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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

Find equation of tangent to a curve

admin — Thu, 19 Nov 2009 14:39:45 +0000

The tangent to a curve is a line which touches the curve at a point without intersecting it at that point so the gradient of the curve at that point and the gradient of the tangent are the same. So we can work out the point the tangent passes though and the gradient of the tangent from the equation of the curve, which will give us enough information to find the equation of the tangent.

Example y=x²
Find the equation of the tangent to the curve when x=4?

To do this we first need to find the gradient of the curve which we can do by differentiating it.

so at the point x=t the gradient is 2t.

From this we can get a general equation for the tangent using the equation for the gradient of a straight line
grad =
to get the general equation for the tangent at the point x=t by substituting x₁=t, y₁=t² and m=2t

Then we can substitute in t=4 to find the equation of the tangent when x=4 to get

which is our final answer.

Also see

First Order Differential Equations

admin — Wed, 29 Apr 2009 14:41:44 +0000

First order differential equations are equations which include a first derivative ie) dy/dx if the equation is in terms of x and y.

Solving these equations involves finding a expression for one of the variables in terms of the other without including a derivative.

To do this usually a general solution is found, one which includes a constant from integration, and then using a given set of conditions (usually initial conditions) the particular solution can be found.

This is done by “splitting the variables” and then integrating the resulting equation. Splitting the variable means that we treat as a fraction so we can “split” the top from the bottom to take the bottom to the other side. What we want is for all the x terms to be on one side and all the y terms on the other. We can then “cross multiply” so that all the x terms are “multiplied” by dx and all the y terms are “multiplied” by dy. We can then add in the integrals and integrate each side (one with respect to x and the other with respect to y).

[Of course this isn't what is actually happening as isn’t a fraction and so we can’t multiply by dx. What happens in reality is that we can rearrange so that all of the y’s are on the side of the and all of the x’s are on the other. Now we have a function of y multiplied by the derivative of y with respect to x. Now we can integrate both sides with respect to x. The side with the x’s using normal rules of integration but on the side with the y we use a change of variable from x to y, to do this we must multiply by or (these can be proven to be the same) which will cancel out the and hence we can simply integrate the expression on the side with the y’s with respect to y giving us the same result as when we “multiply”.]

Then simply re arranging will give the general solution (that is you can rearrange it so that the equation is of the form y=…+C). If you need the particular solution substitute the initial conditions into the equation to find out the value of the constant and then substitute this value for the constant into the general solution.

If all of that was a bit hard to follow here is s a worked example.

Example

Find the general solution to
dy/dx = 3x + 4
and the particular solution in the case y = 4 when x = 0 .
So we can rearrange to get

dy = (3x+4)dx

And then by integrating both sides we get

∫dy = ∫3x+4 dx

y = (3x+4)²/6 + c

Which is the general solution.
We can use boundary conditions, ie) y = 4 when x = 0 to find a particular solution
4 = (0+4)²/6 +c
c = 4-16/6 = 8/6 = 4/3

so the particular solution is
y = (3x+4)²/6 +4/3

Also see

Auxiliary Angle Method for Solving Trigonometry Equations

admin — Sun, 19 Apr 2009 17:05:38 +0000

This is a method of solving equations in the form asinx+bcosx = c where a and b are constants and c is another expression.

It involves rewriting letting asinx + bcosx = rsin(x+y) (or you could use cos(x+y)) where y is acute and then finding values for r and y, then with only one trig function to deal with the equation can be solved more easily.

For example

consider 2sinx + 3cosx = 3

Let 2sinx + 3cosx = rsin(x+y)

Now expand the sin(x+y) to get

2sinx + 3cosx = rsinx cosy + rcosx siny

Since y is constant and therefore cosy and sin y are constant we can compare the coefficients to get

2 = rcosy —–(1)
3 = rsiny ——(2)

We can solve these to find values for r and y.
To find y consider (2)/(1) to get

3/2 = tany
since sin/cos = tan and the r’s cancel
so y = 56.3 °

To find r consider (1)2+(2)2 to get
22+32 = r2
since sin2+cos2 = 1
so r =√13

So we can write

2sinx + 3 cosx = √13 cos(x+56.3) = 3

so x = cos^-1(3/√13) -56.3

so x = cos-1(3/√13) -56.3

since cos-1(3/√13) = 33.7 for solutions between 0° and 90°

x = ±33.7 -56.3 + 180n where n is an integer

In General

asinx + bcosx = √(a2+b2) sin(x+tan-1(b/a))

If you have any questions, comments or corrections please leave them as a comment below

By David Woodford

Also see

Trigonometry Identities (11)
Proof of Cosine Rule (3)
Understand the Sine and Cosine Rules (11)
Area of a Triangle (3)
Proof by Contradiction (0)
Rational and Irrational Numbers (5)

Quadratic Formula

admin — Sun, 24 Feb 2008 19:45:13 +0000

remove_filter(‘the_content’, ‘wpautop’); The quadratic formula is a quick(unless you can factorise) way of solving quadratic equations. You basically take the coefficient’s of x, x² and numbers, put then in the formula, work out the two answers and have your 2 solutions of x. And if that wasn’t easy enough written a console program in c++ that will solve them for you(and gives dodgy answers for complex solution ie) imaginary answers when there are no real roots.

so for the general quadratic
ax²+bx+c=0

So what do you do, well enter a, b and c from the general equation into the formula, work out the answers and they are your solutions.

Whats the plus/minus thingy. You might be wondering what the thing is after the -b, well its a plus or minus sign. Because when you work out a square root it can have to answers, eg)root 9 = +3 and -3, he formula takes this into account by saying you must use the plus and the minus answers. This therefore means you will be 2 solutions to the quadratic, which makes sense as the graph is a curve and it therefore must cut the x-axis twice.

When the root part is negative(before you find the root) there are no real roots, only complex ones. This means that the curve comes down above the x-axis and doesn’t cut it. If you work out the complex roots, using i for root(-1), your pair of answers will be a conjugate pair.

Quadratic Calculator

Enter the coefficients of your quadratic equations into the boxes below and click solve to find the solutions to your quadratic. Will give complex solutions as well if your equation has them.

x²+
x+
=

David Woodfords Quadratic Calculator
This is a console program that i wrote in c++ that will solve a quadratic for you. Just download the file and run it, follow the instructions and it will output the 2 answers for you. I tried to make it work for complex roots as well but somewhere in the decimal data types Ive messed up so only the second parts of the complex answers are correct – though im sure working out -b/2a isn’t too hard for you.
Download quadratic calculator