Solution:

Let the number of 5 rupee notes be x and the number of 10 rupee notes be y.

From the first condition,

5x + 10y = 350 ... (i)

From the second condition,

x = 2y − 10 ... (ii)

Substituting the value of x in (i),

5x + 10y = 350 ... (i)

∴ 5(2y − 10) + 10y = 350

∴ 10y − 50 + 10y = 350

∴ 20y − 50 = 350

∴ 20y = 350 + 50

∴ 20y = 400

\(\displaystyle \therefore y = \frac {400}{20}\)

∴ y = 20 ... (iii)

Substituting the value of y in (ii), we get,

  x = 2y − 10 ... (ii)

∴ x = 2(20) − 10

∴ x = 40 − 10

∴ x = 30 ... (iv)

Hence, there are 30 notes of ₹ 5 and 20 notes of ₹ 10.

Solution:

Let, the numerator of that fraction be x and the denominator be y.

From the first condition,

y = 2x − 1 ... (i)

From the second condition,

\(\displaystyle \frac {x + 1}{y + 1} = \frac {3}{5}\)

By cross multiplication,

   5(x + 1) = 3(y + 1)

∴ 5x + 5 = 3y + 3

∴ 5x − 3y = 5 − 3

∴ 5x − 3y = − 2 ... (ii)

Substituting the value of y from (i) in (ii), we get,

   5x − 3y = −2 ... (ii)

∴ 5x − 3(2x − 1) = − 2

∴ 5x − 6x + 3 = − 2

∴ − x = − 2 − 3

∴ − x = − 5

i.e. x = 5 ... (iii)

Substituting the value of x in (i), we get,

  y = 2x − 1 ... (i)

∴ y = 2(5) − 1

∴ y = 10 − 1

∴ y = 9 ... (iv)

Hence, the required fraction is \(\displaystyle \frac {5}{9}\).

Solution:

Let, the present age of Priyanka be x years and the present age of Deepika be y years.

From the first condition,

x + y = 34 ... (i)

From the second condition,

x − y = 6 ... (ii)

Adding (i) and (ii),

		x	+	y	=		34	... (i)
+		x	−	y	=		6	... (ii)
		2x			=		40

\(\displaystyle \therefore x = \frac {40}{2}\)

∴ x = 20 ... (iii)

Substituting the value of x in (i),

 x + y = 34

∴ 20 + y = 34

∴ y = 34 − 20

∴ y = 14 ... (iv)

∴ Priyanka’s present age is 20 years and Deepika’s present age is 14 years.

Solution:

Let, the number of lions be x and the number of peacocks be y.

From the first condition,

x + y = 50 ... (i)

From the second condition,

4x + 2y = 140 ... 

Dividing both sides by 2, we get:

2x + y = 70 ... (ii)

Subtracting (i) from (ii),

		2x	+	y	=		70	... (ii)
−	⊕	x	⊕	y	=	⊕	50	... (i)
	−		−			−
		x			=		20	... (iii)

Substituting x = 20 in equation (i),

 x + y = 50 ... (i)

∴ 20 + y = 50

∴ y = 50 − 20

∴ y = 30 ... (iv)

∴ There are 20 lions and 30 peacocks in that zoo.

Solution:

Let, Sanjay’s monthly salary be ₹ x and the yearly increment be ₹ y.

From the first condition,

 x + 4y = 4500 ... (i)

From the second condition,

 x + 10y = 5400 ... (ii)

Subtracting equation (i) from equation (ii),

		x	+	10y	=		5400	... (ii)
−	⊕	x	⊕	4y	=	⊕	4500	... (i)
	−		−			−
				6y	=		900

\(\displaystyle \therefore y = \frac {900}{6}\)

∴ y = ₹ 150 ... (iii)

Substituting the value of y in (i), we get,

 x + 4y = 4500 ... (i)

∴ x + 4 × 150 = 4500

∴ x + 600 = 4500

∴ x = 4500 − 600

∴ x = ₹ 3900 ... (iv)

Hence, Sanjay’s monthly salary is ₹ 3900 and his yearly increment is ₹ 150.

Solution:

Let, the price of one chair be ₹ x and the price of one table be ₹ y.

From the first condition,

 3x + 2y = 4500 ... (i)

From the second condition,

 5x + 3y = 7000 ... (ii)

Multiplying equation (i) by 3 and equation (ii) by 2,

 9x + 6y = 13500 ... (iii)

 10x + 6y = 14000 ... (iv)

Subtracting equation (iv) from equation (iii),

		10x	+	6y	=		14000	... (iv)
−	⊕	9x	⊕	6y	=	⊕	13500	... (iii)
	−		−			−
		x			=		500	... (v)

Subtracting equation (i) from equation (ii),

		x	+	10y	=		5400	... (ii)
−	⊕	x	⊕	4y	=	⊕	4500	... (i)
	−		−			−
				6y	=		900

\(\displaystyle \therefore y = \frac {900}{6}\)

∴ y = ₹ 150 ... (iii)

Substituting the value of y in (i), we get,

 x + 4y = 4500 ... (i)

∴ x + 4 × 150 = 4500

∴ x + 600 = 4500

∴ x = 4500 − 600

∴ x = ₹ 3900 ... (iv)

Hence, Sanjay’s monthly salary is ₹ 3900 and his yearly increment is ₹ 150.

Solution:

 2y − x = 0
∴ 2y = x
i.e. x = 2y ... (i)

and 10x + 15y = 105 ... (ii)

Substituting the value of x in (ii),

  10x + 15y = 105 ... (ii)

∴ 10(2y) + 15y = 105

∴ 20y + 15y = 105

∴ 35y = 105

\(\displaystyle \therefore y = \frac {105}{35}\)

∴ y = 3 ... (iii)

Substituting the value of y in (i),

 x = 2y ... (i)

∴ x = 2 × 3

∴ x = 6 ... (iv)

∴ (6, 3) is the solution of the given equations.

(This method is known as method of substitution.)

Solution:

 2x + 3y + 4 = 0
∴ 2x + 3y = − 4 ... (i)

Also, x − 5y = 11
∴ x = 11 + 5y ... (ii)

Substituting the value of x in (i),

2x + 3y = − 4 ... (i)

∴ 2(11 + 5y) = − 4

∴ 22 + 10y + 3y = − 4

∴ 13y = − 4 − 22

∴ 13y = − 26

\(\displaystyle \therefore y = \frac {-26}{13}\)

∴ y = − 2 ... (iii)

Substituting the value of y in (ii),

  x = 11 + 5y ... (ii)

∴ x = 11 + 5 × − 2

∴ x = 11 − 10

∴ x = 1 ... (iv)

∴ (1, − 2) is the solution of the given equations.

Solution:

Let, 2x − 7y = 7 ... (i)
and 3x + y = 22 ... (ii)

Multiplying (ii) by 7,

21x + 7y = 154 ... (iii)

Adding (i) and (iii),

		2x	−	7y	=		7	... (i)
+		21x	+	7y	=		154	... (iii)
		23x			=		161

\(\displaystyle \therefore x = \frac {161}{23}\)

∴ x = 7 ... (iv)

Substituting the value of x in (ii),

  3x + y = 22 ... (ii)

∴ 3(7) + y = 22

∴ 21 + y = 22

∴ y = 22 − 21

∴ y = 1 ... (v)

∴ (7, 1) is the solution of the given equations.