# Distributive Property on Fractions: Swapping Denominators

I'm learning Algebra and am curious about some methodological fundamentals here. One, in particular is why the following equation:

6(2x + 1 / 3) = 6(x + 4 / 2)


results in:

2(2x + 1) = 3(x + 4)


It's obvious that the distributive property swaps the numerators of the fractions and chooses to use another distributive property to complete the equation. Is there a specific formula for this, and why does it work that way specifically?

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Maybe you are mistyping $6(2x+1/3)$ instead of $6((2x+1)/3)$? Perhaps what the author wrote is $6\frac{2x+1}{3} = 6\frac{x+4}{2}$? –  Dilip Sarwate Oct 7 '11 at 18:31
What you wrote is correct if you have $6((2x+1)/3) = 6((x+4)/2)$. But $2x+1/3$ is not the same as $(2x+1)/3)$ and $x+4/2$ is not the same as $(x+4)/2$. –  Michael Hardy Oct 7 '11 at 19:14

HINT $\$ Apply the associative law $\rm\displaystyle\ \ A\ \bigg(\!\frac{1}{B}\ C\bigg)\ =\ \bigg(A\ \frac{1}B\bigg)\ C$

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Also @lhf Taken as typed by the OP, $6(2x + 1/3)$ works out to $12x + 2$, not $2(2x + 2)$ which is what $6((2x + 1)/3)$ works out to be. I believe that the OP is confused by his own mis-typing (and also possibly by how to apply the associative law to fractions..) –  Dilip Sarwate Oct 7 '11 at 19:17
@Dil Surely the OP intends $\:(2\:x+1)/3\:,\:$ else the stated answer would be incorrect. –  Bill Dubuque Oct 7 '11 at 20:21

$$a \left(\frac b c\right) = \frac a1 \frac b c = \frac {ab}{1c} = \frac {ab}{c1} = \frac ac \frac b1 = \frac ac b$$

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May be the following steps help you see how you get the result from the given expression - Swapping is fine as long as you understand the meaning of it so that you don't make mistakes.

Given:

$6 \left ( \frac{2x+1}{3} \right )= 6 \left ( \frac{x+4}{2} \right )$

a-multiply both sides by 1/6 to get:

$\left ( \frac{2x+1}{3} \right )= \left ( \frac{x+4}{2} \right )$

b-multiply both sides by 3 to get:

$3\left ( \frac{2x+1}{3} \right )= 3 \left ( \frac{x+4}{2} \right )$

c-This is equal to:

$\left ( \frac{2x+1}{1} \right )= 3 \left ( \frac{x+4}{2} \right )$

d-Multiply both sides by 2

$2 \left ( \frac{2x+1}{1} \right )= 2 * 3 \left ( \frac{x+4}{2} \right )$

e-Simplifying the right hand side you get

$2 \left ( \frac{2x+1}{1} \right )= 3 \left ( \frac{x+4}{1} \right )$

f-Which is:

$2(2x+1)=3(x+4)$

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