Question

Solve the following systems of equations

(i)`\frac{15}{u} + \frac{2}{v} = 17`

`\frac{1}{u} + \frac{1}{v} = \frac{36}{5}`

(ii) ` \frac{11}{v} – \frac{7}{u} = 1`

`\frac{9}{v} + \frac{4}{u} = 6`

Answer 1

(i) The given system of equation is

`\frac{15}{u} + \frac{2}{v} = 17 ….(1)`

`\frac{1}{u} + \frac{1}{v} = \frac{36}{5} ….(2)`

Considering 1/u = x, 1/v = y, the above system of linear equations can be written as

15x + 2y = 17 ….(3)

`x + y = \frac{36}{5} ….(4)`

Multiplying (4) by 15 and (iii) by 1, we get

15x + 2y = 17 ….(5)

`15x + 15y = \frac{36}{5} × 15 = 108 ….(6)`

Subtracting (6) form (5), we get

–13y = – 91 ⇒ y = 7

Substituting y = 7 in (4), we get

`x + 7 = \frac{36}{5} ⇒ x = \frac{36}{5} – 7 = \frac{1}{5}`

But,

`y = \frac{1}{v} = 7 ⇒ v = \frac{1}{7}`

and, `x = \frac{1}{u} = \frac{1}{5} ⇒ u = 5`

Hence, the required solution of the given system is u = 5, v = 1/7

(ii) The given system of equation is

`\frac{11}{v} – \frac{7}{u} = 1; \frac{9}{v} + \frac{4}{u} = 6`

Taking 1/n = x and 1/u = y, the above system of equations can be written as

11x – 7y = 1 ….(1)

9x – 4y = 6 ….(2)

Multiplying (1) by 4 and (2) by 7, we get,

44x – 28y = 4 ….(3)

63x – 28y = 42 ….(4)

Subtracting (4) from (3) we get,

– 19x = –38 ⇒ x = 2

Substituting the above value of x in (2), we get;

9 × 2 – 4y = 6 ⇒ –4y = – 12

⇒ y = 3

But, `x = \frac{1}{v}= 2 ⇒ v = \frac{1}{2}`

and,

`y = \frac{1}{u} = 3`

`⇒ u = \frac{1}{3}`

Hence, the required solution of the given system of the equation is `v = \frac{1}{2}, u = \frac{1}{3}`