Question

If n(A-B)= 33+x, n(A intersection B) = 3x, n(B-A) = 15+4x and
n(A) = n(B). Then find x and n(AUB).​

Answer 1

Answer:

The value of x is 6

The value of n(AUB) is 96

Step-by-step explanation:

$\text{Given:}$

$n(A-B)= 33+x$

$n(B-A)=15+4x$

$n(A\cap\,B)=15+4x$

$\text{Now }\:n(A)=n(B)$

$\implies\,n(A-B)=n(B-A)$

$\implies\,33+x=15+4x$

$\implies\,33-15=4x-x$

$\implies\,3x=18$

$\implies\,\boxed{\bf\,x=6}$

$n(A-B)=33+6$

$\implies\,n(A-B)=39$

$n(B-A)=15+4(6)$

$\implies\,n(B-A)=39$

$n(A\cap\,B)=3(6)=18$

We know that

$\boxed{\bf\,n(A\cup\,B)=n(A-B)+n(A\cap\,B)+n(B-A)}$

$n(A\cup\,B)=39+39+18$

$\implies\boxed{\bf\,n(A\cup\,B)=96}$

Answer 2

Answer:  The required value of x is 6 and the value of n(AUB) is 96.

Step-by-step explanation:  Given that A and B are two sets with n(A) = n(B), where

$n(A-B)=33+x,~~~n(B-A)=15+4x,~~~n(A\cap B)=3x.$

We are to find the values of x and n(AUB).

Since n(A) = n(B), so we must have

$n(A-B)=n(B-A)\\\\\Rightarrow 33+x=15+4x\\\\\Rightarrow 4x-x=33-15\\\\\Rightarrow 3x=18\\\\\Rightarrow x=\dfrac{18}{3}\\\\\Rightarrow x=6.$

So, we get

$n(A-B)=33+6=39,\\\\n(B-A)=15+4\times6=15+24=39,\\\\n(A\cap B)=3\times6=18.$

From Set Theory, we get

$n(A\cup B)=n(A-B)+n(A\cap B)+n(B-A)=39+18+39=96.$

Thus, the required value of x is 6 and the value of n(AUB) is 96.