Calculations with sums and differences

The difference 2·3 - 9:3 involves the product 2·3 and the quotient 9:3. You must first multiply and divide: 2·3 - 9:3 = 6 - 3 = 3. Rule:

Actions of a higher order precede those of lower order - multiplications and divisions precede additions and subtractions.

Examples:

If calculations are to follow a given path, you must indicate it by means of brackets. For example, if the sum 6 + 3 = 9 is to be multiplied by 5, you must write

(6 + 3)·5 = 9·5 = 45.

You have learned already that you must first compute the expressions inside brackets!.

Examples:

A bracket indicates that the quantities inside belong together and represent a single quantity. The general expression (a + b)x becomes ax + bx.

Addition of sums and differences

Task: Someone has already paid 100 Baht and still owes 50 Baht. He can settle his debt in a single payment or step by step, for example, 25 Baht, 15 Baht and 10 Baht.

Task: Siriporn has 11 Baht and Vichai must give her 3 Baht. He could do so, but does not have the right change and gives her 5 Baht, whence she must return to him 2 Baht.

These relations lead to the Rule:

If a plus sign is in front of a bracket, it can be omitted.

The brackets can contain any number of terms. Also, you know from Arithmetic, how to remove brackets.

Example: Compute 36 + 17.

First add 10 to 36, then 7, that is, decompose 17 into the sum (10 + 7):

Example: Compute 45 + 19, using round-off to 10!

Add first 20 to 45 and then subtract 1:

45 + (20 - 1) = 45 + 20 -1 = 64.

Example: 2a + 3b + (6a + 3b + 2c) + (4a - 3b - 2c) = 2a + 3b + 6a + 3b+ 2c + 4a - 3b - 2c = 2a + 6a + 4a + 3b + 3b + 2c - 2c = 12a + 3b.

Subtraction of sums and differences

Siriporn has 90 Satang and must give one friend 30 Satang, another friend 10 Satang. She can give 40 Satang to the other friend, who will settle her debt; she can also pay her debt without such a help. The final result is the same, whether she pays 40 Satang in one go or separately:

Siriporn has 70 Baht and owes 40 Baht. She has no change and gives a 50 Baht note and retains 20 Baht. She receives 10 Baht change and has then 30 Baht.

Again there is a simple Rule; however, you must keep in mind that in brackets, for example in (3 + 4), the leading term 3 is +3, or in (b - c) the leading term b is +b:

If there is a minus sign in front of brackets and you remove the brackets, you must change the signs of the individual terms.

If you change in a - (b - c) the signs of the terms in brackets, then -(+b) and -(-c) become -b and +c.

Examples from Arithmetic: In order to compute 27 - 13 by counting backwards, you subtract at first 10 from 27 and obtain 17, and then subtract 3 to obtain 14. The number 13 is decomposed into the sum 10 + 3.

27 - (10 + 3) = 27 - 10 - 3.

For example, in order to compute 126 - 99, you can make use round off to 10 and subtract at first 100, which yields 26. You have now subtracted 1 too much and must add it:

126 - (100 - 1) = 126 - 100 + 1 = 27.

Examples:

If an equation contains brackets, you must first open them.

Multiple brackets

2a - [6a - (4a - 4b)] - [(4a - 4b) - (3a - 6b)]

You use brackets to relate the expressions in brackets to other expressions. For this purpose, you use different forms of brackets: ( ), [ ], { }, etc. inward outward in the indicated order. You can open the brackets up in any order, but it is recommended that you start from the inside, that is, in the above example, you open first the round brackets ( ) and then the square brackets [ ]:

2a - [6a - 4a + 4b] - [4a - 4b - 3a + 6b] =
2a - [2a + 4b] - [a + 2b] = 2a - [2a + 4b + a + 2b] = 2a - [3a + 6b]=
= 2a - 3a - 6b = -a - 6b.

Inversions

You often must convert an equation without brackets into one with brackets. In order to add 37 + 25 + 5, you combine 25 + 5 and add 30 to 37. This is the inverse of the earlier Rule.

You know from your lessons in Geometry, that inverse rules are not always valid and must first be proved. The sentence "if it rains, it is wet" does not imply "if it is wet, it rains". The inversion of a sentence into its opposite is often employed by skillful speakers, in order to ridicule opponents in a less critical audience. In Arithmetic, we must also deal with inversion of rules, but the conditions are especially simple.

If numerical relations can be given the form of an equation, the inverse always applies.

The proof employs the fact that the sides of an equation can be interchanged. The preceding rules thus lead to the inverse Rules:

For a positive bracket, its terms retain their original signs,
for a negative bracket, its terms must be given the opposite sign.

Examples

You combine numbers with equal signs by forming their sum and giving it the common sign.

Examples

If there are additions and subtractions, you also have a simple Rule:

Numbers with different signs are combined by subtraction of the smaller one from the larger one and placing outside the brackets the sign of the larger term.

Examples

Multiplication of a sum or difference by a number

Seven children are given pictures, at first 2 and then 3, that is, altogether 2·7+3·7=35. In this distribution, each child could have received directly 2+3=5 pictures.

The bracket on the left hand side is to indicate that there occurs at first the addition 2 + 3 = 5 and then the multiplication by 7. On the right hand side, multiplication occurs before addition.

Each of 7 children has 8 Baht and spends 2 Baht. How many Baht do they still have all of them together? You can compute the answer in different ways:

1. The children own altogether 8·7 = 56 Baht and spend 2·7 = 14 Baht, so that they retain 8·7 - 2·7 = 56 - 14 = 42 Baht.

2. Each child retains (8 - 2) Baht, that is, they altogether retain (8 - 2)·7=42Baht. You obtain the same result, whence

(8 - 2)·7 = 8·7 - 2·7.

These relations can also be derived, based on the definition of multiplication. You have that a·3=a+a+a as well as

(a + b)·3 = a + b + a + b + a + b

or, alternatively,

Similarly,

(a-b)·2 = a-b + a-b = a+a - b-b = 2a - 2b
(a - b)·n = an - bn,
(a-b-c)·3 = a-b+c + a-b+c + a-b+c = a+a+a - b-b-b + c+c+c = 3a - 3b +3c
(a - b + c)·n = an - bn + cn.

Since the sequence of the factors is arbitrary, you also have

n·(a - b + c) = an - bn + cn.

If an expression with many terms is multiplied by a number,
each of its terms is multiplied by that number.

You have used this rule in Arithmetic. In the task 123·3, you compute first 100·3, then 20·3, then 3·3 and add these products.

(100 + 20 + 3)·3 = 100·3 +20·3 +3·3.

In order to compute 79·6, you use round-off to 10 and proceed as follows

(80 - 1)·6 = 80·6 - 1·6 = 480 - 6 = 474.

Examples

1. m - a(b - c)

Operations of the second kind take place before those of the first kind, whence you first form the product a(b - c) = ab - ac and then subtract it from m.

m - a(b - c) = m - [ab - ac] = m - ab + ac.

2. 6a + 2(2b - 2a) - (b - 3a ) ·3 = 6a + [4b - 4a] - [3b - 9a] =
= 6a + 4b - 4a - 3b + 9a = 11a + b.

Products of sums and differences

In order to work out the product (a + b)(c + d), you set provisionally a + b = x and obtain

x(c + d) = xc + xd = (a + b)·c + (a + b)·d = [ac + bc] + [ad + bd]:

(a + b)·(c + d) = ac + bc + ad + bd.

You meet here what you have learned in Arithmetic. The product 32·23 is computed as follows:

(30 + 2)·(20 + 3) = 30·20 + 2·20 + 30·3 + 2·3.

In a similar manner you can convert other products into sums:

in (a - b)·(c + d), set a - b = x:
x(c + d) = xc + xd = (a - b)·c + (a - b)·d = [ac - bc] + [ad - bd]
(a - b)(c + d) = ac - bc + ad - bd.

in (a - b)·(c - d), set a + b = x:
x(c - d) = xc - xd = (a + b)·c - (a - b)·d = [ac - bc] + [ad - bd]
(a - b)·(c - d) = ac - bc - ad + bd.

These formulae can be extended to brackets with more terms:

in (a - b - c)·(d + e), set a - b - c = x:
x·(d + e) = xd + xe = (a - b - c)·d + (a - b - c)·e = ad - bd - cd + ae - be - ce.

If you interpret in (a - b) the number a as the additive term +a, all these multiplications are combined in the Rule:

Two brackets with several terms are multiplied by forming the products of each term with each term and inserting + if the signs of the terms are equal and - if they differ.

During written multiplication, you have followed a definite sequence of steps, in order not to overlook terms; you should also adopt here a sequence which is similar to that rule:

Examples

You can place all factors side by side.

3. (2x + y)·(x - 2y)·(2x + 3y)

You multiply first two factors (in brackets) and then the result by the third factor (in brackets):

(2x + y)(x - 2y) = 2x² + xy - 4xy - 2y² = 2x² - 3xy - 2y²

(2x²-3xy-2y²)(2x+3y)=4x²-6x²y-4xy²+6x²y-9xy²-6y²=4x²-13xy²-6y² .

When you check your computations, you change the order of the operations (cf. Checking results); in this case, you open the brackets in a different manner. Also, evaluation of a task and the result for different values of x provides a check.

Equation: 4x - 11 = (2x - 3)(4x + 4) - (4x + 3)(2x - 3)

4x - 11 = [8x² - 12x + 8x - 12] - [8x² + 6x - 12x - 9]

Be careful and do not take two steps at a time, but retain the brackets!

4x - 11 = 8x² - 12x + 8x - 12 - 8x² - 6x + 12x + 9

Instead of evaluating m - a(b - c) = m - [ab - ac], you can, taking the sign of x into account, write down directly the result. a has the sign -, b the sign +, different signs yield the product -ab; moreover, a has the sign -, c the sign -, equal signs yield +ac.

m - a(b - c) = m - ab + ac.

Examples

Important formulae

If you now want to evaluate (x + 1)², you no longer start from (x + 1)(x + 1), as during evaluations with numbers, but substitute a = x, b = 1 and find

(x + 1)² = x² + 2x + 1² = x² + 2x + 1.

Example: (2x - 3y)²

In (a - b)² = a² - 2ab + b², let a = 2x and b = 3y, then

Equation:

More Formulae

Inversion. Transformation of a sum or difference into a product

c(a + b) = ac + bc or c(a - b) = ac - bc imply inversely ac + bc = c(a + b) and ac - bc = c(a - b), respectively.

The same conclusion applies to multi-term expressions.

If terms in sums or differences have a common factor,
it can be extracted by means of brackets.

Subsequent multiplication confirms whether the task was solved correctly. Watch the sign rules!

Examples:

Multiple brackets

21 + 3y + 7x + xy = 3(7 + y) + x(7 + y) = (7 + y)(3 + x),

also

21 + 3y + 7x + xy = 7(3 + x) + y(3 + x) = (3 + x)(7 + y).

Applications of formulae

Formulae may let you convert a sum into a product. For example, compare x²+4x+4 with the left hand side of a²+2ab+b²=(a+b)². Let x = a and 4x = 2ab, then b must be 2 and therefore x² + 4x + 4 = (x + 2)².

Examples

Equation:

Expansion and combination yields:

(x + a)(x + b) = x² + (a + b)x + ab.

The coefficient of x becomes a sum, the single term the product of a and b. You can now examine a given expression to find out whether these relations are valid with due consideration of the given sign rules.

x² - 7x + 12 = (x - 3)(x - 4).

Division of a sum or difference by a number

During an excursion, Siriporn has spent on behalf of 7 friends 14 Baht and later on yet another 28 Baht. How much must each of them give back to her? Let their shares be equal?

You can compute 42/7 = 6 or 14/7+28/7 = 6, so that (14+28)/7 = 14/7+28/7.

Let a/n denote the number, which on multiplication by n yields a, whence (a+b)/n is a number which yields a+b on multiplication by n. If you now also multiply a/n+b/n by n, you obtain (a/n+b/n)·n=(a/n)n+(b/n)n, which is the same result n = a + b, whence

(a + b)/n = a/n + b/n as well as (a - b)/n = a/n - b/n.

The same derivation applies to expressions with more terms

(a - b - c - d)/n = a/n - b/n - c/n - d/n.

An expression with several terms is divided by a number
by dividing each term by that number.

In Arithmetic you have learned: Compute 69:3 by setting 69 = 60 + 9 and then dividing each term by 3:

69/3 = (60 + 9)/3 = 60/3 + 9/3.

You proceed in a similar manner with the advantage of round-off to 10:

998/2 = (1000 - 2)/2 = 1000/2 - 2/2.

Watch your mode of writing! It may be preferable to use the division sign / instead of the sign :.

(a + b)/c = (a + b) : c.

You must use brackets here, because a + b : c has a different meaning.

Division by a sum or difference

It will be best, if we start here from what you have learned in Arithmetic. How do you compute 682 : 22?

(600 + 80 + 2) : (20 + 2) = 30 + 1				(6·10² + 8·10 + 2):(2 ·10 + 2) = 3·10 + 1
-600 60				-6·10²6·101
		or
20 + 2				2·101 + 2
-20 2				-2·101 2

				dividend	divisor
If you replace the number 10 by z		then		(6z² + 8z + 2)	:(2z + 2)	= 3z + 1
				-6z² 6z
				2z + 2
				-2z 2

You see that you must arrange the dividend and divisor according to decreasing powers of z. The calculation then involves every time the stages:

1. Divide the first term of the dividend (6z²) by the first term of the divisor (2z) and write down the result (3z) next to the equality sign.
2. Multiply this result (3z) by the divisor (2z + 2).
3. Subtract this product (6z² + 6z) from the dividend ab.
4. If there is a remainder, the division proceeds as before.

This method can also be applied to expressions involving subtractions:

(5x4+6x³-4x²+208x):(x+4)	=	5x² - 14x + 52x
-5x4 20x³
		Subtraction of the first result 5x²+20x of the division
-14x³ - 4x²		yields -14x³-4x²; now you must divide by
14x³ 56x²		first term of the divisor, that is, x. You must then
		determine the sign, that the following multiplication
52x² + 208x		does not violate the sign rule by which equal signs
-52x² 208x		yield +, unequal signs -. The result here is -14x²

The check is the same as in normal division. If a/b = c, then cb = a, whence also

(5x³ - 14x² + 52x)·(x + 4) = 5x⁴ + 6x³ - 4x² + 208x.

If you are given expressions which you cannot arrange according to increasing or decreasing powers, you must rearrange according to the letters, indeed the dividend and divisor in the same manner.

Task: (ad + bc + bd + ac) : (b + a)!

For example, you can here arrange the terms alphabetically, so that the task becomes (ac + ad + bc + bd):(a + b) and divide:

Like in division of numbers, a remainder may arise. Division is useful during decomposition of sums into products.

Division allows you to see immediately whether a given expression contains a definite factor.

Task: Is x - 3 a factor of x² - 7x + 12?

The division (x² - 7x + 12):(x - 3) yields x - 4. There is no remainder, whence

(x² - 7x + 12) = (x - 3)·(x - 4).

Tasks

Exercises: Simplify the following expressions:

Answers

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