1. ## Distributing with fractions

I don't understand the rules for distributing with fractions in problems like these:

2/3 (c + 7)

-5/8 (k + 8)

24 (x/3 - 1/8)

I'd greatly appreciate help on these specific problems, and I would also appreciate if someone could explain the rules to me.

2. Originally Posted by endlesst0m
I don't understand the rules for distributing with fractions in problems like these:

2/3 (c + 7)
the distributive law states:

$a(b + c) = ab + ac$

in other words, the term that is multiplying a sum is distributed and multiply each term in the sum.

so here, $\frac 23 (c + 7) = \frac 23 \cdot c + \frac 23 \cdot 7 = \frac {2c}3 + \frac {14}3$

you can combine these, but if tht is how you wanted the answer, the distributive law could be applied differently.

$\frac 23 (c + 7) = \frac {2(c + 7)}3 = \frac {2c + 14}3$

now try the others. if it helps, you can think of 24 as $\frac {24}1$

3. So basically, when you distribute a fraction into a variable, the numerator in the fraction combines with the variable, and the denominator stays the same.

And when you distribute a fraction into a number, you multiply that number times the numerator, and the denominator stays the same.

Right?

4. Originally Posted by endlesst0m
So basically, when you distribute a fraction into a variable, the numerator in the fraction combines with the variable, and the denominator stays the same.

And when you distribute a fraction into a number, you multiply that number times the numerator, and the denominator stays the same.

Right?
no, you multiply each term. look at the rule. it is a coincidence that the denominator stays the same in this case.

recall how to multiply fractions: $\frac ab \cdot \frac cd = \frac {ac}{bd}$

the denominator in this case stays the same, because if you write c + 7 as a fraction, it would be $\frac {c + 7}1$. so when you multiply the denominators, it keeps the denominator of the fraction outside, because you are multiplying it by 1

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# Algebra distributing a fraction into whole numbers

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