Wednesday, 16 December 2015

Unit 5 sec 2.3 collecting like terms

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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