Question

1/3x+y + 1/3x-y= 12/35 ; 1/2(3x+y)- 1/2(3x-y)=2/35
find the value of x and y
solve by simultaneous equation​

Answer 1

$x=\frac{35}{16}$  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)

Put  $\frac{1}{3x+y}$ = u  and $\frac{1}{3x-y}$ = v in (1) and (2) we get

$u+v=\frac{12}{35}$ ........(3)        

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

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

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

Adding equation of (3) and (5), we get

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

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

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

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

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

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

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

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

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

To learn more

i)Solve the following pair of equation

1/3x + y + 1/3x - y = 3/4 , 1/2(3x + y) - 1/2(x-y) = -1/8

https://brainly.in/question/2344036

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

https://brainly.in/question/15263101