(i) 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

∴ 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

∴ x = 9/13, y = -5/13

(ii) 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