Question

set theory

Answer 1

e + t + m + x = 180

e + x = 90
t + x = 120
m + x = 150

Answer 2

how shuld i proceed

Answer 3

four variable , four unknowns

a matrix might be fun


1 0 0 1 90
0 1 0 1 120
0 0 1 1 150
1 1 1 1 180

Answer 4

r4 -> r4 -r1 -r2 -r3

1 0 0 1 90
0 1 0 1 120
0 0 1 1 150
0 0 0 -2 180-90-120-150



r4 -> r4/-2

1 0 0 1 90
0 1 0 1 120
0 0 1 1 150
0 0 0 1 (180-90-120-150)/-2

Answer 5

by the way uncle i dont understand matrix

Answer 6

\[|A\cap B| = |B\cap C| = |A \cap C| = |A \cap B\cap C|\]But\[|A \cup B \cup C| = |A|+|B|+|C|-|A\cap B| - |B\cap C| - |A\cap C| + |A\cap B \cap C|\]\[\Rightarrow 180 = 90+120+150 - 2x\]\[\Rightarrow x = 90\]

Answer 7

from the big equation
e + t + m + x = 180

take away the other equations
e + x = 90
t + x = 120
m + x = 150

and solve for x

Answer 8

parthkohli how did u get 2x


\(\Rightarrow 180 = 90+120+150 - \color{red}{2x}\)

Answer 9

\[|A\cap B| = |B\cap C| = |A \cap C| = |A \cap B\cap C| = \color{red}x \]

Answer 10

but i still don't understand how this become equal with the three

\(|A\cap B| = |B\cap C| = |A \cap C| =\color{red}{ |A \cap B\cap C|}\)

Answer 11

no one was faced with exactly two problems

Answer 12

ok

Answer 13

thnks

Answer 14

i have a question why can't \(x=0\) as
\(|A\cap B| = |B\cap C| = |A \cap C| = 0\)

Answer 15

Exactly two intersections \(\ne\) At least two two intersections

And concluding from the above, we can say that if A and B intersect, C must also intersect there. Which is why I wrote that.