1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The only way the value of the expression will be unchanged is if you both multiply and divide by $3$. $\endgroup$ – saulspatz Jan 7 at 21:14
  • 1
    $\begingroup$ 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 $30$. $\endgroup$ – N. F. Taussig Jan 7 at 21:21
  • $\begingroup$ Please use MathJax. $\endgroup$ – N. F. Taussig Jan 7 at 21:24
0
$\begingroup$

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 $$((6\cdot3)/3+2)\cdot3=(6\cdot3/3)\cdot3+2\cdot3=6\cdot3+6$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.