Unit 5 sec 2.3
collecting like terms
16 December 2015
23:27
Let us look worse. That how it works with numbers. If you
have 2 batches of 4 dots, then altogether you have
2 + 3 batches of 4 dots, that is, 5 batches of 4 dots.
Of course, this does not work, just with batches of dots.
For example
2 × 7 + 3 × 7 = 5 × 7.
In fact, no matter what number a is,
2a +3a =5a.
This gives you a way to simplify expressions that contain a
number of batches of something, added to another number of batches of the same
thing. For example, consider the expression
5bc + 4bc.
Adding 4 batches of bc to find batches of bc gives 9 batches
of bc:
5bc +4bc = (5 + 4)bc = 9bc.
Terms that are “batches of the same thing” I called like terms. 4 turns to be away terms,
the letters and the powers of the letters each term must be the same. So, for
example
7√A and 3√A R White term because they are both terms in √A;
2x² and -0.5x² I like terms because they are both terms in x².
However,
5c and 4c² are not like terms because 5c is a term in c and
4c² is a term in c².
Like terms can always be collected in a similar way to the
examples above: you just add the coefficients (including -ones). You can add
any number of white terms.
Example 2 collecting
like terms
simplify the following
expressions.
(A)
12m +15m -26m
(B)
0.5XY² + 0.1XY²
(C)
5p -p
(D)
1/3 – 2d
Solution
(A)
12m +15m -26m = (12+
15 – 26)m = 1m =m
(B)
0.5XY² + 0.1XY² = (
0.5 +0.1)XY² = 0.6XY²
(C)
5p -p = 5p -1p =(5 –
1)p = 4p
(D)
1/3 – 2d = (1/3 – 2)d
= (1/3 – 6/3)d = -5/3d
The fractional ceifficients were not converted to
approximate decimal values. In algebra, you should work with exact numbers,
such as 1/3 and √5, rather than decimal approximations, wherever possible.
However, if you are using algebra to solve practical problem, then you may have
to use decimal approximations.
It is easier to spot like terms. If you make sure that all
the electors in each term are written in alphabetical order.
Example 3.
Recognising like terms
simplify the following
expressions.
(A)
5st+2st
(B)
-6q²rp + 4prq²
Solution
(A)
5st+2st = 5st +2st =
7st
(B)
-6q²rp + 4prq² = -6q²rp
+ 4prq² = -2pq²r
The lower and upper case versions of the same letter are
different symbols in mathematics. So, for example, 4y and 9Y are not like
terms.
Any to constant terms like terms. They can be collected
using the normal rules of arithmetic.
Often an expression contains some light terms in some unlike
terms. You can simplify the expression by 1st changing the order of
its terms so that the light terms are grouped together, and then collecting the
light terms. This leaves an expression in which all the terms are, which cannot
be simplified any further. Here is an example
Example 4
collecting more like terms
simplify the following
expressions.
(A)
2a +5a – 7a + 3b
(B)
12 – 4pq – 2q +1 –
qp -2
Solution
(A)
the light terms,
then let them
2a +5b – 7a + 3b =2a -7a +5b +3b = -5a +8b
(B)
write qp as pq, group
the like terms, then collect them.
12 – 4pq – 2q +1 – qp -2= 12 – 4pq -2q +1 – pq -2
= 12+ 1 -2 – 4pq -pq – 2q
= 11 – 5pq – 2q
the terms in the final expression can be written in any
order. For example, an alternative answer is 11 – 2q – 5pq.
As you become more used to working
with expressions, you will probably find that you drink lots of white terms without
grouping them together first.
Sometimes when you collect 2 or more like terms, you find that the
result is zero - that is, the terms cancel
each other out.
Example 5 terms that Cancel out
simplify
the expression
M + 2N
+3M – 2N.
Solution
M + 2N
+3M – 2N = 4M +0 = 4M
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