##### Class 10^{th} Mathematics CBSE Solution

**Exercise 3.1**- Aftab tells his daughter, Seven years ago, I was seven times as old as you were…
- The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys…
- The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160.…

**Exercise 3.2**- Form the pair of linear equations in the following problems, and find their…
- On comparing the ratios a_1/a_2 , b_1/b_2 and c_1/c_2 find out whether the lines…
- On comparing the ratios a_1/a_2 , b_1/b_2 and c_1/c_2 find out whether the…
- Which of the following pairs of linear equations are consistent/inconsistent? If…
- Half the perimeter of a rectangular garden, whose length is 4 m more than its…
- Given the linear equation 2x+3y-8 = 0 write another linear equation in two…
- Draw the graphs of the equations and Determine the coordinates of the vertices…

**Exercise 3.3**- Solve the following pair of linear equations by the substitution method. (i)…
- Solve 2x+3y = 11 and 2x-4y = - 24 and hence find the value of m for which y =…
- Form the pair of linear equations for the following problems and find their…

**Exercise 3.4**- Solve the following pair of linear equations by the elimination method and the…
- Form the pair of linear equations in the following problems, and find their…

**Exercise 3.5**- Which of the following pairs of linear equations has unique solution, no…
- For which values of a and b does the following pair of linear equations have an…
- For which value of k will the following pair of linear equations have no…
- Solve the following pair of linear equations by the substitution and…
- Form the pair of linear equations in the following problems and find their…

**Exercise 3.6**- Solve the following pairs of equations by reducing them to a pair of linear…
- Formulate the following problems as a pair of equations, and hence find their…

**Exercise 3.7**- The ages of two friends Ani and Biju differ by 3 years. Anis father Dharam is…
- One says, Give me a hundred, friend! I shall then become twice as rich as you.…
- A train covered a certain distance at a uniform speed. If the train would have…
- The students of a class are made to stand in rows. If 3 students are extra in a…
- In a deltaabc , angle c = 3 angle b = 2 (angle a + angle b) Find the three…
- Draw the graphs of the equations 5x - y = 5 and 3x - y = 3. Determine the…
- Solve the following pair of linear equations. (i) px + qy = p - q qx - py = p +…
- ABCD is a cyclic quadrilateral (see Fig. 3.7). Find the angles of the cyclic…

**Exercise 3.1**

- Aftab tells his daughter, Seven years ago, I was seven times as old as you were…
- The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys…
- The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160.…

**Exercise 3.2**

- Form the pair of linear equations in the following problems, and find their…
- On comparing the ratios a_1/a_2 , b_1/b_2 and c_1/c_2 find out whether the lines…
- On comparing the ratios a_1/a_2 , b_1/b_2 and c_1/c_2 find out whether the…
- Which of the following pairs of linear equations are consistent/inconsistent? If…
- Half the perimeter of a rectangular garden, whose length is 4 m more than its…
- Given the linear equation 2x+3y-8 = 0 write another linear equation in two…
- Draw the graphs of the equations and Determine the coordinates of the vertices…

**Exercise 3.3**

- Solve the following pair of linear equations by the substitution method. (i)…
- Solve 2x+3y = 11 and 2x-4y = - 24 and hence find the value of m for which y =…
- Form the pair of linear equations for the following problems and find their…

**Exercise 3.4**

- Solve the following pair of linear equations by the elimination method and the…
- Form the pair of linear equations in the following problems, and find their…

**Exercise 3.5**

- Which of the following pairs of linear equations has unique solution, no…
- For which values of a and b does the following pair of linear equations have an…
- For which value of k will the following pair of linear equations have no…
- Solve the following pair of linear equations by the substitution and…
- Form the pair of linear equations in the following problems and find their…

**Exercise 3.6**

- Solve the following pairs of equations by reducing them to a pair of linear…
- Formulate the following problems as a pair of equations, and hence find their…

**Exercise 3.7**

- The ages of two friends Ani and Biju differ by 3 years. Anis father Dharam is…
- One says, Give me a hundred, friend! I shall then become twice as rich as you.…
- A train covered a certain distance at a uniform speed. If the train would have…
- The students of a class are made to stand in rows. If 3 students are extra in a…
- In a deltaabc , angle c = 3 angle b = 2 (angle a + angle b) Find the three…
- Draw the graphs of the equations 5x - y = 5 and 3x - y = 3. Determine the…
- Solve the following pair of linear equations. (i) px + qy = p - q qx - py = p +…
- ABCD is a cyclic quadrilateral (see Fig. 3.7). Find the angles of the cyclic…

###### Exercise 3.1

**Question 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.

**Answer:****Given:** Seven years ago, Aftab was seven times as old as his daughter and after 3 years Aftab will be 3 times as old as his daughter.

To Represent the situations algebraically and graphically we need to find the linear equations for these situations.

Let present age of Aftab = x

Let present age of his daughter = y

Then, seven years ago the age of Aftab and his daughter must have been seven less than their present ages,

Age of Aftab seven years ago = x – 7

Age of Daughter seven years ago = y – 7

According to the question,

Seven years ago, Aftab was seven times as old as his daughter, So

x - 7 = 7(y - 7)

⇒ x - 7 = 7y - 49

⇒ x = 7y - 42

Now for finding different points of this equation, we can either take different values of x and put them in the equation to obtain values of y or vice versa

Putting y = 5, 6 and 7 in equation (¡),

we get,

For y = 5,

x = 7 × 5 – 42

= 35 – 42

= -7

For x =6,

x = 7 × 6 – 42

= 42 – 42

= 0

For y = 7,

x = 7 × 7 – 42

= 49 – 42

= 7

Thus we got 3 points to plot on graph for this equation.

Three years from now,

Age of Aftab = x+3

Age of Daughter = y+3

According to the question,

⇒ x + 3 = 3(y + 3)

⇒ x + 3 = 3y + 9

⇒ x = 3y + 6

Now for finding different points of this equation, we can either take different values of x and put them in equation to obtain values of y or vice versa

Putting x = 0, 3 and 6

Thus we got 3 points to plot on graph for this equation.

**Algebraic representation**

x - 7y = -42 (i)

x - 3y = 6 (ii)

**Graphical representation:**

**Question 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.

**Answer:**We need to form linear equations for the situations.

Let cost of one bat = Rs. x

Let cost of one ball = Rs. y

In first case, three bats and 6 balls cost him 3900 rupees. therefore our equation becomes

⇒ 3x + 6y = 3900 ........(i)

Dividing equation by 3 both side

⇒ x + 2y = 1300

⇒ x = 1300 - 2y

For plotting the equation of graph, take different values of y and obtain the value of x from equation or you can do vice versa

At y = 0

⇒ x = 1300 - 2(0)

⇒ x = 1300

Now finding the value at x = 0

⇒ 0 = 1300 - 2y

⇒ 2y = 1300

⇒ y = 650

From above points, we make the graph as follows [Graphic representation]

In second case he buys one bat and 3 balls for 1300, therefore,

⇒ x + 3y = 1300 ......(ii) [Algebraic representation]

⇒ x = 1300 - 3y

At y = 400

⇒ x = 1300 - 3(400)

⇒ x = 100

At y = 300

⇒ x = 1300 - 3(300)

⇒ x = 400

From above points, we make the graph as follows, [Graphical representation]

**Question 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.

**Answer:**Let the cost of each kg of apples = Rs. X

Let the cost of each kg of grapes = Rs. Y

According to the question,

Cost of 2 kg of Apples = 2 x

Cost of 1 kg of grapes = y

Putting y = 20, 40 and 60 we get,

x = (160 - 20)/2 = 70

x = (160 - 40))/2 = 60

x = (160 - 60)/2 = 50

Algebraic Representation:

2 x + y = 160

Graphical Representation:

Now taking another case

Cost of 4 kg of apples and 2 kg of grapes is Rs. 300……………………………(Given)

So,

Dividing the equation by 2 , we get,

Putting x = 70, 75 and 80 we get,

y = 150 – 2×70 = 10

y = 150 – 2(75) = 0

y = 150 – 2(80) = 150 – 160 = -10

Algebraic representation :

4 x + 2 y = 300

Graphical representation :

**Question 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.

**Answer:**

**Given:** Seven years ago, Aftab was seven times as old as his daughter and after 3 years Aftab will be 3 times as old as his daughter.

To Represent the situations algebraically and graphically we need to find the linear equations for these situations.

Let present age of Aftab = x

Let present age of his daughter = y

Then, seven years ago the age of Aftab and his daughter must have been seven less than their present ages,

Age of Aftab seven years ago = x – 7

Age of Daughter seven years ago = y – 7

According to the question,

Seven years ago, Aftab was seven times as old as his daughter, Sox - 7 = 7(y - 7)

⇒ x - 7 = 7y - 49

⇒ x = 7y - 42

Putting y = 5, 6 and 7 in equation (¡),

we get,For y = 5,

x = 7 × 5 – 42

= 35 – 42

= -7

For x =6,

x = 7 × 6 – 42

= 42 – 42

= 0

For y = 7,

x = 7 × 7 – 42

= 49 – 42

= 7

Thus we got 3 points to plot on graph for this equation.

Three years from now,

Age of Aftab = x+3

Age of Daughter = y+3

According to the question,

⇒ x + 3 = 3(y + 3)

⇒ x + 3 = 3y + 9

⇒ x = 3y + 6

Putting x = 0, 3 and 6

Thus we got 3 points to plot on graph for this equation.**Algebraic representation**

x - 7y = -42 (i)

x - 3y = 6 (ii)

**Graphical representation:**

**Question 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.

**Answer:**

We need to form linear equations for the situations.

Let cost of one bat = Rs. x

Let cost of one ball = Rs. y

In first case, three bats and 6 balls cost him 3900 rupees. therefore our equation becomes

⇒ 3x + 6y = 3900 ........(i)

Dividing equation by 3 both side

⇒ x + 2y = 1300

⇒ x = 1300 - 2y

For plotting the equation of graph, take different values of y and obtain the value of x from equation or you can do vice versa

At y = 0

⇒ x = 1300 - 2(0)

⇒ x = 1300

Now finding the value at x = 0

⇒ 0 = 1300 - 2y

⇒ 2y = 1300

⇒ y = 650

From above points, we make the graph as follows [Graphic representation]

In second case he buys one bat and 3 balls for 1300, therefore,

⇒ x + 3y = 1300 ......(ii) [Algebraic representation]

⇒ x = 1300 - 3y

At y = 400

⇒ x = 1300 - 3(400)

⇒ x = 100

At y = 300

⇒ x = 1300 - 3(300)

⇒ x = 400

From above points, we make the graph as follows, [Graphical representation]

**Question 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.

**Answer:**

Let the cost of each kg of apples = Rs. X

Let the cost of each kg of grapes = Rs. Y

According to the question,

Cost of 2 kg of Apples = 2 xCost of 1 kg of grapes = y

Putting y = 20, 40 and 60 we get,

x = (160 - 20)/2 = 70

x = (160 - 40))/2 = 60

x = (160 - 60)/2 = 50

Algebraic Representation:

2 x + y = 160

Graphical Representation:

Now taking another case

Cost of 4 kg of apples and 2 kg of grapes is Rs. 300……………………………(Given)

So,

Dividing the equation by 2 , we get,

Putting x = 70, 75 and 80 we get,

y = 150 – 2×70 = 10

y = 150 – 2(75) = 0

y = 150 – 2(80) = 150 – 160 = -10

Algebraic representation :

4 x + 2 y = 300

Graphical representation :

###### 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.

**Answer:**For representing the situation graphically and algebraically, we need to form linear equations

(i) Let number of girls = x

Let number of boys = y

According to the question, Total no of students is equal to 10,

⇒ ................................eq(i)

Now we will find different points to plot the equation. We can take any value of y and put in eq (i) to obtain the value of x at that point

Putting y = 4, 5 and 6. we get,

at x = 4

X = 10 – 4 = 6

at x = 5

X = 10 – 5 = 5

at x = 6

X = 10 – 6 = 4

Number of girls is 4 more than number of boys …………Given

So,

x = y + 4

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

Now for plotting the points on graph, take any values of x and put them in eq (ii) to obtain values of y

Putting x = 3 ,5 and 7 we get

at x = 3

y = 3 - 4 = -1

at x = 4

y = 5 - 4 = 1

at x = 7

y = 7 - 4 = 3

Graphical representation :

Plotting the points obtain on graph we get,

As, both lines intersect each other at (7, 3)

Solution of this pair of equation is (7, 3)

i.e.

No of girls, x = 7

No of boys, y = 3

(ii)

Let cost of one pencil = Rs. X

Let cost of one pen = Rs. Y

According to the question, 5 pencils and 7 pens together cost Rs 50

5x + 7y = 50

⇒5x = 50 - 7y

⇒

Putting value of y = 0 , 5 , 10 we get,

x = 10 - 0 = 10

Now,

7 pencils and 5 pens together cost Rs. 46

7x + 5y = 46

⇒ 5y = 46 – 7x

Putting x = -2 , 3 , 8 we get,

Graphical Representation:

Plotting the points obtain on graph we get,

As, both lines intersect each other at (3, 5)

Solution of this pair of equation is (3, 5)

i.e.

cost of pencil, x = 3

cost of pen, y = 5

**Question 2.**On comparing the ratios and find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

(i)

(ii)

(iii)

**Answer:**i) Comparing these equation with

=

=

We get,

=

=

Hence,

=

We find that,

Therefore, both lines intersect at one point.

Graph of the lines look like below:

ii) Comparing these equations with ,

=

=

We get,

=

=

Hence,

=

We find that,

=

Therefore , both lines are coincident.

iii) Comparing these equations with,

=

=

We get,

=

=

Hence,

=

We find that,

=

Therefore, both lines are parallel.

**Question 3.**On comparing the ratios and find out whether the following pair of linearequations are consistent, or inconsistent.

(i)

(ii)

(iii)

(iv)

(v)

**Answer:**(i) We get,

=

=

Hence,

=

Therefore , these linear equations will intersect at one point only and have only one possible solution and pair of linear equations is inconsistent.

(ii) We get,

=

=

=

Hence,

=

Therefore, these linear equations are parallel to each other and have no possible solution.

And pair of linear equations is inconsistent.

(iii)

9x – 10y = 14

We get,

=

=

=

Hence,

=

Therefore, these linear equations will intersect each other at one point and have only one possible solution.

And, the pair of linear equations is consistent.

(iv) We get,

=

=

=

Hence,

=

Therefore these pair of lines has the infinite number of solutions. and pair of linear equations is inconsistent

v) We get,

=

=

=

Hence,

=

Therefore these pair of lines have infinite number of solutions and pair of linear equation is consistent.

**Question 4.**Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

(i)

(ii)

(iii)

(iv)

**Answer:**(i) We get,

=

=

=

Hence,

=

Therefore these pair of lines have infinite number of solutions and pair of linear equations is consistent.

x + y = 5

x = 5 - y

putting y = 1,2,3 we get,

x = 5 -1 = 4

x = 5 - 2 = 3

x = 5 - 3 = 2

And, 2x + 2y = 10

x =

(ii) We get,

=

=

Hence,

Therefore, these linear equations are parallel to each other and have no possible solution,

Hence, the pair of linear equations is inconsistent.

(iii) We get,

=

=

=

Hence,

=

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution.

Hence, pair of linear equations is consistent.

=

=

And,

=

Graphical representation

(iv) We get,

=

=

= s

Hence,

=

Therefore, these linear equations are parallel to each other and have no possible solution,

Hence, the pair of linear equations is inconsistent.

**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.

**Answer:**Let the length and breadth of garden be 'x' and 'y' respectively.

Given,

Half the perimeter is 36 m

We know perimeter of Rectangle = 2(length + breadth)

Half the perimeter = x + y

⇒ x + y = 36

⇒ x = 36 - y [1]

Also, Length is 4 m more than width

⇒ x = y + 4 [2]

From [1] and [2], we have

36 - y = y + 4

⇒ 2y = 32

⇒ y = 16

Putting in [1], we have

⇒ x = 36 - 16 = 20

Hence, Length and Breadth of garden are 20 m and 16 m respectively.

**Question 6.**Given the linear equation 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

**Answer:****(i)**__Intersecting lines:__ For this condition,

The second line such that it is intersecting the given line is

2x + 4y - 6 = 0

As

**(ii)**__Parallel lines:__

For this condition,

Hence, the second line can be 4*x* + 6y - 8 = 0

As

So,

**(iii)**__Coincident lines:__ For coincident lines,

Hence,

the second line can be 6*x* + 9*y* - 24 = 0

So,

**Question 7.**Draw the graphs of the equations and Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

**Answer:***Equation: 1*

x - *y* + 1 = 0

*x* = *y* – 1

Let us find the coordinates satisfying the above equation,

Equation 2:

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

Coordinates for equation 2 are:

By taking, coordinates we can plot the both equations in graph.

Hence, the graphic representation is as follows.

**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.

**Answer:**

For representing the situation graphically and algebraically, we need to form linear equations

(i) Let number of girls = x

Let number of boys = y

According to the question, Total no of students is equal to 10,

⇒ ................................eq(i)

Now we will find different points to plot the equation. We can take any value of y and put in eq (i) to obtain the value of x at that pointPutting y = 4, 5 and 6. we get,

at x = 4X = 10 – 4 = 6

at x = 5X = 10 – 5 = 5

at x = 6X = 10 – 6 = 4

Number of girls is 4 more than number of boys …………Given

So,

x = y + 4

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

Putting x = 3 ,5 and 7 we get

at x = 3

y = 3 - 4 = -1

at x = 4

y = 5 - 4 = 1

at x = 7

y = 7 - 4 = 3

Graphical representation :

Plotting the points obtain on graph we get,

Solution of this pair of equation is (7, 3)

i.e.

No of girls, x = 7

No of boys, y = 3

(ii)

Let cost of one pencil = Rs. X

Let cost of one pen = Rs. Y

According to the question, 5 pencils and 7 pens together cost Rs 50

5x + 7y = 50

⇒5x = 50 - 7y

⇒

Putting value of y = 0 , 5 , 10 we get,

x = 10 - 0 = 10

Now,

7 pencils and 5 pens together cost Rs. 46

7x + 5y = 46

⇒ 5y = 46 – 7x

Putting x = -2 , 3 , 8 we get,

Graphical Representation:

Plotting the points obtain on graph we get,As, both lines intersect each other at (3, 5)

Solution of this pair of equation is (3, 5)

i.e.

cost of pencil, x = 3

cost of pen, y = 5

**Question 2.**

On comparing the ratios and find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

(i)

(ii)

(iii)

**Answer:**

i) Comparing these equation with

=

=

We get,

=

=

Hence,

=

We find that,

Therefore, both lines intersect at one point.

Graph of the lines look like below:

ii) Comparing these equations with ,

=

=

We get,

=

=

Hence,

=

We find that,

=

Therefore , both lines are coincident.

iii) Comparing these equations with,

=

=

We get,

=

=

Hence,

=

We find that,

=

Therefore, both lines are parallel.

**Question 3.**

On comparing the ratios and find out whether the following pair of linearequations are consistent, or inconsistent.

(i)

(ii)

(iii)

(iv)

(v)

**Answer:**

(i) We get,

=

=

Hence,

=

Therefore , these linear equations will intersect at one point only and have only one possible solution and pair of linear equations is inconsistent.

(ii) We get,

=

=

=

Hence,

=

Therefore, these linear equations are parallel to each other and have no possible solution.

And pair of linear equations is inconsistent.

(iii)

9x – 10y = 14

We get,

=

=

=

Hence,

=

Therefore, these linear equations will intersect each other at one point and have only one possible solution.

And, the pair of linear equations is consistent.

(iv) We get,

=

=

=

Hence,

=

Therefore these pair of lines has the infinite number of solutions. and pair of linear equations is inconsistent

v) We get,

=

=

=

Hence,

=

Therefore these pair of lines have infinite number of solutions and pair of linear equation is consistent.

**Question 4.**

Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

(i)

(ii)

(iii)

(iv)

**Answer:**

(i) We get,

=

=

=

Hence,

=

Therefore these pair of lines have infinite number of solutions and pair of linear equations is consistent.

x + y = 5

x = 5 - y

putting y = 1,2,3 we get,

x = 5 -1 = 4

x = 5 - 2 = 3

x = 5 - 3 = 2

And, 2x + 2y = 10

x =

(ii) We get,

=

=

Hence,

Therefore, these linear equations are parallel to each other and have no possible solution,

Hence, the pair of linear equations is inconsistent.

(iii) We get,

=

=

=

Hence,

=

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution.

Hence, pair of linear equations is consistent.

=

=

And,

=

Graphical representation

(iv) We get,

=

=

= s

Hence,

=

Therefore, these linear equations are parallel to each other and have no possible solution,

Hence, the pair of linear equations is inconsistent.

**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.

**Answer:**

Let the length and breadth of garden be 'x' and 'y' respectively.

Given,

Half the perimeter is 36 m

We know perimeter of Rectangle = 2(length + breadth)

Half the perimeter = x + y

⇒ x + y = 36

⇒ x = 36 - y [1]

Also, Length is 4 m more than width

⇒ x = y + 4 [2]

From [1] and [2], we have

36 - y = y + 4

⇒ 2y = 32

⇒ y = 16

Putting in [1], we have

⇒ x = 36 - 16 = 20

Hence, Length and Breadth of garden are 20 m and 16 m respectively.

**Question 6.**

Given the linear equation 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

**Answer:**

**(i)**__Intersecting lines:__ For this condition,

The second line such that it is intersecting the given line is

2x + 4y - 6 = 0

As

**(ii)**__Parallel lines:__

For this condition,

Hence, the second line can be 4*x* + 6y - 8 = 0

As

So,

**(iii)**__Coincident lines:__ For coincident lines,

Hence,

the second line can be 6*x* + 9*y* - 24 = 0

So,

**Question 7.**

Draw the graphs of the equations and Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

**Answer:**

*Equation: 1x* -

*y*+ 1 = 0

*x* = *y* – 1

Let us find the coordinates satisfying the above equation,

Equation 2:

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

Coordinates for equation 2 are:

By taking, coordinates we can plot the both equations in graph.

Hence, the graphic representation is as follows.

###### Exercise 3.3

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

**Answer:**i) x + y = 14...............(i)

x - y = 4....................(ii)

From equation (i)

take x on one side and when we take y to the other side its sign changes and we get,

x = 14 - y .........(iii)

Putting value of x in equation (ii) we get,

(14 - y) - y = 4

14 - 2y = 4

2y = 10

Putting value of y in equation (iii) we get,

x = 14 - 5 = 9

Hence, x = 9 and y = 5

ii) s - t = 3......................(i)

and,

From equation (i) we get,

taking t to the other side, the sign of t changes to positive

s = t + 3...............(iii)

Putting value of x from (iii) to (ii)

⇒

⇒ 2t + 6 + 3t = 36

⇒ 5t = 30

⇒ t =

Putting value of t in equation (iii) , we get,

s = 6 + 3 = 9

Hence, s = 9, t = 6

iii) 3x - y = 3..................(i)

9x - 3y = 9......................(ii)

Comparing with general pair of equations i.e.

a_{1}x + b_{1}y + c_{1} = 0

a_{2}x + b_{2}y + c_{2} = 0, we jhave

a_{1} = 3, b_{1} = -1, c_{1} = -3

a_{2} = 9, b_{2} = -2 and c_{2} = -9

Here,

and In this case, the system of linear equation is consistent and has infinite solutions.

iv) 0.2 x + 0.3 y = 1.3 .....................(i)

0.4 x + 0.5 y = 2.3 ........................(ii)

From equation (i) . we get,

Putting value of x in equation (ii) we get,

(6.5 - 1.5 y) × 0.4 + 0.5 y = 2.3

6.5 x 0.4 - 1.5 y x 0.4 + 0.5 y = 2.3

2.6 - 0.6 y + 0.5 y = 2.3

- 0.1 y = -0.3

y =

Putting value of y in equation (iii) we get,

x = 6.5 - 1.5 × 3 = 6.5 - 4.5 = 2

Hence, x = 2 and y = 3

v)

From equation (i) , we get,

**Question 1.**

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

**Answer:**

i) x + y = 14...............(i)

x - y = 4....................(ii)

From equation (i)

take x on one side and when we take y to the other side its sign changes and we get,

x = 14 - y .........(iii)

Putting value of x in equation (ii) we get,

(14 - y) - y = 4

14 - 2y = 4

2y = 10

Putting value of y in equation (iii) we get,

x = 14 - 5 = 9

Hence, x = 9 and y = 5

ii) s - t = 3......................(i)

and,

From equation (i) we get,

taking t to the other side, the sign of t changes to positives = t + 3...............(iii)

Putting value of x from (iii) to (ii)

⇒

⇒ 2t + 6 + 3t = 36

⇒ 5t = 30

⇒ t =

Putting value of t in equation (iii) , we get,

s = 6 + 3 = 9

Hence, s = 9, t = 6

iii) 3x - y = 3..................(i)

9x - 3y = 9......................(ii)

Comparing with general pair of equations i.e.a_{1}x + b_{1}y + c_{1} = 0

a_{2}x + b_{2}y + c_{2} = 0, we jhave

a_{1} = 3, b_{1} = -1, c_{1} = -3

a_{2} = 9, b_{2} = -2 and c_{2} = -9

Here,

and In this case, the system of linear equation is consistent and has infinite solutions.

iv) 0.2 x + 0.3 y = 1.3 .....................(i)

0.4 x + 0.5 y = 2.3 ........................(ii)

From equation (i) . we get,

Putting value of x in equation (ii) we get,

(6.5 - 1.5 y) × 0.4 + 0.5 y = 2.3

6.5 x 0.4 - 1.5 y x 0.4 + 0.5 y = 2.3

2.6 - 0.6 y + 0.5 y = 2.3

y =

Putting value of y in equation (iii) we get,

x = 6.5 - 1.5 × 3 = 6.5 - 4.5 = 2

Hence, x = 2 and y = 3

v)

From equation (i) , we get,