. Evaluate: |4| + |–3|
Rule: To remove absolute value bars,
1. If the number between the bars is positive, simply erase the absolute
value bars
2. If the number between the bars is negative, erase both the absolute
value bars AND the negative sign, making the results a positive number.
3. If the number between the bars is zero, simply erase the absolute
value bars and the result is zero.
|4| + |–3|
Replace |4| by 4
4 + |–3|
Replace |-3| by 3
4 + 3
Answer: 7
--------------------
. Evaluate: |-9| - |–4|
Replace |-9| by 9
9 - |–4|
Replace |-4| by 4
9 - 4
Answer: 5
--------------------------
. Add: 7 + (–9) + (–5) + 6
There are no absolute value bars here. Be sure not to
confuse | | with ( ). The ( ) here are only used around
negative numbers to keep the sign of operation separate from
the sign of the number.
Think of it as $7 + (–$9) + (–$5) + $6
Pretend you are played 4 games of poker.
On the first game you win $7. So you are now $7 to the good
after the first game.
On the second game you lose $9. So you are now $2 in the hole
after the second game.
On the third game you lose $5. So you are now $7 in the hole
after the third game.
On the fourth and last game you win $6. So you are now $1 in
the hole after the fourth game
All in all you finish the four poker games losing $1, so you
indicate that by writing -$1.
So 7 + (–9) + (–5) + 6 = -1
---------------------------
Add: 1/3 + (5/6) + (-1/2)
The dnominators are 3,6, and 2. These will all divide
evenly into 6.
1/3 becomes 2/6
5/6 stays as 5/6
-1/2 becomes (-3/6)
So we rewrite the problem as
2/6 + (5/6) + (-3/6)
Since the denominators are now all the same, we
need only add the numerators.
Think of the first numerator 2 as winning $2 on the first game,
so you're $2 to the good.
Think of the second numerator 5 as winning $5 on the second game,
so you're $7 to the good.
Think of the third numerator -3 as losing $3 on the third game,
so you end up $4 to the good.
Bring the denominator back:
So the answer is 4/6, which reduces to 2/3.
Edwin
