# PSEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2

PSEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2

## PSEB 10th Class Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 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 ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one

pencil and that of one pen.

Solution:

(i) Let the number of boys in the Quiz = X

and the number of girls in the Quiz = y

Total number of students took part in Quiz = 10

x + y = 10

or x + y – 10 = 0

According to Question,

y = x + 4

or x = y – 4

Now, draw the graph of linear equations

x + y = 10

and x – y + 4 = 0

x + y = 10

or x = 10 – y

Putting y = 0 in (1), we get :

x = 10 – 0 = 10

Putting y = 7 in(1), we get:

x = 10 – 7 = 3

Putting y = 10 in (1) we get:

X = 10 – 10 = 0

Plotting the points A (10, 0), B (3, 7), C (0, 10) and drawing a line joining them we get the graph of the equation x + y = 10

Now x – y + 4 = 0

or x = y – 4

Putting y = 0 in (2), we get:

x = 0 – 4 = -4

Putting y = 7 in (2), we get:

x = 7 – 4 = 3

Putting y = 4 in (2), we get:

x = 4 – 4 = 0

Plotting the points D (-4, 0), B (3, 7), E (0, 4) and drawing a line joining them, we get the graph of the equation x – y + 4 = 0

From the graph it is clear that both the linear equations meets at a point B (3, 7).

∴ Point B (3, 7) is the graphic solution.

Hence, number of boys in the Quiz = 3

Number of girls in the Quiz = 7

Table:

Plotting the points E (6.5, 0), B (3, 5), F (9.5. -4) and drawing a line joining them, we get the graph of the equation.

7x + 5y = 46

From the graph, it is clear that both the linear equations meets at a point B (3, 5).

∴ point B (3, 5) is the graphic solution.

Hence, cost of one pencil = ₹ 3

Cost of one pen = ₹ 5

(iv) Given pair of linear equations is

5x – 3y= 11

and -10x + 6y = -22

Or 5x – 3y – 11 = 0

and -10x + 6y + 22 = 0

Here a_{1} = 5, b_{1} = -3, c_{1} = -11

a_{2} = -10, b_{2} = 6, c_{2} = 22

Hence given pair of linear equations is consistent.

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:

(1) Given pair of linear equations is

x + y = 5

and 2x + 2y = 10

Or x + y – 5 = 0

2x + 2y – 10 = 0

Here a_{1} = 1, b_{1} = 1, c_{1} = -5

a_{2} = 2, b_{2} = 2, c_{2} = -10

Now

x + y = 5

x = 5 – y ………….(1)

Putting y = 0 in (1), we get:

x = 5 – 0 = 5

Putting y = 3 in (1), we get

x = 5 – 3 = 2

Putting y = 5 in (1), we get

x = 5 – 5 = 0

Plotting the points A (5, 0), B (2, 3), C (0, 5) and drawing a line joining them, we ge the graph of the equation x + y = 5

2x + 2y = 10 Or 2 (x + y) = 10

Or x + y = 5

Or x = 5 – y

Putting y = 0 in (1), we get :

x = 5 – 0 = 5

Putting y = 2 in (2), we get :

x = 5 – 2 = 3

Putting y = 5 in (2), we get:

x = 5 – 5 = 0

Putting y = 2 in (2), we get :

x = 5 – 2 = 3

Putting y = 5 in (2), we get:

x = 5 – 5 = 0

Plotting the points A (5, 0), D (3, 2), C (0, 5) and drawing a line joining them, we get the graph of the equation 2x + 2y = 10

The graphs of two equations are coincident. Hence system of equations has infinitely many solutions i.e. consistent.

(ii) Given pair of linear equations is :

x – y = 8

and 3x – 3y = 16

Or

x – y – 8 = 0

and 3x – 3y – 16 = 0

Plotting the points D (1, 0), B (2, 2), E (0, -2) and drawing a line joining them, we get the graph of the equation

4x – 2y – 4 = 0

From the graph, it is clear that given system of equations meets at a point B (2, 2).

Hence, given pair of linear equations have unique solution.

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 length of garden = x m

Width of garden =y m

Perimeter of garden = 2 [x + y] m

Half perimeter of garden = (x + y) m

According to 1st condition x = y +4

According to 2nd condition

x + y = 36

∴ Pair of linear equations is

x = y + 4

and x + y = 36

x = y + 4 ………………(1)

Putting y = 0 in (1), we get :

x = 0 + 4 = 4

Putting y = – 4 in (1), we get:

x = -4 + 4 = 0

Putting y = 16 in (1), we get :

x = 16 + 4 = 20

Plotting the points A (4, 0) B (0, -4), C (20, 16) and drawing a line joining them.

we get the graph of the equation.

x = y + 4

Now x + y = 36

x = 36 – y

Putting y = 12 in (2), we get:

x = 36 – 12 = 24

Putting y = 24 in (2), we get :

x = 36 – 24 = 12

Putting y = 16 in (2), we get:

x = 36 – 16 = 20

Plotting the points D (24, 12), E (12, 24) C(20, 16) and drawing a line joining them, we get the raph of the equation.

x + y = 36

From the graph. it is clear that pair of linear equations meet at a point C (20, 16).

∴ C (20, 16) i.e. x = 20 and y = 16 is the solution of linear equations.

Hence, length of garden = 20 m

Width of garden = 16 m

Another Method:

Let width of garden = x m

Lengh of garden = (x + 4) m

Perimeter of garden = 2 [Length + Width]

= 2 [x + x + 4] m

= 2[2x + 4]m

∴ Half perimeter of garden = (2x +4) m

According to Question,

2x + 4 = 36

or 2x = 36 – 4

or 2x = 32

or x = 32/16 = 2 m

Hence, width of garden = 16 m

and length of garden = (16 + 4)m = 20 m

Question 6.

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

(i) intersecting lines

(ii) parallel lines

(iii) coincident lines

Solution:

Case I. For Intersecting Unes

Given linear equation is:

2x + 3y – 8 = 0

There are many another linear equations in two variahes which satisfies the condition of intersecing lines i.e.

Case II:

For Parallel Lines

Given linear equation is

2x + 3y – 8 = 0 ……………(1)

There are many other linear equation in two variables which satisfies the condition of parallel lines i.e.

Plotting the points G (2.5, 0), H (-2, 3), I (7, -3) and drawing a line joining them,

we get the graph of the equation

2 + 3y – 5 = 0.

Case III. For Coincident Lines

Given linear equation is

2x + 3y – 8 = 0

There are many other linear equations ¡n two variables which satisfies the condition of coincident lines i.e.

One of which is as follow :

6x+ 9y – 24 = 0

Now, draw the graph of linear equations (1) and (4).

Consider linear equation (4)

6x + 9y – 24 = o

or 3[2x + 3 – 8] = 0

or 2x + 3y – 8 = 0

∴ The points of both are same and line of both equations are same.

Question 7.

Drab tht graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed b these lines and the x-axis and shade the triangular region. (Pb. 2018 Set I, II, III)

Solution:

Consider the pair of linear equation

x – y + 1 =0

and 3x + 2y – 12 = 0

x – y + 1 = 0

or x = y – 1

Putting y = 0 in (1), we get:

x = 0 – 1 = -1

Puning y = 3 in (1) we get:

x = 3 – 1 = 2

Putting y = 1 in (1) , we get:

x = 1 – 1 = 0

Plotting the points D(4, 0) B (2, 3), E (0, 6) and drawing a line oinin them, we get the graph of the equation 3x – 2y – 12 = 0

The vertices of the triangle formed by pair of linear equations and the x-axis are shaded in the graph. The triangle so formed is ∆ABD.Coordinates of the vertices of ∆ABD are

A(-1, 0), B(2, 3) and D(4, 0).

Now, length of Base AD = AO + OD = 1 + 4 = 5 units

Length of perpendicular BF = 3 units

∴ Area of ∆ABD = 1/2 × Base × altitude

= 1/2 × AD × BF

= (1/2 × 5 × 3) sq. units

= 15/2 = 7.5 sq. units

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