Question

if p = x/ x+y and q= y/x+y find 1/p-q - 2q/ p^2 - q^2​

Answer 1

Step-by-step explanation:

see the answer in attachment

Answer 2

The value of $\frac{1}{p-q}-\frac{2 q}{p^{2}-q^{2}}$ is 1.

Given:

The equations are $p=\frac{x}{x+y} \quad$ and $q &=\frac{y}{x+y}$

To find:

The value of $\frac{1}{p-q}-\frac{2 q}{p^{2}-q^{2}}$

Formula used:

The algebraic identity is $a^2-b^2=(a+b)(a-b)$

Step-by-step explanation:

Since, we have $p=\frac{x}{x+y} \quad$ and

$q &=\frac{y}{x+y}$

Then,

$\frac{1}{p-q}-\frac{2 q}{p^{2}-q^{2}} &=\frac{p+q-2 q}{p^{2}-q^{2}}$

Solve the numerator.

$=\frac{p-q}{p^{2}-q^{2}}$

We know the algebraic identity as  $a^2-b^2=(a+b)(a-b)$

Using the above formula, we get as follows:

$&=\frac{(p-q)}{(p-q)(p+q)}$

$=\frac{1}{p+q}$

Substitute for p and q.

$&=\frac{1}{\frac{x}{x+y}+\frac{y}{x+y}}$

Taking LCM in denominator.

$=\frac{1}{\frac{x+y}{x+y}}$

$&=\frac{1}{1}$

= 1

Hence, the value of $\frac{1}{p-q}-\frac{2 q}{p^{2}-q^{2}}$ is 1.

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