Question

Solve `\frac{2}{x+2y}+\frac{6}{2x-y}=4\text{ ;}\frac{5}{2( x+2y)}+\frac{1}{3( 2x-y)}=1` where, x + 2y ≠ 0 and 2x – y ≠ 0

Answer 1

Taking `\frac { 1 }{ x+2y } = u and \frac { 1 }{ 2x-y } = v` , the above system of equations becomes

2u + 6v = 4 ….(1)

`\frac { 5u }{ 2 } + \frac { v }{ 3 } = 1 ….(2)`

Multiplying equation (2) by 18, we have;

45u + 6v = 18 ….(3)

Now, subtracting equation (3) from equation (1), we get ;

`–43u = – 14 ⇒ u = \frac { 14 }{ 43 }`

Putting u = 14/43 in equation (1), we get

`2 × \frac { 14 }{ 43 } + 6v = 4`

`⇒ 6v = 4 – \frac { 28 }{ 43 } = \frac { 172-28 }{ 43 } ⇒ v = \frac {144 }{ 43 }`

Now,

`u = \frac { 14 }{ 43 } = \frac { 1 }{ x+2y }`

⇒ 14x + 28y = 43 ….(4)

And,

`v = \frac { 144 }{ 43 } = \frac { 1 }{ 2x-y }`

⇒ 288x – 144y = 43 ….(5)

Multiplying equation (4) by 288 and (5) by 14, the system of equations becomes

288 × 14x + 28y × 288 = 43 × 288

288x × 14 – 144y × 14 = 43 × 4

⇒ 4022x + 8064y = 12384 ….(6)

4022x – 2016y = 602 ….(7)

Subtracting equation (7) from (6), we get

10080y = 11782 ⇒ y = 1.6(approx)

Now, putting 1.6 in (4), we get,

14x + 28 × 1.6 = 63

⇒ 14x + 44.8 = 63 ⇒ 14x = 18.2

⇒ x = 1.3 (approx)

Thus, solution of the given system of equation is x = 1.3 (approx), y = 1.6 (approx).