SOLUTION: I hope I am posting this in the correct place. I am really stuck on this and do not know what to do. Please help me understand how to do this, Thank you so much. 2{[3(m-5)+18]-[3(

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: I hope I am posting this in the correct place. I am really stuck on this and do not know what to do. Please help me understand how to do this, Thank you so much. 2{[3(m-5)+18]-[3(      Log On


   

Question 117895: I hope I am posting this in the correct place. I am really stuck on this and do not know what to do. Please help me understand how to do this, Thank you so much.
2{[3(m-5)+18]-[3(5m-3)+4]}
Found 3 solutions by Earlsdon, solver91311, Edwin McCravy:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
2{[3(m-5)+18]-[3(5m-3)+4]} Start by applying the "distributive property" to the inner-most parentheses.
2{[(3m-15)+18]-[(15m-9)+4]} Now you can remove the inner-most parentheses and simplify the inner backets ([...])
2{[3m-15+18}-[15m-9+4]}
2{[3m+3]-[15m-5]} Remove the brackets and simplify inside the curly brackets ({...})
2{3m+3-15m+5}
2{3m-15m+3+5}
2{-12m+8} Finally, apply the distibutive property.
-24m+16

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
2%28%283%28m-5%29%2B18%29-%283%285m-3%29%2B4%29%29

This is just an exercise in the application of the Distributive, Associative, and Commutative Properties:

Distributive Property: a%28b+%2Bc%29=ab%2Bac for all real a, b, and c.
Associative Property: %28a%2Bb%29%2Bc=a%2B%28b%2Bc%29 for all real a, b, and c.
Commutative Property: a%2Bb=b%2Ba for all real a and b.

For this problem you need to work from the inside and go out. Start with the 3%28m-5%29 part. Using the distributive property, you can write that as 3m-15. Same thing for the 3%285m-3%29 part, which can be written as 15m-9. Put these two expressions back into the original expression:

2%28%283m-15%29%2B18-%2815m-9%29%2B4%29%29

Now remove the parentheses around -%2815m-9%29 by distributing the minus sign. Think of the minus sign in front of a set of parentheses as a -1, and then apply the distributive property. You end up with -15m%2B9. Put this back into the expression.

2%28%283m-15%29%2B18-15m%2B9%2B4%29

Next you can remove the parentheses from the %283m-15%29 without changing anything else because there is no minus sign in front of this quantity.

2%283m-15%2B18-15m%2B9%2B4%29

For the expression that is still inside of the parentheses, you have 2 terms that contain the variable m, and 4 terms that are just numbers. Add up all the like terms so that you get -12m and %2B16. This is where we are applying the Associative and Commutative Properties.

2%28-12m%2B16%29

One last application of the Distributive Property and we are done:

-24m%2B32, or it might be more neatly put 32-24m

Hope that helps. I hope you and your family have a joyous holiday season.
John

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

2{[3(m-5)+18]-[3(5m-3)+4]}

First look for the first innermost pair of grouping symbols.
That is, grouping sysmbols which have no grouping symbols 
between them.  So we see (m-5).

Then we ask: "Is there anything which we can do with what is
inside " m-5 ". 

The answer is no.  

Then we ask: "How do we remove those grouping symbols around
" m-5 "?"

The answer is by distributing the 3 just outside the (m-5),
getting 3m-15, so in place of 3(m-5) we write 3m-15, and
copy everything else over, getting

2{[3m-15+18]-[3(5m-3)+4]}

-----------------------------------------

Now we start over. Look for the first innermost pair of 
grouping symbols. That is, grouping sysmbols which have 
no grouping symbols between them.  So we see [3m-15+18].

Then we ask: "Is there anything which we can do with what is
inside "3m-15+18". 

The answer is yes, we can combine the -15 and the +18 getting
"3m+3".  So we replace "3m-15+18" by "3m+3", and copy 
everything else over:

2{[3m+3]-[3(5m-3)+4]}   

---------------------------------------

Now we start over. Look for the first innermost pair of 
grouping symbols. That is, grouping sysmbols which have 
no grouping symbols between them.  So we see [3m+3].

Then we ask: "How do we remove those grouping symbols around
3m+3?"

The answer is by putting the invisible 1 just left of [3m+3],
like this:

2{1[3m+3]-[3(5m-3)+4]}

and distributing the 1 just outside the [3m+3],
getting 3m+3, so in place of [3m+3] we write 3m+3, and
copy everything else over, getting

2{3m+3-[3(5m-3)+4]}

---------------------------------------

Now we start over again. Look for the first innermost pair of 
grouping symbols. That is, grouping symbols which have 
no grouping symbols between them.  So we see (5m-3).

Then we ask: "Is there anything which we can do with what is
inside, that is, "5m-3". 

The answer is no.  

Then we ask: "How do we remove those grouping symbols around
" 5m-3 "?

The answer is by distributing the 3 just outside the (5m-3),
getting 15m-9, so in place of 3(5m-3) we write 15m-9, and
copy everything else over, getting

2{3m+3-[15m-9+4]}

--------------------------------------------

Once more, we start over. Look for the first innermost pair 
of grouping symbols. That is, grouping sysmbols which have 
no grouping symbols between them.  So we see [15m-9+4].

Then we ask: "Is there anything which we can do with what is
inside, namely, "15m-9+4". 

The answer is yes. The answer is yes, we can combine the -9 
and the +4 getting "15m-5".  So we replace "15m-9+4" by "15m-5",
and copy everything else over:

2{3m+3-[15m-5]}   

-----------------------------

Once more, we start over. Look for the first innermost pair 
of grouping symbols. That is, grouping sysmbols which have 
no grouping symbols between them.  So we see [15m-5].

Then we ask: "Is there anything which we can do with what is
inside, namely, "15m-5". 

The answer is no.

Then we ask: "How do we remove those grouping symbols around
15m-5?"

The answer is by putting the invisible 1 just left of [15m-5],
like this:

2{3m+3-1[15m-5]}

and distributing the -1 just outside the [15m-5],
getting -15m+5, so in place of -1[15m-5] we write -15m+5, and
copy everything else over, getting

2{3m+3-15m+5}

--------

Now we start over again. Look for the first innermost pair of 
grouping symbols. That is, grouping symbols which have 
no grouping symbols between them.  So we see {3m+3-15m+5}.

Then we ask: "Is there anything which we can do with what is
inside, namely " 3m+3-15m+5 ". 

The answer is yes, we can combine the 3m and the -15m, and also
the +3 and the +5, getting "-12m+8".  So we replace "3m+3-15m+5"
by "-12m+8", and copy everything else over:

2{-12m+8}   

---------------------------------------

Starting over again, we look for the first innermost pair of 
grouping symbols. That is, grouping sysmbols which have no 
grouping symbols between them.  So we see {-12m+8}.

Then we ask: "Is there anything which we can do with what is
inside, namely, " -12m+8 ". 

The answer is no.  

Then we ask: "How do we remove those grouping symbols around
" -12m+8 "?"

The answer is by distributing the 2 just outside the {-12m+8},
getting -24m+16, so in place of 2{-12m+8} we write -24m+16.

There are no grouping symbols and nothing else can be done with
-24m+16, so we are done.

The answer is -24m+16.

Edwin