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