7
REMOVING
GROUPING SYMBOLS
The rules for removing parentheses
The relationship of a − b to b − a
WE HAVE SEEN that the terms of this sum
a − b + c − d
are
a, −b, c, −d.
We include the minus sign as part of the name of the term. (Lesson 3)
Problem 1. In each of the following, name each term.
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a) 3 −4 + 5 −6 3, −4, 5, −6.
b) x − y + z x, −y, z.
c) −x + y − z −x, y, −z.
The rules for removing parentheses
Parentheses will be preceded either by a plus sign:
a + (b −c + d)
or a minus sign:
a − (b −c + d).
When parentheses are preceded by a plus sign +
simply remove them. Nothing changes.
a + (b −c + d) = a + b − c + d.
When parentheses are preceded by a minus sign −
change the sign of every term within the parentheses.
a − (b −c + d) = a − b + c − d.
The sign of b within the parentheses is understood to be + . Therefore, upon removing the parentheses, it becomes −b.
−c within the parentheses becomes +c. And +d becomes −d.
We can justify this with examples from arithmetic, because algebra is abstracted -- taken from -- arithmetic. If it were not, algebra would not be very useful, would it?
For example,
| 256 + 98 | = | 256 + 100 − 2 | |
| = | 356 − 2 | ||
| = | 354. | ||
| That is, | |||
| 256 + (100 − 2) | = | 256 + 100 − 2. | |
| On the other hand, | |||
| 256 − 98 | = | 256 − 100 + 2 | |
| = | 156 + 2 | ||
| = | 158. | ||
| That is, | |||
| 256 − (100 − 2) | = | 256 − 100 + 2. | |
Problem 2. Remove the parentheses.
a) p + (q −r + s) = p + q − r + s
b) p − (q −r + s) = p − q + r − s
In each of the following problems, remove the parentheses, then simplify
by adding the numbers.
For example,
| (x − 3) − (y − 4) | = | x − 3 − y + 4 |
| = | x − y + 1. | |
It is conventional to write the literal terms, x − y, to the left of the numerical term.
Also, the sign preceding (x − 3) is understood to be + . Therefore the signs within those parentheses do not change.
| Problem 3. (x + 2) + (y + 8) | = | x + 2 + y + 8 |
| = | x + y + 10 | |
| Problem 4. (x + 2) − (y + 8) | = | x + 2 − y − 8 |
| = | x − y − 6 | |
| Problem 5. (x − 2) + (y + 8) | = | x − 2 + y + 8 |
| = | x + y + 6 | |
| Problem 6. (x − 2) − (y + 8) | = | x − 2 − y − 8 |
| = | x − y − 10 | |
| Problem 7. (x − 2) − (y − 8) | = | x − 2 − y + 8 |
| = | x − y + 6 | |
| Problem 8. (x − 2) + (y − 8) | = | x − 2 + y − 8 |
| = | x + y − 10 | |
| Problem 9. (a − 2) + (b + 3) − (c − 7) | = | a − 2 + b + 3 − c + 7 |
| = | a + b − c + 8 | |
| Problem 10. (a − 5) − (b + 6) − (c − 9) | = | a − 5 − b − 6 − c + 9 |
| = | a − b − c − 2 | |
| Problem 11. (a + 2) − (b − 3) + (c − 8) − (d + 1) | ||
| = | a + 2 − b + 3 + c − 8 − d − 1 | |
| = | a − b + c − d − 4 | |
Thus, a minus sign before a parentheses changes every sign within them. We saw that in the rule
−(−x) = x.
| Problem 12. −(−x + y) | = | x − y |
| Problem 13. −(x − y) | = | −x + y |
| Problem 14. −(x + y − 2) | = | −x − y + 2 |
Problem 15. Write the negative of a − b + c − d.
−a + b − c + d
Problem 16. Placing parentheses. Rewrite each of the following by placing parentheses.
a) −x + y = −(x − y)
b) −x − y = −(x + y)
c) −a + b − c + d = −(a − b + c − d)
d) Place parentheses around b and c:
a − b + c − d = a − (b − c) − d
Brackets and braces
Brackets [ ] and braces { } have the same function as parentheses. They are all grouping symbols. After parentheses are used, we use brackets. And after brackets, braces.
Example 1. a − [b − (c − d + e)]
We will remove all the grouping symbols. We will do it by removing the brackets first. Then we will do it again removing the parentheses first. The student should have the skill to do it either way.
So, upon removing the brackets:
| a − [b − (c − d + e)] | = | a − b + (c − d + e) |
Within the brackets, there are two terms The first term is b. The second term is −(c − d + e). Since the brackets are preceded by − , only the sign of each term changes. The signs within the second term do not change.
Finally, we remove the parentheses, which are preceded by + :
| = | a − b + c − d + e. |
Now let us do this same problem by removing the parentheses first:
| a − [b − (c − d + e)] | = | a − [b − c + d − e] |
| = | a − b + c − d + e | |
Since the parentheses are preceded by − , every sign within them changes. And since the brackets are also preceded by − , every sign within them changes.
Problem 17.
a) First remove the brackets, then remove the parentheses.
w + [x − (y + z)] = w + x − (y + z) = w + x − y − z
First remove the parentheses, then remove the brackets.
w + [x − (y + z)] = w + [x − y − z] = w + x − y − z
b) First remove the brackets, then remove the parentheses.
w − [x + (y − z)] = w − x − (y − z) = w − x − y + z
First remove the parentheses, then remove the brackets.
w − [x + (y − z)] = w − [x + y − z] = w − x − y + z
c) First remove the brackets, then remove the parentheses.
w − [x − (y + z)] = w − x + (y + z) = w − x + y + z
First remove the parentheses, then remove the brackets.
w − [x − (y + z)] = w − [x − y − z] = w − x + y + z
d) First remove the brackets, then remove the parentheses.
w + [x − (y − z)] = w + x − (y − z) = w + x − y + z
First remove the parentheses, then remove the brackets.
w + [x − (y − z)] = w + [x − y + z] = w + x − y + z
Problem 18. Remove all the grouping symbols. Simplify as you go by evaluating the numbers. Remove the brackets first.
| a) 5 − [3 − (x − 2)] | = | 5 − 3 + (x − 2) |
| = | 2 + x − 2 | |
| = | x | |
| b) 5 − [3 − (x + 2)] | = | 5 − 3 + (x + 2) |
| = | 2 + x + 2 | |
| = | x + 4 | |
| c) −5 + [3 − (x − 2)] | = | −5 + 3 − (x − 2) |
| = | −2 − x + 2 | |
| = | −x | |
| d) 5 − [−3 − (x + 2)] | = | 5 + 3 + (x + 2) |
| = | 8 + x + 2 | |
| = | x + 10 | |
Problem 19.
a) First remove the braces, then the brackets, then the parentheses.
a) Simplify by adding the numbers.
| 10 − {2 + [3 − (x − 5)]} | = | 10 − 2 − [3 − (x − 5)] |
| = | 8 − 3 + (x − 5) | |
| = | 5 + x − 5 | |
| = | x | |
First remove the parentheses, then the brackets, then the braces.
| 10 − {2 + [3 − (x − 5)]} | = | 10 − {2 + [3 − x + 5]} |
| = | 10 − {2 + 3 − x + 5} | |
| = | 10 − 10 + x | |
| = | x | |
b) First remove the braces, then the brackets, then the parentheses.
| 8 + {2 − [12 + (x − 2)]} | = | 8 + 2 − [12 + (x − 2)] |
| = | 10 − 12 − (x − 2) | |
| = | −2 − x + 2 | |
| = | −x | |
First remove the parentheses, then the brackets, then the braces.
| 8 + {2 − [12 + (x − 2)]} | = | 8 + {2 − [12 + x − 2]} |
| = | 8 + {2 − 12 − x + 2} | |
| = | 8 + 2 − 12 − x + 2 | |
| = | −x | |
c) Remove the grouping symbols. Start with the parentheses.
| 7a − {3a − [4a − (5a − 2a)]} | = | 7a − {3a − [4a − 5a + 2a]} |
| = | 7a − {3a − 4a + 5a − 2a} | |
| = | 7a − 3a + 4a − 5a + 2a} | |
| = | 13a − 8a | |
| = | 5a | |
d) Remove the grouping symbols. Start with the parentheses.
| 3a − {a + b − [a + b + c − (a + b + c + d)]} | ||
| = | 3a − {a + b − [a + b + c − a − b − c − d]} | |
| = | 3a − {a + b − [−d]} | |
| = | 3a − {a + b + d} | |
| = | 3a − a − b − d | |
| = | 2a − b − d | |
Next Lesson: Adding like terms
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