1
$\begingroup$

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?

  • 0
    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 $(x+4)/2$.2011-10-07

3 Answers 3

1

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

  • 0
    @Dil Surely the OP intends $\:(2\:x+1)/3\:,\:$ else the stated answer would be incorrect.2011-10-07
0

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

0

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:

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