X + k y = 1
k x + y = 1 / * ( - k )
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x + k y = 1
- k² x - k y = - k
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x - k² x = 1 - k
x ( 1 - k² ) = 1 - k
x = ( 1 - k ) / ( 1 - k² ) = ( 1 - k ) / ( 1 - k ) ( 1 + k )
y = 1 - k( 1 - k )/( 1 - k² )
y = ( 1 - k ) / ( 1 - k² ) = ( 1 - k ) / ( 1 - k ) ( 1 + k )
a ) For k = - 1 this system has no solution.
b ) For k ≠ - 1 and k ≠ 1, the system has unique solution:
( x , y ) = ( 1/ (1 + k) , 1/( 1 + k ) ).
c ) For k = 1, there are infinitely many solutions.
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Answer:
x = -1, x = 1
Step-by-step explanation:
Factor the equation. (It's a difference of squares, so we know the form will be (a-b)(a+b).)
Test this by FOILing it out, if you're unsure. This is something it can be good to memorize!
Set it equal to 0.
Separate the two parenthetical expressions by the Zero Product Property.
Solve for x!
Answer:
Option A. √(x + 1)
Step-by-step explanation:
Data obtained from the question include:
f(x) = √(x² – 1)
g(x) = √(x – 1)
(f/g) (x) =..?
(x² – 1) => difference of two square
(x² – 1) => (x – 1)(x + 1)
f(x) = √(x² – 1)
f(x) = √(x – 1)(x + 1)
(f/g) (x) = f(x) /g(x)
f(x) = √(x – 1)(x + 1)
g(x) = √(x – 1)
(f/g) (x) = √(x – 1)(x + 1) / √(x – 1)
(f/g) (x) = √[(x – 1)(x + 1) / (x – 1)]
(f/g) (x) = √(x + 1)