Pair of Linear Equations in Two Variables NCERT Solution, study material, CBSE Notes

NCERT Solution: Pair of Linear Equations in Two Variables

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

(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 traveling a distance of 25 km?

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
People have to pay for traveling a distance of 25 km
= x + 25y
= 5 + 25 × 10
= 5 + 250
=255

A person have to pay Rs 255 for 25 Km.

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

(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.

Let Numerator = x
Denominator = y
Fraction will = x/y
A fraction becomes 9/11, if 2 is added to both the numerator and the denominator
(x + 2)/y+2 = 9/11

By Cross multiplication, we get
11x + 22 = 9y + 18
Subtracting 22 both side, we get
11x = 9y – 4
Dividing by 11, we get
x = 9y – 4/11 … (i)
Given that 3 is added to both the numerator and the denominator it becomes 5/6.
If, 3 is added to both the numerator and the denominator it becomes 5/6
(x+3)/y +3 = 5/6 … (ii)
By Cross multiplication, we get
6x + 18 = 5y + 15
Subtracting the value of x, we get
6(9y – 4 )/11 + 18 = 5y + 15
Subtract 18 both side we get
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.

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

Let present age of Jacob = x year
And present Age of his son is = y year
Five years hence,
Age of Jacob will = x + 5 year
Age of his son will = y + 5year
Given that the age of Jacob will be three times that of his son
x + 5 = 3(y + 5)
Adding 5 both side, we get
x = 3y + 15 - 5
x = 3y + 10 … (i)
Five years ago,
Age of Jacob will = x - 5 year
Age of his son will = 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 first we get
x = 3 × 10 + 10
x = 40 years
Hence, Present age of Jacob = 40 years and present age of his son = 10 years.

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

(i) 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 methodx + y = 5 ... (i)
Subtracting y 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

7x + 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 both side 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
Solve it we get
3(4 + 5y ) – 2y = 7
12 + 15y – 2y = 7
13y = - 5
y = -5/13

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

Hence our answer is x = 2 and y = −3.

By substitution method
x – y/3 = 3 ... (ii)
Add y/3 both side, we get
x = 3 + y/3 ... (iv)
Putting this 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, we get
9 + y + 4y = - 6
5y = -15
y = - 3

Hence our answer is x = 2 and y = −3

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?

(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

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

(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?

(ii) Let present age of Nuri = x
and present age of Sonu = y
According to the given information,question,(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

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

(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.

(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.

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

(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.

(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.