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7

REMOVING
GROUPING SYMBOLS

The rules for removing parentheses

2nd level

The relationship of  ab  to  ba

WE HAVE SEEN that the terms of this sum

ab + cd

are

a,  −b,   c,  −d.

The minus sign is part of the name of the term.

Problem 1.   In each of the following, name each term.

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a)   3 −4 + 5 −6.    3, −4,  5, −6.

b)    xy + z.    x, −y,  z.

c)   (a + b) − (cde) − (f + g).

(a + b),  −(cde),  −(f + g).
The parentheses indicate that we must treat what they enclose as one number.

The rules for removing parentheses

Parentheses will be preceded either by a plus sign +

a + (bc + d)

or a minus sign −

a − (bc + d).

When parentheses are preceded by a plus sign +
simply remove them.  Nothing changes.

a + (bc + d) = a + bc + d.

When parentheses are preceded by a minus sign −
change the sign of every term within the parentheses.
Change + to − and − to + .

a − (bc + d) = ab + cd.

The sign of b within the parentheses is understood to be + .  Therefore upon removing the parentheses, that term becomes −b.

c within the parentheses becomes +c.  And +d becomes −d.

We can justify these two possibilities with examples from arithmetic, because algebra is abstracted -- taken from -- arithmetic.

For example, here is how we could calculate 256 + 98:

  256 + 98  =  256 + 100 − 2
 
   =  356 − 2
 
   =  354.
That is,  
 
  256 + (100 − 2)  =  256 + 100 − 2.
 
When we remove those parentheses, nothing changes.
 
And here is how to calculate 256 − 98:
 
  256 − 98  =  256 − 100 + 2
 
   =  156 + 2
 
   =  158.
 
That is,  
 
  256 − (100 − 2)  =  256 − 100 + 2.
 
When we remove those parentheses, the sign of every term
within the parentheses changes.

Problem 2.   Remove the parentheses.

a)   p + (qr + s) = p + qr + s

b)   p − (qr + s) = pq + rs

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
 
    =   xy + 1.

The sign preceding  (x − 3)  is understood to be + . Therefore the signs within those parentheses do not change.

But the sign preceding (y − 4) is minus.  Therefore, y changes to −y, and −4 changes to +4.

Finally, it is the style in algebra to to write the literal terms, xy, to the left of the numerical term.

   Problem 3.  (x + 2) + (y + 8)   =   x + 2 + y + 8
 
    =   x + y + 10.
   Problem 4.  (x + 2) − (y + 8)   =   x + 2 − y − 8
 
    =   xy − 6.
   Problem 5.  (x − 2) + (y + 8)   =   x − 2 + y + 8
 
    =   x + y + 6.
   Problem 6.  (x − 2) − (y + 8)   =   x − 2 − y − 8
 
    =   xy − 10.
   Problem 7.  (x − 2) − (y − 8)   =   x − 2 − y + 8
 
    =   xy + 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 + bc + 8.
   Problem 10.  (a − 5) − (b + 6) − (c − 9)   =   a − 5 − b − 6 − c + 9
 
    =   abc − 2.
  Problem 11.  (a + 2) − (b − 3) + (c − 8) − (d + 1)
 
    =   a + 2 − b + 3 + c − 8 − d − 1
 
    =   ab + cd − 4.

Again, when there is a minus sign before parentheses, every sign within them changes.  We saw that before in the rule of Lesson 3:

a − (−b)  =  a + b.

   Problem 12.   −(−x + y)   =   xy.
   Problem 13.   −(xy)   =   x + y.
   Problem 14.   −(x + y − 2)   =   xy + 2.

Problem 15.   Write the negative of  ab + cd.

a + bc + d.

Example 1.  Placing parentheses.  The rules of algebra go both ways. Therefore, since we may remove parentheses, we may also place them. We may write  ab + cd  in the following ways:

a − (bc + d),  (ab) − (−c + d),  a − (bc) − d.

Problem 16.   Rewrite each of the following by placing parentheses.

a)  −x + y = −(xy).

b)   −xy =   −(x + y)

c)   −a + bc + d =   −(ab + cd).

d)   Place parentheses around b and c:

ab + cd =   a − (bc) − d.

Brackets and braces

Brackets  [   ]  and braces  { }  have the same function as parentheses.  They are all grouping symbols.  After parentheses are used, then for clarity we use brackets.  Ater brackets, braces.

Removing brackets or braces will follow the same rules for removing parentheses.

Example 2.   a − [b − (cd + 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 − (cd + e)] = ab + (cd + e).

Within the brackets, there are two terms.  The first term is b.  The second term is −(cd + e).  (See Problem 1c above.)  Since the brackets are preceded by  − , the sign of each of the two terms changes.  The signs within the term (cd + e) do not change.  

Finally, we remove the parentheses, which are preceded by + :

  = ab + cd + e.

Now let us do this same problem by removing the parentheses first:

a − [b − (cd + e)] = a − [bc + de]
 
  = ab + cd + e.

Since the parentheses are preceded by  − , every sign within them changes.  And since the brackets also are 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 + xyz.

     First remove the parentheses, then remove the brackets.

w + [x − (y + z)]  =   w + [xyz] = w + xyz.

b)   First remove the brackets, then remove the parentheses.

w − [x + (yz)]  =  wx − (yz)  =  wxy + z.

     First remove the parentheses, then remove the brackets.

w − [x + (yz)]  =  w − [x + yz]  = wxy + z.

c)   First remove the brackets, then remove the parentheses.

w − [x − (y + z)]  =  wx + (y + z)  = wx + y + z.

     First remove the parentheses, then remove the brackets.

w − [x − (y + z)]  =  w − [xyz]  = wx + y + z.

d)   First remove the brackets, then remove the parentheses.

w + [x − (yz)]  =  w + x − (yz)  = w + xy + z.

     First remove the parentheses, then remove the brackets.

w + [x − (yz)]  =  w + [xy + z]  = w + xy + 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 + cabcd]}
 
  = 3a − {a + b − [−d]}
 
  = 3a − {a + b + d}
 
  = 3aabd
 
  = 2abd.

2nd Level

Next Lesson:  Adding like terms


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