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?
On
HINT $\ $ Apply the associative law $\rm\displaystyle\ \ A\ \bigg(\!\frac{1}{B}\ C\bigg)\ =\ \bigg(A\ \frac{1}B\bigg)\ C$
On
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:
$2(2x+1)=3(x+4)$
$$a \left(\frac b c\right) = \frac a1 \frac b c = \frac {ab}{1c} = \frac {ab}{c1} = \frac ac \frac b1 = \frac ac b$$