(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) x/2 + 2y/3 = -1 and x – y/3 = 3

**Solution:**

**(i) x + y = 5 and 2x – 3y = 4**

**By the method of elimination.**

x + y = 5 ……………………………….. (i)

2x – 3y = 4 ……………………………..(ii)

When the equation (i) is multiplied by 2, we get;

2x + 2y = 10 ……………………………(iii)

When the equation (ii) is subtracted from (iii) we get;

5y = 6

y = 6/5 ………………………………………(iv)

Substituting the value of y in eq. (i) we get;

x=5−6/5 = 19/5

**∴x = 19/5 , y = 6/5**

**By the method of substitution.**

From the equation (i), we get:

x = 5 – y………………………………….. (v)

When the value is put in equation (ii) we get;

2(5 – y) – 3y = 4

-5y = -6

y = 6/5

When the values are substituted in equation (v), we get;

x =5− 6/5 = 19/5

Therefore, x = 19/5 ,y = 6/5

**(ii) 3x + 4y = 10 and 2x – 2y = 2**

**By the method of elimination.**

3x + 4y = 10……………………….(i)

2x – 2y = 2 ………………………. (ii)

When the equation (i) and (ii) is multiplied by 2, we get;

4x – 4y = 4 ………………………..(iii)

When the Equation (i) and (iii) are added, we get;

7x = 14

x = 2 ……………………………….(iv)

Substituting equation (iv) in (i), we get;

6 + 4y = 10

4y = 4

y = 1

Hence, x = 2 and y = 1

**By the method of Substitution**

From equation (ii) we get,

x = 1 + y……………………………… (v)

Substituting equation (v) in equation (i), we get;

3(1 + y) + 4y = 10

7y = 7

y = 1

When y = 1 is substituted in equation (v), we get;

A = 1 + 1 = 2

Therefore, A = 2 and B = 1

**(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7**

**By the method of elimination:**

3x – 5y – 4 = 0 ………………………………… (i)

9x = 2y + 7

9x – 2y – 7 = 0 …………………………………(ii)

When the equation (i) and (iii) is multiplied, we get;

9x – 15y – 12 = 0 ………………………………(iii)

When the equation (iii) is subtracted from equation (ii), we get;

13y = -5

y = -5/13 ………………………………………….(iv)

When equation (iv) is substituted in equation (i), we get;

3x +25/13 −4=0

3x = 27/13

x =9/13

Therefore, x = 9/13 and y = -5/13

**By the method of Substitution:**

From the equation (i), we get;

x = (5y+4)/3 …………………………………………… (v)

Putting the value (v) in equation (ii), we get;

9(5y+4)/3 −2y −7=0

13y = -5

y = -5/13

Substituting this value in equation (v), we get;

x = (5(-5/13)+4)/3

x = 9/13

Therefore, x = 9/13, y = -5/13

**(iv) x/2 + 2y/3 = -1 and x-y/3 = 3**

**By the method of Elimination.**

3x + 4y = -6 …………………………. (i)

x-y/3 = 3

3x – y = 9 ……………………………. (ii)

When the equation (ii) is subtracted from equation (i), we get;

5y = -15

y = -3 ………………………………….(iii)

When the equation (iii) is substituted in (i), we get;

3x – 12 = -6

3x = 6

x = 2

Hence, x = 2 , y = -3

**By the method of Substitution:**

From the equation (ii) we get,

x = (y+9)/3…………………………………(v)

Putting the value obtained from equation (v) in equation (i), we get;

3(y+9)/3 +4y =−6

5y = -15

y = -3

When y = -3 is substituted in equation (v), we get;

x = (-3+9)/3 = 2

Therefore, x = 2 and y = -3