Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev

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Class 10 : Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev

The document Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.
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Ques 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) Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Sol: (i) x + y = 5   ...(1)
2x - 3y = 4   ...(2)
Multiplying (1) by 3 and adding it to (2),
3 [x + y = 5] ⇒ 3x + 3y = 15
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
⇒ x = (19/5)
Now, putting x = 19/5 in (1), we get
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Thus, x = 19/5 and y = 6/5
(ii) 3x + 4y = 10  ...(1)
2x - 2y = 2  ...(2)
From equation (2), we have:
2 [2x - 2y = 2] ⇒ 4x - 4y = 4 ...(3)
Adding (1) and (3),
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
⇒ x = 14/7 = 2
Putting x = 2 in (1), we get
3 (2) + 4y = 10
⇒ 6 + 4y = 10
⇒ 4y = 10 - 6 = 4
⇒ y = 4/4 = 1
Thus, x = 2 and y = 1
(iii) 3x - 5y - 4 = 0   ...(1)
9x = 2y + 7 or 9x - 2y - 7 = 0   ...(2)
From equation (1), we have:
3 [3x - 5y - 4 = 0]  ⇒ 9x - 15y - 12 = 0 ...(3)
Subtracting (2) from (3),
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
⇒ y = (-5/13)
Now from (1), we get
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Thus, x = 9/13 and Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev

(iv) Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev  ...(1)
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev  ..(2)
Multiplying equation (2) by 2,
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Adding (1) and (3), we have:
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev

Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev ⇒ Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Putting x = 2 in (1), we get
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev ⇒ Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Thus, x =2 and y = - 3

Ques 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 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.
(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.
Sol: (i) Let the numerator = x
And the denominator = y
∴ Fraction = [x/y]

Case I:
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
⇒ x + 1 = y - 1
x - y = - 1 - 1 = - 2
⇒ x - y = - 2   ...(1)
Case II:
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev    ...(2)
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Subtracting (3) from (1), we have:
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
⇒ y = 5
Now, putting y = 5 in (2), we have
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRevThus x = 3 and y = 5
∴The required fraction = 3/5
(ii) Let the present age of Nuri = x years
And the present age of Sonu = y years.
5 years ago:
Age of Nuri = (x - 5) yrs
Age of Sonu = (y - 5) yrs
Condition:
Age of Nuri = 3 [Age of Sonu]
⇒ x - 5 = 3 [y - 5]
⇒ x - 5 = 3y - 15
⇒ x - 3y - 5 + 15 = 0
⇒ x - 3y + 10 = 0 ...(1)
10 years later:
Age of Nuri = (x + 10) years
Age of Sonu = (y + 10) yrs
Condition:
Age of Nuri = 2 [Age of Sonu]
⇒ x + 10 = 2 (y + 10)
⇒ x + 10 = 2y + 20
⇒ x - 2y + 10 - 20 = 0
⇒ x - 2y - 10 = 0 ...(2)
Subtracting (1) from (2),
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
⇒ y = 20
Putting y = 20 in (1), we get
x - 3 (20) + 10 = 0
⇒ x - 60 + 10 = 0
⇒ x - 50 = 0 ⇒ x = 50
Thus, x = 50 and y = 20
∴ Age of Nuri = 50 years
Age of Sonu = 20 years
(iii) Let the digit at unit place = x
And the digit at tens place = y
∴ The number = 10y + x
The number obtained by reversing the digits = 10x + y
∵ 9 [The number] = 2 [Number obtained by reversing the digits]
∴ 9 [10y + x] = 2 [10x + y]
⇒ 90y + 9x = 20x + 2y
⇒ 90y + 9x - 20x - 2y = 0
⇒ - 11x + 88y = 0
⇒ x - 8y = 0   ...(1)
Also Sum of the digits = 9
∴ x + y = 9   ...(2)
Subtracting (1) from (2), we have :
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
⇒ y = 9/9  = 1
Putting y = 1 in (2), we get

x + 1 = 9
⇒ x = 9 - 1 = 8
Thus, x = 8 and y = 1
∴ The required number = (10 × 1) + 8
= 10 + 8
= 18
(iv) Let the number of 50 rupee notes = x
And the number of 100 rupee notes = y
According to the conditions
Total number of notes = 25
∴ x + y = 25    ...(1)
∵ The value of all the notes = ₹ 2000
∴ 50x + 100y = 2000
⇒ x + 2y = 40   ...(2)
Subtracting equation (1) from (2),
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Putting y = 15 in (1),
x + 15 = 25
⇒ x = 25 - 15 = 10
Thus, x = 10 and y = 15
⇒ Number of 50 rupee notes = 10
Number of 100 rupee notes = 15
(v) Let the fixed charge (for first three days) = ₹ x
And the additional charge for each extra day = ₹ y
First Condition,
Charge for 7 days = ₹ 27
∴  x + 4y = 27    ...(1)
[∵ Extra days = 7 - 3 = 4]
Charge for 5 days = ₹ 21
⇒ x + 2y = 21  ...(2)
[∵ Extra days = 5 - 3 = 2]
Subtracting (2) from (1), we get
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
⇒ y = 6/2 = 3
Putting y = 3 in (2), we have:

x + 2 (3) = 21
x + 6 = 21
x = 21 - 6 = 15
Since x = 15  and  y = 3
∴Fixed charge = ₹ 15
 Additional per day charge = ₹ 3

l Cross-Multiplication Method
In cross-multiplication method, we use the following steps:
I. Write down the given pair of equations in the form of:
a1 x + b1 y + c1 = 0
a2 x + b2 y + c2 = 0
II. Write the coefficients with the column y, constant terms, coefficients of x and repeating with that of y as shown below:
b1  c1  a1  b1
b2  c2  a2  b2
III. Now, write the coefficients of x, y and 1 in the following manner.
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev 

Note:
(i) Below x, we write coefficients of y and constant terms.
(ii) Below y, we write constant terms and coefficients of x.
(iii) Below 1, we write coefficients of x and y.

IV. The expressions in the denominator of x, y and 1 are obtained by multiplying the number written at the end of arrows such that:
(i) Numbers with downward arrows are multiplied first.
(ii) Then the products of numbers with upwards arrows are subtracted from the above products.
i.e.,
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev   Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev


Note:
I. The cross-multiplication method can be used only for the consistent systems of linear equations. We cannot use it for the inconsistent and dependent systems.
II. In case of Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRevthe system has no solution because
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRevand division by zero is not defined.
III. In case Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev, the values of x and y are reduced to  form i.e.,
Ex 3.4 NCERT Solutions- Pair of Linear Equations in Two Variables Class 10 Notes | EduRev
Such systems are dependent and pair of given equations have unlimited solutions.

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