Algebra Tutorial

The first topic to be reviewed is fractions. A fraction is expressed as \(\frac{a}{b}\), where \(a\) is called the numerator and \(b\), the denominator.

The addition or subtraction of fractions can be expressed generally as \(\frac{a}{b}+/- \frac{c}{d}\). In order to carry out the operation indicated it is necessary that the expression have a common denominator, \(b \times d\). This is accomplished by multiplying both terms by \(bd\), and dividing both terms by \(bd\). That is, \(\frac{a}{b} +/- \frac{c}{d}\) is equal to \(\frac{ad +/- bc}{bd}\).

As an example, what is \(1/2 + 1/3\)? Using the previous equation , where now a=1, b=2, c=1 and d=3, we obtain \(\frac{1\times 3 + 1\times 2}{2\times 3} = \frac {5}{6}\).

The product of two fractions is the product of their numerators divided by the product of their denominators. That is, \(\frac{a}{b} \times \frac {c}{d} = \frac {a\times c}{b\times d}\) .

Now we will investigate power laws. A number raised to a power is written as \(a^x\), where a is the base and \(x\) is the power or exponent.

When multiplying \(a^x\) by \(a^y\) the rule is to add the exponents, thus, \(a^x\) times \(a^y\) equals \(a^{x + y}\). This is obvious since \(a^x\) times \(a^y\) is \((x+y)\) \(a's\) multiplied together.

When dividing \(a^x\) by \(a^y\), the rule is to subtract the exponents. Thus \(a^x\) divided by \(a^y\) is equal to \(a^{x - y}\).

When raising a number to a power, to a further power, the exponents are multiplied. Thus \((a^x )^y\) is equal to \(a^{xy}\).

How do we handle expressions like \(a^x\) times \(b^y\)? Now the bases are different and the above rules no longer apply, however, if we can express \(b\) as a multiple or a power of \(a\) then the problem can be solved, as before. Suppose \(b\) can be expressed as a quantity \(n\) times \(a\). Then \(a^x\) times \(b^y\) equals \(a^x \times (na)^y\), which is equal to \(n^y\) times \(a^x\) times \(a^y\), which is equal to \(n^y\) times \(a^{x + y}\).

Alternatively it could be done this way, suppose that \(b\) could be written as \(a^n\). Then we would write \(a^x\) times \(b^y\) is equal to \(a^x\) times \((a^n)^y\) which is equal to \(a^x\) times \(a^{ny}\), which is equal to \(a^{x + ny}\).

On the other hand, if b is neither a simple multiple nor a simple power of a, then the solution is best handled by resorting to the use of logarithms.

It should be mentioned that the above rules for handling powers are valid for the exponent being ± an integer or a fraction.

Note however that any base raised to the power of 0, is 1, and that 0 can not be used as a base.

Now let's look at proportionality. Often in physical problems, one deals with parameters that vary as certain other parameters are varied. These parameters are called variables.

Let's consider two cases. The case where the parameter , x , varies directly with y, or sometimes we say x is proportional to y, \(x\propto y\) . And the case where x varies inversely with y, or sometimes we say x is inversely proportional to y, \(x \propto \frac{1}{y}\) . These expressions can be written as equalities by inserting constants of proportionality, for example \(x = ky\) and \(x= \frac{k^1}{y}\). Where k and k1 are unique constants, expressed in the appropriate units, that characterize the particular problem.

As an example, Newton's Law of Gravitation states, that the force of attraction between two bodies varies directly as the product of their masses and inversely as the square of the distance between them. That is \(F \propto \frac{m_1m_2}{r^2}\) . Or written as an equality \(F = G \frac{m_1m_2}{r^2}\) , where G is the Universal Gravitation Constant.

We will now examine the solutions of a linear equation in one unknown, remember that an equation is simply a statement of equality.

Let us suppose that we are confronted with an equation of the following form, \(ad = by + c\). Where we know values for a, b, c, and d, and wish to solve for the unknown designated y. First put the expression involving the unknown y on the left hand side of the equation, and everything else on the right hand side. In other words, transpose terms from one side of the equation to the other, remembering that terms change signs when they change sides. Thus we obtain, \(by = ad - c\). To obtain y alone on the left hand side, it is necessary to divide the left hand side by b, but to retain the equality, the right hand side must also be divided by b, giving \(y = \frac{ad-c}{b}\).

The rule is that you can do anything you want to an equation in terms of adding and subtracting terms, multiplying and dividing by terms etc, as long as you do the same thing to both sides of the equation. One word of caution however, division by 0 is not allowed.

Now how do we deal with common factors? Often the solution of an equation can be simplified by removing factors common to all terms.

For example, suppose we wish to solve the following equation for x, \(x^2-a^2 = 2(x - a)^2\). Factor both sides of the equation, giving \((x - a)(x + a) = 2(x - a)(x - a)\). Divide both sides of the equation by \((x - a)\), and we get \((x + a) = 2(x - a)\). Or finally, \(x = 3a\).

Let's now turn our attention to the dimensionality of the equations. All terms in an equation, connected by a plus or minus sign must have the same dimensions and be in the same system of units.

Consider for example, the equation for uniform motion in a straight line, from kinematics, \(v = u + at\), where v is the final, and u is the initial velocity, a is the acceleration and t is the time interval. There are three basic dimensions, M, L and T, for mass, length and time into which most physical quantities can be reduced.

For example, velocities have the dimension L/T, acceleration has the dimensions L/T2, and time of course has the dimension of T. Thus, as shown, v has the dimensions L T-1 , u has the dimensions L T-1 and, a times t has the dimensions L T-2 times t, which is L T-1. Therefore the equation \(v = u + at\) is dimensionally correct.

If you wish to substitute in numerical values for the parameters, first choose a system of units, for example the mks system (where mass is in kilograms, distance in metres and time in seconds), and then v and u have units of m/s, a has units of m/s2 and t, of course, has units of seconds.

Now let's look at the solutions of two simultaneous equations.

Two equations and two unknowns can be solved in a straight forward fashion using the techniques already developed.

Let us consider the two equations \(x + y = 3\) and \(4x - y = 2\). We wish to solve for both x and y. The first step is to eliminate one of the unknowns, this can be accomplished in several ways, I will demonstrate only one. Solve either one of the equations, for one of the unknowns in terms of the other. Consider the equation \(x + y = 3\), which can be written, \(x = 3 - y\), substitute this into \(4x - y = 2\) to get \(4(3-y) - y = 2\). Solve to get \(12 -4y -y = 2\) or \(5y = 10\) and finally \(y = 2\). Now substitute this value into the previous equation \((x=3-y)\) to get \(x = 3-2\), which equals 1. Therefore the solutions to the equation is \(x = 1\) and \(y = 2\).

Finally let us look at the binomial expansion. Often it is necessary to calculate expressions of the following form, \((1 + x)^n\), where n may be an integer or a fraction.

Let us write down the expression for several values of n, and expand.

For \(n = 0\), we would have \((1 + x)^0 = 1\),

for \(n = 1\) we have \((1 + x)^1 = 1 + x\),

and for \(n = 2\) we have \((1 + x)^2 = 1 + 2x + x^2\),

for \(n = 3\) we have \((1 + x)^3 = 1 + 3x + 3x^2 + x^3\) and so on.

For \(n = 0\), 1 is still 1, for \(n = 1\), I have written the answer as \(1 + \frac{1x}{1}\), for \(n = 2\), I have written the answer as \(1 + \frac{2x}{1}=\frac{2 \times 1 \times(x^2)}{2\times1}\).

Let's do one more, for \(n = 3\), I have written the answer as \(1+\frac{3x}{1} + \frac{3 \times 2 \times (x^2)}{2 \times 1} +\frac{3 \times 2 \times 1 \times (x^3)}{3 \times 2 \times 1}\).

You will note that there is a general format becoming evident.

The format that we saw emerging, we can write in a general way as follows,

\(n = n\) 

\((1 +x)^n = 1 +\frac{nx}{1}+\frac {n(n-1)}{2 \times 1} x^2 + \frac {n(n-1)(n-2)x^3}{3 \times 2 \times 1} + etc.\).

This equation is called the binomial expansion, if x is a number less than 1 many of the higher order terms in the expansion can be deleted, because they become negligible.

For example, suppose x = 0.01 and n = 7. If we substitute n = 7 into the binomial expansion, we'll find \((1+x)^7 = 1 + \frac {7x}{1} + \frac {7 \times 6(x^2)}{2 \times 1} + etc.\). And now with x as 0.01 these terms become \(1 + .07 + (2 \times .0001)\) and so on. If we were working to only three significant figures, only the first two terms in the expansion would contribute, and we could say that \(1 + 7x = 1.07\).

The binomial expansion can also be used if the exponent is negative, and then the general expression takes the following form ( the expansion converges for   \(x^2 < 1\)), \((1 +x)^{-n} = 1 + \frac{(-n)x}{1}+ \frac {(-n)(-n-1)x^2}{2 \times 1} + \frac {(-n)(-n-1)(-n-2)x^3}{3 \times 2 \times 1} + etc.\) .

And this becomes,

\(= 1 - \frac{nx}{1} + \frac {n(n+1)x^2}{2 \times 1}- \frac {n(n+1)(n+2)x^3}{3 \times 2 \times 1}\)  and so on.

Again, n may be an integer or a fraction.