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How do you evaluate #\frac{1}{x^{2}-16}=-\frac{1}{x^{2}-4x}#?

Answer 1

See a solution process below:

First, factor the denominators of each fraction as:

#1/((x + 4)(x - 4)) = -1/(x (x - 4)#

Next, multiply each side of the equation by #color(red)((x - 4))# to eliminate a common factor while keeping the equation balanced:

#color(red)((x - 4)) xx 1/((x + 4)(x - 4)) = color(red)((x - 4)) xx -1/(x(x - 4)#

#cancel(color(red)((x - 4))) xx 1/((x + 4)color(red)(cancel(color(black)((x - 4))))) = cancel(color(red)((x - 4))) xx -1/(xcolor(red)(cancel(color(black)((x - 4)))))#

#1/(x + 4) = -1/x#

Then, because we have a pure fraction on each side of the equation we can flip the fractions giving:

#(x + 4)/1 = -x/1#

#x + 4 = -x#

Next, we can subtract #color(red)(4)# and add #color(blue)(x)# to each side of the equation to isolate the #x# term while keeping the equation balanced:

#color(blue)(x) + x + 4 - color(red)(4) = color(blue)(x) - x - color(red)(4)#

#1color(blue)(x) + 1x + 0 = 0 - 4#

#(1 + 1)x = -4#

#2x = -4#

Now, divide both sides of the equation by #color(red)(2)# to solve for #x# while keeping the equation balanced:

#(2x)/color(red)(2) = -4/color(red)(2)#

#(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = -2#

#x = -2#

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Answer 2

Hazel Anglin

#x=-2#

#1/(x^2-16)=-1/(x^2-4x)#

We can multiply both sides of the equation by #(x^2-16)(x^2-4x)#

#color(red)(cancel((x^2-16))(x^2-4x))/cancel(x^2-16)=-color(red)((x^2-16)cancel((x^2-4x)))/cancel(x^2-4x)#

#x^2-4x=-(x^2-16)# #x^2-4x=-x^2+16#

Solve like a quadratic:

#2x^2-4x-16=0# #x^2-2x-8=0# #(x-4)(x+2)=0#

#x=-2# or #x=4#

We need to check our answers however. This is because we can't divide by zero, this is undefined

Because of this, if we let #x=4#, we get denominators of 0. This cannot happen, so #x!=4# This means that the only solution is #x=-2#

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Answer 3

Julia Adams

To evaluate the equation (\frac{1}{x^{2}-16}=-\frac{1}{x^{2}-4x}), first, find a common denominator and then equate the numerators.

(\frac{1}{x^{2}-16}=-\frac{1}{x^{2}-4x})

Cross multiply:

(1(x^2 - 4x) = -1(x^2 - 16))

Expand and rearrange terms:

(x^2 - 4x = -x^2 + 16)

Add (x^2) to both sides:

(2x^2 - 4x = 16)

Divide both sides by 2:

(x^2 - 2x = 8)

Complete the square on the left side:

(x^2 - 2x + 1 = 8 + 1)

Factor the left side:

((x - 1)^2 = 9)

Take the square root of both sides:

(x - 1 = \pm 3)

Add 1 to both sides:

(x = 1 \pm 3)

Solve for (x):

(x = 4) or (x = -2)

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.