194-NCERT New Syllabus Grade 10 Pair of Linear Equations in Two Variables Ex-3.3

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NCERT New Syllabus Mathematics

Class: 10

Exercise 3.3

Topic: Pair of Linear Equations in Two Variables 

Exploring the Concept of Pair of Linear Equations in Two Variables: NCERT Class 10

Linear equations are everywhere in our daily lives, from calculating expenses to solving practical problems. In this chapter, we dive deeper into understanding how pairs of linear equations work in two variables. Whether finding the meeting point of two straight lines or determining the solution to complex problems, this topic opens doors to various real-world applications.

This chapter focuses on exploring how two linear equations with two variables can be represented graphically and algebraically and how to find their solutions using different methods. We will explore everything from graphical representations to algebraic solutions such as substitution, elimination, and cross-multiplication.

By this topic's end, you'll understand how to interpret, solve, and apply linear equations in two variables effectively!

Elimination Method Steps:

Step 1: Equalize the Coefficients

Multiply both equations by appropriate constants to make the coefficients of either x or y the same.

Step 2: Eliminate One Variable

Add or subtract the equations to cancel out one variable. Depending on the result: 

If you obtain an equation in one variable, move to Step 3.

If a true statement without variables is obtained (e.g., 0 = 0), the

system has infinitely many solutions.

If a false statement is obtained (e.g., 0 = 5), the system has no

solution and is inconsistent.

Step 3: Solve for One Variable

Solve the resulting equation to find the value of x or y.

 

EXERCISE 3.3

Q1. 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, x – (y/3) = 3

Explanation:

1) Two equations in two variables are provided.

2) Equalize the coefficients of either x or y by multiplying both equations by

suitable non-zero constants.

3) Eliminate the variable by adding or subtracting one equation from the other.

4) This process is called the Elimination Method.

Solution:

(i) x + y = 5 and 2x – 3y = 4

a) Elimination method

1) Given equations are

x + y = 5 ---------------equation 1

2x – 3y = 4 ---------------equation 2

2) Here multiply equation 1 by 2 to get the coefficient of x same.

2(x + y) = 2(5)

2x + 2y = 10 ---------------equation 3

3) Subtracting equation 2 from equation 3, we get,

2x + 2y = 10

2x – 3y  =  4

    (–)    (+)      (–)

---------------------------

5y = 6

y = 6/5 ---------------equation 4

4) Put the value of y =6/5 from equation 4 in equation 1, we get,

x + y = 5

x + (6/5) = 5

x = 5 – (6/5)

x = (25 – 6)/5

x = (19)/5 

x = 19/5 -------------------------- equation 5. 

5) Therefore, x = 19/5 and y = 6/5.

b) Substitution method

6) x + y = 5 ------------ equation 6

7) 2x – 3y = 4 ------------ equation 7

8) Simplify equation 6, and we get

x + y = 5

y = 5 – x ------------ equation 8

9) Substitute the value of y = (5 – x) from equation 8 in equation 7, we get

2x – 3y = 4

2x – 3(5 – x) = 4

2x – 15 + 3x = 4

5x – 15 = 4

5x = 15 + 4

5x = 19

x = 19/5 ------------ equation 9

10) Put the value of x = 19/5 from equation 9 in equation 8, and we get

y = 5 – x

y = 5 – (19/5)

y = (25 – 19)/5  

y = 6/5 ------------ equation 10

11) The value of x = 19/5 and the value of y = 6/5.

 

(ii) 3x + 4y = 10 and 2x – 2y = 2

a) Elimination method

1) Given equations are

3x + 4y = 10 ---------------equation 1

2x – 2y = 2 ---------------equation 2

2) Here multiply equation 2 by 2 to get the coefficient of y same.

2(2x – 2y) = 2(2)

4x – 4y = 4 ---------------equation 3

3) Adding equation 1 and equation 3, we get,

3x + 4y = 10

4x – 4y  =  4

    ------------------------

7x = 14

x = 14/7

x = 2 ---------------equation 4

4) Put the value of x = 2 from equation 4 in equation 2, we get,

2x – 2y = 2

2(2) – 2y = 2

2y = 4 – 2

2y = 2

y = 1 -------------------------- equation 5. 

5) Therefore, x = 2 and y = 1.

b) Substitution method

6) 3x + 4y = 10 ------------ equation 6

7) 2x – 2y = 2 ------------ equation 7

8) Simplify equation 7, we get

2x – 2y = 2

    x – y = 1 

x = y + 1 ------------ equation 8

9) Substitute the value of x =(y + 1) from equation 8 in equation 6, we get

3x + 4y = 10

3(y + 1) + 4y = 10

3y + 3 + 4y = 10

7y + 3 = 10

7y = 10 – 3

7y = 7

y = 1 ------------ equation 9

10) Put the value of y = 1 from equation 9 in equation 8, we get

x = y + 1

x = 1 + 1

x = 2 ------------ equation 10

11) Therefore, x = 2 and y = 1.

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

a) Elimination method

1) Given equations are

3x – 5y – 4 = 0

3x – 5y = 4 ---------------equation 1

9x = 2y + 7

9x – 2y = 7 ---------------equation 2

2) Here multiply equation 1 by 3 to get the coefficient of x as same.

3(3x – 5y) = 3(4)

9x – 15y = 12 ---------------equation 3

3) Subtracting equation 2 from equation 3, we get,

9x – 15y = 12

9x –  2y  =  7

     (–)    (+)      (–)

    ------------------------

     –13 y = 5

    y = – (5/13) ---------------equation 4

4) Put the value of y = – (5/13) from equation 4 in equation 1, we get,

3x – 5y = 4

3x – 5(– 5/13) = 4

3x + (25/13) = 4 

3x = 4 – (25/13)

3x = (52 – 25)/13

3x = (27)/13 

x = (27)/(13 x 3)

x = 9/13 -------------------------- equation 5. 

5) Therefore, x = 9/13 and y = - (5/13).

b) Substitution method

6) 3x – 5y = 4 ------------ equation 6

7) 9x – 2y = 7 ------------ equation 7

8) Simplify equation 7, we get

3x – 5y = 4

3x = 5y +4

x = (5y + 4)/3 ------------ equation 8

9) Substitute the value of x = (5y + 4)/3 from equation 8 in equation 7, we get

9x – 2y = 7

9(5y + 4)/3 – 2y = 7

3(5y + 4) – 2y = 7 

15y + 12 – 2y = 7

13y + 12 = 7

13y = 7 – 12

13y = – 5

y = – (5/13) ------------ equation 9

10) Put the value of y = – (5/13) from equation 9 in equation 8, we get

x = (5(– 5/13) + 4)/3

x = (– 25/13) + 4)/3

x = (– 25 + 52)/(13(3))

x = (27)/(13(3))

x = (9)/(13) 

x = 9/13 ------------ equation 10

11) Therefore, x = 9/13 and y = – (5/13).

(iv) (x/2) + (2y/3) = – 1, x – (y/3) = 3

a) Elimination method

1) Given equations are

(x/2) + (2y/3) = – 1 ---------------equation 1

x – (y/3) = 3 ---------------equation 2

2) Here multiply equation 1 by 2 to get the coefficient of x as same.

2(x/2) + 2(2y/3) = 2(– 1)

x + (4y/3) = – 2 ---------------equation 3

3) Subtracting equation 2 from equation 3, we get,

x + 4y/3 = – 2

x –  y/3  =    3

     (–)  (+)        (–)

    --------------------------

    5 y/3 = – 5

y/3 = – 1

y = – 3 ---------------equation 4

4) Put the value of y = – 3 from equation 4 in equation 1, we get,

(x/2) + (2y/3) = – 1

(x/2) + (2(– 3)/3) = – 1

(x/2) + (– 2) = – 1

(x/2) = – 1 + 2

(x/2) = 1

x = 2 -------------------------- equation 5. 

5) Therefore, x = 2 and y = – 3.

b) Substitution method

6) 3x – 5y = 4 ------------ equation 6

7) 9x – 2y = 7 ------------ equation 7

8) Simplify equation 7, we get

3x – 5y = 4

3x = 5y +4

x = (5y + 4)/3 ------------ equation 8

9) Substitute the value of x = (5y + 4)/3 from equation 8 in equation 7, we get

9x – 2y = 7

9(5y + 4)/3 – 2y = 7

3(5y + 4) – 2y = 7 

15y + 12 – 2y = 7

13y + 12 = 7

13y = 7 – 12

13y = – 5

y = – (5/13) ------------ equation 9

10) Put the value of y = – (5/13) from equation 9 in equation 8, we get

x = (5(– 5/13) + 4)/3

x = (– 25/13) + 4)/3

x = (– 25 + 52)/(13(3))

x = (27)/(13(3))

x = (9)/(13) 

x = 9/13 ------------ equation 10

11) Therefore, x = 9/13 and y = – (5/13).

Q2. 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 a 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.

Explanation:

1) Two equations in two variables are provided.

2) Equalize the coefficients of either x or y by multiplying both equations by

suitable non-zero constants.

3) Eliminate the variable by adding or subtracting one equation from the other.

4) This process is called the Elimination Method.

Solution:

(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?

1) Let the numerator be x and the denominator be y.

2) According to the first condition, we have,

(x + 1)/(y – 1) = 1

(x + 1) = (y – 1) 

x – y = – 2 ------------ equation 1

3) According to the second condition, we have,

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

2x = y + 1

2x – y = 1 ------------ equation 2

4) Subtract equation 1 from equation 2, and we get,

2x –  y =   1

  x –  y = – 2

    (–)   (+)       (–)

----------------------------

  x         = 3

x = 3 ---------------equation 3

5) Put the value of x = 3 from equation 3 in equation 1, we get,

x – y = – 2

3 – y = – 2

y = 3 + 2 

y = 5 -------------------------- equation 4.

6) The equations are x – y = – 2 and 2x – y = 1, where x and y are the numerator

and denominator of the fraction is 3/5.

(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?

1) Let Nuri's present age be x and Sonu's present age be y.

2) 5 years ago their ages were (x – 5) and (y – 5).

3) According to the first relation given in the problem,

(x – 5) = 3(y – 5)

(x – 5) = 3y – 15

x – 3y = – 15 + 5
x – 3y = – 10  ---------------------- equation 1

4) 10 years later, their ages will be (x + 10) and (y + 10). 

5) According to the relation given in the problem,

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

(x + 10) = 2y + 20

x – 2y = 20 – 10
x – 2y = 10  ---------------------- equation 2

6) Subtract equation 1 from equation 2, and we get,

x – 2y =    10

x – 3y = – 10

    (–)  (+)      (+)

---------------------------- 

y = 20

y = 20 ---------------equation 3

7) Put the value of y = 20 from equation 3 in equation 2, we get,

x – 2y = 10

x – 2(20) = 10

x = 10 + 40 

x = 50 -------------------------- equation 4.

8) The equations are x – 3y = – 10 and x – 2y = 10, where x and y are the ages 

(in years) of Nuri and Sonu respectively. So Nuri's present age is 50 years and Sonu's present age is 20 years.

(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.

1) Let the ones place digit be x and the tens place digit be y.

2) According to the first condition, we have,

x + y = 9

x + y = 9 ------------ equation 1

3) The original number: 10 x + y

4) The number obtained by reversing the digits: 10 y + x

5) According to the second condition, we have,

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

90 x + 9 y = 20 y + 2 x 

90 x – 2 x + 9 y – 20 y = 0

88 x – 11 y = 0

11(8 x – y) = 0

8 x – y = 0 ------------ equation 2

4) Add equation 1 to equation 2, and we get,

8 x – y = 0

  x + y = 9

  ----------------------------

9x      = 9

x = 1 ---------------equation 3

5) Put the value of x = 1 from equation 3 in equation 1, we get,

x + y = 9

1 + y = 9

y = 9 – 1 

y = 8 -------------------------- equation 4.

6) The equations are x + y = 9 and 8 x – y = 0, Where x and y represent the tens

and units digits of the number, respectively. Solving these, we get 

y = 8, which means the number is 18.

(iv) Meena went to a 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.

1) Let the number of Rs 50 be x and the number of Rs 100 be y.

2) According to the first condition, we have,

x + y = 25

x + y = 25 ------------ equation 1

3) According to the second condition, we have,

50 x + 100 y = 2000

50 (x + 2 y) = 50 (40) 

x + 2 y = 40

x + 2 y = 40 ------------ equation 2

4) Subtract equation 1 from equation 2, and we get,

x + 2 y = 40

x +    y = 25

     (–)  (–)      (–)

  ----------------------------

y = 15

y = 15 ---------------equation 3

5) Put the value of y = 15 from equation 3 in equation 1, we get,

x + y = 25

x + 15 = 25

x = 25 – 15 

x = 10 -------------------------- equation 4.

6) The equations are x + y = 25 and x + 2 y = 40, Where x and y represent the

number of Rs 50 notes and Rs 100 notes

(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 a book she kept for five days. Find the fixed charge and the charge for each extra day.

1) Let the fixed charges for the first 3 days be Rs x.

2) Let the additional charges per extra day be Rs y.

3) Saritha paid Rs 27 for 7 days, so she paid Rs x for the first 3 days and the remaining for

4 days, so we have,

x + 4y = 27

x + 4y = 27 ------------ equation 1

4) Susy paid Rs 21 for 5 days, so she paid Rs x for the first 3 days and the remaining for

2 days, so we have,

x + 2y = 21

x + 2y = 21 ------------ equation 2

5) Subtract equation 2 from equation 1, and we get,

x + 4 y = 27

x + 2 y = 21

     (–)  (–)      (–)

  ----------------------------

2 y = 6

   y = 3 ---------------equation 3

5) Put the value of y = 3 from equation 3 in equation 2, we get,

x + 2 y = 21

x + 2(3) = 21

x + 6 = 21 

x = 21 – 6

x = 15 -------------------------- equation 4.

6) The equations are x + 4y = 27 and x + 2y = 21, where x is the fixed charge 

So, 

Conclusion: Mastering Pair of Linear Equations in Two Variables

As we reach the end of this fascinating topic on Pair of Linear Equations in Two Variables, it’s clear how these equations form the foundation for solving real-world problems involving relationships between two variables. By learning how to graph, calculate, and interpret their solutions, you’re honing your mathematical skills and building your problem-solving mindset. Remember, mastering this concept will open doors to more advanced mathematical techniques in the future. So keep practicing, explore different substitution, elimination, and graphical representation methods, and enjoy the learning journey!

Stay curious and keep solving!

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NCERT New Syllabus Class 10 - Quadratic Equations. Exercise 4.1

Anil Satpute

193-NCERT New Syllabus Grade 10 Pair of Linear Equations in Two Variables Ex-3.2

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NCERT New Syllabus Mathematics

Class: 10

Exercise 3.2

Topic: Pair of Linear Equations in Two Variables 

Exploring the Concept of Pair of Linear Equations in Two Variables: NCERT Class 10

Linear equations are everywhere in our daily lives, from calculating expenses to solving practical problems. In this chapter, we dive deeper into understanding how pairs of linear equations work in two variables. Whether finding the meeting point of two straight lines or determining the solution to complex problems, this topic opens doors to various real-world applications.

This chapter explores how two linear equations with two variables can be represented graphically and algebraically and how to find their solutions using different methods. We will explore everything from graphical representations to algebraic solutions such as substitution, elimination, and cross-multiplication.

By the end of this topic, you will understand how to interpret, solve, and apply linear equations in two variables effectively!

EXERCISE 3.2

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) √2 x + √3 y = 0,  √3 x – √8 y = 0

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

Explanation:

1) Start with a simple linear equation.

2) Solve for one variable (say x or y) using straightforward steps.

3) Substitute the value of the solved variable into the other equation to find the

value of the remaining variable.

Solution:

(i) x + y = 14, x – y = 4

1) x + y = 14 ------------ equation 1

2) x – y = 4 ------------ equation 2

3) Simplify equation 2,  we get

x – y = 4

y = x – 4 ------------ equation 3

4) Substitute the value of y from equation 3 in equation 1, we get

x + y = 14

x + (x – 4) = 14

2x – 4 = 14

2x = 14 + 4

2x = 18

x = 9 ------------ equation 4

5) Put the value of x from equation 4 in equation 3, we get

y = x – 4

y = 9 – 4 

y = 5

6) The value of x = 9 and the value of y = 5. 

(ii) s – t = 3, (s/3) + (t/2) = 6

1) s – t = 3 ------------ equation 1

2) (s/3) + (t/2) = 6 ------------ equation 2

3) Simplify equation 1, we get

s – t = 3

s = t + 3 ------------ equation 3

4) Substitute the value of s from equation 3 in equation 2, we get

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

((t + 3)/3) + (t/2) = 6

[(2t + 6) + (3t)]/6 = 6

(5t + 6)/6 = 6

(5t + 6) = 36

5t = 36 – 6

5t = 30 

t = 6 ------------ equation 4

5) Put the value of t from equation 4 in equation 3, we get

s = t + 3

s = 6 + 3

s = 9

6) The value of s = 9 and the value of t = 6. 

(iii) 3x – y = 3, 9x – 3y = 9

1) 3x – y = 3 ------------ equation 1

2) 9x – 3y = 9 ------------ equation 2

3) Simplify equation 1, we get

3x – y = 3

y = 3x – 3 ------------ equation 3

4) Substitute the value of y from equation 3 in equation 2, we get

9x – 3y = 9

9x – 3(3x – 3) = 9

9x – 9x – 9 = 9

– 9 = 9

5) This shows that the lines are coincident with infinitely many solutions.

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

1) 0.2x + 0.3y = 1.3 ------------ equation 1

2) 0.4x + 0.5y = 2.3 ------------ equation 2

3) Multiplying by 10 to equation 1 and equation 2, we get,

2x + 3y = 13 ------------ equation 3

4x + 5y = 23 ------------ equation 4

4) Simplify equation 3, we get

2x + 3y = 13

2x = 13 – 3y

x = (13 – 3y)/2 ------------ equation 5

5) Substitute the value of x from equation 5 in equation 4, we get

4x + 5y = 23

4[(13 – 3y)/2] + 5y = 23

2(13 – 3y) + 5y = 23

26 – 6y + 5y = 23

26 – y = 23 

y = 26 – 23

y = 3 ------------ equation 6

5) Put the value of y from equation 6 in equation 5, we get

x = (13 – 3y)/2

x = (13 – 3(3))/2

x = (13 – 9)/2

x = 4/2

x = 2 

6) The value of x = 2 and the value of y = 3.

(v) √2 x + √3 y = 0,  √3 x – √8 y = 0

1) √2x + √3y = 0 ------------ equation 1

2) √3x – √8y = 0 ------------ equation 2

3) Simplify equation 2, we get

√3x – √8y = 0

√3x = √8y

x = √(8/3)y ------------ equation 3

4) Substitute the value of x from equation 3 in equation 1, we get

√2x + √3y = 0

√2(8/3)y + √3y = 0

[√2(8/3) + √3]y = 0

[√16/3) + √3]y = 0

y = 0 ------------ equation 4

5) Put the value of y from equation 4 in equation 3, we get

x = √(8/3)y

x = √(8/3)x0 

x = 0

6) The value of x = 0 and the value of y = 0.

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

1) (3x/2) – (5y/3) = – 2 ------------ equation 1

2) (x/3) + (y/2) = 13/6 ------------ equation 2

3) Simplify equation 1 and equation 2, and we get,

9x – 10y = – 12 ------------ equation 3

2x + 3y = 13 ------------ equation 4

4) Simplify equation 3, we get

9x – 10y = – 12

9x = 10y – 12 

x = (10y – 12)/9 ------------ equation 5

5) Substitute the value of x from equation 5 in equation 4, we get

2x + 3y = 13

2[(10y – 12)/9] + 3y = 13

(20y – 24)/9 + 3y = 13

(20y – 24 + 27y)/9 = 13

(47y – 24)/9 = 13 

47y – 24 = 13 x 9

47y = 24 + 117

47y = 141

y = 141/47 

y = 3 ------------ equation 6

6) Put the value of y from equation 6 in equation 4, we get

2x + 3y = 13 

2x + (3 x 3) = 13

2x = 13 – 9 = 4

x = 4/2

x = 2 

7) The value of x = 2 and the value of y = 3.

Q2. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

1) 2x + 3y = 11 ------------ equation 1

2) 2x – 4y = – 24 ------------ equation 2

3) Simplify equation 2, we get

2x – 4y = – 24

2x = 4y – 24

x = (4y – 24)/2 

x = 2y – 12 ------------ equation 3

4) Substitute the value of x from equation 3 in equation 1, and we get

2x + 3y = 11

2(2y – 12) + 3y = 11

4y – 24 + 3y = 11

7y – 24 = 11

7y = 24 + 11

7y = 35

y = 35/7

y = 5 ------------ equation 4

5) Put the value of y = 5, from equation 4 in equation 3, and we get

x = 2y – 12

x = (2 x 5) – 12

x = 10 – 12  

x = – 2

6) The value of x = – 2 and the value of y = 5.

7) Put x = – 2 and y = 5 in the equation y = mx + 3, we get

y = mx + 3

5 = m(– 2) + 3

– 2m = 5 – 3

– 2m = 2

m = 2/(– 2)

m = – 1

8) Therefore, x = – 2, y = 5, and m = – 1. 

Q3. 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, the 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?

Explanation:

1) Let x and y represent the two variables.

2) Based on the given conditions, formulate the corresponding equations.

3) Solve the two equations obtained from the conditions to determine the values of

x and y. 

Solution:

(i) The difference between two numbers is 26 and one number is three times

the other. Find them.

1) Let the greater number be x, and the smaller number be y.

2) According to the first condition, 

x – y = 26 -------------------------- equation 1.

3) According to the second condition, 

x = 3y -------------------------- equation 2.

4) Put the value of x from equation 2 in equation 1, we get,

x – y = 26

3y – y = 26

2y = 26

y = 26/2

y = 13 -------------------------- equation 3.

5) Put the value of y from equation 3 in equation 2, we get,

x = 3y

x = 3 x 13

x = 39 -------------------------- equation 4.

6) The numbers are 13 and 39.

7) The equations are x – y = 26, and x = 3y, 

where x and y are two numbers (x > y); x = 39, y = 13.

(ii) The larger of two supplementary angles exceeds the smaller by 18

degrees. Find them.

1) Let the greater angle be x, and the smaller angle be y.

2) According to the first condition, 

x = y + 18 -------------------------- equation 1.

3) We know that the sum of the supplementary angles is 180 degrees, 

x + y = 180 -------------------------- equation 2.

4) Put the value of x = y + 18 from equation 1 in equation 2, we get,

x + y = 180

(y + 18) + y = 180

2y + 18 = 180

2y = 180 – 18

2y = 162 

y = 81⁰ -------------------------- equation 3.

5) Put the value of y = 81 from equation 3 in equation 1, we get,

x = y + 18

x = 81 + 18

x = 99⁰ -------------------------- equation 4.

6) The angles are 81⁰ and 99⁰. 

7) The equations are x = y + 18, x + y = 180, where x and y are the measures of

the two angles in degrees; x = 99⁰, y = 81⁰.

(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.

1) Let the cost of the bat be Rs x and the cost of the ball be Rs y.

2) As 7 bats and 6 balls cost Rs 3800, we have,

7x + 6y = 3800

7x = 3800 – 6y

x = (3800 – 6y)/7 ------------ equation 1

3) As 3 bats and 5 balls cost Rs 1750, we have,

3x + 5y = 1750 ------------ equation 2

4) Put the value of x = (3800 – 6y)/7 from equation 1 in equation 2, we get,

3x + 5y = 1750

3(3800 – 6y)/7 + 5y = 1750

[(11400 – 18y) + 35y]/7 = 1750

[(11400 – 18y) + 35y] = 1750 x 7

11400 + 17y = 1750 x 7

11400 + 17y = 12250

17y = 12250 – 11400

17y = 850

y = 850/17

y = 50 -------------------------- equation 3.

5) Put the value of y = 50 from equation 3 in equation 2, we get,

3x + 5y = 1750

3x + 5(50) = 1750

3x + 250 = 1750

3x = 1750 – 250 

3x = 1500

x = 1500/3 

x = 500 -------------------------- equation 4.

6) The equations are 7x + 6y = 3800, and 3x + 5y = 1750, where x and y are the

costs (in Rs) of one bat and one ball respectively; and the cost of the bat is Rs 500 and the cost of the ball is Rs 50.

(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, the 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?

1) Let the fixed charges be Rs x and the charges per km be Rs y.

2) According to the first condition, we have,

x + 10y = 105

x = 105 – 10y ------------ equation 1

3) According to the second condition, we have,

x + 15y = 155 ------------ equation 2

4) Put the value of x = 105 – 10y from equation 1 in equation 2, we get,

x + 15y = 155

(105 – 10y) + 15y = 155

105 + 5y = 155

5y = 155 – 105

5y = 50

y = 10 -------------------------- equation 3.

5) Put the value of y = 10 from equation 3 in equation 1, we get,

x = 105 – 10y

x = 105 – 10(10)

x = 105 – 100

x = 5  -------------------------- equation 4.

6) The charges for 25 km travel will be x + 25y.

7) So the total charges for 25 km travel will be,

x + 25y = 5 + 25(10)

= 5 + 250

= 255

8)  The equations are x + 10y = 105 and x + 15y = 155, where x is the fixed charges (in Rs), y is the charge (in Rs per km), and the fixed charges are x = Rs 5, the charge per kilometer is y = Rs 10, and the total travel cost for 25 km will be Rs 255.

 

(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.

1) Let the numerator be x and the denominator be y.

2) According to the first condition, we have,

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

11x + 22 = 9y +18

9y = 11x + 22 – 18

9y = 11x + 4

y = (11x + 4)/9 ------------ equation 1

3) According to the second condition, we have,

(x + 3)/(y + 3) = 5/6

6x + 18 = 5y +15

6x – 5y = 15 – 18

6x – 5y = – 3 ------------ equation 2

4) Put the value of y = (11x + 4)/9 from equation 1 in equation 2, we get,

6x – 5y = – 3

6x – 5(11x + 4)/9 = – 3

[54x – 5(11x + 4)]/9 = – 3

[54x – 5(11x + 4)] = – 3 (9)

[54x – 55x – 20] = – 27 

54x – 55x = – 27 + 20

– x = – 7 

x = 7 -------------------------- equation 3.

5) Put the value of x = 7 from equation 3 in equation 1, we get,

y = (11x + 4)/9

y = (11(7) + 4)/9

y = (77 + 4)/9 

y = 81/9

y = 9  -------------------------- equation 4.

6) The equations are 9y = 11x + 4, and 6x – 5y = – 3, where x and y are the

numerator and denominator of the fraction; the numerator is x = 7 and the denominator is y = 9. So the fraction is 7/9.

(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?

1) Let Jacob's present age be x and his son's present age be y.

2) 5 years hence their ages will be (x + 5) and (y + 5).

3) According to the first relation given in the problem,

(x + 5) = 3(y + 5)

(x + 5) = 3y + 15

x = 3y + 15 – 5

x = 3y + 10  ---------------------- equation 1

4) 5 years ago, their ages were (x – 5) and (y – 5). 

5) According to the relation given in the problem,

(x – 5) = 7(y – 5)

(x – 5) = 7y – 35

x = 7y – 35 + 5

x = 7y – 30  ---------------------- equation 2

6) Put the value of x = 3y + 10 from equation 1 in equation 2, and we get,

x = 7y – 30

3y + 10 = 7y – 30

3y – 7y = – 30 – 10

     – 4y = – 40

       4y = 40 

y = 10 -------------------------- equation 3.

7) Put the value of y = 10 from equation 3 in equation 1, we get,

x = 3y + 10

x = 3(10) + 10

x = 30 + 10 

x = 40  -------------------------- equation 4.

8) The equations are x = 3y + 10 and x = 7y – 30, where x and y are the ages in

years of Jacob and his son. So Jacob's present age is x = 40 years and his son's present age is y = 10 years. 

Conclusion: Mastering Pair of Linear Equations in Two Variables

As we reach the end of this fascinating topic on Pair of Linear Equations in Two Variables, it’s clear how these equations form the foundation for solving real-world problems involving relationships between two variables. By learning how to graph, calculate, and interpret their solutions, you’re honing your mathematical skills and building your problem-solving mindset. Remember, mastering this concept will open doors to more advanced mathematical techniques in the future. So keep practicing, explore different substitution, elimination, and graphical representation methods, and enjoy the learning journey!

Stay curious and keep solving!

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Anil Satpute