Xpert Solution For class 10 chapter 3: Pair of Linear Equations in Two Variables

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.

Solution:

Let present age of Aftab be x
And, present age of daughter is represented by y
Then Seven years ago,
Age of Aftab = -7
Age of daughter = y-7
According to the question,
(– 7)  = 7 (– 7 )
– 7 = 7 – 49
x– 7= – 49 + 7
– 7y = – 42 …(i)
x = 7y – 42
Putting y = 5, 6 and 7, we get
x = 7 × 5 – 42 = 35 – 42 = – 7
x = 7 × 6 – 42 = 42 – 42 = 0
x = 7 × 7 – 42 = 49 – 42 = 7

x-707
y567

Three years from now ,
Age of Aftab = +3
Age of daughter = +3
According to the question,
(+ 3) = 3 (+ 3)
+ 3 = 3+ 9
-3= 9-3
-3= 6 …(ii)
= 3+ 6
Putting, = -2,-1 and 0, we get
= 3 × – 2 + 6 = -6 + 6 =0
= 3 × – 1 + 6 = -3 + 6 = 3
= 3 × 0 + 6 = 0 + 6 = 6

 

x 036
y-2-1 0

Algebraic representation
From equation (i) and (ii)
– 7= – 42 …(i)
– 3= 6 …(ii)
Graphical representation

Linear_Equ1

  1. 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.

Solution:

Let cost of one bat = Rs x
Cost of one ball = Rs y
3 bats and 6 balls for Rs 3900 So that
3+ 6y = 3900 … (i)
Dividing equation by 3, we get
+ 2y = 1300
Subtracting 2y both side we get
x = 1300 – 2
Putting y = -1300, 0 and 1300 we get
x = 1300 – 2 (-1300) = 1300 + 2600 = 3900
= 1300 -2(0) = 1300 – 0 = 1300
x = 1300 – 2(1300) = 1300 – 2600 = – 1300

x39001300-1300
y-1300 01300

Given that she buys another bat and 2 more balls of the same kind for Rs 1300
So, we get
x + 2= 1300 … (ii)
Subtracting 2y both side we get
= 1300 – 2y
Putting y = – 1300, 0 and 1300 we get
x = 1300 – 2 (-1300) = 1300 + 2600 = 3900
= 1300 – 2 (0) = 1300 – 0 = 1300
= 1300 – 2(1300) = 1300 – 2600 = -1300

x39001300-1300
y-1300 01300

Algebraic representation
3+ 6y = 3900 … (i)
+ 2= 1300 … (ii)
Graphical representation,

Linear_Equation_2

  1. 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.

Solution:

Let cost each kg of apples = Rs x
Cost of each kg of grapes = Rs y
Given that the cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160
So that
= 160 … (i)
2x = 160 – y
x = (160 – y)/2
Let y = 0 , 80 and 160,  we get
x = (160 – ( 0 )/2 = 80
x = (160- 80 )/2 = 40
x = (160 – 2 × 80)/2 = 0

x8040 0
y 080160

Given that the cost of 4 kg of apples and 2 kg of grapes is Rs 300
So we get
4x + 2= 300 … (ii)
Dividing by 2 we get
2x + y = 150
Subtracting 2x both side, we get
= 150 – 2x
Putting x = 0 , 50 , 100 we get
= 150 – 2 × 0 = 150
= 150 – 2 ×  50 = 50
= 150 – 2 × (100) = -50

x 050100
y15050-50

Algebraic representation,
2y = 160 … (i)
4x + 2y = 300 … (ii)
Graphical representation,

Linear_Equation_3

 Exercise 3.2

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

(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.

Solution:

Let number of boys = x
Number of girls = y
Given that total number of student is 10 so that
= 10
Subtract y both side we get
= 10 – y
Putting = 0 , 5, 10 we get
= 10 – 0 = 10
= 10 – 5 = 5
= 10 – 10 = 0

x105
y 05

Given that If the number of girls is 4 more than the number of boys
So that
+ 4
Putting x = -4, 0, 4, and we get
= – 4 + 4 = 0
= 0 + 4 = 4
= 4 + 4 = 8

x-4 04
y 048

Graphical representation

Linear_Equation_4

Therefore, number of boys = 3 and number of girls = 7.

(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.

Solution:

Let cost of pencil = Rs x
Cost of pens = Rs y
5 pencils and 7 pens together cost Rs 50,
So we get
5x + 7y = 50
Subtracting 7y both sides we get
5x = 50 – 7y
Dividing by 5 we get
x = 10 – 7 y /5
Putting value of y = 5 , 10 and 15 we get
x = 10 – 7 × 5/5 = 10 – 7 = 3
x = 10 – 7 × 10/5 = 10 – 14 = – 4

x = 10 – 7 × 15/5 = 10 – 21 = – 11

x3-4-11
y51015

Given that 7 pencils and 5 pens together cost Rs 46
7x + 5y = 46
Subtracting 7x both side we get
5y = 46 – 7x
Dividing by 5 we get
y = 46/5 – 7x/5
y = 9.2 – 1.4x
Putting x = 0 , 2 and 4 we get
y = 9.2 – 1.4 × 0 = 9.2 – 0 = 9.2
y = 9.2 – 1.4 (2) = 9.2 – 2.8 = 6.4
y = 9.2 – 1.4 (4) = 9.2 – 5.6 = 3.6

x 024
y9.26.43.6

Graphical representation

Linear_Equation_5

Therefore, cost of one pencil = Rs 3 and cost of one pen = Rs 5.

  1. On comparing the ratios a/ab/b₂and C/c₂, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident.

Solution:

(i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0

Comparing these equation with
ax + by + C = 0
ax + by + C= 0

We get
a = 5, b= -4, and C = 8
a=7, b₂ = 6 and C₂= -9
a₁/a₂ = 5/7,
b₁/b₂ = -4/6 and
C/C= 8/-9
Hence, a/a ≠ b₁/b₂

Therefore, both are intersecting lines at one point.

(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
Comparing these equations with

ax + by + C = 0
ax + by + C= 0

We get
a = 9, b= 3, and C = 12
a = 18, b₂ = 6 and C= 24
a/a = 9/18 = 1/2
b/b₂ = 3/6 = 1/2 and
C/C= 12/24 = 1/2
Hence, a/a = b/b₂ = C/C
Therefore, both lines are coincident

(iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
Comparing these equations with

ax + by + c = 0
ax + by + c= 0

We get
a = 6, b= -3, and C = 10
a = 2, b₂ = -1 and C₂= 9
a/a = 6/2 = 3/1
b/b₂ = -3/-1 = 3/1 and
C/C= 12/24 = 1/2
Hence, a/a₂ = b/b₂  C/c₂
Therefore, both lines are parallel

  1. On comparing the ratios a/ab/b₂and C/Cfind out whether the following pair of linear equations are consistent, or inconsistent.
    (i) 3x + 2y = 5 ; 2x – 3y = 7
    (ii) 2x – 3y = 8 ; 4x – 6y = 9
    (iii) 3/2x + 5/3y = 7 ; 9– 10y = 14
    (iv) 5x – 3y = 11 ; – 10x + 6y = –22
    (v) 4/3x + 2y =8 ; 2x + 3y = 12

Solution:

(i) 3x + 2y = 5 ; 2x – 3y = 7
a/a = 3/2
b/b₂ = -2/3 and
C/C= 5/7
Hence, a/a ≠ b₁/b₂
These linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

(ii) 2x – 3y = 8 ; 4x – 6y = 9
a/a = 2/4 = 1/2
b/b₂ = -3/-6 = 1/2 and
C/C= 8/9
Hence, a/a₂ = b/b C/c2

Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.

(iii) 3/2x + 5/3y = 7 ; 9– 10y = 14
a/a = 3/2/9 = 1/6
b/b₂ = 5/3/-10 = -1/6 and
C/C= 7/14 = 1/2
Hence, a/a₂ ≠ b/b₂

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

(iv) 5x – 3y = 11 ; – 10x + 6y = –22
a/a = 5/-10 = -1/2
b/b₂ = -3/6 = -1/2 and
C/C= 11/-22 = -1/2
Hence, a/a₂ = b/b₂ = C/cb₂

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

(v) 4/3x + 2y =8 ; 2x + 3y = 12
a/a = 4/3/2 = 2/3
b/b₂ = /3 and
C/C= 8/12 = 2/3
Hence, a/a₂ = b/b₂C/C

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

  1. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
    (i)xy = 5, 2x + 2y = 10

(ii) x – y = 8, 3x – 3y = 16

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

Solution:

(i) x + y = 5; 2x + 2y = 10
a/a = 1/2
b/b₂ = 1/2 and
C/C= 5/10 = 1/2
Hence, a/a₂ = b/b₂ = C/c₂
Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

x + y = 5

x = 5 – y

x432
y123

And, 2x + 2y = 10
x = 10-2y/2

x432
y123

Graphical representation

Linear_Equation_3.2(4)

From the figure, it can be observed that these lines are overlapping each other. Therefore, infinite solutions are possible for the given pair of equations.

(ii) x – y = 8, 3x – 3y = 16
a/a = 1/3
b/b₂ = -1/-3 = 1/3 and
C/C= 8/16 = 1/2
Hence, a/a₂ = b/b₂  C/C₂  Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
a/a = 2/4 = 1/2
b/b₂ = -1/2 and
C/C= -6/-4 = 3/2
Hence, a/a ≠ b₁/b₂
Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

2x + y – 6 = 0
y = 6 – 2x

x 012
y642

And, 4x – 2y -4 = 0
y = 4x – 4/2

x123
y 024

Graphical representation

Linear_Equation_3.2 _4(iii)

From the figure, it can be observed that these lines are intersecting each other at the only one point i.e., (2,2) which is the solution for the given pair of equations.

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
a/a = 2/4 = 1/2
b/b₂= -2/-4 = 1/2 and
C/C= 2/5
Hence, a/a₂ = b/b₂  C/C

Therefore, these linear equations are parallel to each other and thus, have no possible solution. Hence, the pair of linear equations is inconsistent.

  1. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Solution:

Let length of rectangle = x m
Width of the rectangle = m
According to the question,
y – x = 4 … (i)
y + x = 36 … (ii)
y – x = 4
y = x + 4

x812
y41216

y + x = 36

x3616
y3620

Graphical representation

Linear_Equation_3.2 _4(v)

From the figure, it can be observed that these lines are intersecting each other at only point i.e., (16, 20). Therefore, the length and width of the given garden is 20 m and 16 m respectively.

  1. Given the linear equation 2x+ 3y– 8 = 0, write another linear equations in two variables such that the geometrical representation of the pair so formed is:

(i) intersecting lines
(ii) parallel lines
(iii) coincident lines

Solution:

(i) Intersecting lines:
For this condition,
a/a ≠ b₁/b
The second line such that it is intersecting the given line is
2x + 4y – 6 = 0 as
a/a = 2/2 = 1
b/b₂ = 3/4 and
a/a ≠ b₁/b

(ii) Parallel lines

For this condition,

a/a₂ = b/b₂  C/c₂
Hence, the second line can be
4x + 6y – 8 = 0 as
a/a = 2/4 = 1/2
b/b₂ = 3/6 = 1/2 and
C/C= -8/-8 = 1
and a/a₂ = b/b₂  C/c₂

(iii) Coincident lines
For coincident lines,
a/a₂ = b/b₂C/c₂
Hence, the second line can be
6x + 9y – 24 = 0 as
a/a = 2/6 = 1/3
b/b₂ = 3/9 = 1/3 and
C/C= -8/-24 = 1/3
and a/a₂ = b/b₂ = C/c₂

7. Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Solution:

x – y + 1 = 0
x = y – 1

x12
y123

3x + 2y – 12 = 0

x = 12 – 2y/3

x42
y36

Graphical representation

Linear_Equation_3.2(7)

From the figure, it can be observed that these lines are intersecting each other at point (2, 3) and x-axis at ( – 1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), ( – 1, 0), and (4, 0).

Exercise 3.3

  1. Solve the following pair of linear equations by the substitution method.

(i) = 14 ; – = 4
(ii) – = 3 ; s/3 + t/2 = 6
(iii) 3x – y = 3 ; 9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3 ; 0.4x + 0.5y = 2.3
(v) √2x+ √3y = 0 ; √3x – √8y = 0
(vi) 3/2x – 5/3y = -2 ; x/3 + y/2 = 13/6

Solution:

(i) x + y = 14 … (i)

x – y = 4 … (ii)
From equation (i), we get

x = 14 – y … (iii)
Putting this value in equation (ii), we get

(14 – y) – y = 4

14 – 2y = 4

10 = 2y

= 5 … (iv)

Putting this in equation (iii), we get

= 9

∴ = 9 and y = 5

(ii) – = 3 … (i)
s/3 + t/2 = 6 … (ii)
From equation (i), we gett + 3
Putting this value in equation (ii), we get
t+3/3 + t/2 = 6
2t + 6 + 3t = 36
5t = 30
t = 30/5 … (iv)
Putting in equation (iii), we obtain
s = 9
∴ s = 9, t = 6

(iii) 3x – = 3 … (i)
9x – 3y = 9 … (ii)
From equation (i), we get
y = 3x – 3 … (iii)
Putting this value in equation (ii), we get
9x – 3(3x – 3) = 9
9x – 9x + 9 = 9
9 = 9
This is always true.
Hence, the given pair of equations has infinite possible solutions and the relation between these variables can be given by
y = 3x – 3
Therefore, one of its possible solutions is x = 1, y = 0.

(iv) 0.2x + 0.3y = 1.3 … (i) 
0.4x + 0.5y = 2.3 … (ii)
0.2x + 0.3y = 1.3
Solving equation (i), we get
0.2x = 1.3 – 0.3y
Dividing by 0.2, we get
x = 1.3/0.2 – 0.3/0.2
x = 6.5 – 1.5 y …(iii)
Putting the value in equation (ii), we get
0.4x + 0.5y = 2.3
(6.5 – 1.5y) × 0.4x + 0.5y = 2.3
2.6 – 0.6y + 0.5y = 2.3
-0.1y = 2.3 – 2.6
= -0.3/-0.1
y = 3
Putting this value in equation (iii) we get
x = 6.5 – 1.5 y
x = 6.5 – 1.5(3)
x = 6.5 – 4.5
x = 2
∴ = 2 and y = 3

Linear_Equation_3.3_2

  1. Solve 2x+ 3y= 11 and 2x – 4y = – 24 and hence find the value of ‘m‘ for which y =mx + 3.

Solution:

2x + 3= 11 … (i)
Subtracting 3y both side we get
2x = 11 – 3y … (ii)
Putting this value in equation second we get
2x – 4y = – 24 … (iii)
11- 3y – 4y = – 24
7y = – 24 – 11
-7y = – 35
y = – 35/-7
y = 5
Putting this value in equation (iii) we get
2x = 11 – 3 × 5
2x = 11- 15
2x = – 4
Dividing by 2 we get
x = – 2
Putting the value of x and y
y = mx + 3.
5 = -2m +3
2m = 3 – 5
m = -2/2
m = -1

  1. Form the pair of linear equations for the following problems and find their solution by substitution method
    (i) The difference between two numbers is 26 and one number is three times the other. Find them.

Solution:

Let larger number = x

Smaller number = y
The difference between two numbers is 26
x – y = 26
x = 26 + y
Given that one number is three times the other
So x = 3y
Putting the value of x we get
26y = 3y
-2y = – 2 6
y = 13
So value of x = 3y
Putting value of y, we get
x = 3 × 13 = 39
Hence the numbers are 13 and 39.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Solution:

Let first angle = x
And second number = y
As both angles are supplementary so that sum will 180
x + y = 180
x = 180 – y … (i)
Difference is 18 degree so that
x – y = 18
Putting the value of we get
180 – y – y = 18
– 2y = -162
y = -162/-2
y = 81
Putting the value back in equation (i), we get
x = 180 – 81 = 99Hence, the angles are 99º and 81º.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

Solution:

Let cost of each bat = Rs x
Cost of each ball = Rs y

Given that coach of a cricket team buys 7 bats and 6 balls for Rs 3800.
7x + 6y = 3800
6y = 3800 – 7x
Dividing by 6, we get
y = (3800 – 7x)/6 … (i)

Given that she buys 3 bats and 5 balls for Rs 1750 later.
3x + 5= 1750
Putting the value of y
3x + 5 ((3800 – 7x)/6) = 1750
Multiplying by 6, we get
18x + 19000 – 35x = 10500
-17x =10500 – 19000
-17x = -8500
x = – 8500/- 17
= 500
Putting this value in equation (i) we get
= ( 3800 – 7 × 500)/6
= 300/6
= 50

Hence cost of each bat = Rs 500 and cost of each balls = Rs 50.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for traveling a distance of 25 km?

Solution:

Let the fixed charge for taxi = Rs x
And variable cost per km = Rs y
Total cost = fixed charge + variable charge
Given that for a distance of 10 km, the charge paid is Rs 105
x + 10y = 105 … (i)
x = 105 – 10y
Given that for a journey of 15 km, the charge paid is Rs 155
x + 15y = 155
Putting the value of x we get
105 – 10y + 15y = 155
5y = 155 – 105
5y = 50
Dividing by 5, we get
y = 50/5 = 10
Putting this value in equation (i) we get
= 105 – 10 × 10
= 5
People have to pay for traveling a distance of 25 km
x + 25y
= 5 + 25 × 10
= 5 + 250
=255

A person have to pay Rs 255 for 25 Km.

(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6 . Find the fraction.

Solution:

Let Numerator = x
Denominator = y
Fraction will = x/y
A fraction becomes 9/11, if 2 is added to both the numerator and the denominator
(x + 2)/y+2 = 9/11

By Cross multiplication, we get
11x + 22 = 9y + 18
Subtracting 22 both side, we get
11x = 9y – 4
Dividing by 11, we get
= 9y – 4/11 … (i)
Given that 3 is added to both the numerator and the denominator it becomes 5/6.
If, 3 is added to both the numerator and the denominator it becomes 5/6
(x+3)/+3  = 5/6 … (ii)
By Cross multiplication, we get
6x + 18 = 5y + 15
Subtracting the value of x, we get
6(9y – 4 )/11 + 18 = 5y + 15
Subtract 18 both side we get
6(9y – 4 )/11 = 5y – 3
54 – 24 = 55– 33
y = -9
y = 9
Putting this value of y in equation (i), we get
= 9y – 4
11 … (i)
x = (81 – 4)/77
x = 77/11
x = 7

Hence our fraction is 7/9.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Solution:

Let present age of Jacob = x year
And present Age of his son is = y year
Five years hence,
Age of Jacob will = x + 5 year
Age of his son will = y + 5year
Given that the age of Jacob will be three times that of his son
x + 5 = 3(+ 5)
Adding 5 both side, we get
x = 3y + 15 – 5
x = 3y + 10 … (i)
Five years ago,
Age of Jacob will = x – 5 year
Age of his son will = y – 5 year
Jacob’s age was seven times that of his son
x – 5 = 7(y -5)
Putting the value of x from equation (i) we get
3+ 10 – 5 = 7y – 35
3y + 5 = 7y – 35
3y – 7y = -35 – 5
-4y = – 40
y = – 40/- 4
y = 10 year
Putting the value of y in equation first we get
x = 3 × 10 + 10
x = 40 years
Hence, Present age of Jacob = 40 years and present age of his son = 10 years.

Exercise 3.4

  1. Solve the following pair of linear equations by the elimination method and the substitution method:
    (i)xy =5 and 2x –3y = 4
    (ii) 3x + 4y = 10 and 2x – 2y = 2
    (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
    (iv) x/2 + 2y/3 = – 1 and x – y/3 = 3

Solution:

(i) x + y =5 and 2x –3y = 4
By elimination method
x + y =5 … (i)
2x –3y = 4 … (ii)
Multiplying equation (i) by (ii), we get
2x + 2y = 10 … (iii)
2x –3y = 4 … (ii)
Subtracting equation (ii) from equation (iii), we get
5y = 6
y = 6/5
Putting the value in equation (i), we get
x = 5 – (6/5) = 19/5

Hence, x = 19/5 and y = 6/5

By substitution methodx + y = 5 … (i)
Subtracting y both side, we get
x = 5 – y … (iv)
Putting the value of x in equation (ii) we get
2(5 – y) – 3y = 4
-5y = – 6
y = -6/-5 = 6/5
Putting the value of y in equation (iv) we get
x = 5 – 6/5
x = 19/5
Hence, x = 19/5 and y = 6/5 again

(ii) 3x + 4y = 10 and 2x – 2y = 2
By elimination method
3x + 4y = 10 …. (i)
2x – 2y = 2 … (ii)
Multiplying equation (ii) by 2, we get
4x – 4y = 4 … (iii)
3x + 4y = 10 … (i)
Adding equation (i) and (iii), we get
7x + 0 = 14
Dividing both side by 7, we get
x = 14/7 = 2
Putting in equation (i), we get
3x + 4y = 10
3(2) + 4y = 10
6 + 4y = 10
4y = 10 – 6
4y = 4
y = 4/4 = 1

Hence, answer is x = 2, y = 1

By substitution method
3x + 4y = 10 … (i)
Subtract 3x both side, we get
4y = 10 – 3x
Divide by 4 we get
y = (10 – 3x )/4
Putting this value in equation (ii), we get
2x – 2y = 2 … (i)
2x – 2(10 – 3x )/4) = 2
Multiply by 4 we get
8x – 2(10 – 3x) = 8
8x – 20 + 6x = 8
14x = 28
x = 28/14 = 2
= (10 – 3x)/4

= 4/4 = 1

Hence, answer is x = 2, y = 1 again.

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
By elimination method
3x – 5y – 4 = 0
3x – 5y = 4 …(i)
9x = 2y + 7
9x – 2= 7 … (ii)
Multiplying equation (i) by 3, we get
9 x – 15 y = 11 … (iii)
9x – 2y = 7 … (ii)
Subtracting equation (ii) from equation (iii), we get
-13y = 5
y = -5/13
Putting value in equation (i), we get
3x – 5y = 4 … (i)
3x – 5(-5/13) = 4
Multiplying by 13 we get
39x + 25 = 52
39x = 27
x =27/39 = 9/13
Hence our answer is x = 9/13 and y = – 5/13

By substitution method
3x – 5y = 4 … (i)
Adding 5y both side we get
3x = 4 + 5y
Dividing by 3 we get
x = (4 + 5y )/3 … (iv)
Putting this value in equation (ii) we get
9x – 2y = 7 … (ii)
9 ((4 + 5)/3) – 2y = 7
Solve it we get
3(4 + 5y ) – 2y = 7
12 + 15y – 2y = 7
13y = – 5
y = -5/13

Hence we get x = 9/13 and y = – 5/13 again.

(iv) x/2 + 2y/3 = – 1 and x – y/3 = 3
By elimination method
x/2 + 2y/3 = -1 … (i)
x – y/3 = 3 … (ii)
Multiplying equation (i) by 2, we get
x + 4y/3 = – 2 … (iii)
x – y/3 = 3 … (ii)
Subtracting equation (ii) from equation (iii), we get
5y/3 = -5
Dividing by 5 and multiplying by 3, we get
= -15/5
= – 3
Putting this value in equation (ii), we get
x – y/3 = 3 … (ii)
x – (-3)/3 = 3
x + 1 = 3
x = 2

Hence our answer is x = 2 and y = −3.

By substitution method
x – y/3 = 3 … (ii)
Add y/3 both side, we get
= 3 + y/3 … (iv)
Putting this value in equation (i) we get
x/2 + 2y/3 = – 1 … (i)
(3+ y/3)/2 + 2y/3 = -1
3/2 + y/6 + 2y/3 = – 1
Multiplying by 6, we get
9 + y + 4= – 6
5y = -15
y = – 3

Hence our answer is x = 2 and y = −3.

  1. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?

(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?

(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.

(iv) Meena went to 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.

(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:

(i) Let the fraction be x/y
According to the question,x + 1/y – 1 = 1
⇒ – = -2 … (i)x/y+1 = 1/2
⇒ 2x – = 1 … (ii)
Subtracting equation (i) from equation (ii), we get
x = 3 … (iii)
Putting this value in equation (i), we get
3 – y = -2
y = -5
y = 5
Hence, the fraction is 3/5

(ii) Let present age of Nuri = x
and present age of Sonu = y
According to the given information,question,(x – 5) = 3(y – 5)
x – 3y = -10 … (i)
(x + 10y) = 2(y + 10)
x – 2y = 10 … (ii)
Subtracting equation (i) from equation (ii), we get
y = 20 … (iii)
Putting this value in equation (i), we get
x – 60 = -10
= 50
Hence, age of Nuri = 50 years and age of Sonu = 20 years.

(iii) Let the unit digit and tens digits of the number be and yrespectively.
Then, number = 10y + x
Number after reversing the digits = 10x + y
According to the question,
x + y = 9 … (i)
9(10x) = 2(10x + y)
88y – 11x = 0
– x + 8y =0 … (ii)
Adding equation (i) and (ii), we get
9y = 9
y = 1 … (iii)
Putting the value in equation (i), we get
x = 8
Hence, the number is 10y + x = 10 × 1 + 8 = 18.

(iv) Let the number of Rs 50 notes and Rs 100 notes be x and yrespectively.
According to the question,
= 25 … (i)
50x + 100y = 2000 … (ii)
Multiplying equation (i) by 50, we get
50x + 50y = 1250 … (iii)
Subtracting equation (iii) from equation (ii), we get
50y = 750
y = 15
Putting this value in equation (i), we have x = 10
Hence, Meena has 10 notes of Rs 50 and 15 notes of Rs 100.

(v) Let the fixed charge for first three days and each day charge thereafter be Rs and Rs respectively.
According to the question,
+ 4y = 27 … (i)
+ 2y = 21 … (ii)
Subtracting equation (ii) from equation (i), we get
2y = 6
y = 3 … (iii)
Putting in equation (i), we get
x + 12 =27
x = 15
Hence, fixed charge = Rs 15 and Charge per day = Rs 3.

Exercise 3.5  

  1. Which of the following pairs of linear equations has unique solution, no solution or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method.

(i) x – 3y – 3 = 0 ; 3x – 9y – 2 =0
(ii) 2x + y = 5 ; 3x +2y =8
(iii) 3x – 5y = 20 ; 6x – 10y =40
(iv) x – 3y – 7 = 0 ; 3x – 3y – 15= 0

Solution:

(i) x – 3y – 3 = 0
3x – 9y – 2 =0
a/a= 1/3
b/b= -3/-9 = 1/3 and
C/C = -3/-2 = 3/2
a/a₂= b/b C/C

 Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect each other and thus, there will not be any solution for these equations.

(ii) 2x + y = 5
3x +2y = 8
a/a= 2/3
b/b= 1/2 and
C/C = -5/-8 = 5/8
a/a₂≠ b/b2

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication method,
x/bCb2C = y/Ca2Ca = 1/ab2a2b
x/-8-(-10) = y/-15+16 = 1/4-3
x/2 = y/1 = 1
x/2 = 1, y/1 = 1
∴ = 2, = 1.

(iii) 3x – 5y = 20
6x – 10y = 40
a/a= 3/6 = 1/2
b/b= -5/-10 = 1/2 and
C/C = -20/-40 = 1/2
a/a₂= b/b= C/C

Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite solutions possible for these equations.

(iv) x – 3y – 7 = 0
3x – 3y – 15= 0
a/a= 1/3
b/b= -3/-3 = 1 and
C/C = -7/-15 = 7/15
a/a₂≠ b/b2

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication,

x/45-(21) = y/-21-(-15) = 1/-3-(-9)
x/24 = y/-6 = 1/6
x/24 = 1/6 and y/-6 = 1/6
= 4 and = -1
∴ = 4, = -1.

  1. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?
    2x+ 3y=7

(a – b)x + (a + b)y = 3a +b –2

Solution:

2x + 3y -7 = 0

(a – b)x + (a + b)y – (3a +b –2) = 0

a/a= 2/ab = 1/2
b/b= -7/a+b and
C/C = -7/-(3a+b-2) = 7/(3a+b-2)
For infinitely many solutions,a/ab/bC/C

2/a= 7/3a+b-26a + 2b – 4 = 7a – 7b
a – 9b = -4 … (i)

2/a= 3/a+b
2a + 2b = 3a – 3b
a – 5b = 0 … (ii)

Subtracting equation (i) from (ii), we get
4b = 4
b = 1
Putting this value in equation (ii), we get
a – 5 × 1 = 0
a = 5
Hence, a = 5 and b = 1 are the values for which the given equations give infinitely many solutions.

(ii) For which value of k will the following pair of linear equations have no solution?

3x + = 1

(2k –1)x + (k –1)y = 2k + 1

Solution:

3x + -1 = 0

(2k –1)x + (k –1)y – (2k + 1) = 0

a/a= 3/2k-1
b/b= 1/k-1 and
C/C = -1/-2k-1 = 1/2k+1
For no solutions,
a/ab/b C/C
3/2k-1 = 1/k-1 ≠ 1/2k+1

3/2k-1 = 1/k-1
3k – 3 = 2k – 1
k = 2
Hence, for k = 2, the given equation has no solution.

  1. Solve the following pair of linear equations by the substitution and cross-multiplication methods:
    8x+5y= 9
    3x +2y = 4

Solution:

8x +5y = 9 … (i)
3x +2y = 4 … (ii)
From equation (ii), we get
x = 4-2y/3 … (iii)
Putting this value in equation (i), we get
8(4-2y/3) + 5y = 9
32 – 16y +15y = 27
y = -5
y = 5 … (iv)
Putting this value in equation (ii), we get
3x + 10 = 4
x = -2
Hence, x = -2, = 5
By cross multiplication again, we get

8x + 5y -9 = 0

3x + 2y – 4 = 0

x/-20-(-18) = y/-27-(-32) = 1/16-15
x/-2 = y/5 = 1/1
x/-2 = 1 and y/5 = 1
x = -2 and y = 5

  1. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.

Solution:Let be the fixed charge of the food and be the charge for food per day.

According to the question,

x + 20y = 1000 … (i)

x + 26y = 1180 … (ii)
Subtracting equation (i) from equation (ii), we get

6y = 180

y = 180/6 = 30
Putting this value in equation (i), we get
x + 20 × 30 = 1000
x = 1000 – 600
x = 400
Hence, fixed charge = Rs 400 and charge per day = Rs 30

(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. Find the fraction.

Solution:
Let the fraction be x/y
According to the question,

x-1/y = 1/3
⇒ 3x – y = 3… (i)
x/y+8 = 1/4
⇒ 4x – y = 8 … (ii)
Subtracting equation (i) from equation (ii), we get

x = 5 … (iii)

Putting this value in equation (i), we get

15 – = 3

= 12
Hence, the fraction is 5/12.

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

Solution: Let the number of right answers and wrong answers be and respectively.

According to the question,

3xy = 40 … (i)
4x – 2y = 50
⇒ 2xy = 25 … (ii)

Subtracting equation (ii) from equation (i), we get
x = 15 … (iii)
Putting this value in equation (ii), we get

30 – y = 25
y = 5
Therefore, number of right answers = 15
And number of wrong answers = 5
Total number of questions = 20

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?Solution:

Let the speed of 1st car and 2nd car be u km/h and v km/h.
Respective speed of both cars while they are travelling in same direction = (u – v) km/h

Respective speed of both cars while they are travelling in opposite directions i.e., travelling towards each other = (+ v) km/h

According to the question,
5(u v) = 100
uv = 20 … (i)
1(u + v) = 100 … (ii)

Adding both the equations, we get

2u = 120
u = 60 km/h … (iii)

Putting this value in equation (ii), we obtain
v = 40 km/h
Hence, speed of one car = 60 km/h and speed of other car = 40 km/h

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Solution:

Let length and breadth of rectangle be x unit and y unit respectively.
Area = xy
According to the question,
(x – 5) (y + 3) = xy – 9
⇒ 3x – 5y – 6 = 0 … (i)
(x + 3) (y + 2) = xy + 67
⇒ 2x – 3y – 61 = 0 … (ii)
By cross multiplication, we get
x/305-(-18) = y/-12-(-183) = 1/9-(-10)
x/323 = y/171 = 1/19
x = 17, y = 9
Hence, the length of the rectangle = 17 units and breadth of the rectangle = 9 units.

Exercise 3.6

1. Solve the following pairs of equations by reducing them to a pair of linear equations:
(i) 1/2x + 1/3y = 2
1/3x + 1/2y = 13/6

(ii) 2/√x +3/√y = 2
4/√x – 9/√y = -1

(iii) 4/x + 3y = 14
3/x – 4y = 23

(iv) 5/x-1 + 1/y-2 = 2
6/x-1 – 3/y-2 = 1

(v) 7x-2y/xy = 5
8x + 7y/xy = 15

(vi) 6x + 3y = 6xy
2x + 4y = 5xy

(vii) 10/x+y + 2/xy = 4
15/x+y – 5/xy = -2

(viii) 1/3x+y + 1/3xy = 3/4
1/2(3xy) – 1/2(3xy) = -1/8

Solution:

(i) 1/2x + 1/3y = 2
1/3x + 1/2= 13/6
Let 1/x = p and 1/y = q, then the equations changes as below:
p/2 + q/3 = 2
⇒ 3p + 2q -12 = 0 … (i)
p/3 + q/2 = 13/6
⇒ 2p + 3q -13 = 0 … (ii)

By cross-multiplication method, we get

p/-26-(-36) = q/-24-(-39) = 1/9-4
p/10 = q/15 = 1/5
p/10 = 1/5 and q/15 = 1/5
p = 2 and q = 3
1/x = 2 and 1/y = 3
Hence, x = 1/2 and y = 1/3

(ii) 2/√x +3/√y = 2
4/√x – 9/√y = -1
Let 1/√= p and 1/√y = q, then the equations changes as below:
2p + 3q = 2 … (i)
4p – 9q = -1 … (ii)
Multiplying equation (i) by 3, we get
6p + 9q = 6 … (iii)
Adding equation (ii) and (iii), we get

10p = 5
p = 1/2 … (iv)
Putting in equation (i), we get
2 × 1/2 + 3q = 2
3q = 1
q = 1/3p = 1/√x = 1/2
x = 2
x = 4
and
q = 1/√y = 1/3
y = 3
y = 9
Hence, x = 4, y = 9

(iii) 4/x + 3y = 14
3/x – 4y = 23
Putting 1/x = p in the given equations, we get
4p + 3y = 14
⇒ 4p + 3y – 14 = 0
3p – 4y = 23
⇒ 3p – 4y -23 = 0
By cross-multiplication, we get
p/-69-56 = y/-42-(-92) = 1/-16-9
⇒ –p/125 = y/50 = -1/25
Now,
p/125 = -1/25 and y/50 = -1/25
p = 5 and y = -2
Also, p = 1/x = 5
x = 1/5
So, x = 1/5 and y = -2 is the solution.

(iv) 5/x-1 + 1/y-2 = 2
6/x-1 – 3/y-2 = 1
Putting 1/x-1 = p and 1/y-2 = q in the given equations, we obtain
5p + q = 2 … (i)
6p – 3q = 1 … (ii)
Now, by multiplying equation (i) by 3 we get
15p + 3q = 6 … (iii)
Now, adding equation (ii) and (iii)
21p = 7
p = 1/3
Putting this value in equation (ii) we get,
6×1/3 – 3q =1
⇒ 2-3q = 1
⇒ -3q = 1-2
⇒ -3q = -1
q = 1/3
Now,
p = 1/x-1 = 1/3
⇒1/x-1 = 1/3
⇒ 3 = – 1
x = 4
Also,
q = 1/y-2 = 1/3
⇒ 1/y-2 = 1/3
⇒ 3 = y-2
y = 5
Hence, x = 4 and y = 5 is the solution.

(v) 7x-2y/xy = 5
⇒ 7x/xy – 2y/xy = 5
⇒ 7/y – 2/x = 5 … (i)
8x+7y/xy = 15
⇒ 8x/xy + 7y/xy = 15
⇒ 8/y + 7/x = 15 … (ii)
Putting 1/x = p and 1/y = q in (i) and (ii) we get,
7q – 2p = 5 … (iii)
8q + 7p = 15 … (iv)
Multiplying equation (iii) by 7 and multiplying equation (iv) by 2 we get,
49q – 14p = 35 … (v)
16q + 14p = 30 … (vi)
Now, adding equation (v) and (vi) we get,
49q – 14p + 16q + 14p = 35 + 30
⇒ 65q = 65
q = 1
Putting the value of q in equation (iv)
8 + 7p = 15
⇒ 7p = 7
p = 1
Now,
p = 1/x = 1
⇒ 1/x = 1
x = 1
also, q = 1 = 1/y
⇒ 1/y = 1
y = 1
Hence, =1 and y = 1 is the solution.

(vi) 6x + 3y = 6xy
⇒ 6x/xy + 3y/xy = 6
⇒ 6/y + 3/x = 6 … (i)
2x + 4y = 5xy
⇒ 2x/xy + 4y/xy = 5
⇒ 2/y + 4/x = 5 … (ii)
Putting 1/x = p and 1/y = q in (i) and (ii) we get,
6q + 3p – 6 = 0
2q + 4p – 5 = 0
By cross multiplication method, we get
p/-30-(-12) = q/-24-(-15) = 1/6-24
p/-18 = q/-9 = 1/-18
p/-18 = 1/-18 and q/-9 = 1/-18
p = 1 and q = 1/2
p = 1/x = 1 and q = 1/y = 1/2
x = 1, y = 2
Hence, x = 1 and y = 2

(vii) 10/x+y + 2/xy = 4
15/x+y – 5/xy = -2
Putting 1/x+y = p and 1/xy = q in the given equations, we get:
10p + 2q = 4
⇒ 10p + 2q – 4 = 0 … (i)
15p – 5q = -2
⇒ 15p – 5q + 2 = 0 … (ii)
Using cross multiplication, we get
p/4-20 = q/-60-(-20) = 1/-50-30
p/-16 = q/-80 = 1/-80
p/-16 = 1/-80 and q/-80 = 1/-80
p = 1/5 and q = 1
p = 1/x+y = 1/5 and q = 1/xy = 1
x + y = 5 … (iii)
and xy = 1 … (iv)
Adding equation (iii) and (iv), we get
2x = 6
x = 3 …. (v)
Putting value of x in equation (iii), we get
y = 2
Hence, x = 3 and y = 2

(viii) 1/3x+y + 1/3xy = 3/4
1/2(3xy) – 1/2(3xy) = -1/8
Putting 1/3x+y = p and 1/3xy = q in the given equations, we get
p + q = 3/4 … (i)
p/2 – q/2 = -1/8
p q = -1/4 … (ii)
Adding (i) and (ii), we get
2p = 3/4 – 1/4
2p = 1/2
p = 1/4
Putting the value in equation (ii), we get
1/4 – q = -1/4
q = 1/4 + 1/4 = 1/2
p = 1/3x+y = 1/4
3x + y = 4 … (iii)
q = 1/3xy = 1/2
3x – y = 2 … (iv)
Adding equations (iii) and (iv), we get
6x = 6
x = 1 … (v)
Putting the value in equation (iii), we get
3(1) + y = 4
y = 1
Hence, x = 1 and y = 1

2. Formulate the following problems as a pair of equations, and hence find their solutions:

(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

Solution:

Let the speed of Ritu in still water and the speed of stream be xkm/h
and y km/h respectively.
Speed of Ritu while rowing
Upstream = (xy) km/h

Downstream = (x + y) km/h

According to question,

2(x + y) = 20
x + y = 10 … (i)
2(xy) = 4
xy = 2 … (ii)
Adding equation (i) and (ii), we get
Putting this equation in (i), we get
y = 4

Hence, Ritu’s speed in still water is 6 km/h and the speed of the current is 4 km/h.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

Solution:

Let the number of days taken by a woman and a man be x and yrespectively.
Therefore, work done by a woman in 1 day = 1/x

According to the question,
4(2/x + 5/y) = 1
2/x + 5/y = 1/4
3(3/x + 6/y) = 1
3/x + 6/y = 1/3
Putting 1/x = p and 1/= q in these equations, we get
2p + 5q = 1/4
By cross multiplication, we get
p/-20-(-18) = q/-9-(-18) = 1/144-180
p/-2 = q/-1 = 1/-36
p/-2 = -1/36 and q/-1 = 1/-36
p = 1/18 and q = 1/36
p = 1/x = 1/18 and q = 1/y = 1/36
x = 18 and y = 36
Hence, number of days taken by a woman = 18 and number of days taken by a man = 36

(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.Answer

Let the speed of train and bus be u km/h and v km/h respectively.
According to the given information,

60/u + 240/v = 4 … (i)
100/u + 200/v = 25/6 … (ii)
Putting 1/u = p and 1/v = q in the equations, we get
60p + 240q = 4 … (iii)
100p + 200q = 25/6
600p + 1200q = 25 … (iv)
Multiplying equation (iii) by 10, we get
600p + 2400q = 40 …. (v)
Subtracting equation (iv) from (v), we get1200q = 15
q = 15/200 = 1/80 … (vi)
Putting equation (iii), we get
60p + 3 = 4
60p = 1
p = 1/60
p = 1/u = 1/60 and q = 1/v = 1/80
u = 60 and v = 80
Hence, speed of train = 60 km/h and speed of bus = 80 km/h.

 

Xpert Solutions For Class X Math Chapter 2: Polynomials

Excercise 2.1

  1. The graphs of y = p(x) are given in following figure, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

Sol:

graph

(i) The number of zeroes is 0 as the graph does not cut the x-axis at any point.
(ii) The number of zeroes is 1 as the graph intersects the x-axis at only 1 point.
(iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.
(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points.
(v) The number of zeroes is 4 as the graph intersects the x-axis at 4 points.
(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

Exercise 2.2

  1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
    (i) x²– 2x– 8
    (ii) 4s² – 4s + 1
    (iii) 6x² – 3 – 7x
    (iv) 4u² + 8u
    (v) t² – 15
    (vi) 3x² – x – 4

Sol:

(i) x² – 2x – 8

To make factors given below the method:
The Product of  Factors should be: 1 × -8  = -8   (Coefficient of  X²  × constant value)
And   The sum of factors using above value should be   -2 (Coefficient of X)

fac

2, 2, 2       (Therefore    -4 + 2 = -2,   -4 × 2 = -8)
= (x – 4) (x + 2)
The value of x² – 2x – 8 is zero when x – 4 = 0 or x + 2 = 0, i.e., when x = 4 or x = -2
Therefore, the zeroes of x² – 2x – 8 are 4 and -2.
Sum of zeroes = 4 + (-2) = 2 = -(-2)/1 = -(Coefficient of x)/Coefficient of x²
Product of zeroes = 4 × (-2) = -8 = -8/1 = Constant term/Coefficient of x²

(ii) 4s² – 4s + 1
= (2s-1)²                  (Using Formula Here:  (a-b)² =  a² – 2ab + b² )

The value of 4s² – 4s + 1 is zero when 2s – 1 = 0, i.e., s = 1/2

Therefore, the zeroes of 4s² – 4s + 1 are 1/2 and 1/2.
Sum of zeroes = 1/2 + 1/2 = 1 = -(-4)/4 = -(Coefficient ofs)/Coefficient of s²
Product of zeroes = 1/2 × 1/2 = 1/4 = Constant term/Coefficient of s² .

(iii) 6 – 3 – 7x
6 – 7– 3
= 6 +2x -9x – 3
[ i.e 6 × -3 = -18, L.C.M = 2 × 3 × 3, Product of -9 and 2 is -18 & Sum is -7]
= 2x(3x+1 ) – 3x(3x+1)
= (3x + 1) (2x – 3)
The value of 6 – 3 – 7x is zero when 3x + 1 = 0 or 2x – 3 = 0, i.e.,x = -1/3 or x = 3/2

Therefore, the zeroes of 6 – 3 – 7x are -1/3 and 3/2.
Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(Coefficient of x)/Coefficient of 
Product of zeroes = -1/3 × 3/2 = -1/2 = -3/6 = Constant term/Coefficient of 

(iv) 4 + 8u
4 + 8
= 4u(u + 2)
The value of 4 + 8u is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = – 2
Therefore, the zeroes of 4 + 8u are 0 and – 2.
Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of u)/Coefficient of 
Product of zeroes = 0 × (-2) = 0 = 0/4 = Constant term/Coefficient of .

(v)  – 15
 – 0.t – 15
= (– √15) (t + √15)
The value of  – 15 is zero when t – √15 = 0 or t + √15 = 0, i.e., when t = √15 or = -√15
Therefore, the zeroes of  – 15 are √15 and -√15.Sum of zeroes = √15 + -√15 = 0 = -0/1 = -(Coefficient of t)/Coefficient of t2
Product of zeroes = (√15) (-√15) = -15 = -15/1 = Constant term/Coefficient of .

(vi) 3 – x – 4
= (3x – 4) (x + 1)
The value of 3 – x – 4 is zero when 3x – 4 = 0 and x + 1 = 0,i.e., when x = 4/3 or x = -1
Therefore, the zeroes of 3 – x – 4 are 4/3 and -1.

Sum of zeroes = 4/3 + (-1) = 1/3 = -(-1)/3 = -(Coefficient of x)/Coefficient of 

Product of zeroes = 4/3 × (-1) = -4/3 = Constant term/Coefficient of .

          1. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) 1/4 , -1
(ii) √2 , 1/3
(iii) 0, √5
(iv) 1,1
(v) -1/4 ,1/4
(vi) 4,1
Sol:
(i) 1/4 , -1
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = 1/4 = –b/a
αß = -1 = -4/4 = c/a
If a = 4, then b = -1, c = -4
Therefore, the quadratic polynomial is 4 – x -4.

(ii) √2 , 1/3
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = √2 = -3√2/3 = –b/a  [multiply value  by 3/3 to equal value of a with αß  ]
αß = 1/3 = c/a
If a = 3, then b = -3√2, c = 1
Therefore, the quadratic polynomial is 3 -3√2x +1.
or

k { x² – (sum of zeros)x + Product of zeroes }
k { x² – (√2)x + 1/3 }
k{ 3x² – 3√2x + 1 }

(iii) 0, √5
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = 0 = 0/1 = –b/a
αß = √5 = √5/1 = c/a
If a = 1, then b = 0, c = √5
Therefore, the quadratic polynomial is  + √5.

(iv) 1, 1
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = 1 = 1/1 = –b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -1, c = 1
Therefore, the quadratic polynomial is  – x +1.

(v) -1/4 ,1/4
Let the polynomial be ax² bx + c, and its zeroes be α and ß
α + ß = -1/4 = –b/a
αß = 1/4 = c/a
If a = 4, then b = 1, c = 1
Therefore, the quadratic polynomial is 4 + x +1.

(vi) 4,1
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = 4 = 4/1 = –b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -4, c = 1
Therefore, the quadratic polynomial is  – 4x +1.

Exercise 2.3

  1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

Sol
(i) p(x) =  – 3x² + 5x – 3, g(x) = x² – 2

EX2.3(1)

Quotient = x-3 and remainder 7x – 9

(ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x

2.3(II)

Quotient = x² + – 3 and remainder 8

(iii) p(x) = x⁴ – 5x + 6, g(x) = 2 – x²

ex2.3(iii)

Quotient = –x² -2 and remainder -5x +10

  1. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

Sol

(i) t² – 3,  2t⁴ + 3 – 2t² – 9t – 12

t² – 3 exactly divides 2t⁴ + 3 – 2t² – 9t – 12 leaving no remainder. Hence, it is a factor of 2t⁴ + 3 – 2t² – 9t – 12.

(ii) x² + 3x + 1, 3x⁴ + 5 – 7x² + 2x + 2

ex2.3(iv)

x² + 3x + 1 exactly divides 3x⁴ + 5 – 7x² + 2x + 2 leaving no remainder. Hence, it is factor of 3x⁴ + 5 – 7x² + 2x + 2.

(iii)  – 3x + 1, x⁵ – 4 + x² + 3x + 1

 – 3x + 1 didn’t divides exactly  x⁵ – 4 + x² + 3x + 1 and leaves 2 as remainder. Hence, it not a factor of  x⁵ – 4 + x² + 3x + 1

  1. Obtain all other zeroes of 3x+ 6x³– 2x² – 10x – 5, if two of its zeroes are √(5/3) and – √(5/3).

Sol:

p(x) = 3x+ 6x³– 2x² – 10x – 5
Since the two zeroes are √(5/3) and – √(5/3).

find other zero polynomial

We factorize x² + 2+ 1 = (+ 1)²     [ From the other equ ( x²-5/3) factor are  √(5/3), – √(5/3) already given]

Therefore, its zero is given by x + 1 = 0   or   x = -1

As it has the term (+ 1)² , therefore, there will be 2 zeroes at x = – 1.
Hence,the zeroes of the given polynomial are √(5/3), – √(5/3), – 1 and – 1.

  1.  On dividing x³– 3x²x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x + 4, respectively. Find g(x).

Sol:

Here in the given question,

Dividend = x³– 3x²x + 2

Quotient = x – 2

Remainder = -2x + 4

Divisor = g(x)

We know that,

Dividend = Quotient × Divisor + Remainder

⇒ x³– 3x²x + 2 = (x – 2) × g(x) + (-2x + 4)
⇒ x³– 3x²x + 2 – (-2x + 4) = (x – 2) × g(x)
⇒ x3 – 3x2 + 3x – 2 = (x – 2) × g(x)
⇒ g(x) =  (x3 – 3x2 + 3x – 2)/(x – 2)

polynomial

∴ g(x) = (x2 – x + 1)

5.Give examples of polynomial p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0

Sol:

(i) Let us assume the division of 6x² + 2x + 2 by 2
Here, p(x) = 6x² + 2x + 2
g(x) = 2
q(x) = 3x² + x + 1
r(x) = 0
Degree of p(x) and q(x) is same i.e. 2.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
Or, 6x² + 2x + 2 = 2x (3x² + + 1)
Hence, division algorithm is satisfied.
(ii) Let us assume the division of x³x by x²,
Here, p(x) = x³ + x
g(x) = x²
q(x) = x and r(x) = x
Clearly, the degree of q(x) and r(x) is the same i.e., 1.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
x³ + x = (x² ) × x + x
x³ + x = x³ + x
Thus, the division algorithm is satisfied.

(iii) Let us assume the division of x³+ 1 by x²
Here, p(x) = x³ + 1
g(x) = x²
q(x) = x and r(x) = 1
Clearly, the degree of r(x) is 0.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
x³ + 1 = (x² ) × + 1
x³ + 1 = x³ + 1
Thus, the division algorithm is satisfied.