Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths

RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q 15 : Draw the graph of y = | x |.

Ans. We are given,

y = |x|

Substituting x = 1 , we get y = 1

Substituting x = -1, we get y = 1

Substituting x = 2, we get y = 2

Substituting x = -2, we get y = 2

RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

For every value of x, whether positive or negative, we get y as a positive number.


Q 16: Draw the graph of  y = |x| + 2.

Ans. We are given,

Y = |x|+2

Substituting x = 0 we get y=2

Substituting x = I , we get y=3

Substituting x =-1 , we get Y = 3

Substituting x = 2, we get y = 4

Substituting x = -2, we get y = 4

RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

For every value of x, whether positive or negative, we get y as a positive number and the minimum value of y is equal to 2 units.

Q 17 : Draw the graphs of the following linear equations on the same graph paper: 2x + 3y = 12, x – y = 1 Find the coordinates of the vertices of the triangle formed by the two straight lines and the y-axis. Also, find the area of the triangle.

Ans. We are given,

2x +3y  = 12

We get,  RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Now, substituting x = 0 in y RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics , we get

 

Substituting x = 6 in  RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics we get

y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X06
Y40


Plotting A(0,4) and E(6,0) on the graph and by joining the points , we obtain the graph of equation

2x+3y = 12.

We are given,

x – y = 1

We get, y = x – 1

Now, substituting x = 0 in y = x-1,

we get y = -1

Substituting x in y = x-1,

we get y = -2

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X0-1
Y-1-2


Plotting D(0,) and E(-1,0) on the graph and by joining the points , we obtain the graph of equation  x— y = I .

RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

By the intersection of lines formed by 2x + 3y = 12 and x—y=1 on the graph, triangle ABC is formed on y axis.

Therefore, AC at y axis is the base of triangle ABC having AC = 5 units on y axis.

Draw FE perpendicular from F on y axis. FE parallel to x axis is the height of triangle ABC having FE = 3 units on x axis.

Therefore, Area of triangle ABC, say A is given by A = (Base x Height)/2 =(AC x FE)/2 = (5×3)/2    ⇒15/2 = 7.5 sq. units


Q 18 : Draw the graphs of the linear equations 4x – 3y + 4 = 0 and 4x + 3y – 20 = 0. Find the area bounded by these lines and x-axis.

Ans. We are given, 4x – 3y + 4 = 0

We get, RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Now, substituting x = 0 in RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics, we get

Substituting x = -I in RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

we get y =0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

x0-1
y4/30


Plotting E(0, 4/3) and A (-1, 0) on the graph and by joining the points, we obtain the graph of equation

4x – 3y + 4 = 0.

We are given, 4x + 3y – 20 = 0

We get,

RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Now, substituting x = 0 in RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics, we get

y = 7

Substituting x = 5 in RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics,we get

y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X05
Y20/30


Plotting D (0,20/3) and B(5,0) on the graph and by joining the points , we obtain the graph of equation 4x +3y – 20 = 0.

RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

By the intersection of lines formed by 4x-3y + 4 = 0 and 4x+ 3y – 20 = 0 on the graph,

Triangle ABC is formed on x axis. Therefore, AB at x axis is the base of triangle ABC having AB = 6 units on x axis.

Draw CF perpendicular from C on x axis.

CF parallel to y axis is the height of triangle ABC having CF = 4 units on y axis.

Therefore, Area of triangle ABC, say A is given by

A = (Base x Height)/2

A = (AB x CF)/2

A = (6 x 4)/2

k = 12 sq. units


Q19 : The path of a train A is given by the equation 3x + 4y -12 = 0 and the path of another train B is given by the equation 6x + 8y – 48 = 0. Represent this situation graphically.

Ans. We are given the path of train A, 3x + 4y – 12 = 0

We get,

RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Now, substituting x = 0 in RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics,  we get

Y = 3

Substituting x = 4 in RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics, we get

y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X04
Y30


Plotting A(4,0) and E(0,3) on the graph and by joining the points , we obtain the graph of equation 3x+4y-12 = 0.

We are given the path of train B,

6x + 8y – 48 = 0

We get,  RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Now, substituting x = 0 in RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics,we get

y = 6

Substituting x = 8 in  RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics,we get

y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X08
Y60


Plotting C(0,6) and D(8,0) on the graph and by joining the points , we obtain the graph of equation 6x+8y-48 = 0

RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Q 20 : Ravish tells his daughter Aarushi, “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”. If present ages of Aarushi and Ravish are x and y years respectively, represent this situation algebraically as well as graphically.

Ans. We are given the present age of Ravish as y years and Aarushi as x years.

Age of Ravish seven years ago = y – 7

Age of Aarushi seven years ago = x – 7

It has already been said by Ravish that seven years ago he was seven times old then Aarushi was then

So, y – 7 = 7 (x – 7)

Y – 7 = 7 x – 49

7x  – y = – 7 + 49

7x – y – 42 = 0                                    ——(1)

Age of Ravish three years from now = y + 3

Age of Aarushi three years from now = x+3

It has already been said by Ravish that three years from now he will be three times older then Aarushi will be then So,

Y + 3 = 3 (x + 3)

y + 3 = 3x + 9

3x + 9 – y – 3 = 0

3x – y + 6 = 0                                      —–(2)

(1) and (2) are the algebraic representation of the given statement.

We are given,

7x – y- 42 = 0

We get,

Y = 7x – 42

Now, substituting x = 0 in y = 7x —42,

we get y = -42

Substituting x = 6 in y = 7x – 42,

we get y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X06
Y-420

We are given,

3x – y + 6 =0

We get,

Y = 3x + 6

Now, substituting x=0 in y = 3x + 6,

We get y = 6

Substituting x= —2 in y = 3x + 6,

We get y = 0

Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

X0-2
Y60

RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

The red -line represents the equation 7x—y-42 =0.

The blue-line represents the equation 3x —y +6 =0.

 

Q21: Aarushi was driving a car with uniform speed of 60 km/h. Draw distance-time graph From the graph, find the distance travelled by Aarushi in  

(i) 212 Hours
 (ii) 12 Hour

Ans. Aarushi is driving the car with the uniform speed of 60 km/h. We represent time on X-axis and distance on Y-axis Now, graphically

RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

We are given that the car is travelling with a uniform speed 60 km/hr. This means car travels 60 km distance each hour. Thus the graph we get is of a straight line.

Also, we know when the car is at rest, the distance travelled is 0 km, speed is 0 km/hr and the time is also 0 hr. Thus, the given straight line will pass through O (0 , 0) and M (1, 60).

Join the points 0 and M and extend the line in both directions.

Now, we draw a dotted line parallel to y-axis from x = 12 that meets the straight line graph at L from which we draw a line parallel to x-axis that crosses y-axis at 30. Thus, in 12hr, distance travelled by the car is 30 km.

Now, we draw a dotted line parallel to y-axis from x = 212 that meets the straight line graph at N from which we draw a line parallel to x-axis that crosses y-axis at 150. Thus, in 212hr, distance travelled by the car is 150 km.

(i) Distance = Speed x Time Distance travelled in RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics hours is given by

Distance=60 x RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Distance = 60 x 5/2

Distance = 150 Km

(ii) Distance = Speed x Time Distance travelled in 12 hour is given by

Distance=60 x RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Distance = 30 km

The document RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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FAQs on RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths - RD Sharma Solutions for Class 9 Mathematics

1. What are linear equations in two variables?
Ans. Linear equations in two variables are equations that involve two variables, usually represented as x and y, and have a degree of 1. They can be expressed in the form ax + by = c, where a, b, and c are constants.
2. How can linear equations in two variables be solved?
Ans. Linear equations in two variables can be solved using various methods such as the substitution method, elimination method, and graphical method. These methods involve manipulating the equations to eliminate one variable and solve for the other.
3. Can linear equations in two variables have multiple solutions?
Ans. Yes, linear equations in two variables can have multiple solutions. This means that there can be more than one pair of values for x and y that satisfy the equation. These solutions can be represented as ordered pairs (x, y) on a coordinate plane.
4. What is the importance of linear equations in two variables?
Ans. Linear equations in two variables have various applications in real-life situations. They are used to model and solve problems related to geometry, physics, economics, and many other fields. They provide a mathematical framework for analyzing and understanding relationships between two variables.
5. Are linear equations in two variables limited to straight lines?
Ans. No, linear equations in two variables are not limited to straight lines. While the graphical representation of a linear equation is a straight line, the equations themselves can represent various relationships between the variables. The linearity refers to the power of the variables, not the shape of the graph.
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