# How does rearrangement work for pulling out factors

Given 10(6*3+2) we can rearrange the equation by pulling out the 3 and getting 10((6*3)/3+(2/3))3

Okay I understand that it works. But I don't understand how. I would have assumed to do 10((6*3)/3+2)3 since we are only removing the 3 as a factor to 6. Why does 3 need to be divided by the 2 as well? 3 has nothing to do with the 2.

• The only way the value of the expression will be unchanged is if you both multiply and divide by $3$. – saulspatz Jan 7 at 21:14
• Notice that everything in the parentheses is being multiplied by $10$. If you extract a $3$ from the parentheses, each term left in the parentheses is being multiplied by $30$. – N. F. Taussig Jan 7 at 21:21
• Please use MathJax. – N. F. Taussig Jan 7 at 21:24

According to the distributive law $$a(b+c)=ab+ac$$ In your problem $$a=3,~b=((6\cdot3)/3),~c=(2/3)$$ $$((6\cdot3)/3+(2/3))\cdot3=((6\cdot3)/3) \cdot3 +(2/3)\cdot3=6\cdot3+2$$ On the other hand, you are wrong to say that $$((6\cdot3)/3+2)\cdot3=6\cdot3+2$$ Since according to the distributive law $$((6\cdot3)/3+2)\cdot3=(6\cdot3/3)\cdot3+2\cdot3=6\cdot3+6$$