Trevor Pythagoras Maths » Algebra

Simplifying and Solving equations with Algebraic Fractions

admin — Sun, 20 Jun 2010 13:08:34 +0000

Algebraic fractions are fractions which include variables. The objective of simplifying an algebraic fraction is to have a single fraction (ie one denominator) for the whole expression or equation and to cancel as many terms as possible. To solve equations with algebraic fractions we must first simplify them and then cross multiply (ie times the equation by the denominator so it is cancelled from one side and multiplies the other).

Example 1: simplifying single fractions

Consider

We want to find things that divide both the numerator (top) and denominator (bottom). In this case 2x does. We can then divide both the top and bottom by 2x (known as cancelling out the 2x) to gain the fraction

Note that you can only cancel things that divide the whole of the top and bottom not part of the top and bottom or things that are added to the top and bottom. For example

Sometimes the common factor that we are cancelling isn’t immediately obvious, for example if we have quadratic expressions. In these cases you should try to factorise the quadratic to see if there is anything you should cancel.

Note on quadratics: If a fraction has quadratics that can be factorised you should always leave the denominator factorised whilst it is common to multiply out the numerator (unless there is a need to leave it factorised or it is very complex)

Example 2: adding fractions

Consider the algebraic expression

Before we consider adding them we should make each of the fractions in the sum into its simplest form.

Firstly we want to have one denominator. To do this we need to find the simplest common denominator and then make the denominator of both these fractions this. This is done in a similar way in which we find the lowest common denominator when we are adding normal fractions. The easiest way to do this is to multiply numerator and the denominator of both factions by the denominator of the other so the expression becomes:

Now, as when adding normal fractions, because the denominator of both fractions is the same we can add their numerators together and put them over a single denominator to get

Example 3: Solving Equations

To solve an equation involving an algebraic fraction you should first multiply out the fractions as explained above.

Consider the equation

which is already simplified. To solve this we need to cross multiply, this involves multiplying both sides of the equation by the both the denominators and then simplifying both sides. However, because after the multiplication the denominator of each fraction will be multiplied by the numerator they will cancel out with the result that we have multiplied each side of the equation by the denominator of the other side and have removed the denominators from the fractions. This is most easily show by example

ie)

becomes

with intermediate step

We can then solve the equation using normal methods (note that if the result is a quadratic you may have to use the quadratic formula)

Also see

Finding the Inverse of a Function

admin — Wed, 20 Jan 2010 18:01:30 +0000

Before you can try and find the inverse of a function you need to determine if one exists. For an inverse to exist exactly one element of the domain must map to each element of the co-domain (though you can always re-define the co-domain to only include elements that are mapped to).

So for example f(x) = x+2 (from reals to reals) has an inverse as every element y is mapped to only by y-2 but f(x) = x² (from reals to reals) doesnt because both -1 and 1 map to 1 so how would we decide which of these would be the unique result of the inverse applied to 1 (we could of course define the domain to be the positive reals in which case it would have an inverse).

When finding the inverse of a function you are really looking to see what maps to each element of the range or codomain, to find the inverse of f you are looking for the element x in the domain for a given y in the range such that f(x) = y. This basically means you are reversing the process of the function.

Finding the Inverse when the function is a formula

When a function is given by a formula what you need to try and do is apply the operations of that formula backwards. This easiest way of doing this is to let the function f(x) = y. Now you know the formula to get from x to y so substitue this in for f(x). All that now needs to be done is to rearrange this equation so that x is the subject and the resulting rexpression on the otherside is only in terms of y. This function is the inverse of f, to show this lets denote it g so we g(y) = x.

Then g(f(x)) = g(y) = x since we started by letting f(x) = y and created g such that g(y) =x.
and f(g(y)) = f(x) = y

so g is indeed the inverse of f.

Example

This is most easily demostrated through an example.
Let
Then to find the inverse of f we first write

and rearrange as follows

So the inverse of f is given by the formula so we can write

Note that we have replaced the y’s with x’s, this doesnt matter as we can put any variable we like into the function but it important to make sure that you use the same variable as the parameter of the function and in the formula that defines it.

Also see

Why the proof 2=1 is wrong

admin — Fri, 18 Sep 2009 10:26:35 +0000

There are a number of apparent proof that 2=1. An example of one such proof is below

Suppose a=b then
a² = ab
2a²=a²+ab
2a²-2ab = a²+ab-2ab
2a²-2ab=a²-ab
2(a²-ab) = 1(a²-ab)
2 = 1

However this is clearly not true which mean there must be a step in the above algebra which isnt valid.

The invalid step

The invalid step is in fact the very last step in the proof, cancelling both sides by a²-ab. This is because a²-ab = 0 since we initially assumed a=b hence

ab = aa = a²

This then means that the second last line states

2 x 0 = 1 x 0

which is obviously true but you cant “cancel” both sides by 0 since any number divided by zero is meaningless and doesnt have a values. The trick involved in this proof is hiding the divide by zero which most people would recognise as invalid with a more complicated looking expression.

What this means

This example demonstrates the dangers of allowing divide by zeros to occur in your calculations even if its in the form of algebra. It also demonstrates the problems of considering infinty (x/0) as a number that can be used in calculations.

Also see

Proof by Mathematical Induction

admin — Wed, 27 May 2009 15:50:10 +0000

Proof by induction involves proving that if a statement is true in one case then it must also be true for the next case. Then by showing that it is true for a base case, eg 0, you can conclude that it is true for all positive integral cases, since if it is true for case 0 it must be true for case 1 and if it is true for case 1 it must also be true for case 2 etc

To do this it is normal to let Pn be the statement you are trying to prove in a general case n. Let us use the example Pn: 9ⁿ – 1 is divisible by 8 so we may right 9ⁿ-1 = 8a where a is an integer

Then we must prove that if Pk is true Pk+1 is also true. This will require using the expression Pk otherwise it isn’t induction since you are not showing that the truth of each statement can be found from the previous one.

so consider 9^(k+1) – 1
This is equal to 9x9k -1=9x9k – 9+ 8=9(9k-1)+8
but from Pk we know 9k-1 = 8a so
9^k+1-1 = 9(8a)+8=8(9a+1)
which is of the form 8b so is divisible by 8
Hence if Pk “9^k-1 is divisible by 8″ is true then Pk+1 is also true

Finally we must prove that P1 is true and we then know that Pn is true for all k >=1 where k is an integer.

P1 is the statement 9¹ – 1 is divisible by eight
9¹-1=9-1=8 so is true

therefore Pn is true for all positive integral value of n

Also see

Transformations of Graphs

admin — Sun, 11 Jan 2009 21:41:34 +0000

This looks at how a given graph will change when the the function is changed slightly eg how the graph y=x will change when it becomes y = 2x.

y = af(x)

The graph f(x) will “steeper” as the y value of each point is multiplied by a. It will appear like a “stretched” version of the graph y=f(x)

y =f(x) + a

The graph f(x) will move up by the amount a as a is added to each y value. This means that the points of intersection of the graph and the y axis will increase by the amount a. The intersection of the graph and the x-axis will depend upon the function of the graph.

y = f(ax)

This will make the graph appear “narrower” beacuse y is taking the value of f(x) a time across. So if the value of f(x) is 3 when x =12 then the value of f(4x) = 3 when x=3 as 12/4=3.

y=f(x+a)

This will shift the graph to the left by a. This is because the value of f(x) at x+a is displayed at the point x so effectively the graph occurs a earlier and therefore shifts to the left.