Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions Ex-13.3, (Part -1), Linear Equation In Two Variables, Class 9, Maths

RD Sharma Solutions Ex-13.3, (Part -1), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q 1: Draw the graph of each of the following linear equations in two variables:

(i) x + y = 4
 (ii) x – y = 2
 (iii) -x + y = 6
 (iv) y = 2x
 (v) 3x + 5y = 15

RD Sharma Solutions Ex-13.3, (Part -1), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics
RD Sharma Solutions Ex-13.3, (Part -1), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Ans. (i) We are given, x + y = 4

We get, y = 4 – x,

Now, substituting x = 0 in y = 4 – x,

we get y = 4

Substituting x = 4 in y = 4 — x, we get y = 0

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

X04
Y40

RD Sharma Solutions Ex-13.3, (Part -1), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

(ii) We are given, x – y = 2

We get, y = x – 2

Now, substituting x = 0 in y= x – 2, we get y = – 2

Substituting x = 2 in y = x – 2, 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

X02
Y-20

RD Sharma Solutions Ex-13.3, (Part -1), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

(iii) We are given, – x + y = 6

We get, y = 6 + x

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

We get y =6

Substituting x = -6 in y = 6+ x, 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.

X0-6
Y60

RD Sharma Solutions Ex-13.3, (Part -1), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

(iv) We are given, y = 2x

Now, substituting x = 1 in y = 2x

We get y = 2

Substituting x = 3 in y = 2x

We get y = 6

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

X13
Y26

RD Sharma Solutions Ex-13.3, (Part -1), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

(v) We are given, 3x + 5y = 15

We get, 15 – 3x = 5y

Now, substituting x = 0 in 5y = 15 – 3x,

We get; 5y = 15

y =3

Substituting x = 5 in 5y = 15 – 3x

we get 5 y = 0

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

X05
Y30

RD Sharma Solutions Ex-13.3, (Part -1), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

(vi) we are given.

RD Sharma Solutions Ex-13.3, (Part -1), 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 -1), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

We get y = -6

Substituting x = 4 inRD Sharma Solutions Ex-13.3, (Part -1), 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
Y-60

RD Sharma Solutions Ex-13.3, (Part -1), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

(vii) We are given,

RD Sharma Solutions Ex-13.3, (Part -1), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

We get, x-2 = 3(Y-3)

x – 2 = 3y – 9

x = 3y – 7

Now, substituting x = 5 in x = 3y – 7,

We get; y = 4

Substituting x = 8 in x  = 3y – 7 ,

We get; y = 5

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

X58
Y45

RD Sharma Solutions Ex-13.3, (Part -1), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

(viii) We are given, 2y = – x +1

We get, 1 – x = 2Y

Now, substituting x =1 in1 – x = 2Y, we get

y = 0

Substituting x = 5 in 1 – x = 2Y , 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

X15
Y0-2

RD Sharma Solutions Ex-13.3, (Part -1), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics


Q 2: Give the equations of two lines passing through ( 3, 12). How many more such lines are there, and why?

Ans. We observe that x = 3 and y = 12 is the solution of the following equations

4x – y = 0 and 3x – y + 3 = 0

So, we get the equations of two lines passing through (3, 12) are, 4x – y = 0 and 3x – y + 3 = 0.

We know that passing through the given point infinitely many lines can be drawn.

So, there are infinitely many lines passing through (3, 12)


Q 3 : A three-wheeler scooter charges Rs 15 for first kilometer and Rs 8 each for every subsequent kilometer. For a distance of x km, an amount of Rs y is paid. Write the linear equation representing the above information.

Ans. Total fare of Rs y for covering the distance of x km is given by

y = 15 + 8(x – 1)

y = 15 + 8x – 8

y = 8x + 7

Where, Rs y is the total fare (x – 1) is taken as the cost of first kilometer is already given Rs 15 and 1 has to subtracted from the total distance travelled to deduct the cost of first Kilometer.


Q 4 : A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Aarushi paid Rs 27 for a book kept for seven days. If fixed charges are Rs x and per day charges are Rs y. Write the linear equation representing the above information.

Ans. Total charges of Rs 27 of which Rs x for first three days and Rs y per day for 4 more days is given by

x + y ( 7 – 3 ) = 27

x + 4y = 27

Here, (7 —3) is taken as the charges for the first three days are already given at Rs x and we have to find the charges for the remaining four days as the book is kept for the total of 7 days.


Q5: A number is 27 more than the number obtained by reversing its digits. lf its unit’s and ten’s digit are x and y respectively, write the linear equation representing the statement.

Ans. The number given to us is in the form of  ‘ yx ‘,

Where y represents the ten’s place of the number

And x represents the unit’s place of the number.

Now, the given number is 10y + x

Number obtained by reversing the digits of the number is 10x + y

It is given to us that the original number is 27 more than the number obtained by reversing its digits

So, 10y + x = 10x + y + 27

10y – y + x – 10x = 27

9y – 9x = 27

9 ( y – x ) = 27

y – x  = 27/9 = 3

x – y + 3 = 0


Q6: The Sum of a two digit number and the number obtained by reversing the order of its digits is 121. If units and tens digit of the number are x and y respectively, then write the linear equation representing the above statement.

Ans. The number given to us is in the form of ‘ yx’ ,

Where y represents the ten’s place of the number and x represents the units place of the number

Now, the given number is 10y + x

Number obtained by reversing the digits of the number is 10x+ y

It is given to us that the sum of these two numbers is 121

So, (10y + x)+ (10x + y) = 121

10y + y + x + 10x = 121

11y + 11x = 121

11 (y + x) = 121

x + y = 121/11 = 11

x + y = 11


Q7 : Plot the Points (3,5) and (-1,3) on a graph paper and verify that the straight line passing through the points, also passes through the point (1,4)

Ans.

RD Sharma Solutions Ex-13.3, (Part -1), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

By plotting the given points (3, 5) and (-1, 3) on a graph paper, we get the line BC.

We have already plotted the point A (1, 4) on the given plane by the intersecting lines.

Therefore, it is proved that the straight line passing through (3, 5) and (-1, 3) also passes through A (1, 4).

The document RD Sharma Solutions Ex-13.3, (Part -1), 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 -1), Linear Equation In Two Variables, Class 9, Maths - RD Sharma Solutions for Class 9 Mathematics

1. What is the importance of studying linear equations in two variables in Class 9 Mathematics?
Ans. Studying linear equations in two variables in Class 9 Mathematics is important because it helps in understanding the relationship between two variables and how they change with respect to each other. It also helps in solving real-life problems involving two unknown quantities. Moreover, it serves as a foundation for higher-level mathematics and other subjects like physics and economics.
2. How can we solve linear equations in two variables using the elimination method?
Ans. The elimination method is used to solve linear equations in two variables. In this method, we eliminate one variable by adding or subtracting the equations to get a new equation with only one variable. Then, we solve the resulting equation to find the value of the remaining variable. Finally, we substitute this value back into one of the original equations to find the value of the other variable.
3. Can linear equations in two variables have more than one solution?
Ans. Yes, linear equations in two variables can have more than one solution. If the two equations represent the same line, they have infinitely many solutions because all points on the line satisfy both equations. However, if the two equations represent parallel lines, they have no common solution and are said to be inconsistent.
4. What is the graphical method of solving linear equations in two variables?
Ans. The graphical method of solving linear equations in two variables involves plotting the given equations on a graph and finding the point of intersection, if any. The point of intersection represents the solution to the system of equations. If the lines are parallel, there is no point of intersection and the system of equations has no solution.
5. How can linear equations in two variables be applied in real-life situations?
Ans. Linear equations in two variables can be applied in various real-life situations. For example, they can be used to determine the cost of items based on their quantity, to calculate the time and distance traveled by an object, to analyze the relationship between two variables in scientific experiments, and to solve optimization problems in economics and business.
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