Question

if x = (p+q)/(p-q), then x - 1 = ?

Answer 1

Answer:

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

Step-by-step explanation:

x=

$\frac{p + q}{p - q}$

or, x-1 =

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

or, x-1=

$\frac{p + q - p + q}{p - q}$

or, x-1 =

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

Answer 2

Answer:

Required value of (x-1) is $\frac{2q}{p - q}$

Step-by-step explanation:

Let,

$x = \frac{p + q}{p - q}$

Here we want to find value of $x - 1$

We are putting value of $x = \frac{p + q}{p - q}$ in $(x - 1)$

So,

$x - 1 \\ = \frac{p + q}{p - q} - 1 \\ = \frac{p +q - (p - q)}{p - q} \\ = \frac{p + q - p + q}{p - q} \\ = \frac{2q}{p - q}$

This is a problem of Algebra.

Some important Algebra formulas:

${(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \\ {(x + y)}^{3} = {x}^{3} + 3 {x}^{2} y + 3x {y}^{2} + {y}^{3} \\ {(x - y)}^{3} = {x}^{3} - 3 {x}^{2} y + 3x {y}^{2} - {y}^{3} \\ {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y) \\ {x}^{3} - {y}^{3} = {(x - y)}^{3} + 3xy(x - y) \\ {x}^{2} - {y}^{2} = (x + y)(x - y) \\ {x}^{2} + {y}^{2} = {(x - y)}^{2} + 2xy \\ {x}^{2} - {y}^{2} = {(x + y)}^{2} - 2xy \\ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} ) \\ {x}^{3} + {y}^{3} = (x - + y)( {x}^{2} - xy + {y}^{2} )$

Know more about Algebra,

1) https://brainly.in/question/13024124

2) https://brainly.in/question/1169549

#SPJ5