Solution:

1. x + y = 10
2. 2x − y = 5
3. 3x + 4y = 12
4. x − y = 3
5. 5x + 2y = 15

Solution:

1. x = 1, y = 6 (since 1 + 6 = 7)
2. x = 2, y = 5 (since 2 + 5 = 7)
3. x = 3, y = 4 (since 3 + 4 = 7)
4. x = 4, y = 3 (since 4 + 3 = 7)
5. x = 5, y = 2 (since 5 + 2 = 7)

Solution:

Let, x + y = 4 ... (i)
and 2x − 5y = 1 ... (ii)

Multiplying (i) by 5,

5x + 5y = 20 ... (iii)

Adding (ii) and (iii),

		2x	−	5y	=		1	... (ii)
+		5x	+	5y	=		20	... (iii)
		7x			=		21

\(\displaystyle \therefore x = \frac {21}{7}\)

∴ x = 3 ... (iv)

Substituting the value of x in (i),

 x + y = 4

∴ 3 + y = 4

∴ y = 4 − 3

∴ y = 1 ... (v)

∴ (3, 1) is the solution of the given equations.

(This method is known as method of elimination.)

Solution:

Let, 2x + y = 5 ... (i)
and 3x − y = 5 ... (ii)

Adding (i) and (ii),

		2x	+	y	=		5	... (i)
+		3x	−	y	=		5	... (ii)
		5x			=		10

\(\displaystyle \therefore x = \frac {10}{5}\)

∴ x = 2 ... (iv)

Substituting the value of x in (i),

 2x + y = 5

∴ 2 × 2 + y = 5

∴ 4 + y = 5

∴ y = 5 − 4

∴ y = 1 ... (v)

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

Solution:

Let, 3x − 5y = 16 ... (i)
and x − 3y = 8 ... (ii)

Multiplying (ii) by 3,

 3x − 9y = 24 ... (iii)

Subtracting (iii) from (i),

		3x	−	5y	=		16	... (i)
−	⊕	3x	⊖	9y	=	⊕	24	... (iii)
	−		+			−
				4y	=	−	8

\(\displaystyle \therefore y = \frac {- 8}{4}\)

∴ y = − 2 ... (iv)

Substituting the value of y in (ii),

 x − 3y = 8 ... (ii)

∴ x − 3 × − 2 = 8

∴ x + 6 = 8

∴ x = 8 − 6

∴ x = 2 ... (v)

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

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.