Pair Of Linear Equations In Two Variables Class 10th Mathematics CBSE Solution

Class 10^th Mathematics CBSE Solution

Exercise 3.1

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

Exercise 3.2

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

Exercise 3.3

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

Exercise 3.4

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

Exercise 3.5

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

Exercise 3.6

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

Exercise 3.7

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

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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 :

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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 4x + 6y - 8 = 0

As 

So,

(iii)Coincident lines: For coincident lines,

Hence,

the second line can be 6x + 9y - 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:
3x + 2y - 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.

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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. 

a₁x + b_1­y + c₁ = 0

a₂x + b₂y + c₂ = 0, we jhave

a₁ = 3, b₁ = -1, c₁ = -3

a₂ = 9, b₂ = -2 and c₂ = -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,