Math  /  Algebra

QuestionSimplify the expression: 6x12x+1xx1+1x+1=\frac{\frac{6}{x-1}-\frac{2}{x+1}}{\frac{x}{x-1}+\frac{1}{x+1}}=

Studdy Solution

STEP 1

What is this asking? We're dealing with a big fraction of fractions, and we need to simplify it into something much nicer! Watch out! Don't forget about those sneaky common denominators when adding or subtracting fractions!

STEP 2

1. Simplify the numerator
2. Simplify the denominator
3. Combine the simplified numerator and denominator

STEP 3

To subtract the fractions 6x1\frac{6}{x-1} and 2x+1\frac{2}{x+1}, we need a common denominator.
Let's use (x1)(x+1)(x-1) \cdot (x+1).
We multiply the first fraction by x+1x+1\frac{x+1}{x+1} (which is just multiplying by one!) and the second fraction by x1x1\frac{x-1}{x-1} (again, just one!).

STEP 4

This gives us: 6(x+1)(x1)(x+1)2(x1)(x+1)(x1) \frac{6(x+1)}{(x-1)(x+1)} - \frac{2(x-1)}{(x+1)(x-1)}

STEP 5

Now that we have a common denominator, we can subtract the numerators: 6(x+1)2(x1)(x1)(x+1)=6x+62x+2(x1)(x+1) \frac{6(x+1) - 2(x-1)}{(x-1)(x+1)} = \frac{6x+6-2x+2}{(x-1)(x+1)}

STEP 6

Combining like terms in the numerator, we get: 4x+8(x1)(x+1) \frac{4x+8}{(x-1)(x+1)} So, the simplified numerator is 4x+8(x1)(x+1)\frac{4x+8}{(x-1)(x+1)}!

STEP 7

Just like we did for the numerator, we need a common denominator for xx1\frac{x}{x-1} and 1x+1\frac{1}{x+1}.
We'll use (x1)(x+1)(x-1) \cdot (x+1) again.
Multiply the first fraction by x+1x+1\frac{x+1}{x+1} and the second by x1x1\frac{x-1}{x-1}.

STEP 8

This gives us: x(x+1)(x1)(x+1)+1(x1)(x+1)(x1) \frac{x(x+1)}{(x-1)(x+1)} + \frac{1(x-1)}{(x+1)(x-1)}

STEP 9

Now, add the numerators: x(x+1)+(x1)(x1)(x+1)=x2+x+x1(x1)(x+1) \frac{x(x+1) + (x-1)}{(x-1)(x+1)} = \frac{x^2 + x + x - 1}{(x-1)(x+1)}

STEP 10

Combining like terms in the denominator, we get: x2+2x1(x1)(x+1) \frac{x^2 + 2x - 1}{(x-1)(x+1)} The simplified denominator is x2+2x1(x1)(x+1)\frac{x^2 + 2x - 1}{(x-1)(x+1)}!

STEP 11

We now have: 4x+8(x1)(x+1)x2+2x1(x1)(x+1) \frac{\frac{4x+8}{(x-1)(x+1)}}{\frac{x^2 + 2x - 1}{(x-1)(x+1)}}

STEP 12

Remember, dividing by a fraction is the same as multiplying by its reciprocal: 4x+8(x1)(x+1)(x1)(x+1)x2+2x1 \frac{4x+8}{(x-1)(x+1)} \cdot \frac{(x-1)(x+1)}{x^2 + 2x - 1}

STEP 13

Notice that (x1)(x+1)(x-1)(x+1) appears in both the numerator and the denominator.
We can multiply by 1(x1)(x+1)\frac{1}{(x-1)(x+1)} in the numerator and denominator, which simplifies things greatly: 4x+8x2+2x1 \frac{4x+8}{x^2 + 2x - 1}

STEP 14

The simplified expression is 4x+8x2+2x1\frac{4x+8}{x^2 + 2x - 1}!

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