Class 10 - Maths - Linear Equations in 2 Variables

                                                                      Exercise 3.1

Question 1:

Aftab tells his daughter, “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.” (Isn’t this interesting?)

Represent this situation algebraically and graphically.

Answer:

Let the present age of Aftab = x

and present age of his daughter is represented as = y

Seven years ago,

Aftab’s age = x – 7

Age of Aftab’s daughter = y – 7

According to the question,

      (x – 7) = 7 (y – 7 )

=> x – 7 = 7 y – 49

=> x – 7y = – 49 + 7

=> x – 7y = – 42 …           (i)

=> x = 7y – 42

=> Putting y = 5, 6 and 7, we get

=> x = 7 * 5 – 42 = 35 – 42 = – 7

=> x = 7 * 6 – 42 = 42 – 42 = 0

=> x = 7 * 7 – 42 = 49 – 42 = 7

x

-7

0

7

y

5

6

7

 

Three years from now,

Aftab’s age = x + 3

Age of Aftab’s daughter = y + 3

According to the question,

     (x + 3) = 3 (y + 3)

=> x + 3 = 3y + 9

=> x – 3y = 9 – 3

=> x – 3y = 6 …           (ii)

=> x = 3y + 6

=> Putting, y = – 2, –1 and 0, we get

=> x = 3 * (-2) + 6 = – 6 + 6 =0

=> x = 3 * (-1) + 6 = – 3 + 6 = 3

=> x = 3 * 0 + 6 = 0 + 6 = 6    

x

0

3

6

y

-2

-1

0

 

From equation (i) and (ii)

x – 7y = – 42 …           (i)

x – 3y = 6 …                  (ii)

Graphical representation is given as above.

Class_10_Maths_LinearEquations_In_2_Variables_Graph8

 

 

Question 2:

The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 2 more balls of the same kind for Rs 1300.

Represent this situation algebraically and geometrically.

Answer:

Let the cost of one bat = Rs x

Let the cost of one ball = Rs y

According to first condition,

3x + 6y = 3900   …………..1

According to second condition,

x + 2y = 1300   …………….2

Hence, the following is the algebraic

representation of the situation:

3x + 6y = 3900

x + 2y = 1300

Now, for graphical representation, the three solutions of each equation are as follows:

From equation 1, we get

y = (3900 – 3x)/6

x

300

100

-100

y

500

600

700

 

From equation 2, we get

y = (1300 – x)/2

x

300

100

-100

y

500

600

700

 

Class_10_Maths_LinearEquations_In_2_Variables_Graph7

Question 3:

The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160.

After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300.

Represent the situation algebraically and geometrically.

Answer:

Let the cost of 1 kg of apple = Rs x

Let the cost of 1 kg of grapes = Rs y

According to first condition,

2x + y = 160   …………..1

According to second condition,

4x + 2y = 300   …………….2

Hence, the following is the algebraic

representation of the situation:

2x + y = 160

4x + 2y = 300

Now, for graphical representation, the three solutions of each equation are as follows:

From equation 1, we get

y = 160 – 2x

x

50

60

70

y

60

40

20

 

From equation 2, we get

y = (300 – 4x)/2

x

70

80

75

y

10

-10

0

Class_10_Maths_LinearEquations_In_2_Variables_Graph6

                                                                        Exercise 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 Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.

Answer:

(i) Let the number of boys be = x  

Let the number of girls be = y

Given that total number of students is 10

Therefore, x + y = 10

=> x = 10 – y

Putting y = 0, 5, 10 we get

x = 10 – 0 = 10

x = 10 – 5 = 5

x = 10 – 10 = 0

x

10

5

0

y

0

5

10

                                                         

     

Given: the number of girls is 4 more than the number of boys

=> y = x + 4

Putting x = – 4, 0, 4, and we get

y = – 4 + 4 = 0

y = 0 + 4 = 4

y = 4 + 4 = 8

x

-4

0

4

y

0

4

8

 

 

Graphical representation is given below:

 

 Class_10_Maths_LinearEquations_In_2_Variables_Graph5

 

From the graph, we get

Number of boys = 3 and number of girls = 7

(ii) Let the cost of one pencil = Rs x

Let the cost of one pen = Rs y

5 pencils and 7 pens together cost = Rs 50

=> 5x + 7y = 50

=> 5x = 50 – 7y

=> x = (10 – 7y) /5

Putting value of y = 5, 10 and 15 we get

x = 10 – 7 * 5/5 = 10 – 7 = 3

x = 10 – 7 * 10/5 = 10 – 14 = – 4

x = 10 – 7 * 15/5 = 10 – 21 = – 11

x

3

-4

11

y

5

10

15

 

 

Given: 7 pencils and 5 pens together cost Rs 46

=> 7x + 5y = 46

=> 5y = 46 – 7x

=> y = 46/5 – 7x/5

=> y = 9.2 – 1.4x

Putting x = 0, 2 and 4 we get

y = 9.2 – 1.4 * 0 = 9.2 – 0 = 9.2

y = 9.2 – 1.4 * 2 = 9.2 – 2.8 = 6.4

y = 9.2 – 1.4 * 2 = 9.2 – 5.6 = 3.6

x

0

2

4

y

9.2

6.4

3.6

       

 

Graphical representation is given below

From the graph, we get

Cost of one pencil = Rs 3 and cost of one pen = Rs 5.

 

 Class_10_Maths_LinearEquations_In_2_Variables_Graph4

 

 

Question 2:

On comparing the ratios a1/a2, b1/b2 and c1/c2, find out whether the lines representing the following pairs of linear equations intersect at a point,

are parallel or coincident:

(i) 5x – 4y + 8 = 0                           (ii) 9x + 3y + 12 = 0                   (iii) 6x – 3y + 10 = 0       

    7x + 6y – 9 = 0                                 18x + 6y + 24 = 0                        2x – y + 9 = 0

Answer:

(i) 5x – 4y + 8 = 0

     7x + 6y – 9 = 0

On comparing these equation with

a1 x + b1 y + c1 = 0

a2 x + b2 y + c2 = 0

We get

a1 = 5, b1 = – 4, and c1 = 8

a2 =7, b2 = 6 and c2 = – 9

a1/a2 = 5/7, b1/b2 = – 4/6 and c1/c2 = 8/(-9)

Hence, a1/a2 ≠ b1/b2

Therefore, both the lines intersect at one point.

(ii)  9x + 3y + 12 = 0

      18x + 6y + 24 = 0

Comparing these equations with

a1 x + b1 y + c1 = 0

a2 x + b2 y + c2 = 0

We get

a1 = 9, b1 = 3, and c1 = 12

a2 =18, b2 = 6 and c2 = 24

a1/a2 = 9/18 = 1/2

b1/b2 = 3/6 = 1/2

and c1/c2 = 12/24 = 1/2

Hence, a1/a2 = b1/b2 = c1/c2

Therefore, both the lines are coincident

(iii) 6x – 3y + 10 = 0

       2x – y + 9 = 0

Comparing these equations with

a1 x + b1 y + c1 = 0

a2 x + b2 y + c2 = 0

We get

a1 = 6, b1 = -3, and c1 = 10

a2 =2, b2 = -1 and c2 = 9

a1/a2 = 6/2 = 3/1

b1/b2 = -3/(-1) = 3/1

and c1/c2 = 12/24 = 1/2

Hence, a1/a2 = b1/b2 ≠ c1/c2

Therefore, both lines are parallel.

Question 3:

On comparing the ratios a1/a2, b1/b2 and c1/c2, find out whether the following pair of linear equations are consistent, or inconsistent.

(i) 3x + 2y = 5; 2x – 3y = 7

(ii) 2x – 3y = 8; 4x – 6y = 9

(iii) 3/2x + 5/3y = 7; 9x – 10y = 14

(iv) 5x – 3y = 11; -10x + 6y = –22

(v) 4/3x + 2y =8; 2x + 3y = 12

Answer:

(i)   3x + 2y = 5; 2x – 3y = 7

a1 /a2 = 3/2

b1 /b2 = -2/3 and

c1 /c2 = 5/7

Hence, a1 /a2 ≠ b1 /b2

These linear equations intersect each other at one point and therefore have only one possible

solution. Hence, the pair of linear equations is consistent.

(ii)  2x – 3y = 8; 4x – 6y = 9

a1 /a2 = 2/4 = 1/2

b1 /b2 = -3/(-6) = 1/2

c1 /c2 = 8/9

Hence, a1 /a2 = b1 /b2 ≠ c1 /c2

Therefore, these linear equations are parallel to each other and thus have no possible

solution. Hence, the pair of linear equations is inconsistent.

(iii) 3/2x + 5/3y = 7; 9x – 10y = 14

a1 /a2 = (3/2)/9 = 1/6

b1 /b2 = (5/3)/(-10) = – 1/6

c1 /c2 = 7/14 = 1/2

Hence, a1 /a2 ≠ b1 /b2

Therefore, these linear equations intersect each other at one point and thus have only one

possible solution. Hence, the pair of linear equations is consistent.

(iv) 5x – 3y = 11; – 10x + 6y = –22

 a1 /a2 = 5/–10 = – 1/2

b1 /b2 = – 3/6 = – 1/2

c1 /c2 = 11/–22 = – 1/2

Hence, a1 /a2 = b1 /b2 = c1 /c2

Therefore, these linear equations are coincident and have infinite number of possible

solutions. Therefore, the pair of linear equations is consistent.

(v) 4/3x + 2y =8; 2x + 3y = 12

a1 /a2 = 4/3/2 = 2/3

b1 /b2 = 2 /3 and

c1 /c2 = 8/12 = 2/3

Hence, a1 /a2 = b1 /b2 = c1 /c2

Therefore, these linear equations are coincident and have infinite number of possible

solutions. Therefore, the 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

Answer:

(i)   x + y = 5; 2x + 2y = 10

a1 /a2 = 1/2

b1 /b2 = 1/2 and

c1 /c2 = 5/10 = 1/2

Hence, a1 /a2 = b1 /b2 = c1 /c2

Therefore, these linear equations are coincident and have infinite number of possible

solutions. Therefore, the pair of linear equations is consistent.

 x + y = 5

x

4

3

2

y

1

2

3

 x = 5 – y

       

 

 and, 2x + 2y = 10

 x = (10 – 2y)/2

     

x

4

3

2

y

1

2

3

 

 

Graphical representation

        Class_10_Maths_LinearEquations_In_2_Variables_Graph3  

From the figure, it can be observed that these lines are overlapping each other. Therefore,

infinite number of solutions are possible for the given pair of equations.

(ii)  x – y = 8, 3x – 3y = 16

a1 /a2 = 1/3

b1 /b2 = – 1/–3 = 1/3 and

c1 /c2 = 8/16 = 1/2

Hence, a1 /a2 = b1 /b2 ≠ c1 /c2

Therefore, these linear equations are parallel to each other and thus have no possible

solution. Hence, the pair of linear equations is inconsistent.

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0

a1 /a2 = 2/4 = 1/2

b1 /b2 = – 1/2 and

c1 /c2 = – 6/–4 = 3/2

Hence, a1 /a2 ≠ b1 /b2

Therefore, these linear equations are intersecting each other at one point and thus have only

one possible solution. Hence, the pair of linear equations is consistent.

2x + y – 6 = 0

y = 6 – 2x

       

x

0

1

2

y

6

4

2

 

 

and, 4x – 2y – 4 = 0

 y = (4x – 4)/2

x

1

2

3

y

0

2

4

      

 

Graphical representation is given below:

From the figure, it can be observed that these lines are intersecting each other at the only one

point i.e., (2,2) which is the solution for the given pair of equations.

 Class_10_Maths_LinearEquations_In_2_Variables_Graph2

             

(iv)  2x – 2y – 2 = 0, 4x – 4y – 5 = 0

a1 /a2 = 2/4 = 1/2

b1 /b2 = – 2/–4 = 1/2 and

c1 /c2 = 2/5

Hence, a1 /a2 = b1 /b2 ≠ c1 /c2

Therefore, these linear equations are parallel to each other and thus, have no possible

solution. Hence, the pair of linear equations is inconsistent.

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.

Answer:

Let length of the rectangle be = x m

Let Width of the rectangle be = y m

According to the question,

y – x = 4 ...              (i)

y + x = 36 ...           (ii)

y – x = 4

y = x + 4

x

0

8

12

y

4

12

16

       

 

y + x = 36

x

0

36

16

y

36

0

20

       

 

Graphical representation is given below:

        Class_10_Maths_LinearEquations_In_2_Variables_Graph1 

 

From the figure, it can be seen that these lines intersect each other at only one point i.e.,    

(16, 20). Therefore, the length and width of the given garden is 20 m and 16 m respectively.

 

Question 6:

Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the

geometrical representation of the pair so formed is:

(i) intersecting lines    (ii) parallel lines      (iii) coincident lines

Answer:

(i) Intersecting lines:

Condition,

a1 /a2 ≠ b1 /b2

The second line such that it is intersecting the given line is

2x + 4y – 6 = 0 as

a1 /a2 = 2/2 = 1

b1 /b2 = 3/4 and

a1 /a2 ≠ b1 /b2

(ii) Parallel lines

Condition,

a1 /a2 = b1 /b2 ≠ c1 /c2

Hence, the second line can be

4x + 6y – 8 = 0 as

a1 /a2 = 2/4 = 1/2

b1 /b2 = 3/6 = 1/2 and

c1 /c2 = – 8/–8 = 1

and a1 /a2 = b1 /b2 ≠ c1 /c2

(iii) Coincident lines

Condition,

a1 /a2 = b1 /b2 = c1 /c2

Hence, the second line can be

6x + 9y – 24 = 0 as

a1 /a2 = 2/6 = 1/3

b1 /b2 = 3/9 = 1/3 and

c1 /c2 = – 8/–24 = 1/3

and a1 /a2 = b1 /b2 = c1 /c2

Question 7:

Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0.

Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Answer:

From equation 1, we get

x = y - 1

x

0

1

2

y

1

2

3

 

 

From equation 2, we get

x = (12 – 2y)/3

x

4

2

0

y

0

3

6

 

 

The coordinate of the vertices of the triangle formed by these lines and the x-axis are:

(-1, 0), (4, 0) and (2, 3).

 Class_10_Maths_LinearEquations_In_2_Variables_Graph

 

                                                                       Exercise 3.3

Question 1:

Solve the following pair of linear equations by the substitution method.

(i)    x + y = 14; x – y = 4    

(ii)   s – t = 3; s/3 + t/2 = 6               

(iii)  3x – y = 3; 9x – 3y = 9          

(iv)  0.2x + 0.3y = 1.3; 0.4x + 0.5y = 2.3       

(v) √2x + √3y = 0; √3x – √8y = 0

(vi) 3x/2 – 5y/3 = -2; x/3 + y/2 = 13/6

Answer:

(i)   x + y = 14 ... (i)

      x – y = 4 ... (ii)

From equation (i), we get

x = 14 – y ... (iii)

Putting this value in equation (ii), we get

(14 – y) – y = 4

14 – 2y = 4

10 = 2y

y = 5 ... (iv)

Putting this in equation (iii), we get

x = 9

So, x = 9 and y = 5

(ii)  s – t = 3 ... (i)

s/3 + t/2 = 6 ... (ii)

From equation (i), we get s = t + 3

Putting this value in equation (ii), we get

t + 3/3 + t/2 = 6

2t + 6 + 3t = 36

5t = 30

t = 30/5 ... (iv)

Putting in equation (iii), we get

s = 9

So, s = 9, t = 6

(iii) 3x – y = 3 ... (i)

9x – 3y = 9 ... (ii)

From equation (i), we get

y = 3x – 3 ... (iii)

Putting this value in equation (ii), we get

9x – 3(3x – 3) = 9

9x – 9x + 9 = 9

9 = 9

This is always true.

Hence, the given pair of equations has infinite possible solutions and the relation between these variables can be given by y = 3x – 3

Therefore, one of its possible solutions is x = 1, y = 0.

(iv) 0.2x + 0.3y = 1.3 ... (i)

    0.4x + 0.5y = 2.3 ... (ii)

0.2x + 0.3y = 1.3

Solving equation (i), we get

0.2x = 1.3 – 0.3y

Dividing by 0.2, we get

x = 1.3/0.2 – 0.3/0.2

x = 6.5 – 1.5 y ... (iii)

Putting the value in equation (ii), we get

0.4x + 0.5y = 2.3

(6.5 – 1.5y) × 0.4x + 0.5y = 2.3

2.6 – 0.6y + 0.5y = 2.3

–0.1y = 2.3 – 2.6

y = – 0.3/–0.1

y = 3

Putting this value in equation (iii) we get

x = 6.5 – 1.5 y

x = 6.5 – 1.5(3)

x = 6.5 – 4.5

x = 2

So, x = 2 and y = 3

(v) √2x + √3y = 0  …………..1

 √3x – √8y = 0   ………….2

From equation 1, we get

y = -√2x/√3   ………….3

Put value of y in equation 2, we get

√3x - √2(-√2x/√3) = 0

3x – 2x = 0

x = 0

Put value ox in equation 3, we get

y = 0

Hence, x = 0, y = 0

(vi) 3x/2 – 5y/3 = -2    ………1

x/3 + y/2 = 13/6   …………….2

From equation 1, we get

5y/3 = 3x/2 + 2

5y/3 = (3x + 4)/2

y = 3(3x + 4)/(2 * 5)

y = (9x + 12)/10          ………3

Put value of y in equation 2, we get

      x/3 + (1/2) * (9x + 12)/10 = 13/6

=> (20x + 27x + 36)/60 = 13/6

=> (47x + 36)/60 = 13/6

=> (47x + 36)/10 = 13

=> 47x + 36 = 13 * 10

=> 47x + 36 = 130

=> 47x = 130 – 36

=> 47x = 94     => x = 94/47        => x = 2

Put value of x in equation 3, we get

      y = (9 * 2 + 12)/10

=> y = (18 + 12)/10

=> y = 30/10

=> y = 3          Hence, x = 2, y = 3

Question 2:

Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

Answer:

Given, 2x + 3y = 11 ... (i)

Subtracting 3y both side we get

2x = 11 – 3y … (ii)

Putting this value in equation second we get

2x – 4y = – 24 … (iii)

11- 3y – 4y = – 24

7y = – 24 – 11

–7y = – 35

y = – 35/–7

y = 5

Putting this value in equation (iii) we get

2x = 11 – 3 * 5

2x = 11 – 15

2x = – 4

Dividing by 2 we get

x = – 2

Putting the value of x and y

y = mx + 3

5 = – 2m + 3

2m = 3 – 5

m = – 2/2              or, m = – 1

Question 3:

Form the pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other. Find them.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750.

Find the cost of each bat and each ball.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km,

he charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155.

What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator.

If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.

(vi) Five years hence, the age of Jacob will be three times that of his son.

Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Answer:

(i) Let the larger number be = x

 Let the smaller number be = y

 The difference between the two numbers is 26

 x – y = 26

 x = 26 + y

 Given that one number is three times the other

 So, x = 3y

 Putting the value of x we get

 26 + y = 3y

 –2y = – 2 6

 y = 13

Putting value of y, we get

 x = 3 * 13 = 39

 Therefore the numbers are 13 and 39.

(ii)Let the first angle be= x

and the second angle be = y

As both angles are supplementary so that sum will 180

x + y = 180

x = 180 – y ... (i)

Given; difference is 18 degree

Therefore, x – y = 18

Putting the value of x we get

180 – y – y = 18

 – 2y = – 162

y = – 162/–2

y = 81

Putting the value back in equation (i), we get

x = 180 – 81 = 99

Hence, the angles are 99° and 81°.

(iii) Let the cost of each bat be = Rs x

Let the cost of each ball be = Rs y

Given: coach of a cricket team buys 7 bats and 6 balls for Rs 3800.

7x + 6y = 3800

6y = 3800 – 7x

Dividing by 6, we get

y = (3800 – 7x)/6 ……….. (i)

Given that she buys 3 bats and 5 balls for Rs 1750 later.

3x + 5y = 1750

Putting the value of y

3x + 5 ((3800 – 7x)/6) = 1750

Multiplying by 6, we get

18x + 19000 – 35x = 10500

 –17x =10500 – 19000

–17x = – 8500

x = – 8500/–17

x = 500

Putting this value in equation (i) we get

y = (3800 – 7 * 500)/6

y = 300/6

y = 50

Hence the cost of each bat = Rs 500 and the cost of each balls = Rs 50.

(iv) Let the fixed charge for taxi = Rs x

and variable cost per km = Rs y

Total cost = fixed charge + variable charge

Given that for a distance of 10 km, the charge paid is Rs 105

x + 10y = 105 … (i)

x = 105 – 10y

Given that for a journey of 15 km, the charge paid is Rs 155

x + 15y = 155

Putting the value of x we get

105 – 10y + 15y = 155

5y = 155 – 105

5y = 50

Dividing by 5, we get

y = 50/5 = 10

Putting this value in equation (i) we get

x = 105 – 10 * 10

x = 5

Cost for traveling a distance of 25 km = x + 25y

                                                                    = 5 + 25 * 10

                                                                    = 5 + 250

                                                                    =255

A person has to pay Rs 255 for 25 Km.

(v)  Let the Numerator be = x

Let the Denominator be = y

Fraction = x/y

A fraction becomes 9/11, if 2 is added to both the numerator and the denominator

(x + 2)/y + 2 = 9/11

On Cross multiplying,

11x + 22 = 9y + 18

Subtracting 22 from both sides,

11x = 9y – 4

Dividing by 11, we get

x = (9y – 4)/11 … (i)

Given: if 3 is added to both the numerator and the denominator it becomes 5/6.

(x + 3)/(y + 3) = 5/6 … (ii)

On Cross multiplying,

6x + 18 = 5y + 15

Subtracting the value of x, we get

6(9y – 4 )/11 + 18 = 5y + 15

Subtracting 18 from both the sides

6(9y – 4 )/11 = 5y – 3

54 – 24 = 55y – 33

 –y = – 9

y = 9

Putting this value of y in equation (i), we get

x = (9y – 4)/11 … (i)

x = (81 – 4)/77

x = 77/11

x = 7

Hence our fraction is 7/9.

(vi) Let the present age of Jacob be = x year

and the present Age of his son be = y year

Five years from now,

Jacob’s age will be = x + 5 year

Age of his son will be = y + 5year

Given; the age of Jacob will be three times that of his son

x + 5 = 3(y + 5)

Adding 5 to both side,

x = 3y + 15 – 5

x = 3y + 10 … (i)

Five years ago,

Jacob’s age= x – 5 year

His son’s age = y – 5 year

Jacob’s age was seven times that of his son

x – 5 = 7(y – 5)

Putting the value of x from equation (i) we get

3y + 10 – 5 = 7y – 35

3y + 5 = 7y – 35

3y – 7y = – 35 – 5

 –4y = – 40

y = – 40/–4

y = 10 year

Putting the value of y in equation we get

x = 3 * 10 + 10

x = 40 years

Hence, Present age of Jacob = 40 years and present age of his son = 10 years.

 

                                                                       Exercise 3.4

Question 1:

Solve the following pair of linear equations by the elimination method and the substitution method:

(i) x + y = 5 and 2x – 3y = 4                                (ii) 3x + 4y = 10 and 2x – 2y = 2

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7                  (iv) x/2 + 2y/3 = -1 and x – y/3 = 3

Answer:

x + y =5 and 2x –3y = 4

By elimination method

x + y =5 ... (i)

2x –3y = 4 ... (ii)

Multiplying equation (i) by (ii), we get

2x + 2y = 10 ... (iii)

2x – 3y = 4 ... (ii)

Subtracting equation (ii) from equation (iii), we get

5y = 6

y = 6/5

Putting the value in equation (i), we get

x = 5 – (6/5) = 19/5

Hence, x = 19/5 and y = 6/5

By substitution method

x + y = 5 ... (i)

Subtracting y from both side, we get

x = 5 – y ... (iv)

Putting the value of x in equation (ii) we get

2(5 – y) – 3y = 4

–5y = – 6

y = – 6/–5 = 6/5

Putting the value of y in equation (iv) we get

x = 5 – 6/5

x = 19/5

Hence, x = 19/5 and y = 6/5 again

(ii)  3x + 4y = 10 and 2x – 2y = 2

By elimination method

3x + 4y = 10 .... (i)

2x – 2y = 2 ... (ii)

Multiplying equation (ii) by 2, we get

4x – 4y = 4 ... (iii)

3x + 4y = 10 ... (i)

Adding equation (i) and (iii), we get

x + 0 = 14

Dividing both side by 7, we get

x = 14/7 = 2

Putting in equation (i), we get

3x + 4y = 10

3(2) + 4y = 10

6 + 4y = 10

4y = 10 – 6

4y = 4

y = 4/4 = 1

Hence, answer is x = 2, y = 1

By substitution method

3x + 4y = 10 ... (i)

Subtract 3x both side, we get

4y = 10 – 3x

Divide by 4 we get

y = (10 – 3x )/4

Putting this value in equation (ii), we get

2x – 2y = 2 ... (i)

2x – 2(10 – 3x )/4) = 2

Multiply by 4 we get

8x – 2(10 – 3x) = 8

8x – 20 + 6x = 8

14x = 28

x = 28/14 = 2

y = (10 – 3x)/4

y = 4/4 = 1

Hence, answer is x = 2, y = 1 again.

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

By elimination method

3x – 5y – 4 = 0

3x – 5y = 4 ...(i)

9x = 2y + 7

9x – 2y = 7 ... (ii)

Multiplying equation (i) by 3, we get

9 x – 15 y = 11 ... (iii)

9x – 2y = 7 ... (ii)

Subtracting equation (ii) from equation (iii), we get

–13y = 5

y = – 5/13

Putting value in equation (i), we get

3x – 5y = 4 ... (i)

3x – 5(–5/13) = 4

Multiplying by 13 we get

39x + 25 = 52

39x = 27

x =27/39 = 9/13

Hence our answer is x = 9/13 and y = – 5/13

By substitution method

3x – 5y = 4 ... (i)

Adding 5y on both sides we get

3x = 4 + 5y

Dividing by 3 we get

x = (4 + 5y )/3 ... (iv)

Putting this value in equation (ii) we get

9x – 2y = 7 ... (ii)

9 ((4 + 5y )/3) – 2y = 7

On solve we get

3(4 + 5y ) – 2y = 7

12 + 15y – 2y = 7

13y = – 5

y = – 5/13

 x = {4 + 5 * (-5/13)}/3

x = (4 – 25/13)/3

x = {(52 - 25)/13}/3

x = 27/(13 * 3)

x = 9/13

Hence, we get x = 9/13 and y = -5/13 again.

(iv) x/2 + 2y/3 = – 1 and x – y/3 = 3

By elimination method

x/2 + 2y/3 = – 1 ... (i)

x – y/3 = 3 ... (ii)

Multiplying equation (i) by 2, we get

x + 4y/3 = – 2 ... (iii)

x – y/3 = 3 ... (ii)

Subtracting equation (ii) from equation (iii), we get

5y/3 = – 5

Dividing by 5 and multiplying by 3, we get

y = – 15/5

y = – 3

Putting this value in equation (ii), we get

x – y/3 = 3 ... (ii)

x – (–3)/3 = 3

x + 1 = 3

x = 2

Therefore x = 2 and y = – 3.

By substitution method

x – y/3 = 3 ... (ii)

Adding y/3 on both sides, we get

x = 3 + y/3 ... (iv)

Putting the value in equation (i) we get

x/2 + 2y/3 = – 1 ... (i)

(3 + y/3)/2 + 2y/3 = – 1

3/2 + y/6 + 2y/3 = – 1

Multiplying by 6,

9 + y + 4y = – 6

5y = – 15

y = – 3

Therefore x = 2 and y = – 3

Question 2:

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

(i)    If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1.

It becomes 1/2 if we only add 1 to the denominator. What is the fraction?

(ii)   Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu.

How old are Nuri and Sonu?

(iii)  The sum of the digits of a two-digit number is 9.

Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

(iv)  Meena went to bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only.

Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.

(v)   A lending library has a fixed charge for the first three days and an additional charge for each day thereafter.

Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days.

Find the fixed charge and the charge for each extra day.

Answer:

(i) Let the fraction be x/y

According to the question,

    (x + 1)/(y – 1) = 1

=> x – y = – 2 ... (i)

    x/y + 1 = 1/2

=> 2x – y = 1 ... (ii)

Subtracting equation (i) from equation (ii), we get

x = 3 ... (iii)

Putting this value in equation (i), we get

3 – y = – 2

 –y = – 5

 y = 5

Hence, the fraction is 3/5

(ii)  Let the present age of Nuri be = x

Let the present age of Sonu be = y

According to the given information,

(x – 5) = 3(y – 5)

x – 3y = – 10 ... (i)

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

x – 2y = 10 ... (ii)

Subtracting equation (i) from equation (ii), we get

y = 20 ... (iii)

Putting this value in equation (i), we get

x – 60 = – 10

x = 50

Hence, age of Nuri = 50 years and age of Sonu = 20 years.

(iii) Let the unit digit and tens digits of the number be x and y respectively.

Then, number = 10y + x

Number after reversing the digits = 10x + y

According to the question,

x + y = 9 ... (i)

9(10y + x) = 2(10x + y)

88y – 11x = 0

 –x + 8y =0 ... (ii)

Adding equation (i) and (ii), we get

9y = 9

y = 1 ... (iii)

Putting the value in equation (i), we get

x = 8

Hence, the number is 10y + x = 10 × 1 + 8 = 18.

(iv)  Let the number of Rs 50 notes and Rs 100 notes be x and y respectively.

According to the question,

x + y = 25 ... (i)

50x + 100y = 2000 ... (ii)

Multiplying equation (i) by 50, we get

50x + 50y = 1250 ... (iii)

Subtracting equation (iii) from equation (ii), we get

50y = 750

y = 15

Putting this value in equation (i), we have x = 10

Hence, Meena has 10 notes of Rs 50 and 15 notes of Rs 100.

(v)   Let the fixed charge for first three days and each day charge thereafter be Rs x and Rs y

respectively.

According to the question,

x + 4y = 27 ... (i)

x + 2y = 21 ... (ii)

Subtracting equation (ii) from equation (i), we get

2y = 6

y = 3 ... (iii)

Putting in equation (i), we get

x + 12 =27

x = 15

Hence, fixed charge = Rs 15 and Charge per day = Rs 3

                                                  Exercise 3.5

Question 1:

Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions.

In case there is a unique solution, find it by using cross multiplication method.

(i) x – 3y – 3 = 0                                     (ii) 2x + y = 5

    3x – 9y – 2 = 0                                         3x + 2y = 8

(iii) 3x – 5y = 20                                   (iv) x – 3y – 7 = 0

      6x – 10y = 40                                        3x – 3y – 15 = 0

Answer:

(i) x – 3y – 3 = 0

    3x – 9y – 2 =0

a1 /a2 = 1/3

b1 /b2 = – 3/-9 = 1/3 and

c1 /c2 = – 3/-2 = 3/2

a1 /a2 = b1 /b2 ≠ c1 /c2

Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect

each other and thus, there will not be any solution for these equations.

(ii)  2x + y = 5

       3x + 2y = 8

a1 /a2 = 2/3

b1 /b2 = 1/2 and

c1 /c2 = – 5/–8 = 5/8

a1 /a2 ≠ b1 /b2

Therefore, they will intersect each other at a unique point and thus, there will be a unique

solution for these equations.

By cross-multiplication method,

x/(b1 c2 – b2 c1) = y/(c1 a2 – c2 a1) = 1/(a1 b2 – a2 b1)

x/{-8–(–10)} = y/{–15 + 16} = 1/{4 – 3}

x/2 = y/1 = 1

x/2 = 1, y/1 = 1

So, x = 2, y = 1

(iii) 3x – 5y = 20

       6x – 10y = 40

a1 /a2 = 3/6 = 1/2

b1 /b2 = – 5/–10 = 1/2 and

c1 /c2 = – 20/–40 = 1/2

a1 /a2 = b1 /b2 = c1 /c2

Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident

to each other and thus, there are infinite solutions possible for these equations.

(iv) x – 3y – 7 = 0

      3x – 3y – 15= 0

a1 /a2 = 1/3

b1 /b2 = – 3/–3 = 1 and

c1 /c2 = – 7/–15 = 7/15

a1 /a2 ≠ b1 /b2

Therefore, they will intersect each other at a unique point and thus, there will be a unique

solution for these equations.

By cross-multiplication,

x/{45 – (21)} = y/{–21 – (–15)} = 1/{–3 – (–9)}

x/24 = y/(–6) = 1/6

x/24 = 1/6 and y/(–6) = 1/6

x = 24/6 and y = (-6)/6

x = 4 and y = – 1

So, x = 4, y = – 1

Question 2:

(i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?

2x + 3y = 7

(a – b) x + (a + b) y = 3a + b – 2

(ii) For which value of k will the following pair of linear equations have no solution?

3x + y = 1

(2k – 1) x + (k – 1) y = 2k + 1

Answer:

(i) 2x + 3y –7 = 0

    (a – b)x + (a + b)y – (3a + b –2) = 0

a1 /a2 = 2/(a – b) = 1/2

b1 /b2 = – 7/(a + b) and

c1 /c2 = – 7/–(3a + b – 2) = 7/(3a + b – 2)

For infinitely many solutions, a1 /a2 = b1 /b2 = c1 /c2

2/(a – b) = 7/(3a + b – 2)

6a + 2b – 4 = 7a – 7b

a – 9b = – 4 ... (i)

2/(a – b) = 3/(a + b)

2(a + b) = 3(a - b)

2a + 2b = 3a – 3b

a – 5b = 0 ... (ii)

Subtracting equation (i) from (ii), we get

4b = 4

b = 1

Putting this value in equation (ii), we get

a – 5 * 1 = 0

a = 5

Hence, a = 5 and b = 1 are the values for which the given equations give infinitely many

solutions.

(ii) 3x + y –1 = 0

      (2k –1)x + (k –1)y – (2k + 1) = 0

a1 /a2 = 3/(2k – 1)

b1 /b2 = 1/(k – 1) and

c1 /c2 = – 1/(–2k – 1) = 1/(2k + 1)

For no solutions,

a1 /a2 = b1 /b2 ≠ c1 /c2

3/(2k – 1) = 1/(k – 1) ≠ 1/(2k + 1)

3/(2k – 1) = 1/(k – 1)

3k – 3 = 2k – 1

k = 2

Hence, for k = 2, the given equation has no solution.

Question 3:

Solve the following pair of linear equations by the substitution and cross-multiplication methods :

8x + 5y = 9

3x + 2y = 4

Answer:

8x + 5y = 9 ... (i)

3x + 2y = 4 ... (ii)

From equation (ii), we get

x = (4 – 2y)/3 ... (iii)

Putting this value in equation (i), we get

8{(4 – 2y)/3} + 5y = 9

32 – 16y + 15y = 27

 –y = – 5

y = 5 ... (iv)

Putting this value in equation (ii), we get

3x + 10 = 4

x = – 2

Hence, x = – 2, y = 5

By cross multiplication again, we get

8x + 5y – 9 = 0

3x + 2y – 4 = 0

x/{–20 – (–18)} = y/{–27 – (–32)} = 1/(16 – 15)

x/(–2) = y/5 = 1/1

x/(–2) = 1 and y/5 = 1

x = – 2 and y = 5

Question 4:

Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess.

When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B,

who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.

(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator.

Find the fraction.

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer.

Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks.

How many questions were there in the test?

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time.

If the cars travel in the same direction at different speeds, they meet in 5 hours.

If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units.

If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Answer:

(i) Let x be the fixed charge of the food and y be the charge for food per day.

According to the question,

x + 20y = 1000 ... (i)

x + 26y = 1180 ... (ii)

Subtracting equation (i) from equation (ii), we get

6y = 180

y = 180/6 = 30

Putting this value in equation (i), we get

x + 20 × 30 = 1000

x = 1000 – 600

x = 400

Hence, fixed charge = Rs 400 and charge per day = Rs 30

(ii) Let the fraction be x/y

 According to the question,

      (x – 1)/y = 1/3

=> 3(x - 1) = y

=> 3x – y = 3... (i)

       x/(y + 8) = 1/4

=> 4x = y + 8

=> 4x – y = 8 ... (ii)

Subtracting equation (i) from equation (ii), we get

x = 5 ... (iii)

Putting this value in equation (i), we get

15 – y = 3

y = 12

Hence, the fraction is 5/12.

(iii) Let the number of right answers and wrong answers be x and y respectively.

According to the question,

     3x – y = 40 ... (i)

     4x – 2y = 50

=> 2(2x - y) = 50

=> 2x – y = 50/2

=> 2x – y = 25 ... (ii)

Subtracting equation (ii) from equation (i), we get

x = 15 ... (iii)

Putting this value in equation (ii), we get

30 – y = 25

y = 5

Therefore, number of right answers = 15

And number of wrong answers = 5

Total number of questions = 20

(iv) Let the speed of 1st car and 2nd car be u km/h and v km/h.

Respective speed of both cars while they are travelling in same direction = (u – v) km/h

Respective speed of both cars while they are travelling in opposite directions i.e., travelling

towards each other = (u + v) km/h

According to the question,

     5(u – v) = 100

=> u – v = 20 ... (i)

1(u + v) = 100 ... (ii)

Adding both the equations, we get

2u = 120

u = 60 km/h ... (iii)

Putting this value in equation (ii), we obtain

v = 40 km/h

Hence, speed of one car = 60 km/h and speed of other car = 40 km/h

(v) Let length and breadth of rectangle be x unit and y unit respectively.

Area = xy

According to the question,

     (x – 5) (y + 3) = xy – 9

=> xy + 3x – 5y – 15 = xy - 9

=> 3x – 5y – 6 = 0 ... (i)

      (x + 3) (y + 2) = xy + 67

=> xy + 2x + 3y + 6 = xy + 67

=> 2x – 3y – 61 = 0 ... (ii)

By cross multiplication, we get

     x/{305 – (–18)} = y/{–12 – (–183)} = 1/{9 – (–10)}

=> x/(305 + 18) = y/(–12 + 183) = 1/(9 + 10)

=> x/323 = y/171 = 1/19

=> x/323 = 1/19 and y/171 = 1/19

=> x = 323/19 and y = 171/19

=> x = 17, y = 9

Hence, the length of the rectangle = 17 units and breadth of the rectangle = 9 units.

                                                                         Exercise 3.6

Question 1:

Solve the following pairs of equations by reducing them to a pair of linear equations:

(i) 1/2x + 1/3y = 2                                        (ii) 2/√x + 3/√y = 2

     1/3x + 1/2y = 13/6                                       4/√x – 9/√y = -1

(iii) 4/x + 3y = 14                                          (iv) 5/(x - 1) + 1/(y – 2) = 2

       3/x – 4y = 23                                                 6/(x - 1) - 3/(y – 2) = 1

(v) (7x – 2y)/xy =   5                                      (vi) 6x + 3y = 6xy

      (8x + 7y)/xy = 15                                            2x + 4y = 5xy

(vii) 10/(x + y) + 2/(x – y) = 4                       (viii) 1/(3x + y) + 1/(3x – y) = 3/4

        15/(x + y) - 5/(x – y) = -2                               1/2(3x + y) - 1/2(3x – y) = -1/8

Answer:

(i) 1/2x + 1/3y = 2

     1/3x + 1/2y = 13/6

Let 1/x = p and 1/y = q, then the equations changes as below:

p/2 + q/3 = 2

3p + 2q – 12 = 0 ... (i)

p/3 + q/2 = 13/6

2p + 3q – 13 = 0 ... (ii)

By cross-multiplication method, we get

p/{–26 – (–36)} = q/{–24 – (–39)} = 1/{9 – 4}

p/10 = q/15 = 1/5

p/10 = 1/5 and q/15 = 1/5

p = 2 and q = 3

1/x = 2 and 1/y = 3

Hence, x = 1/2 and y = 1/3

(ii) 2/√x + 3/√y = 2

      4/√x + 9/√y = -1    

Let 1/√x = p and 1/√y = q, then the equation changes as below:   

2p + 3q = 2 ... (i)

4p – 9q = – 1 ... (ii)

Multiplying equation (i) by 3, we get

6p + 9q = 6 ... (iii)

Adding equation (ii) and (iii), we get

10p = 5

p = 1/2 ... (iv)

Putting value of P in equation (i), we get

2 * 1/2 + 3q = 2

3q = 1

q = 1/3

p = 1/√x = 1/2

√x = 2

x = 4 and

q = 1/√y = 1/3

√y = 3

y = 9

Hence, x = 4, y = 9

(iii) 4/x + 3y = 14

       3/x – 4y = 23

Putting 1/x = p in the given equations, we get

      4p + 3y = 14

=> 4p + 3y – 14 = 0

     3p – 4y = 23

=> 3p – 4y – 23 = 0

By cross-multiplication, we get

p/{–69 – 56} = y/{–42 – (–92)} = 1/{–16 – 9}

=> -p/125 = y/50 = – 1/25

Now, -p/125 = – 1/25 and y/50 = – 1/25

=> p = 5 and y = – 2

Also, p = 1/x = 5

=> x = 1/5

So, x = 1/5 and y = – 2 is the solution.

(v) 5/(x – 1) + 1/(y – 2) = 2

      6/(x – 1) – 3/(y – 2) = 1

Putting 1/(x – 1) = p and 1/(y – 2) = q in the given equations, we obtain

5p + q = 2 ... (i)

6p – 3q = 1 ... (ii)

Now, by multiplying equation (i) by 3 we get

15p + 3q = 6 ... (iii)

Now, adding equation (ii) and (iii)

21p = 7

⇒ p = 1/3

Putting this value in equation (ii) we get,

6 * 1/3 – 3q =1

⇒ 2 – 3q = 1

⇒ – 3q = 1 – 2

⇒ – 3q = – 1

⇒ q = 1/3

Now,

p = 1/(x – 1) = 1/3

⇒ 1/(x – 1) = 1/3

⇒ 3 = x – 1

⇒ x = 4

Also,

q = 1/(y – 2) = 1/3

⇒ 1/(y – 2) = 1/3

⇒ 3 = y – 2

⇒ y = 5

Hence, x = 4 and y = 5 is the solution.

(v) (7x – 2y)/xy = 5

⇒ 7x/xy – 2y/xy = 5

⇒ 7/y – 2/x = 5 ... (i)

(8x + 7y)/xy = 15

⇒ 8x/xy + 7y/xy = 15

⇒ 8/y + 7/x = 15 ... (ii)

Putting 1/x = p and 1/y = q in (i) and (ii) we get,

7q – 2p = 5 ... (iii)

8q + 7p = 15 ... (iv)

Multiplying equation (iii) by 7 and multiplying equation (iv) by 2 we get,

49q – 14p = 35 ... (v)

16q + 14p = 30 ... (vi)

Now, adding equation (v) and (vi) we get,

49q – 14p + 16q + 14p = 35 + 30

⇒ 65q = 65

⇒ q = 1

Putting the value of q in equation (iv)

8 + 7p = 15

⇒ 7p = 7

⇒ p = 1

Now,

p = 1/x = 1

⇒ 1/x = 1

⇒ x = 1

also, q = 1 = 1/y

⇒ 1/y = 1

⇒ y = 1

Hence, x =1 and y = 1 is the solution.

(vi) 6x + 3y = 6xy

⇒ 6x/xy + 3y/xy = 6

⇒ 6/y + 3/x = 6 ... (i)

2x + 4y = 5xy

⇒ 2x/xy + 4y/xy = 5

⇒ 2/y + 4/x = 5 ... (ii)

Putting 1/x = p and 1/y = q in (i) and (ii) we get,

6q + 3p – 6 = 0

2q + 4p – 5 = 0

By cross multiplication method, we get

p/{–30 – (–12)} = q/{–24 – (–15)} = 1/{6 – 24)

p/(–18) = q/(–9) = 1/(–18)

p/(–18) = 1/(–18) and q/(–9) = 1/(–18)

p = 1 and q = 1/2

p = 1/x = 1 and q = 1/y = 1/2

x = 1, y = 2

Hence, x = 1 and y = 2

(vii) 10/(x + y) + 2/(x – y) = 4

        15/(x + y) - 5/(x – y) = – 2

Putting 1/(x + y) = p and 1/(x – y) = q in the given equations, we get:

10p + 2q = 4

⇒ 10p + 2q – 4 = 0 ... (i)

15p – 5q = – 2

⇒ 15p – 5q + 2 = 0 ... (ii)

Using cross multiplication, we get

p/(4 – 20) = q/{–60 – (–20)} = 1/(–50 – 30)

p/(–16) = q/(–80) = 1/(–80)

p/(–16) = 1/(–80) and q/(–80) = 1/(–80)

p = 1/5 and q = 1

p = 1/(x + y) = 1/5 and q = 1/(x – y) = 1

x + y = 5 ... (iii)

and x – y = 1 ... (iv)

Adding equation (iii) and (iv), we get

2x = 6

x = 3 .... (v)

Putting value of x in equation (iii), we get

y = 2

Hence, x = 3 and y = 2

(viii) 1/(3x + y) + 1/(3x – y) = 3/4

         1/2(3x – y) – 1/2(3x – y) = – 1/8

Putting 1/(3x + y) = p and 1/(3x – y) = q in the given equations, we get

p + q = 3/4 ... (i)

p/2 – q/2 = – 1/8

p – q = – 1/4 ... (ii)

Adding equation (i) and (ii), we get

2p = 3/4 – 1/4

2p = 1/2

p = 1/4

Putting the value in equation (ii), we get

1/4 – q = – 1/4

q = 1/4 + 1/4 = 1/2

p = 1/(3x + y) = 1/4

3x + y = 4 ... (iii)

q = 1/(3x – y) = 1/2

3x – y = 2 ... (iv)

Adding equations (iii) and (iv), we get

6x = 6

x = 1 ... (v)

Putting the value in equation (iii), we get

3(1) + y = 4

y = 1

Hence, x = 1 and y = 1   

 

Question 2:

Formulate the following problems as a pair of equations, and hence find their solutions:

(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days.

Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.

(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus.

If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Answer:

Let the speed of Ritu in still water and the speed of stream be x km/h and y km/h respectively.

Speed of Ritu while rowing

Upstream = (x – y) km/h

Downstream = (x + y) km/h

According to question,

2(x + y) = 20

⇒ x + y = 10 ... (i)

2(x – y) = 4

⇒ x – y = 2 ... (ii)

Adding equation (i) and (ii), we get

Putting this equation in (i), we get

y = 4

Hence, Ritu’s speed in still water is 6 km/h and the speed of the current is 4 km/h.

(ii)  Let the number of days taken by a woman and a man be x and y respectively.

Therefore, work done by a woman in 1 day = 1/x

According to the question,

4(2/x + 5/y) = 1

2/x + 5/y = 1/4

3(3/x + 6/y) = 1

3/x + 6/y = 1/3

Putting 1/x = p and 1/y = q in these equations, we get

2p + 5q = 1/4

By cross multiplication, we get

p/{–20 – (–18)} = q/{–9 – (–18)} = 1/(144 – 180)

p/(–2) = q/(–1) = 1/(–36)

p/(–2) = -1/36 and q/(–1) = 1/(–36)

p = 1/18 and q = 1/36

p = 1/x = 1/18 and q = 1/y = 1/36

x = 18 and y = 36

Hence, number of days taken by a woman = 18 and number of days taken by a man = 36

(iii) Let the speed of train and bus be u km/h and v km/h respectively.

According to the given information,

60/u + 240/v = 4 ... (i)

100/u + 200/v = 25/6 ... (ii)

Putting 1/u = p and 1/v = q in the equations, we get

60p + 240q = 4 ... (iii)

100p + 200q = 25/6

600p + 1200q = 25 ... (iv)

Multiplying equation (iii) by 10, we get

600p + 2400q = 40 .... (v)

Subtracting equation (iv) from (v), we get

1200q = 15

q = 15/200 = 1/80 ... (vi)

Putting equation (iii), we get

60p + 3 = 4

60p = 1

p = 1/60

p = 1/u = 1/60 and q = 1/v = 1/80

u = 60 and v = 80

Hence, speed of train = 60 km/h and speed of bus = 80 km/h.

                                              Exercise 3.7 (Optional)*

Question 1:

The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy.

The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

Answer:

The difference between the ages of Biju and Ani is 3 years. Either Biju is 3 years older than Ani

or Ani is 3 years older than Biju. However, it is obvious that in both cases, Ani’s father’s age

will be 30 years more than that of Cathy’s age.

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

Therefore, age of Ani’s father, Dharam = 2 × x = 2x years

And age of Biju’s sister Cathy years

By using the information given in the question,

Case 1: When Ani is older than Biju by 3 years,

x − y = 3 ……….(i)

      2x – y/2 = 30

=> (4x - y) = 30 * 2

=> 4x − y = 60 ………(ii)

Subtracting (i) from (ii), we obtain

     3x = 60 − 3 = 57

=> x = 57/3

=> x = 19

Therefore, age of Ani = 19 years

And age of Biju = 19 − 3 = 16 years

Case 2: When Biju is older than Ani,

y − x = 3 (i)

4x − y = 60 (ii)

Adding (i) and (ii), we obtain

3x = 63

x = 21

Therefore, age of Ani = 21 years

And age of Biju = 21 + 3 = 24 years

Question 2:

One says, “Give me a hundred, friend! I shall then become twice as rich as you”.

The other replies, “If you give me ten, I shall be six times as rich as you”.

Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II] 

   [Hint: x + 100 = 2(y – 100), y + 10 = 6(x – 10)]

Answer:

Let those friends were having Rs x and y with them.

Using the information given in the question, we obtain

      x + 100 = 2(y − 100)

=> x + 100 = 2y − 200

=> x − 2y = −300 ……….(i)

and 6(x – 10) = y + 10

=> 6x – 60 = y + 10

=> 6x – y = 70  …………(ii)

Multiplying equation (ii) by 2, we obtain

12x − 2y = 140 ………..(iii)

Subtracting equation (i) from equation (iii), we obtain

      11x = 140 + 300

=> 11x = 440

=> x = 40

Using this in equation (i), we obtain

     40 − 2y = −300

=> 40 + 300 = 2y

=> 2y = 340

=> y = 170

Therefore, those friends had Rs 40 and Rs 170 with them respectively.

Question 3:

A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster,

it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h;

it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Answer:

Let the speed of the train be x km/h and the time taken by train to travel the given distance be

t hours and the distance to travel was d km. We know that,

     Speed = Distance travelled/Time taken to travel that distance

=> x = d/t

=> d = xt ……….(i)

Using the information given in the question, we obtain

By using equation (i), we obtain

       (x + 10) = d/(t - 2)

=> (x + 10)(t - 2) = d

=> tx – 2x + 10t – 20 = d

By using equation (i), we get

− 2x + 10t = 20 ………(ii)

       (x - 10) = d/(t + 3)

=> (x - 10)(t + 3) = d

=> tx + 3x - 10t – 30 = d

By using equation (i), we obtain

3x − 10t = 30 (iii)

Adding equations (ii) and (iii), we obtain

x = 50

Using equation (ii), we obtain

      (−2) * (50) + 10t = 20

=> −100 + 10t = 20

=> 10t = 120

=> t = 12 hours

From equation (i), we obtain

Distance to travel = d = xt = 50 * 12 = 600 km

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

 

Question 4:

The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less.

If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Answer:

Let the number of rows be x and number of students in a row be y.

Total students of the class = Number of rows * Number of students in a row

                                               = xy

Using the information given in the question,

Condition 1

Total number of students = (x − 1) (y + 3)

=> xy = (x − 1) (y + 3)

=> x y = xy − y + 3x – 3

=> 3x − y − 3 = 0

=> 3x − y = 3 ……….(i)

Condition 2

Total number of students = (x + 2) (y − 3)

=> xy = xy + 2y − 3x − 6

=> 3x − 2y = −6 ……….(ii)

Subtracting equation (ii) from (i), we get

      (3x − y) − (3x − 2y) = 3 − (−6)

=> -y + 2y = 3 + 6

=> y = 9

By using equation (i), we obtain

      3x − 9 = 3

=> 3x = 9 + 3

=> 3x = 12

=> x = 4

Number of rows = x = 4

Number of students in a row = y = 9

Total number of students in the class = xy = 4 * 9 = 36

Question 5:

In a ΔABC, ∠C = 3 ∠B = 2(∠A + ∠B). Find the three angles.

Answer:

Given that,

     ∠C = 3∠B = 2(∠A + ∠B)

=> 3∠B = 2(∠A + ∠B)

=> 3∠B = 2∠A + 2∠B

=> ∠B = 2∠A

=> 2 ∠A − ∠B = 0 …………(i)

We know that the sum of the measures of all angles of a triangle is 180°. Therefore,

=> ∠A + ∠B + ∠C = 180°

=> ∠A + ∠B + 3 ∠B = 180°

=> ∠A + 4 ∠B = 180° … (ii)

Multiplying equation (i) by 4, we obtain

8 ∠A − 4 ∠B = 0 … (iii)

Adding equations (ii) and (iii), we obtain

    9 ∠A = 180°

=> ∠A = 180°/9

=> ∠A = 20°

From equation (ii), we obtain

      20° + 4 ∠B = 180°

=> 4 ∠B = 160°

=> ∠B = 40°

∠C = 3 ∠B

      = 3 * 40°

      = 120°

Hence, ∠A, ∠B, ∠C are 20°, 40°, 120° respectively.  

Question 6:

Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y axis.

Answer:

    5x − y = 5

=> y = 5x – 5

The solution table will be as follows.

x

0

1

2

y

-5

0

5

    

     3x − y = 3

=> y = 3x − 3

The solution table will be as follows.

x

0

1

2

y

-5

0

5

 

The graphical representation of these lines will be as above.

It can be observed that the required triangle is ΔABC formed by these lines and y-axis.

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

Class_10_Maths_LinearEquations_In_2_Variables_Polynomials

Question 7:

Solve the following pair of linear equations:

(i) px + qy = p – q                                            (ii) ax + by = c

     qx – py = p + q                                                  bx + ay = 1 + c

(iii) x/a – y/b = 0                                             (iv) (a – b)x + (a + b) y = a2 – 2ab – b2

      ax + by = a2 + b2                                              (a + b)(x + y) = a2 + b2

(v) 152x – 378y = – 74

      -378x + 152y = – 604

Answer:

(i) px + qy = p − q ……… (1)

    qx − py = p + q ………. (2)

Multiplying equation (1) by p and equation (2) by q, we obtain

p2 x + pqy = p2 − pq ………… (3)

q2 x − pqy = pq + q2 ………… (4)

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

       p2 x + q2 x = p2 + q2

=> (p2 + q2) x = p2 + q2

=> x = (p2 + q2)/ (p2 + q2)

=> x = 1

From equation (1), we obtain

      p * 1 + qy = p – q

=> p + qy = p – q

=> qy = − q

=> y = -q/q

=> y = -1

(ii) ax + by = c ……….. (1)

     bx + ay = 1 + c …….. (2)

Multiplying equation (1) by a and equation (2) by b, we obtain

a2 x + aby = ac ………….. (3)

b2 x + aby = b + bc ……. (4)

Subtracting equation (4) from equation (3),

          (a2 − b2) x = ac − bc – b

=> x = (ac − bc – b)/ (a2 − b2)

=> x = {c(a – b) – b}/ (a2 − b2)

From equation (1), we obtain

      ax + by = c

=> a{c(a – b) – b}/ (a2 − b2) + by = c

=> {ac(a – b) – b}/ (a2 − b2) + by = c

=> by = c - {ac(a – b) – ab}/ (a2 − b2)

=> by = {c(a2 − b2) - ac(a – b) + ab}/ (a2 − b2)

=> by = {a2c − b2 c – a2 c + abc) + ab}/ (a2 − b2)

=> by = {abc − b2 c + ab}/ (a2 − b2)

=> by = {bc(a – b) + ab}/ (a2 − b2)

=> by = [b{c(a – b) + a}]/ (a2 − b2)

=> y = {c(a – b) + a}/ (a2 − b2)

(iii) x/a – y/b = 0

=> bx – ay = 0  ………..1

ax + by = a2 + b2  ………2

Multiplying equation (1) and (2) by b and a respectively, we obtain

b2 x − aby = 0 ………… (3)

a2 x + aby = a3 + ab2 …….. (4)

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

      b2 x + a2 x = a3 + ab2

=> x(b2 + a2) = a(a2 + b2)

=> x = a(a2 + b2)/ (b2 + a2)

=> x = a

By using (1), we obtain

     b * a − ay = 0

=> ab − ay = 0

=> ay = ab

=> y = b

(iv) (a − b)x + (a + b)y = a2 − 2ab − b2 ……….. (1)

(a + b) (x + y) = a2 + b2

(a + b)x + (a + b)y = a2 + b2 ……….. (2)

Subtracting equation (2) from (1), we obtain

      (a − b)x − (a + b)x = (a2 − 2ab − b2) − (a2 + b2)

=> (a − b − a − b) x = − 2ab − 2b2

=> -2bx = − 2b (a + b)

=> x = a + b

Using equation (1), we obtain

     (a − b)(a + b) + (a + b) y = a2 − 2ab − b2

=> a2 − b2 + (a + b) y = a2 − 2ab − b2

=> (a + b) y = − 2ab

=> y = -2ab/(a + b)

(v) 152x − 378y = − 74

      76x − 189y = − 37     …….… (1)

− 378x + 152y = − 604

− 189x + 76y = − 302     ………. (2)

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

      -189{(189y - 37)/76} + 76y = -302

=> -(189)2 y + 189 * 37 + (76) 2 y = − 302 * 76

=> 189 * 37 + 302 * 76 = (189) 2 y − (76) 2 y

=> 6993 + 22952 = (189 - 76)(198 + 76)y

=> 29945 = 113 * 265 * y

=> 29945y = 29945

=> y = 1

From equation (1), w get

      76x – 189 * 1 = − 37

=> 76x – 189 = − 37

=> 76x = 189 – 37

=> 76x = 152

=> x = 152/76

=> x = 2

Question 8:

ABCD is a cyclic quadrilateral (see Fig. 3.7).

Find the angles of the cyclic quadrilateral.

Class_10_Maths_LinearEquations_In_2_Variables_Cyclic_Quadrilateral

Answer:

We know that the sum of the measures of opposite angles in

a cyclic quadrilateral is 180°.

Therefore, ∠A + ∠C = 180°

       4y + 20 − 4x = 180°

=> -4x + 4y = 160°

=> x − y = − 40° …………(i)

Also, ∠B + ∠D = 180°

       3y − 5 − 7x + 5 = 180°

=> -7x + 3y = 180° ………(ii)

Multiplying equation (i) by 3, we obtain

3x − 3y = − 120° (iii)

Adding equations (ii) and (iii), we obtain

     -7x + 3x = 180° − 120°

=> -4x = 60°

=> x = −15°

By using equation (i), we obtain

      x − y = − 40°

=> -15 − y = − 40°

=> y = −15° + 40° = 25°

∠A = 4y + 20° = 4(25°) + 20° = 120°

∠B = 3y − 5° = 3(25°) − 5° = 70°

∠C = − 4x = − 4(− 15°) = 60°

∠D = -7x + 20° = -7(-15°) + 20° = 110°

 

 

 

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