Video: US-SAT03S3-Q16-284160561086

If π‘₯/(π‘₯Β² βˆ’ 4) βˆ’ (1/(π‘₯ + 2)) = 1/(4π‘₯ βˆ’ 8), what is the value of π‘₯?

04:29

Video Transcript

If π‘₯ over π‘₯ squared minus four minus one over π‘₯ plus two equals one over four π‘₯ minus eight, what is the value of π‘₯?

The first step in solving something like this is to take these three fractions and find a common denominator. At first, this might seem a little bit tricky. And so, what we’ll do is see if we can simplify any of the denominators. π‘₯ squared minus four is in the form π‘Ž squared minus 𝑏 squared. We could rewrite it as π‘₯ squared minus two squared. And we call this the difference of squares. We rewrite the difference of squares as π‘Ž plus 𝑏 times π‘Ž minus 𝑏. And in our case, we’ll have π‘₯ plus two times π‘₯ minus two. One over π‘₯ plus two can’t be simplified any further. So we’ll leave it for now.

What about one over four π‘₯ minus eight? We can’t do anything with the numerator. But our denominator has a factor of four. And that means we could rewrite the denominator as four times π‘₯ minus two. Now we see that two of our fractions have a denominator with a factor of π‘₯ plus two. Two of them have a denominator with the factor π‘₯ minus two and one of them has a factor of four. And we need to give all three of these terms all three factors. We multiply our first term by four over four. Four over four equals one. So we haven’t changed the value of this fraction. But we’ve rewritten it. And now, its denominator has a factor of four, a factor of π‘₯ plus two, and a factor of π‘₯ minus two.

Our next term one over π‘₯ plus two needs to be multiplied by π‘₯ minus two over π‘₯ minus two and by four over four. It needs two new terms. So we multiply the numerator one by π‘₯ minus two and four. One times four times π‘₯ minus two equals four times π‘₯ minus two. At this point, our denominators are now the same. Our final term is missing a factor of π‘₯ plus two. So we multiply it by π‘₯ plus two over π‘₯ plus two. One times π‘₯ plus two equals π‘₯ plus two. And now, it too has a denominator of four times π‘₯ plus two times π‘₯ minus two.

We can now do something else to simplify this problem. We have a denominator of four times π‘₯ plus two times π‘₯ minus two on both sides of this equation. If we multiplied both sides of the equation by four times π‘₯ plus two times π‘₯ minus two, the denominators cancel out. And we’re left with four π‘₯ minus four times π‘₯ minus two equals π‘₯ plus two. So we can distribute this multiplication so that we have four π‘₯ minus four π‘₯ plus eight. Negative four times negative two equals positive eight. All of that equal two π‘₯ plus two. Positive four π‘₯ minus four π‘₯ equals zero. We have eight equals π‘₯ plus two. We’ll give ourselves a little more room and then subtract two from both sides. Eight minus two is six. π‘₯ is equal to six. When π‘₯ equals six, this statement is true.

We can also plug six in to our original equation to make sure this is true. We would have six over six squared which is 36 minus four minus one-eighth equals one over 24 minus eight. Six over 32 minus one-eighth equals one sixteenth. If we rewrite them all with the denominator of 32, we see six over 32 minus four over 32 equals two over 32, which is true and confirms that π‘₯ equals six.

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