# RD SHARMA Solutions for Class 10 Maths Chapter 3 - Pairs of Linear Equations in Two Variables

## Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.1

Seven years ago,

Age of Aftab = x - 7

Age of his daughter = y - 7

According to the given condition,

Three years hence,

Age of Aftab = x + 3

Age of his daughter = y + 3

According to the given condition,

Thus, the given conditions can be algebraically represented as:

x - 7y = -42

x - 3y = 6

Three solutions of this equation can be written in a table as follows:

x | -7 | 0 | 7 |

y | 5 | 6 | 7 |

Three solutions of this equation can be written in a table as follows:

x | 6 | 3 | 0 |

y | 0 | -1 | -2 |

The graphical representation is as follows:

**Concept insight:**In order to represent a given situation mathematically, first see what we need to find out in the problem. Here, Aftab and his daughter's present age needs to be found so, so the ages will be represented by variables x and y. The problem talks about their ages seven years ago and three years from now. Here, the words 'seven years ago' means we have to subtract 7 from their present ages, and 'three years from now' or 'three years hence' means we have to add 3 to their present ages. Remember in order to represent the algebraic equations graphically the solution set of equations must be taken as whole numbers only for the accuracy. Graph of the two linear equations will be represented by a straight line.

The given conditions can be algebraically represented as:

Three solutions of this equation can be written in a table as follows:

x |
50 | 60 | 70 |

y | 60 | 40 | 20 |

Three solutions of this equation can be written in a table as follows:

x |
70 | 80 | 75 |

y | 10 |
-10 |
0 |

The graphical representation is as follows:

**Concept insight:**cost of apples and grapes needs to be found so the cost of 1 kg apples and 1 kg grapes will be taken as the variables. From the given conditions of collective cost of apples and grapes, a pair of linear equations in two variables will be obtained. Then, in order to represent the obtained equations graphically, take the values of variables as whole numbers only. Since these values are large so take the suitable scale.

## Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.2

Three solutions of this equation can be written in a table as follows:

x | 0 | 1 | 2 |

y | -5 | 0 | 5 |

x | 0 | 1 | 2 |

y | -3 | 0 | 3 |

The graphical representation of the two lines will be as follows:

It can be observed that the required triangle is ABC.

The coordinates of its vertices are A (1, 0), B (0, -3), C (0, -5).

**Concept insight: **In order to find the coordinates of the vertices of the triangle so formed, find the points where the two lines intersects the y-axis and also where the two lines intersect each other. Here, note that the coordinates of the intersection of lines with y-axis is taken and not with x-axis, this is because the question says to find the triangle formed by the two lines and the y-axis.

(i) Let the number of girls and boys in the class be x and y respectively.

According to the given conditions, we have:

x + y = 10

x - y = 4

x + y = 10 x = 10 - y

Three solutions of this equation can be written in a table as follows:

x | 4 | 5 | 6 |

y | 6 | 5 | 4 |

x - y = 4 x = 4 + y

Three solutions of this equation can be written in a table as follows:

x | 5 | 4 | 3 |

y | 1 | 0 | -1 |

The graphical representation is as follows:

From the graph, it can be observed that the two lines intersect each other at the point (7, 3).

So, x = 7 and y = 3.

Thus, the number of girls and boys in the class are 7 and 3 respectively.

(ii) Let the cost of one pencil and one pen be Rs x and Rs y respectively.

According to the given conditions, we have:

5x + 7y = 50

7x + 5y = 46

Three solutions of this equation can be written in a table as follows:

x | 3 | 10 | -4 |

y | 5 | 0 | 10 |

Three solutions of this equation can be written in a table as follows:

x | 8 | 3 | -2 |

y | -2 | 5 | 12 |

The graphical representation is as follows:

From the graph, it can be observed that the two lines intersect each other at the point (3, 5).

So, x = 3 and y = 5.

Therefore, the cost of one pencil and one pen are Rs 3 and Rs 5 respectively.

(iii)

Let us denote the number of pants by x and the number of skirts by y. Then the equations formed are:

y = 2x - 2 ...(1)

and y = 4x - 4 ...(2)

Let us draw the graphs of Equations (1) and (2) by finding two solutions for each of the equations.

They are given in Table

x | 2 | 0 |

y = 2x - 2 | 2 | -2 |

x | 0 | 1 |

y = 4x - 4 | -4 | 0 |

Plot the points and draw the lines passing through them to represent the equations, as shown in fig.,

The two lines intersect at the point (1,0). So, x = 1, y = 0 is the required solution of the pair of linear equations, i.e., the number of pants she purchased is 1 and she did not buy any skirt.**Concept insight: **Read the question carefully and examine what are the unknowns. Represent the given conditions with the help of equations by taking the unknowns quantities as variables. Also carefully state the variables as whole solution is based on it. On the graph paper, mark the points accurately and neatly using a sharp pencil. Also, take at least three points satisfying the two equations in order to obtain the correct straight line of the equation. Since joining any two points gives a straight line and if one of the points is computed incorrect will give a wrong line and taking third point will give a correct line. The point where the two straight lines will intersect will give the values of the two variables, i.e., the solution of the two linear equations. State the solution point.

(i) For the two lines a_{1}x + b_{1}x + c_{1} = 0 and a_{2}x + b_{2}x + c_{2} = 0, to be intersecting, we must have

So, the other linear equation can be 5x + 6y - 16 = 0

(ii) For the two lines a_{1}x + b_{1}x + c_{1} = 0 and a_{2}x + b_{2}x + c_{2} = 0, to be parallel, we must have

So, the other linear equation can be 6x + 9y + 24 = 0,

(iii) For the two lines a_{1}x + b_{1}x + c_{1} = 0 and a_{2}x + b_{2}x + c_{2} = 0 to be coincident, we must have

So, the other linear equation can be 8x + 12y - 32 = 0,

**Concept insight: **In order to answer such type of problems, just remember the conditions for two lines to be intersecting, parallel, and coincident. This problem will have multiple answers as their can be many equations satisfying the required conditions.

The lines AB and CD intersect at point R(1, 4). Hence, the solution of the given pair of linear equations is x = 1, y = 4.

From R, draw RM ⊥ X-axis and RN ⊥ Y-axis.

Then, from graph, we have

RM = 4 units, RN = 1 unit, AP = 4 units, BQ = 4 units

From the graph, the vertices of the triangle AOP formed by the given lines are A(4, 4), O(0, 0) and P(6, 2).

The graph of x = 3 is a straight line parallel to Y-axis at a distance of 3 units to the right of Y-axis.

The graph of x = 5 is a straight line parallel to Y-axis at a distance of 5 units to the right of Y-axis.

The graph of x = -2 is a straight line parallel to Y-axis at a distance of 2 units to the left of Y-axis.

The graph of y = 3 is a straight line parallel to X-axis at a distance of 3 units above X-axis.

Given equations are:

x - y + 1 = 0 … (i)

3x + 2y - 12 = 0 … (ii)

From (i) we get, x = y - 1

When x = 0, y = 1

When x = -1, y = 0

When x = 1, y = 2

We have the following table:

x |
0 |
-1 |
1 |

y |
1 |
0 |
2 |

From (ii) we get,

When x = 0, y = 6

When x = 4, y = 0

When x = 2, y = 3

We have the following table:

x |
0 |
4 |
2 |

y |
6 |
0 |
3 |

Graph of the given equations is:

As the two lines intersect at (2, 3).

Hence, x = 2, y = 3 is the solution of the given equations.

## Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.3

## Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.5

The given system of equations will have infinite number of solutions if

## Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.8

Let the fraction be

According to the given conditions, we have

Subtracting (ii) from (i), we get x = 7

Substituting the value of x in (ii), we get

y = 15

## Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.4

## Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.6

## Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.7

## Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.9

The difference between the ages of Ani and Biju is given as 3 years. So, either Biju is 3 years older than Ani or Ani is 3 years older than Biju.

Let the age of Ani and Biju be x years and y years respectively.

Age of Dharam = 2 × x = 2x years

x - y = 3 ... (1)

Age of Biju = 19 - 3 = 16 years

Case II: Biju is older than Ani by 3 years

y - x = 3 ... (3)

4x - y = 60 ... (4)

Adding (3) and (4), we obtain:

3x = 63

x = 21

Age of Ani = 21 years

Age of Biju = 21 + 3 = 24 years**Concept Insight: **In this problem, ages of Ani and Biju are the unknown quantities. So, we represent them by variables x and y. Now, note that here it is given that the ages of Ani and Biju differ by 3 years. So, it is not mentioned that which one is older. So, the most important point in this question is to consider both cases Ani is older than Biju and Biju is older than Ani. For second condition the relation on the ages of Dharam and Cathy can be implemented . Pair of linear equations can be solved using a suitable algebraic method.

## Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.10

According to the question,

By using equation (1), we obtain:

3x - 10t = 30 ... (3)

Adding equations (2) and (3), we obtain:

x = 50

Substituting the value of x in equation (2), we obtain:

(-2) x (50) + 10t = 20

-100 + 10t = 20

10t = 120

t = 12

From equation (1), we obtain:

d = xt = 50 x 12 = 600

Thus, the distance covered by the train is 600 km.

**Concept insight:**To solve this problem, it is very important to remember the relation . Now, all these three quantities are unknown. So, we will represent these

by three different

variables. By using the given conditions, a pair of equations will be obtained. Mind one thing that the equations obtained will not be linear. But they can be reduced to linear form by using the fact that . Then two linear equations can be formed which can

be solved easily by elimination method.

## Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.11

We know that the sum of the measures of opposite angles in a cyclic quadrilateral is 180°.

A + C = 180

4y + 20 - 4x = 180

-4x + 4y = 160

x - y = -40 ... (1)

Also, B + D = 180

3y - 5 - 7x + 5 = 180

-7x + 3y = 180 ... (2)

Multiplying equation (1) by 3, we obtain:

3x - 3y = -120 ... (3)

Adding equations (2) and (3), we obtain:

-4x = 60

x = -15

Substituting the value of x in equation (1), we obtain:

-15 - y = -40

y = -15 + 40 = 25

A = 4y + 20 = 4(25) + 20 = 120^{o}

B = 3y - 5 = 3(25) - 5 = 70^{o}

C = -4x = -4(-15) = 60^{o}

D = -7x + 5 = -7(-15) + 5 = 110^{o}**Concept insight: **The most important idea to solve this problem is by using the fact that the sum of the measures of opposite angles in a cyclic quadrilateral is 180^{o}. By using this relation, two linear equations can be obtained which can be solved easily by eliminating a suitable variable.

Let the money with the first person and second person be Rs x and Rs y respectively.

According to the question,

x + 100 = 2(y - 100)

x + 100 = 2y - 200

x - 2y = -300 ... (1)

6(x - 10) = (y + 10)

6x - 60 = y + 10

6x - y = 70 ... (2)

Multiplying equation (2) by 2, we obtain:

12x - 2y = 140 ... (3)

Subtracting equation (1) from equation (3), we obtain:

11x = 140 + 300

11x = 440

x = 40

Putting the value of x in equation (1), we obtain:

40 - 2y = -300

40 + 300 = 2y

2y = 340

y = 170

Thus, the two friends had Rs 40 and Rs 170 with them.**Concept insight: **This problem talks about the amount of capital with two friends. So, we will represent them by variables x and y respectively. Now, using the given conditions, a pair of linear equations can be formed which can then be solved easily using elimination method.

## Chapter 3 - Pairs of Linear Equations in Two Variables Exercise 3.114

So, the correct option is (b).

## Chapter 3 - Pairs of Linear Equations in Two Variables Exercise 3.115

So, the correct option is (a).

So, the correct option is (b).

So, the correct option is (a).

So, the correct option is (c).

Consistent solution means either linear equations have unique solutions or infinite solutions.

⇒ In case of unique solution; lines are intersecting

⇒ If solutions are infinite, lines are coincident.

So, lines are either intersecting or coincident

So, the correct option is (d).

So, the correct option is (c).

So, the correct option is (b).

So, the correct option is (d).

So, the correct option is (d).

So, the correct option is (a).

## Chapter 3 - Pairs of Linear Equations in Two Variables Exercise 3.116

So, the correct option is (b).

Since x = a and y = b is the solution of given system of equations x - y = 2 and x + y = 4, we have

a - b = 2 ….(i)

a + b = 4 ….(ii)

Adding (i) and (ii), we have

2a = 6 ⇒ a = 3

⇒ b = 4 - 3 = 1

Hence, correct option is (a).

### Other Chapters for CBSE Class 10 Mathematics

Chapter 1- Real Numbers Chapter 2- Polynomials Chapter 4- Quadratic Equations Chapter 5- Arithmetic Progressions Chapter 6- Co-ordinate Geometry Chapter 7- Triangles Chapter 8- Circles Chapter 9- Constructions Chapter 10- Trigonometric Ratios Chapter 11- Trigonometric Identities Chapter 12- Heights and Distances Chapter 13- Areas Related to Circles Chapter 14- Surface Areas and Volumes Chapter 15- Statistics Chapter 16- Probability### RD SHARMA Solutions for CBSE Class 10 Subjects

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