Question

$\sf \large{a + b = a \times b = a \div b}$
Find the value of a and b.

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Answer 1

ANSWER:

Given:

• a + b = a × b = a ÷ b

To Find:

• Value of a and b

Solution:

We are given that,

$\implies\sf a+b=a\times b=a\div b$

So,

$\implies\sf a\!\!\!/:\times b=\dfrac{a\!\!\!/:\}{b}$

$\implies\sf b=\dfrac{1}{b}$

Transposing b to LHS,

$\implies\sf b^2=1$

Taking square root,

$\implies\sf \sqrt{b^2}=\sqrt1$

$\implies\sf b=\pm1$

So, now we will put the value of b, in any 2 of those equations(one of them will be a + b).

We had,

$\implies\sf a+b=a\times b$

Taking b = 1,

$\implies\sf a+1=a\times 1$

$\implies\sf a+1=a$

Cancelling a,

$\implies\sf a\!\!\!/\:+1=a\!\!\!/\:$

$\implies\sf 1=0$

As, the staterment is false, b ≠ 1.

Taking b = -1,

$\implies\sf a-1=a\times (-1)$

$\implies\sf a-1=-a$

Transposing -a to LHS,

$\implies\sf a+a=1$

$\implies\sf 2a=1$

Transposing 2 to RHS,

$\implies\sf a=\dfrac{1}{2}$

Hence, for b = -1,  a = 1/2.

Therefore value of a = 1/2 and value of b = -1.

Answer 2

Step-by-step explanation:

Given A x B= {(a, x), (a, y), (b, x), (b, y)}

A is the set of all first elements

i.e. A = {a, b} (Since first element contains only a and b)

and

B is the set of all second elements.

B = {x,y}

(Since second element contains only x and y)