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Troyanec [42]
3 years ago
12

Two problems here I need solved! I need every step, so please have that with your answers!!

Mathematics
1 answer:
Free_Kalibri [48]3 years ago
7 0
QUESTION 1

We want to solve,

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}

We factor the denominator of the fraction on the right hand side to get,

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.

This implies
\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.

\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}

We multiply through by LCM of
(x-4)(x - 2)

(x - 2) + x(x-4) = 2

We expand to get,

x - 2 + {x}^{2} - 4x= 2

We group like terms and equate everything to zero,

{x}^{2} + x - 4x - 2 - 2 = 0

We split the middle term,

{x}^{2} + - 3x - 4 = 0

We factor to get,

{x}^{2} + x - 4x- 4 = 0

x(x + 1) - 4(x + 1) = 0

(x + 1)(x - 4) = 0

x + 1 = 0 \: or \: x - 4 = 0

x = - 1 \: or \: x = 4

But
x = 4
is not in the domain of the given equation.

It is an extraneous solution.

\therefore \: x = - 1
is the only solution.

QUESTION 2

\sqrt{x+11} -x=-1

We add x to both sides,

\sqrt{x+11} =x-1

We square both sides,

x + 11 = (x - 1)^{2}

We expand to get,

x + 11 = {x}^{2} - 2x + 1

This implies,

{x}^{2} - 3x - 10 = 0

We solve this quadratic equation by factorization,

{x}^{2} - 5x + 2x - 10 = 0

x(x - 5) + 2(x - 5) = 0

(x + 2)(x - 5) = 0

x + 2 = 0 \: or \: x - 5 = 0

x = - 2 \: or \: x = 5

But
x = - 2
is an extraneous solution

\therefore \: x = 5
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