Pair of linear equations in two variables Class 10th
Que. 1. Solve the following pair of equations by elimination method and substitution method.
SOLUTION 1. x + y = 5 – (1)
2x – 3y = 4 – (2)
Multiplying equation 1 by 2,
2 (x + y) = 5
2x + 2y = 10 – (3)
Equation (2) – (3)
2x – 3y = 4
2x + 2y = 10
-5y = 6
2. 3x + 4y = 10 – (1)
2x – 2y = 2 – (2)
Multiplying equation 2 by 2,
2 (2x – 2y) = 2
4x – 4y = 2 – (3)
Adding equations (1) and (3)
3x + 4y = 10
4x – 2y = 4
7x = 14
X = 2
Substituting x into equation 2,
2x – 2y = 2
2 ( 2 ) – 2y = 2
4 – 2y = 2
-2y = -2
Y = 1
3. 3x – 5y – 4 = 0
9x = 2y + 7
3x – 5y = 4 – (1)
9x – 2y = 7 – (2)
Multiplying equation (1) by (3),
3 (3x – 5y) = 4
9x – 15y = 12 – (3)
Equation (2) – (3)
9x – 2y = 7
9x – (15y) = -12
13y = -5
Substituting Equations (1) – (2),
3x + 4y = -6
-3x – (+y) = -9
5y = -15
Y = -3
Substituting the value of y into Equation 1,
3x + 4y = -6
3x + 4 x (– 3) = – 6
3x – 12 = -6
3x = -6 + 12
3x = 6
x = 2
Que. 2. Form the pair of linear equations in the following problems and find their solutions (if they exist) by elimination method.
Let the numerator of the fraction = x
Let the denominator of the fraction = y
As per the question,
Width multiplication,
2x = 1 (y – 1)
2x = y + 1
2x – y = 1 – (2)
Substituting equations (1) – (2),
X – y = -2
-2x – (+y) = -1
-x = -3
X = 3
Substituting the value of x into equation 1,
X – y = -2
3 – y = -2
-y = – 2 – 3
-y = -5
Y = 5
Required fraction
(ii) Five years ago, Noori was three times Sonu’s age. Ten years from now, Noori will be twice Sonu’s age. What are the ages of Noori and Sonu?
Let Noorie’s age = x years
Let Sonu’s age = y years
According to the question,
Five years ago,
X – 5 y – 5
X – 5 = 3 (y – 5)
X – 5 = 3y – 15
X – 3y = -15 + 5
X – 3y = -10 – (1)
Ten years later,
X + 10 y = 10
X + 10 = 2 (y + 10)
X + 10 = 2y + 20
X – 2y = 20 – 10
X – 2y = 10 – (2)
Substituting equations 1 and 2,
X – 3y = -10
-X – (+2y) = 10
-y = -20
Y = 20
Substituting the value of y into equation 1,
X – 3y = -10
-X – 3(20) = -10
X – 60 = -10
X = -10 + 60
X = 50
(iii) The sum of the digits of a two-digit number is 9. Nine times this number is twice the number formed by reversing the digits. Find that number.
Let the unit digit of the number = y
Let the tens digit of the number = x
As per the question
X + y = 9 – (1)
9 (10x + y) = 2 (10y + x)
90x – 2x – 20y + 9y = 0
88x – 11y = 0
11 (8x – y) = 0
8x – y = 0
8x – y = 0 – (2)
Substituting equations 1 and 2,
X + y = 9
8x – y = 0
9x = 9
X = 1
Substituting the value of x into equation 1,
X + y = 9
1 + y = 9
Y = 9 – 1
Y = 8
number = xy
= 18
(iv) Meena went to a bank to withdraw ₹2000. She asked the cashier for ₹50 and ₹100 notes. Meena received a total of 25 notes. Find how many ₹50 and ₹100 notes she received.
Let the number of 50 rupee notes = x
Let the number of 100 rupee notes = y
As per the question,
X + y = 25 – (1)
50x + 100y = 2000
50 (x + 2y) = 2000
X + 2y = 40 – (2)
Substituting equations 1 and 2,
X + y = 25
-x + 2y = 40
-y = -15
Y = 15
Substituting the value of y in equation 1,
X + y = 25
X + 15 = 25
X = 25 – 15
X = 10
Therefore, the number of 50 rupee notes is 10 and the number of 100 rupee notes is 10. The number is 15.
(v) A library that rents books has a fixed rent for the first three days and a different rent for each additional day. Sarita paid ₹27 for a book for seven days, while Susie paid ₹21 for a book for five days. Find the fixed rent and the rent for each additional day.
Let the fixed library rent = Rs. x.
Let the additional library rent per day = Rs. y.
For Sarita,
X + 4y = 27 – (1)
For Susie,
X + 2y = 21 – (2)
Equations (1) – (2),
X + 4y = 27
X + (-2y) = -21
2y = 6
Y = 3
Substituting the value of y into Equation 1,
X + 4y = 27
X + 4 (3) = 27
X + 12 = 27
X = 27 – 12
X = 15 Rs.
