# Pair of Linear Equations in two Variables Exercise 3.4

Pair of Linear Equations in two Variables Exercise 3.4

Page No: 56

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

(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
= (10 - 3x)/4

= 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 – 2= 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 + 5)/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
= -15/5
= - 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
= 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 + 4= - 6
5y = -15
y = - 3

Hence our answer is  x = 2 and y = −3.

Page No: 57

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.

(i) Let the fraction be x/y
According to the question,x + 1/y - 1 = 1
⇒ = -2 ... (i)x/y+1 = 1/2
⇒ 2x - = 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 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
= 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 and respectively.
Then, number = 10y + x
Number after reversing the digits = 10x + y
According to the question,
x + y = 9 ... (i)
9(10x) = 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,
= 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 and Rs respectively.
According to the question,
+ 4y = 27 ... (i)
+ 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.