Same Coefficients
Solve the following system for distinct x , y , and z .
⎩ ⎪ ⎨ ⎪ ⎧ 5 7 3 2 x + 2 1 3 4 y + 2 1 3 4 z = 7 8 6 6 2 1 3 4 x + 5 7 3 2 y + 2 1 3 4 z = 6 7 0 2 1 3 4 x + 2 1 3 4 y + 5 7 3 2 z = 1 1 4 6 4
What is x + y + z = ?
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2 solutions
That is the solution I wanted to see. Nice. Thank you.
⎩ ⎪ ⎨ ⎪ ⎧ 5 7 3 2 x + 2 1 3 4 y + 2 1 3 4 z = 7 8 6 6 2 1 3 4 x + 5 7 3 2 y + 2 1 3 4 z = 6 7 0 2 1 3 4 x + 2 1 3 4 y + 5 7 3 2 z = 1 1 4 6 4 . . . ( 1 ) . . . ( 2 ) . . . ( 3 )
( 1 ) + ( 2 ) + ( 3 ) : 1 0 0 0 0 x + 1 0 0 0 0 y + 1 0 0 0 0 z ⟹ x + y + z = 2 0 0 0 0 = 2
That is the solution I wanted to see. Nice. Thank you.
Adding the three equations, we have, 1 0 0 0 0 x + 1 0 0 0 0 y + 1 0 0 0 0 z = 2 0 0 0 0 Dividing both L.H.S and R.H.S by 1 0 0 0 0 , x + y + z = 2
Note: It is possible to solve for x , y , z as we have three equations and three unknowns but this is more efficient since we are only asked for the value of their sum