Can someone help me with these two problems?

Question

1. Given f(x) = x-5 and g(x) =sqrt(x-7), I need to find the domain of the quotient function f/g
2. I have to solve x-1/x+2 = x/x+1

Any help would be greatly appreciated.

Answer 1

1.
f(x)/g(x) = (x-5)/sqrt(x-7)

The domain is the interval on x for which f(x)/g(x) gives real values.

When x is < 7, the argument of the sqrt in g(x) is negative, which means that g(x) is complex/imaginary. Therefore, f(x)/g(x) is also imaginary.

When x is = 7, g(x) = 0. f(0)/g(0) is undefined.

Alas, the domain is (7, inf)

2.
(x-1)/(x+2) = x/(x+1)
(x-1)(x+1)/(x+2) = x
(x-1)(x+1)/(x+2) - x = 0
(x-1)(x+1)/(x+2) - x(x+2)/(x+2) = 0
[(x-1)(x+1) - x(x+2)]/(x+2) = 0
(x^2 - 1 - x^2 - 2x)/(x+2) = 0
(-1 - 2x)/(x+2) = 0
-1 - 2x = 0
-2x = 1
x = -1/2

Answer 2

1. The expression under an even root (a square root is also an even one) must be non-negative. Also, because we divide by g, g must not be zero. f is a polynomial (of the first degree), so its domain is R. The domain of f/g is:

x - 7 > 0 <=> x > 7 <=> D = (7; ∞)

2. (x - 1)/(x + 2) = x/(x+1) /∙(x+1)(x+2) ≠ 0 => x ≠ -1, -2

(x - 1)(x + 1) = x(x + 2)

x² - 1 = x² + 2x

-1 = 2x /:2

x = -1/2 - > the only solution

Answer 3

1)
Inside the root can not be negative
and te denominator must not be zero. Hence

x-7>0

x>7

2)
(x-1)(x+1)=x(x+2)

x^2-1=x^2 +2x

2x=-1

x=-1/2

Answer 4

1.

f/g = (x-5)/sqrt(x-7)

x ∈R, x ≠ 7


2.

(x-1)(x+1)=x(x+2)
x²-1=x²+2x
x=-1/2