Class10 NCRT Pair of Linear Equations in Two Variables Exercise – 3.4 pdf || UP Board
Pair of Linear Equations in Two
Variables
Exercise – 3.4
1. Solve the following pair of linear
equations by the elimination method and the substitution method:
(i) x + y = 5 and 2x – 3y = 4
Solution:
x + y = 5…………….(1)
2x – 3y = 4……………….(2)
In equation (1), multiplying by 2 and subtracting equation (2),
Substituting the value of y in equation (1),
x + 6/5 = 5
x = 5 – (6/5)
x = 19/5
Hence, x = 19/5 and y = 6/5.
(ii) 3x + 4y = 10 and 2x – 2y = 2
Solution:
3x + 4y = 10…………….(1)
2x – 2y = 2……………….(2)
Multiplying 2 in equation (2) and adding it to equation (1),
Substituting the value of x in equation (2),
4 – 2y = 2
4 - 2 = 2y
y = 2/2 = 1
Hence, x = 2 and y = 1.
(iii) 3x - 5y – 4 = 0 and 9x = 2y + 7
Solution:
3x - 5y = 4…………….(1)
9x – 2y = 7……………….(2)
In equation (1), multiplying by 3 and subtracting equation (2),
Substituting the value of y in equation (1),
3x – 5×(-5/13) = 4
3x + 25/13 = 4
3x = 4 - 25/13
3x = 27/13
x = 9/13
Hence, x = 9/13 and y = –5/13.
(iv) x/2 + 2y/3 = -1 and x – y/3 = 3
Solution:
3x + 4y = -6…………….(1)
3x – y = 9……………….(2)
On subtracting equation (2) from equation (1),
Substituting the value of y in equation (2),
3x – (-3) = 9
3x = 6
x = 6/3 = 2
Hence, x = 2 and y = -3.
2. From 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 ½ if we only add 1 to the
denominator. What is the fraction?
Solution:
Let numerator = x
and denominator = y
Hence fractional number = x/y
According to the first condition,
x + 1 = y – 1
x – y = -2……………(1)
according to the second condition,
2x = y + 1
2x – y = 1…………………(2)
On multiplying equation (1) by 2 and subtracting equation (2),
Substituting the value of y in equation (2),
2x – 5 = 1
2x = 6
x = 3
Hence the fractional number = x/y = 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?
Solution:
Let Nuri's present age = x years
And present age of Sonu = y years
five years ago,
Nuri's age = x - 5 years
Sonu's age = y - 5 years
According to Question,
x – 5 = 3(y – 5)
x – 5 = 3y - 15
x – 3y = -10……………..(1)
Ten years later,
Nuri's age = x + 10 years
Sonu's age = y + 10 years
According to Question,
x + 10 = 2(y + 10)
x + 10 = 2y + 20
x – 2y = 10………………(2)
Subtracting equation (2) from equation (1),
Substituting the value of y in equation (2),
x – 2×20 = 10
x = 50
Hence present age of Nuri = 50 years and present age of Sonu = 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.
Solution:
Let unit's digit = x
tens digit = y
So number = 10y + x (Multiplyed y by 10 because the number is two digit e.g.– if y = 1, x = 0 then the number 10 + 0 = 10 which is two digit)
According to the question the sum of the digits of the
number is 9. so,
x + y = 9………….(1)
Number formed by reversing the digits of the number = 10x + y
According to Question,
9(number)
= 2(number
formed by reversing)
9(10y + x) = 2(10x + y)
90y + 9x = 20x + 2y
11x – 88y = 0
x – 8y = 0…………………(2)
Subtracting equation (2) from equation (1),
Substituting the value of y in equation (1),
x + 1 = 9
x = 8
Hence the number = 10y + x = 10×1 + 8 = 18.
(iv) Meena went to a bank to withdraw ₹2000. She asked the cashier to give her ₹50 and ₹100 notes
only. Meena
got 25 notes in all. Find how many notes of ₹50 and ₹100 she
received.
Solution:
Let the number of ₹50 notes = x
and number of ₹100 notes = y
The total number of notes is 25.
Therefore,
x + y = 25………….(1)
Total amount of ₹50 notes and ₹100 notes = ₹2000 therefore,
50x + 100y = 2000
x + 2y = 40……………..(2)
Subtracting equation (2) from equation (1),
Substituting the value of y in equation (1),
x + 15 = 25
x = 10
Hence, number of ₹50 notes = 10 and ₹100 notes = 15.
(v) A lending library has a fixed
charge for the first three days and an additional charge for each day
thereafter. Saritha paid ₹27 for
a book kept for seven days, while Susy paid ₹21 for the book she kept for five days. Find the fixed
charge and the charge for each extra day.
Solution:
Let the fixed charge for the first three days = x
and rent for each additional day = y
Sarita paid ₹27 to keep a book for seven days so,
x + 4y = 27……………(1)
Susy paid ₹21 to keep a book for five days so,
x + 2y = 21……………..(2)
Subtracting equation (2) from equation (1),
Substituting the value of y in equation (1),
x + 4×3 = 27
x = 15
Hence, fixed charge for first three days = ₹ 15 and for each additional day = ₹ 3.
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