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Consider the addition of 2 fractions:

[tex]\displaystyle{\frac{7}{x - 3} + \frac{1}{3 - x}}[/tex]

We know we must find a least common denominator in order to add the fractions, so let's consider them:

We have x - 3, and we also have 3 - x.

Suppose we rewrote the first denominator as -(3 - x). In other words: x - 3 = -(3 - x)

Our denominators would then be -(3 - x) & (3 - x). Note we haven't changed the value of the denominators, we have simply rewritten them.

The fraction problem would then become:

[tex]\displaystyle{\frac{7}{-(3 - x)} + \frac{1}{3 - x}}[/tex]

Note that when a minus sign is placed in a fraction, it can be placed in either the numberator or the denominator, as it does not matter. So, we can have identical denominators for the purpose of addition by simply moving the minus sign to the numberator in the first fraction:

[tex]\displaystyle{\frac{-7}{3 - x} + \frac{1}{3 - x} = \frac{-6}{3 - x}}[/tex]

My question is, why is this wrong?

The book instead rewrites the OTHER denominator, and the final problem looks like this:

[tex]\displaystyle{\frac{7}{x - 3} + \frac{1}{3 - x} = \frac{7}{x - 3} + \frac{1}{-(x - 3)} = \frac{7}{x - 3} + \frac{-1}{x - 3} = \frac{6}{x - 3}}[/tex]

What am I doing wrong?

[tex]\displaystyle{\frac{7}{x - 3} + \frac{1}{3 - x}}[/tex]

We know we must find a least common denominator in order to add the fractions, so let's consider them:

We have x - 3, and we also have 3 - x.

Suppose we rewrote the first denominator as -(3 - x). In other words: x - 3 = -(3 - x)

Our denominators would then be -(3 - x) & (3 - x). Note we haven't changed the value of the denominators, we have simply rewritten them.

The fraction problem would then become:

[tex]\displaystyle{\frac{7}{-(3 - x)} + \frac{1}{3 - x}}[/tex]

Note that when a minus sign is placed in a fraction, it can be placed in either the numberator or the denominator, as it does not matter. So, we can have identical denominators for the purpose of addition by simply moving the minus sign to the numberator in the first fraction:

[tex]\displaystyle{\frac{-7}{3 - x} + \frac{1}{3 - x} = \frac{-6}{3 - x}}[/tex]

My question is, why is this wrong?

The book instead rewrites the OTHER denominator, and the final problem looks like this:

[tex]\displaystyle{\frac{7}{x - 3} + \frac{1}{3 - x} = \frac{7}{x - 3} + \frac{1}{-(x - 3)} = \frac{7}{x - 3} + \frac{-1}{x - 3} = \frac{6}{x - 3}}[/tex]

What am I doing wrong?

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