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A L G E B R A

8

WHEN NUMBERS ARE ADDED OR SUBTRACTED, we call them terms. Like terms are terms that we could combine. Here, for example, is a sum of like terms:

5x + 3x − 2x.

How many x 's are there? 5 + 3 − 2 equals 6 of them. We write:

5x + 3x − 2x = 6x.

We call 5, 3, and −2 the coefficients respectively of each term. Recall that in naming terms we include the minus sign.

Here, on the other hand, is a sum of unlike terms:

x² − 2xy + y²

There is no way that we could combine them.

The most elementary terms that we could combine are a and −a.

5 + (−5) = 0.

In this sum—

2x − 3y + 4x + 5y

—the like terms are 2x and 4x,  y and −5y.

2x + 3y + 4x − 5y = 6x − 2y.

That has the effect of reducing the number of terms. Which is what we like.

We say that there are two terms "in" x and two "in" y. The preposition "in" indicates which are the like terms.

See Problem 12.

To add like terms, add their coefficients. The order in which the terms are written does not matter.

Problem 1 .   6a − 3b + cd.

a)  What number is the coefficient of a ?

6

b)  What number is the coefficient of b ?  −3

c)  What number is the coefficient of c ?   1.  For c  =  1·c.

d)  What number is the coefficient of d ?   −1.  −d  =  −1·d.

See Lesson 5.

It is the style in algebra not to write the coefficients 1 or −1.

Actually, the coefficient of any factor is all the remaining factors. Thus in the term 4ab, the coefficient of a is 4b; the coefficient of 4a is b; and so on. In this term, x(x − 1), the coefficient of (x − 1) is x.

Problem 2.   In the expression 5ayx, name the coefficient of

a)  x  5ay          b)  y  5ax          c)  yx  5a

d)  5a  xy          e)  5  ayx

Problem 3.   In this product  2(x + y)z

a)  name each factor.   2, (x + y), z

b)  name the coefficient of z.   2(x + y)

c)  name the coefficient of (x + y).   2z

Problem 4.   What number is the coefficient of x?

 a) x2 12 Compare Lesson 6, Problem 7b.
 b) 3x 4 34 3x 4 = 34 · x Lesson 4.

Problem 5.   How do we add like terms?

 a)  6x + 2x = 8x b)  6x − 2x = 4x c)  5x + x = 6x d)  5x − x = 4x e)  −4x + 5x = x f)  4x − 5x = −x Again, we do not to write the coefficients 1 or −1. g)  −5x − 3x = −8x h)  −x − x = −2x

i)  −3x − 4 + 2x + 6  = −x + 2

j)  x − 2 − 4x − 5  = −3x − 7

k)  4x + y − 2x + y = 2x + 2y

l)  3xy − 8x + 2y  = −5x + y

m)  4x² − 5x² + x² = 0

a)   2a + 3b   These are not like terms. The literals are different.

b)   2a + 3b + 4a − 5ab  = 6a + 3b − 5ab.
Terms that you cannot combine, simply rewrite.

Problem 8.    Remove parentheses and add like terms.

 a)   (2a − 3b + c) + (5a − 6b + c) = 2a − 3b + c + 5a − 6b + c = 7a − 9b + 2c
 b)    (a + 2b + 4c − 3d) − (3a − 8b − 2c + d) = a + 2b + 4c − 3d −3a + 8b + 2c − d = −2a + 10b + 6c − 4d
 c)    (4x − 3y) + (3y − 5x) + (5z − 4x) = 4x − 3y + 3y − 5x + 5z − 4x = −5x + 5z
 d)    (5xy − 3x + 2y − 1) − (2xy − 7x − 8y + 6) = 5xy − 3x + 2y − 1 −2xy + 7x + 8y − 6 = 3xy + 4x + 10y − 7
 e)   (x − y) − (y + xy − x) − (2x − 4xy − 2y) = x − y −y − xy + x − 2x + 4xy + 2y = 3xy
 f)   (4x² − 7x − 3) − (x² − 4x + 1) = 4x² − 7x − 3 − x² + 4x − 1 = 3x² − 3x − 4
 g)   (6x3 + 4x² − 2x − 6) − (2x3 − 8x² + x − 2) = 6x3 + 4x² − 2x − 6 − 2x3 + 8x² − x + 2 = 4x3 + 12x² − 3x − 4
 h)   (x² + x + 1) + (2x² + 2x + 2) − (x² − x − 1) = x² + x + 1 + 2x² + 2x + 2 − x² + x + 1 = 2x² + 4x + 4

Problem 9.   5abc + 2cba.  Are those like terms?

Yes.  The order of factors does not matter.
Upon combining them, we get 7abc.
When writing the final answer, it is conventional to preserve the alphabetical order.

 a) 4xy − 9yx  = −5xy b) 8x − 5xy − 4x + 4yx  = 4x − xy

c)   9xyz + 3yzx + 5zxy  = 17xyz

d)   3xy − 4xyz + 3x − 8yx + 5yzx − 9x  = −5xy + xyz − 6x

a)   2n + 2 − n =  n + 2

b)   n − 2 − 3n + 1 =  −2n − 1

c)   2n + 4 − 2n − 2 = 2

Problem 12.   Add like terms, which are in (x + 2). Do not remove parentheses.

a)  3(x + 2) + 7(x + 2) =  10(x + 2).

b)  2(x + 2) − 5(x + 2) =  −3(x + 2).

c)  x(x + 2) + 4(x + 2) =  (x + 4)(x + 2).

d)  x(x + 2) − (x + 2) =  (x − 1)(x + 2).

Problem 13.   Add like terms, which are in x or y. Add the coefficients.

a)   px + qx = (p + q)x.

b)   ax + bycx + dy = (ac)x + (b +d)y.

c)   x + ax = (1 + a)x.      d)   axx = (a − 1)x.

e)   (a + b)x + cx = (a + b + c)x.

f)   (ab)xcx = (abc)x.

f)   (a + b)x − (b + a)x = 0.

a)  3a2b3 − 2ab2 + a3b2 − 5b2a + b3a2 = 4a2b3 − 7ab2 + a3b2.

b)   xy2xy + x2yy2x + 2yx2 + yx = 3x2y.

*

In calculus, the student will not see any problem in the form "Subtract a from b." However, in certain standard exams that form tends to come up. Hence, the following rule.

The rule for subtraction

"Subtract a from b."  Is that  ab  or  ba ?

It is  ba.   a is the number being subtracted.  It is called the subtrahend.  The subtrahend appears to the right of the minus sign -- before the word "from."

Example.   Subtract  2x − 3  from  5x − 4

Solution.   2x − 3 is the subtrahend.

 (5x − 4) − (2x − 3) = 5x − 4 − 2x + 3 = 3x − 1.

Notice:  The signs of the subtrahend change.

2x − 3  changes to  −2x + 3.

We can therefore state the following rule for subtraction.

Change the signs of all the terms in the subtrahend.

Problem 15.   Subtract  4a − 2b  from  a + 3b.

Change the signs of the subtrahend, and add:

a + 3b − 4a + 2b = −3a + 5b.

Problem 16.   Subtract  x² − 5x + 7  from  3x² − 8x − 2.

 3x² − 8x − 2 − x² + 5x − 7 = 2x² − 3x − 9. Next Lesson:  Linear equations

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