⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ x + y + z + 1 8 a + b = 1 3 5 8 0 2 4 5 8 x + y + 1 8 z + a + b = 1 3 5 8 0 2 4 5 8 1 8 x + y + z + a + b = 1 3 5 8 0 2 4 5 8 x + 1 8 y + z + a + b = 1 3 5 8 0 2 4 5 8 x + y + z + a + 1 8 b = 1 3 5 8 0 2 4 5 8
Solve the system of equations above for real values of x , y , z , a , b .
Enter the value of x as your answer.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Yay! Same method!
Log in to reply
Thanks for your comment :) ... No matter how complex math we know and understand... These are the things I love the most.... How did you find it?
Same way dude!!!
Log in to reply
@abc xyz cool :) ... Thanks for your comment
Problem Loading...
Note Loading...
Set Loading...
Observing the symmetry in the equation, we see that interchanging the values of x , y , z , a , b cyclically in the first equation will yield b + x + y + 1 8 z + a = 1 3 5 8 0 2 4 5 8 which is the second equation. Similarly, interchanging the variables in cyclical manner would yield all the 5 equations in some order. This shows that all the variables have to be equal. putting x = y = z = a = b
2 2 x = 1 3 5 8 0 2 4 5 8 x = 6 1 7 2 8 3 9