x + y = 25 …. (1)
and  100x + 50y = 2000
    ⇒ 2x + y = 40 … (2)
Subtracting equation (2) from (1)
     x + y = 25
    2x + y = 40
   -     -       - 
   −x = −15
⇒ x = 15
Putting the value of x in equation (1)
    15 + y = 2
⇒ y = 25 – 15
⇒  y = 10
Therefore, number of Rs 100 notes = 15 and number of Rs 50 notes = 10

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution
Let the fixed charge for 3 days be Rs x and the additional charge for each day thereafter be Rs y
According to given condition
x + 4y = 27 ……. (1)
x + 2y = 21 … ….(2)
Subtracting equation (2) from (1)
 x + 4y = 27
 x + 2y = 21
-    -        -   
      2y = 6
  ⇒ y = 3
Putting the value of y in equation  (1)

    x + 4 (3) = 27
⇒ x = 27 – 12 = 15
Therefore, fixed charge for 3 days = Rs 15 and additional charge for each day after 3 days = Rs 3