Question

1/3x+y + 1/3x-y =12/35 ; 1/2 (3x+y) - 1/2(3x -y) =2/35.​

Answer 1

Therefore $x=\frac{35}{16}$   and $y=\frac{-35}{16}$

Step-by-step explanation:

Given,$\frac{1}{3x+y} +\frac{1}{3x-y} =\frac{12}{35}$ .........(1)

and  $\frac{1}{2(3x+y)} -\frac{1}{2(3x-y)} =\frac{2}{35}$.............(2)

The equation (1) and (2) are 2 variables equation. Because here two unknown quantity present i.e x and y.

To solve the problem we have to let  $\frac{1}{3x+y}=u$  and $\frac{1}{3x-y}=v$

so above two equations become

$u+v=\frac{12}{35}$ ........(3)        and  $\frac{1}{2} u-\frac{1}{2} v=\frac{2}{35}$........(4)

Next step, we have to equate the coefficient of either u or v.

For this equation (4) multiply by 2 , then the equation (4) becomes

$u-v=\frac{4}{35}$ ........(5)

Adding equation of (3) and (5)

$u+v+u-v=\frac{12}{35} +\frac{4}{35}$              

$\Rightarrow 2u=\frac{16}{35}$

$\Rightarrow u=\frac{16}{35\times 2}$

$\Rightarrow u=\frac{8}{35}$

Putting the value of $u=\frac{8}{35}$ in equation (3)

$\frac{8}{35}+v=\frac{12}{35}$

$\Rightarrow v=\frac{12}{35}-\frac{8}{35}$

$\Rightarrow v=\frac{4}{35}$

Therefore $u=\frac{8}{35}$     and    $v=\frac{4}{35}$

Therefore $\frac{1}{3x+y} = \frac{8}{35}$     and     $\frac{1}{3x-y} = \frac{4}{35}$     [ since $\frac{1}{3x+y} =u$  and  $\frac{1}{3x-y} = v$]

⇒$3x+y=\frac{35}{8}$  ..........(6)   and      $}{3x-y} = \frac{35}{4}$................(7)

Adding equation (6) and (7)

$3x +y+3x-y=\frac{35}{8} +\frac{35}{4}$

$\Rightarrow 6x=\frac{105}{8}$

$\Rightarrow x=\frac{105}{8\times 6}$

$\Rightarrow x=\frac{105}{48} =\frac{35}{16}$

Putting the value of x in equation (6) we get

$(3\times \frac{35}{16} )+y=\frac{35}{8}$

$\Rightarrow \frac{105}{16} +y=\frac{35}{8}$

$\Rightarrow y=\frac{35}{8}-\frac{105}{16}$

$\Rightarrow y=\frac{-35}{16}$

Therefore $x=\frac{35}{16}$   and $y=\frac{-35}{16}$.

Answer 2

Answer:

Step-by-step explanation: