Help with distributive property and clearing parentheses

J

J_Barber

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Jun 7, 2013
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As the title says I have to use distributive properties and clear the parentheses.

-2/3(x-6)
-2/3(x) here's where I get stuck. not sure how to disttriute the -2/3 to the -6. Do the two negatives then turn into a positive?

-2/3(x)+(2/3-6/1)? don't think this is correct. Can I get a hint?

JB
 
J_Barber said:
As the title says I have to use distributive properties and clear the parentheses.

-2/3(x-6)
-2/3(x) here's where I get stuck. not sure how to disttriute the -2/3 to the -6. Do the two negatives then turn into a positive?

-2/3(x)+(2/3-6/1)? don't think this is correct. It's not. Can I get a hint?

JB
Click to expand...
A negative number times a negative number is a positive number. It makes no difference whether the numbers are fractions or integers or irrationals.

\(\displaystyle u< 0\ and\ v < 0 \implies u * v > 0\ for\ all\ real\ numbers\ u\ and\ v.\)

Now let's understand the distributive property.

\(\displaystyle a(b \pm c) = ab \pm ac\ for\ all\ real\ numbers\ a,\ b, \ and\ c.\)

So here is a hint. Let a = - 2/3. Restate the expression using a. Clear the parentheses using a. Then substitute - 2/3 for a. This trick of a temporary substitution is a good one to have in your toolbox.
 
-2/3(x-6)
-2/3(x) + 2/3(6/1)
-2/3x+12/3
= -2/3x+4

Great explanation, once I read it through it made sense and I was able to zip right through the problem.

Thanks again!

JB
 
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