﻿ Class 10 NCERT Math Solution
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TOPICS
Exercise - 3.2

Question-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.
(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 :-
```(i) Let the no. of girls be x and no. of boys be y.
The algebraic representation is given by the following equations:

x + y = 10 ......(i)
x – y = 4  ......(ii)

Lets represent these equations in graphically:
x + y = 10  .......(i)

x010

y100

x – y = 4  ......(ii)

x04

y-40

Plot the points A(0, 10), B(10, 0) and P(0, -4), Q(4, 0), corresponding to the solutions in Table.
Now draw the lines AB and PQ, representing the equations x + y = 10  and x – y = 4, as shown in following Figure:

Now, Observe that the two lines representing the two equations are intersecting at the point (7, 3).
Therefore, the no. of girls and no. of boys in the class are 7 and 3 respectively.
So, it is consistent.
```
```(ii) Let the cost of pencil be x and cost of pen be y.
The algebraic representation is given by the following equations:

5x + 7y = 50 ......(i)
7x + 4y = 46 ......(ii)

Lets represent these equations in graphically:
5x + 7y = 50 ......(i)

x310

y50

7x + 4y = 46  ......(ii)

x38

y5-2

Plot the points A(3, 5), B(10, 0) and P(3, 5), Q(8, -2), corresponding to the solutions in Table.
Now draw the lines AB and PQ, representing the equations 5x + 7y = 50  and 7x + 4y = 46, as shown in following Figure:

Now, Observe that the two lines representing the two equations are intersecting at the point (3, 5).
Therefore, the cost of one pencil and cost of one pen are 3 and 5 respectively.
So, it is consistent.
```

Question-2 :-  On comparing the ratios a₁/a₂, b₁/b₂ and c₁/c₂ find out whether lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) 5x – 4y + 8 = 0; 7x + 6y – 9 = 0
(ii) 9x + 3y + 12 = 0 ; 18x + 6y + 24 = 0
(iii) 6x – 3y + 10 = 0; 2x – y + 9 = 0

Solution :-
```(i) 5x – 4y + 8 = 0; 7x + 6y – 9 = 0

a₁ = 5, b₁ = -4, c₁ = 8
a₂ = 7, b₂ = 6, c₂ = -9

Here, a₁/a₂ = 5/7,
Also, b₁/b₂ = -4/6 and
c₁/c₂ = 8/(-9)
Therefore, a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂
Now, these linear equations are intersecting each other at one point and its have only one solution.
```
```(ii) 9x + 3y + 12 = 0; 18x + 6y + 24 = 0

a₁ = 9, b₁ = 3, c₁ = 12
a₂ = 18, b₂ = 6, c₂ = 24

Here, a₁/a₂ = 9/18 = 1/2,
Also, b₁/b₂ = 3/6 = 1/2 and
c₁/c₂ = 12/24 = 1/2

Therefore, a₁/a₂ = b₁/b₂ = c₁/c₂
Now, these linear equations are coincident and its have many solution.
```
```(iii) 6x – 3y + 10 = 0; 2x – y + 9 = 0

a₁ = 6, b₁ = -3, c₁ = 10
a₂ = 2, b₂ = -1, c₂ = 9

Here, a₁/a₂ = 6/2 = 3,
Also, b₁/b₂ = -3/(-1) = 3 and
c₁/c₂ = 10/9

Therefore, a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Now, these linear equations are parallel each other and its have no solution.
```

Question-3 :-  On comparing the ratios a₁/a₂, b₁/b₂ and c₁/c₂ find 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) 3x/2 + 5y/3 = 7; 9x - 10y = 14
(iv) 5x – 3y = 11 ; – 10x + 6y = –22
(v) 4x/3 + 2y = 8; 2x + 3y = 12

Solution :-
```(i) 3x + 2y = 5 ; 2x – 3y = 7
3x + 2y - 5 = 0 ; 2x – 3y - 7 = 0

a₁ = 3, b₁ = 2, c₁ = -5
a₂ = 2, b₂ = -3, c₂ = -7

Here, a₁/a₂ = 3/2,
Also, b₁/b₂ = 2/(-3) and
c₁/c₂ = -5/(-7)= 5/7
Therefore, a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂
Now, these linear equations are intersecting each other at one point and its have only one solution.
So, It is consitent.
```
```(ii) 2x – 3y = 8 ; 4x – 6y = 9
2x – 3y - 8 = 0 ; 4x – 6y - 9 = 0

a₁ = 2, b₁ = -3, c₁ = -8
a₂ = 4, b₂ = -6, c₂ = -9

Here, a₁/a₂ = 2/4 = 1/2,
Also, b₁/b₂ = -3/(-6) = 1/2 and
c₁/c₂ = -8/(-9) = 8/9

Therefore, a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Now, these linear equations are parallel to each other and its have only no possible solution.
So, It is inconsitent.
```
```(iii) 3x/2 + 5y/3 = 7; 9x - 10y = 14
3x/2 + 5y/3 - 7 = 0; 9x - 10y - 14 = 0

a₁ = 3/2, b₁ = 5/3, c₁ = -7
a₂ = 9, b₂ = -10, c₂ = -14

Here, a₁/a₂ = (3/2)/9 = 1/6,
Also, b₁/b₂ = (5/3)/(-10) = 1/(-6) and
c₁/c₂ = -7/(-14) = 1/2

Therefore, a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂
Now, these linear equations are intersecting each other at one point and its have only one solution.
So, It is consitent.
```
```(iv) 5x – 3y = 11 ; – 10x + 6y = –22
5x – 3y - 11 = 0 ; – 10x + 6y + 22 = 0

a₁ = 5, b₁ = -3, c₁ = -11
a₂ = -10, b₂ = 6, c₂ = 22

Here, a₁/a₂ = 5/(-10) = 1/(-2),
Also, b₁/b₂ = (-3)/6 = 1/(-2) and
c₁/c₂ = -11/22 = 1/(-2)

Therefore, a₁/a₂ = b₁/b₂ = c₁/c₂
Now, these linear equations are coincident and its have many solution.
So, It is consitent.
```
```(v) 4x/3 + 2y = 8; 2x + 3y = 12
4x/3 + 2y - 8 = 0; 2x + 3y - 12 = 0

a₁ = 4/3, b₁ = 2, c₁ = -8
a₂ = 2, b₂ = 3, c₂ = -12

Here, a₁/a₂ = (4/3)/2 = 2/3,
Also, b₁/b₂ = 2/3 and
c₁/c₂ = -8/(-12) = 2/3

Therefore, a₁/a₂ = b₁/b₂ = c₁/c₂
Now, these linear equations are coincident and its have many solution.
So, It is consitent.
```

Question-4 :-  Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) x + y = 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
x + y - 5 = 0 ; 2x + 2y - 10 = 0

a₁ = 1, b₁ = 1, c₁ = -5
a₂ = 2, b₂ = 2, c₂ = -10

Here, a₁/a₂ = 1/2,
Also, b₁/b₂ = 1/2 and
c₁/c₂ = -5/(-10)= 1/2
Therefore, a₁/a₂ = b₁/b₂ = c₁/c₂
Now, these linear equations are coincident and its have many solution.
So, It is consitent.

Lets represent these equations in graphically:
x + y = 5  ...........(i)

x05

y50

2x + 2y = 10 ...........(ii)

x05

y50

Plot the points A(0, 5), B(5, 0) and P(0, 5), Q(5, 0), corresponding to the solutions in Table.
Now draw the lines AB and PQ, representing the equations  x + y = 5 and 2x + 2y = 10, as shown in following Figure:

```
```
(ii) x – y = 8, 3x – 3y = 16
x – y - 8 = 0 ; 3x – 3y - 16 = 0

a₁ = 1, b₁ = -1, c₁ = -8
a₂ = 3, b₂ = -3, c₂ = -16

Here, a₁/a₂ = 1/3,
Also, b₁/b₂ = -1/(-3) = 1/3 and
c₁/c₂ = -8/(-16) = 1/2

Therefore, a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Now, these linear equations are parallel to each other and its have only no possible solution.
So, It is inconsitent.
```
```(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0

a₁ = 2, b₁ = 1, c₁ = -6
a₂ = 4, b₂ = -2, c₂ = -4

Here, a₁/a₂ = 2/4 = 1/2,
Also, b₁/b₂ = 1/(-2) and
c₁/c₂ = -6/(-4) = 3/2

Therefore, a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂
Now, these linear equations are intersecting each other at one point and its have only one solution.
So, It is consitent.

Lets represent these equations in graphically:
2x + y = 6  ...........(i)

x03

y60

4x – 2y = 4 ...........(ii)

x01

y-20

Plot the points A(0, 6), B(3, 0) and P(0, -2), Q(1, 0), corresponding to the solutions in Table.
Now draw the lines AB and PQ, representing the equations 2x + y = 6  and 4x – 2y = 4, as shown in following Figure:

```
```(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

a₁ = 2, b₁ = -2, c₁ = -2
a₂ = 4, b₂ = -4, c₂ = -5

Here, a₁/a₂ = 2/4 = 1/2,
Also, b₁/b₂ = (-2)/(-4) = 1/2 and
c₁/c₂ = -2/(-5) = 2/5

Therefore, a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Now, these linear equations are parallel to each other and its have only no possible solution.
So, It is inconsitent.
```

Question-5 :-  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 the width of garden and length of garden be araxand y respectively.
According to question linear equations are:

y = x + 4
x - y = -4 ........(i)

y + x = 36
x + y = 36 ........(ii)

Lets represent these equations in graphically:
x - y = -4  ...........(i)

x0-4

y40

x + y = 36 ...........(ii)

x036

y360

Plot the points A(0, 4), B(-4, 0) and P(0, 36), Q(36, 0), corresponding to the solutions in Table.
Now draw the lines AB and PQ, representing the equations  x - y = -4 and x + y = 36, as shown in following Figure:

```

Question-6 :-  Given the linear equation 2x + 3y – 8 = 0, write another linear equation 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) For intersecting lines,
a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂

Given : 2x + 3y - 8 = 0 ....(i)
According to condition the second line is :
3x + 5y - 15 = 0 ....(ii)

a₁ = 2, b₁ = 3, c₁ = -8
a₂ = 3, b₂ = 5, c₂ = -15

Here, a₁/a₂ = 2/3,
Also, b₁/b₂ = 3/5 and
c₁/c₂ = -8/(-15) = 8/15

Hence, a₁/a₂ ≠ b₁/b₂ ≠ c₁/c₂.
```
```(ii) For parallel lines,
a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Given : 2x + 3y - 8 = 0 ....(i)
According to condition the second line is :
4x + 6y - 18 = 0 ....(ii)

a₁ = 2, b₁ = 3, c₁ = -8
a₂ = 4, b₂ = 6, c₂ = -18

Here, a₁/a₂ = 2/4,
Also, b₁/b₂ = 3/6 and
c₁/c₂ = -8/(-18) = 4/9

Hence, a₁/a₂ = b₁/b₂ ≠ c₁/c₂.
```
```(iii) For coincident lines,
a₁/a₂ = b₁/b₂ = c₁/c₂

Given : 2x + 3y - 8 = 0 ....(i)
According to condition the second line is :
6x + 9y - 24 = 0 ....(ii)

a₁ = 2, b₁ = 3, c₁ = -8
a₂ = 6, b₂ = 9, c₂ = -24

Here, a₁/a₂ = 2/6 = 1/3,
Also, b₁/b₂ = 3/9 = 1/3 and
c₁/c₂ = -8/(-24) = 1/3

Hence, a₁/a₂ = b₁/b₂ = c₁/c₂.
```

Question-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 ........(i)

3x + 2y – 12 = 0
3x + 2y = 12 ........(ii)

Lets represent these equations in graphically:
x - y = -1 ........(i)

x0-1

y10

3x + 2y = 12 ........(ii)

x04

y60

Plot the points A(0, 1), B(-1, 0) and P(0, 6), Q(4, 0), corresponding to the solutions in Table.
Now draw the lines AB and PQ, representing the equations x - y = -1 and 3x + 2y = 12, as shown in following Figure:

```
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