Question

a+b=ab=a/b then a-b=?​

Answer 1

Answer:

Step-by-step explanation:

a+b=ab

(which means a and b are not equal to 1 because everything multiplied by one equals to itself while a number added by 1 doesn't)

ab=a/b

$ab^{2} =a$

$ab^{2}=a \\ab^{2}-a=0\\a (b^{2}-1 )=0$

a=0 or b= ± 1

since b≠1 ; b= -1

a=0 --> a+b=ab; therefore b=0 for this function

a+b=a/b

1. For (a,b) = (0,0)

0+0≠0/0 because 0/0 is undefined, therefore it is not true

2. For b=-1

$a+b=\frac{a}{b}\\a-1=\frac{a}{-1} \\a-1=-a\\2a=1\\a=\frac{1}{2}$

therefore (a,b) = (1/2,-1)

apply to a+b=ab

1/2 - 1 = 1/2 x -1 = -1/2 (the equation if correct)

a-b= 1/2-1 = -1/2

Answer 2

Answer:

The value of a-b is

$\frac{3}{2}$

Step-by-step explanation:

The given relation is

$a + b = ab = \frac{a}{b}$

Let us fix some conditions by looking at the relations.

Assume a=b=1 .Then,

$a + b = 1 + 1 = 2 \\ ab = 1.1 = 1$

Both are not equal.Therefore,

$a≠b≠1$

As there is term a/b therefore,

$b≠0$

$ab = \frac{a}{b} \\ = > ab {}^{2} = a = > a( {b}^{2} - 1) = 0 \\ = > a = 0 \: and \: b = - 1$

Substitute a=0 we get

$a + b = ab \\ = > 0 + b = 0 = > b = 0 \\ \: but \: we \: know \: this \: not \: possible$

Hence this case is ruled out.Therefore we go with b=-1

$a - 1 = - a = > a = \frac{1}{2}$

Therefore,

$a - b = \frac{1}{2} - ( - 1) = \frac{3}{2}$

#SPJ3