Question

please help me

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Math's
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please help me please
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Answer 1

★Given:-

• $\sf \frac{6}{x+2y} +\frac{5}{x-2y} =-3$
• $\sf \frac{3}{x+2y} +\frac{7}{x-2y} =-6$
• And,x+2y≠0;x-2y≠0.

★To find:-

• Value of x & y.

★Solution:-

Let 1/x+2y = p & 1/x-2y = q.

Then,

The given equations will be:

→ 6p+5q = -3....(1)

→ 3p+7q = -6.....(2)

Multiplying eqn(2) with 2,

→ 2×(3p+7q=-6)

→ 6p + 14q = -12 .....(3)

Substracting eqn(1) from (3),

6p+14q = -12

6p+5q = -3

_________

     9q = -9

        q = -1

Substituting value of 'q' in eqn(1):

→ 6p+5q = -3

→ 6p+5(-1) = -3

→ 6p - 5 = -3

→ 6p = 5 - 3 = 2

→ p = 2/6

→ p = 1/3

We have,

$\mapsto \sf \frac{1}{x+2y} =p \ ;\ \frac{1}{x-2y} =q$

Now,

$:\implies \sf \frac{1}{x+2y} = p\\\\:\implies \sf \frac{1}{x+2y} =\frac{1}{3} \\\\:\implies \sf x+2y = 3....(4)$

And,

$:\implies \sf \frac{1}{x-2y} = q \\\\:\implies \sf \frac{1}{x-2y} =-1\\\\:\implies 1=-x+2y....(5)$

Substracting eqn(4) from (5),

x+2y = 3

-x+2y = 1

________

     4y = 4

        y = 1

$\mapsto \boxed{\underline{\sf y = 1}}$

Now,

$:\implies \sf x+2y=3\\\\:\implies \sf x+2(1)=3\\\\:\implies \sf x =3-2 =1\\\\:\implies \boxed{\underline{\sf x=1}}$

Therefore,

Values of x is 1 and y is 1.

___________

Answer 2

★Given:-

$\sf \frac{6}{x+2y} +\frac{5}{x-2y} =-3$

$\sf \frac{3}{x+2y} +\frac{7}{x-2y} =-6$

And,x+2y≠0;x-2y≠0.

★To find:-

Value of x & y.

★Solution:-

Let 1/x+2y = p & 1/x-2y = q.

Then,

The given equations will be:

→ 6p+5q = -3....(1)

→ 3p+7q = -6.....(2)

Multiplying eqn(2) with 2,

→ 2×(3p+7q=-6)

→ 6p + 14q = -12 .....(3)

Substracting eqn(1) from (3),

6p+14q = -12

6p+5q = -3

_________

     9q = -9

        q = -1

Substituting value of 'q' in eqn(1):

→ 6p+5q = -3

→ 6p+5(-1) = -3

→ 6p - 5 = -3

→ 6p = 5 - 3 = 2

→ p = 2/6

→ p = 1/3

We have,

$\mapsto \sf \frac{1}{x+2y} =p \ ;\ \frac{1}{x-2y} =q$

Now,

$:\implies \sf \frac{1}{x+2y} = p\\\\:\implies \sf \frac{1}{x+2y} =\frac{1}{3} \\\\:\implies \sf x+2y = 3....(4)$

And,

$:\implies \sf \frac{1}{x-2y} = q \\\\:\implies \sf \frac{1}{x-2y} =-1\\\\:\implies 1=-x+2y....(5)$

Substracting eqn(4) from (5),

x+2y = 3

-x+2y = 1

________

     4y = 4

        y = 1

$\mapsto \boxed{\underline{\sf y = 1}}$

Now,

$:\implies \sf x+2y=3\\\\:\implies \sf x+2(1)=3\\\\:\implies \sf x =3-2 =1\\\\:\implies \boxed{\underline{\sf x=1}}$

Therefore,

Values of x is 1 and y is 1.

___________