Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions Ex-13.3, (Part -2), Linear Equation In Two Variables, Class 9, Maths

RD Sharma Solutions Ex-13.3, (Part -2), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q8: From the choices given below, choose the equations whose graph is given in fig

(i) y = x
 (ii) x + y = 0
 (iii) y = 2x
 (iv) 2 + 3y = 7x

Ans: We are given co-ordinates (1, – 1) and (-1, 1) as the solution of one of the following equations.

We will substitute the value of both co-ordinates in each of the equation and find the equation which satisfies the given co-ordinates.

RD Sharma Solutions Ex-13.3, (Part -2), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

(i) We are given, y = x

Substituting x =I and y = -1 ,

we get; 1 ≠ -1

L.H.S ≠ R.H.S

Substituting x = -1  and y = 1 ,

we get; -1 ≠1

L.H.S ≠R.H.S

Therefore, the given equation y = x does not represent the graph in the figure.

(ii) We are given,

x + y = 0

Substituting x =1 and y = -1 , we get

⇒1 + (-1) = 0

⇒ 0 = 0

L.H.S = R.H.S

Substituting x = —1 and y = 1 ,we get

(-1)+ 1 = 0

0 = 0

L.H.S = R.H.S

Therefore, the given solutions satisfy this equation.

Thus, it is the equation whose graph is given.


Q9: From the choices given below, choose the equation whose graph is given fig:

(i) y = x + 2
 (ii) y = x – 2
 (iii)y = – x + 2
 (iv) x + 2y = 6

Ans. We are given co-ordinates (-1, 3) and (2, 0) as the solution of one of the following equations.

We will substitute the value of both co-ordinates in each of the equation and find the equation which satisfies the given co-ordinates.

RD Sharma Solutions Ex-13.3, (Part -2), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

(i) We are given, y = x+2

Substituting x = – 1 and y = 3 ,we get

3 ≠ – 1 + 2

L.H.S ≠ R.H.S

Substituting x = 2 and y = 0 ,we get

0 ≠ 4

L.H.S ≠R.H.S

Therefore, the given solution does not satisfy this equation.

(ii) We are given, y = x – 2

Substituting x = —1 and y = 3 ,we get

3 = – 1 – 2

L.H.S ≠ R.H.S

Substituting x = 2 and y = 0 ,we get

0 = 0

L.H.S = R.H.S

Therefore, the given solutions does not completely satisfy this equation.

(iii) We are given, y = – x + 2

Substituting x = – 1 and y = 3,we get

3 = – (– 1) + 2

L.H.S = R.H.S

Substituting x = 2 and y = 0 ,we get

0 = -2 + 2

0 = 0

L.H.S = R.H.S

Therefore, the given solutions satisfy this equation.

Thus, it is the equation whose graph is given.


Q 10 : If the point (2, -2) lies on the graph of linear equation, 5x + 4y = 4, find the value of k.

Ans. It is given that the point (2,-2) lies on the given equation,

5x + ky = 4

Clearly, the given point is the solution of the given equation.

Now, Substituting x = 2 and y = – 2 in the given equation, we get 5x + ky = 4

5 x 2 + (– 2) k = 4

2k = 10 – 4

2k = 6

k = 6/2

k = 3


Q 11 : Draw the graph of equation 2x + 3y = 12. From the graph, find the co ordinates of the point:

(i) whose y-coordinate is 3
 (ii) whose x coordinate is -3

Ans. We are given,

2x +3y =12

RD Sharma Solutions Ex-13.3, (Part -2), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Substituting, x = 0 in  RD Sharma Solutions Ex-13.3, (Part -2), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-13.3, (Part -2), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

y = 4

Substituting x = 6 in RD Sharma Solutions Ex-13.3, (Part -2), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

RD Sharma Solutions Ex-13.3, (Part -2), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

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

RD Sharma Solutions Ex-13.3, (Part -2), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

By plotting the given equation on the graph, we get the point B (0, 4) and C (6,0).

(i) Co-ordinates of the point whose y axis is 3 are A (3/2, 3)

(ii) Co-ordinates of the point whose x -coordinate is —3 are D (-3, 6)


Q 12: Draw the graph of each of the equations given below. Also, find the coordinates of the points where the graph cuts the coordinate axes:

(i) 6x – 3y = 12
 (ii) – x + 4y = 8
 (iii) 2x + y = 6
 (iv) 3x + 2y + 6 = 0

Ans. (i) We are given,

6x – 3y = 12 We get,

y = (6x —12) /3

Now, substituting x = 0 in y = – (6x – 12)/3 we get

y =- 4

Substituting x = 2 in y = (- 6x —12)/3, 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-40

RD Sharma Solutions Ex-13.3, (Part -2), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Co-ordinates of the points where graph cuts the co-ordinate axes are y = – 4 at y axis and x = 2 at x axis. (ii) We are given,

– x + 4y = 8

We get,

Now, substituting x = 0 in  RD Sharma Solutions Ex-13.3, (Part -2), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics we get

y = 2

Substituting x = -8 in RD Sharma Solutions Ex-13.3, (Part -2), 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

X0-8
Y20

RD Sharma Solutions Ex-13.3, (Part -2), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Co-ordinates of the points where graph cuts the co-ordinate axes are y = 2 at y axis and x = —8 at x axis.

(iii) We are given,

2x + y = 6

We get, y = 6 – 2x

Now, substituting x = 0 in y = 6 -2x we get

y = 6

Substituting x = 3 in y = 6-2x, 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

X03
Y60

RD Sharma Solutions Ex-13.3, (Part -2), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Co-ordinates of the points where graph cuts the co-ordinate axes are y = 6 at y axis and x =3 at x axis.

(iv) We are given,

3x+2y+6 = 0

We get,RD Sharma Solutions Ex-13.3, (Part -2), 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 -2), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

y= – 3

Substituting x = —2 in RD Sharma Solutions Ex-13.3, (Part -2), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

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
y-30

  RD Sharma Solutions Ex-13.3, (Part -2), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

Co-ordinates of the points where graph cuts the co-ordinate axes are y = – 3 at y axis and x = – 2 at x axis.


Q 13 : Draw the graph of the equation 2x + y = 6. Shade the region bounded by the graph and the coordinate axes. Also, find the area of the shaded region.

Ans. We are given,

2x + y = 6

We get,

y = 6 – 2x

Now, substituting x = 0 in y = 6 – 2x,

we get y = 6

Substituting x =3 in y = 6— 2x,

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

X03
Y60

 

RD Sharma Solutions Ex-13.3, (Part -2), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

The region bounded by the graph is ABC which forms a triangle.

AC at y axis is the base of triangle having AC = 6 units on y axis.

BC at x axis is the height of triangle having BC = 3 units on x axis.

Therefore, Area of triangle ABC, say A is given by A = (Base x Height)/2

A = (AC x BC)/2

A = (6 x 3)/2

A = 9 sq. units

Q 14 : Draw the graph of the equation RD Sharma Solutions Ex-13.3, (Part -2), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 MathematicsAlso, find the area of the triangle formed by 3 4 the line and the coordinates axes.

Ans. We are given.

RD Sharma Solutions Ex-13.3, (Part -2), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

4x +3y = 12

We get,

RD Sharma Solutions Ex-13.3, (Part -2), 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 -2), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics,we get 

y = 4

Substituting x = 3 in  RD Sharma Solutions Ex-13.3, (Part -2), 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

X03
Y40

RD Sharma Solutions Ex-13.3, (Part -2), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics

The region bounded by the graph is ABC which forms a triangle.

AC at y axis is the base of triangle having AC = 4 units on y axis.

BC at x axis is the height of triangle having BC = 3 units on x axis.

Therefore,

Area of triangle ABC, say A is given by

A = (Base x Height)/2

A= (AC x BC)/2

A = (4 x 3)/2

A = 6 sq. units

The document RD Sharma Solutions Ex-13.3, (Part -2), 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 -2), Linear Equation In Two Variables, Class 9, Maths - RD Sharma Solutions for Class 9 Mathematics

1. What is the importance of solving linear equations in two variables?
Ans. Solving linear equations in two variables is important as it helps in finding the relationship between two unknown quantities. It allows us to determine the values of the variables that satisfy the given conditions or constraints. This is useful in various real-life situations such as determining the cost and quantity of items, analyzing data, predicting outcomes, and solving optimization problems.
2. How do we solve linear equations in two variables graphically?
Ans. To solve linear equations in two variables graphically, we plot the given equations on a coordinate plane. The point of intersection of the two lines represents the solution to the equations. If the lines are parallel, it means there is no solution, and if the lines coincide, it means there are infinitely many solutions. By visually analyzing the graph, we can determine the values of the variables that satisfy both equations.
3. What are the different methods to solve linear equations in two variables algebraically?
Ans. There are three main methods to solve linear equations in two variables algebraically: 1. Substitution Method: In this method, we solve one equation for one variable and substitute it into the other equation. This helps us find the value of one variable, which can then be substituted back into either of the original equations to find the value of the other variable. 2. Elimination Method: In this method, we manipulate the equations so that the coefficients of one variable in both equations are equal in magnitude but opposite in sign. By adding or subtracting the equations, we can eliminate one variable and solve for the other. 3. Cross Multiplication Method: This method is used when the equations are in the form of fractions. By cross multiplying the fractions and simplifying the resulting equation, we can find the values of the variables.
4. 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 lines representing the equations coincide, it means that every point on the line is a solution to the equations. In this case, there are infinitely many solutions. However, if the lines are parallel, it means that they do not intersect and there is no common solution.
5. How are linear equations in two variables used in real-life situations?
Ans. Linear equations in two variables are used in various real-life situations such as: 1. Cost and Quantity Analysis: They can be used to determine the cost and quantity of items based on given conditions or constraints. 2. Data Analysis: They can be used to analyze data and find relationships between two variables. 3. Optimization Problems: They can be used to solve optimization problems, such as finding the maximum or minimum value of a quantity. 4. Predicting Outcomes: They can be used to make predictions and forecasts based on the relationship between two variables.
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