Question

Solve the following system of equations `\frac { 1 }{ 2x } – \frac { 1 }{ y } = – 1; \frac { 1 }{ x } + \frac { 1}{ 2y } = 8`

Answer 1

We have,

`\frac { 1 }{ 2x } – \frac { 1 }{ y } = – 1 ….(1)`

`\frac { 1 }{ x } + \frac { 1 }{ 2y } = 8 ….(2)`

Let us consider 1/x = u and 1/y = v.

Putting 1/x = u and 1/y = v in the above equations, we get;

`\frac { u }{ 2 } – v = – 1 ….(3)`

`u + \frac { v }{ 2 } = 8 ….(4)`

Let us eliminate v from the system of equations. So, multiplying equation (3) with 1/2 and (4) with 1, we get

`\frac { u }{ 4 } – \frac { v }{ 2 } = -\frac { 1 }{ 2 } ….(5)`

`u + \frac { v }{ 2 } = 8 ….(6)`

Adding equation (5) and (6), we get ;

`\frac { u }{ 4 } + u = -\frac { 1 }{ 2 } + 8`

⇒ `\frac { 5u }{ 4 } = \frac { 15 }{ 2 }`

⇒ `u = \frac { 15 }{ 2 } × \frac { 4 }{ 5 }`

⇒ u = 6

We know,

`\frac { 1 }{ x } = u ⇒ \frac { 1 }{ x } = 6`

`⇒ x = \frac { 1 }{ 6 }`

Putting 1/x = 6 in equation (2), we get ;

`6 + \frac { 1 }{ 2y } = 8 ⇒ \frac { 1 }{ 2y } = 2`

`⇒ \frac { 1 }{ y } = 4 ⇒ y = \frac { 1 }{ 4 }`

Hence, the solution of the system is,

`x = \frac { 1 }{ 6 } , y = \frac { 1 }{ 4 }`