It is normal when teaching mathematics to start with Arithmetic and to leave Algebra until later as an advanced study. In truth this is a curious approach because Algebra explores the Rules of Mathematics and it is the basis of arithmetical techniques.
When "solving" a problem using Algebra the purpose is not to obtain a numerical answer but to derive a method for solving that problem. Consider for example the problem: A train leaves London for Bristol; how fast does it travel?
Those skilled only in Arithmetic might well reply with: “How long is a piece of string?”. It is not possible to provide an answer unless we are given:
(a) the distance between London and Bristol
(b) the time taken to make the journey.
Algebra can provide a solution in the form of a formula; when the missing information becomes available the figures can be entered into the formula and thc answer is quickly available.
The first step in solving the above problem is to provide substitutes for the missing items and this is the source of that much-parodied phrase: “Let x be the unknown quantity”.
Let the distance from London to Bristol be d miles
Let tim time taken for the journey be T hours
Speed is expressed as distance traveled in a unit of time; for example miles per hour, metres per second
(Note: it is standard practice to use d to represent distance and T to represent time but the units - miles, metres, hours, seconds - must be given.)
It is now possible to write:
If in T hours the train travels d miles
Then in 1 hour (T/T hours) it must travel d/T miles (divide both quantities by T )
i.e. (that is) the speed of the train is d/T miles per hour (m.p.h.).
Although the speed is expressed here as m.p.h. it is, in fact, a general solution; for example if we write:
" in T seconds the train travels d metres" then the speed is again given by d/T BUT the units are metres per second (m.p.s).
To be really scientific about it there is one more step:
Let the speed of the train (in m.p.h.) be represented by v
Thus v = d/T ....................................... (1)
(Note: (i) v is specified in m.p.h. because d is
in miles and T is in hours.
(ii) d/T is
often read as "d by T".)
This final result is known as a formula albeit a very simple one: its usefulness lies in the fact that it is a general solution. In this instance it was derived by considering a hypothetical train on a somewhat facetious journey but it is equally applicable to any object in motion.
For example: If a man walks a distance of 15 miles in three hours then the quantity d becomes 15 miles and the quantity T becomes 3 hours; it is now possible to apply Arithmetic to the formula to give his walking speed as
v = d/T = 15/3 = 5 m.p.h.
These formulae are called equalities because they contain an "equals” sign; the term means simply that the expression on the left of the equals sign is equal to the expression on the right of the equals sign.
NOTE that, in deriving the above formula, it is important to put on your thinking cap and to set out the propositions in the form that will enable you to derive the quantity you seek. The question asked was ‘How fast does the train travel?’; the train's speed is distance/time and so we started with the distance, moved on to the time and then derived speed as distance divided by time.
The above discussion derived a simple formula which expressed speed in terms of a distance and the time required to cover that distance but, in the practical world, things are not necessarily done in that order.
In the first example above the train driver had a speedometer which tells him his speed in m.p.h. and he will be well aware of the distance to be covered; his problem more likely is to work out how long will be required to complete the journey because he wishes to estimate. his arrival time. In other words he does not want speed in terms of time and distance: he wants time in terms of speed and distance. The purpose here is to understand the methods which are used to manipulate equations to obtain any one of the quantities in terms of the others.
The formula far speed involves dividing one quantity by another but another formula might be concerned with adding one quantity to another. For example: a man starts 5 miles from his Home and walks x-miles directly away from Home; what is his final distance from that home?
Let the final distance be D (miles) : then
D = 5 + x .................. (2)
Alternatively, if he walked toward his Home, his final distance D must be less than his starting distance (5 miles) and the equation becomes
D = 5 - x ............... (3)
By comparing equations 2 and 3 you might well deduce that reversing the direction of walking is shown by changing the sign of x ? (If you really want to stir the pudding this can be expressed by saying that the man is walking away from Home by a distance of -5 miles ?)
We can do anything we like to equalities provided that we do the same thing to both sides of the equation thereby preserving the equalness. Thus, by subtracting 5 from each side of equation-2:
D - 5 = 5 - 5 + x
or D - 5 = x ...........................(4)
This is a formula which gives the distance walked (x) in terms of the starting distance (5 miles) and the final distance (D miles); x is expressed as a difference of two distances.
The above is a very simple manipulation in that it involves only three small quantities; where the expression on one side (or on both sides) of the equals sign is complicated it is possible to make errors and so it is useful to develop a Rule-of-thumb which will help to eliminate such errors.
To obtain x from equation-2 it was necessary to get rid of the number 5 from the right-hand side (the r.h.s.) of the equality. By inspecting equation-4 it is seen that, in effect, the figure 5 was taken across to the other side of the equality and its sign changed; i.e. +5 on the r.h.s. has become -5 on the l.h.s.
RULE Quantities to be added or subtracted can be simply transferred across the equality provided their sign is changed.
Equation-1 involves dividing one quantity by another and the above Rule-of-thumb is not applicable; the basic rule still applies however that, whatever operation is performed on one side of an equality, the same operation must be performed on the other.
It was proposed that the Train Driver wishes to find the time T which is on the bottom line of the fraction d/t in equation-1. A fraction is a way of writing down a portion of some whole quantity.
The bottom line of a fraction is known as the denominator because it denominates (or names) the size of the portions into which a whole unit must be divided; for example a 2 on the bottom line tells you to divide it into two equal parts (halves), a 5 tells you to divide it into 5 equal parts.
The top line of a fraction is known as the numerator because it specifies (enumerates) the number of those portions that must be taken; for example 3/4 means divide it into four equal quarters and select a total of three quarters.
RULE Divide by the expression on the bottom line and multiply by the expression on the top line.
The Train Driver's problem is to extract T from the equation for speed:
v = d/T ............................ (1)
Multiply both sides by T:
v x T = d x T/T .......... (5)
If you multiply by T and then divide by T you should not expect to get anywhere; we say that the two T 's, one on the top line and one on the bottom line, cancel. Put another way T divided by T equals 1. Thus
v x T = d ...................... (6)
However the train Driver knows the value of the distance d; it is T that he needs. The process is now repeated as it were in reverse by dividing each side by v:
T x v/v = d/v
or
T (x 1) = d/v .............. (7)
Once again if this manipulation is attempted on an equation with very complex expressions there is a considerable possibility of error; a second rule-of-thumb can be obtained by comparing equation-1 with equation-7:
RULE Quantities to be multiplied or divided can be transferred across the equality provided that they are interchanged between the top line and bottom line of a fraction.
(Remember that a quantity such as T can be represented as a fraction by writing T / 1 ; i.e. divide the whole into 1-only equal parts.)
Thus to obtain equation-7 from equation-1 it is much simpler (and therefore safer) just to interchange the quantities T and v; across the equals-sign the T goes up to the top line and the v goes down to the bottom line.
This rule-of-thumb method of manipulating an equation which involves multiplications and divisions is known as cross-multiplication. If it sounds like cheating remember that it is legitimate because it helps to avoid errors and also because it is soundly based in that it omits only the more difficult "middle bit".
It is tiresome to write, and also difficult to read, a script that constantly spells out words such as “multiplication” or “divided by” and a notation is necessary. In mathematics there are several different notations in use for the same operations; sometimes they duplicate each other but each does have its own particular niche.
In multiplication the symbol x is used universally between the two quantities involved such as
156 x 27
This is satisfactory enough in arithmetic but leads to a deal of confusion in algebra where the symbol x might also be used to represent an unknown quantity. One method used to overcome this is to reserve the lower-case written x (which appears as two c’s back to back) for the unknown.
Additionally the x may be replaced by a full-stop, by round brackets or simply
omitted altogether. Thus the expression "a multiplied by b" may be
written as any of the following:
a x b a(b) b(a) (a)b (b)a a.b ab
The brackets are seldom used in the simple form shown here. The form a(b) means multiply whatever is within the brackets by a. Consider for example a(p + q - s) ; this is read as “a into p plus q minus s”; to evaluate this when given values for p, q and s work out first the additions and differences within the brackets and then perform the multiplication.
In the computer world there is one other common variation because computers are unable to differentiate between “x” meaning a letter, “x” meaning multiply-by and “x” used in place of an unknown quantity. The symbol most often used to represent the operation of multiplying is the asterisk; thus, in a computer program, you find expressions such as 5*L. It is worth noting perhaps that, to get a computer to do pure algebra, is an extremely complicated business that quickly reveals how much we take for granted.
The standard symbol for division is not very convenient in use and it is most often replaced by the solidus (or slash in the USA). The problem however is one of order; there is no ambiguity in
a x b/c (or ab/c or even a*b/c)
but how should we interpret
a/c x b ?
It is in this kind of problem that the correct use of order and (where necessary) brackets becomes important; the possibilities are:
a/bc a/(b x c) a/c x b/1 = a x b/c (or ab/c)
Confusions such as those shown above are obviated in computer programming by adopting a strict order of precedence. However, unless you are well practised in programming that particular computer, it is not easy to remember the relevant order and so the proper use of brackets is safer and quicker.
[The word program is well established in the computer world where it refers to so-called “software” or the written instructions that tell a computer exactly what it is required to do.]
(i) If p + q = 7 what is the value of q?
Take the p to the r.h.s of the equation and change its sign:
q = 7 - p (or q = -p + 7)
Check this result by using the full step-by-step technique.
(ii) If x - ay = p - q what is the value of x?
Take ay to the r.h.s. with a change of sign:
x = p - q + ay
(iii) If x - ay = p - q what is the value of y? ( Watch those minus signs!)
Take x to the r.h.s. with a change of sign:
- ay = p - q - x
Take -a to the r.h.s. and on to the bottom line:
y = (p - q - x) / (-a)
(Note the use of brackets to keep together all the symbols which belong together; without the brackets around -a there is the possibility of error in that the symbol a might be subtracted from the expression instead of being divided into it.)
(iv) If x/y = p/q what is the value of x/p and of q/p ?
Cross-multiply with y and then with p:
x / p = y / q
Alternatively cross-multiply with all four symbols:
x p q y
- = - becomes -- = —
y q p x
(This introduces a useful extension of the Rule above that what is done to one side must also be done to the other; simply turn the whole equation upside down.)
(v) If 3a - b + c = 2p what is the value of a ?
Take both b and c to the r.h.s. with a change of sign
3a = 2p + b - c
Divide everything by 3 (divide throughout by 3) ... take the 3 across to the bottom line:
a = (2p + b - c)/3
(vi) If 3(a - b + c) = 2p what is the value of a?
First remove the bracket so that a is released from b and c:
the expression 3(a - b + c) means three lots of (a - b + c) which is the same as 3a - 3b + 3c (multiply everything within the bracket by 3) hence
3a - 3b + 3c = 2p
Take the terms in b and c across to the r.h.s.:
3a = 2p + 3b - 3c
a = (2p+3b-3c) / 3
or, if you prefer,
a = 2p/3 + b - c
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