## Simultaneous Equations

Simultaneous equations are two or more equations that are all related because they contain the same variables. To solve these equations you must use information from both equations to find a set of values for the variables that hold true for both equations.

You can only solve simultaneous equations if there at least as many different equations (being different meaning that the two equations cant be re-arranged or simplified into each other) as there are variables in the equations. In order to solve the equations we want to eliminate all the variables but one so that we are left with a single equation that can be solved normally.

There are several ways of solving these equations. The method of solving using substitution is shown below using an example.

### Solve using substitution – with an example

Lets consider the equations

(1)- y + 3x = 12 + 4y

(2)- 3x = 7y – 6 + 4x

First of all we must simplify the equations by gathering all the similar terms together and cancelling through any common factors. So we get

(1)=> 3x = 12 + 3y => x = 4 + y

(2)=> 6 = 7y + x

We would now normally arrange one of these equations so that either x or y is the subject of the equation however in this example equation 1 naturally has x as the subject.

We can now substitute the expression for x in equation 1 (4+y) into equation 2 to get

6 = 7y + 4 +y

Again we must now simplify this to get

2 = 8y

and then we can solve it as a single equation to find

y=1/4

We can now substitute this value for y into either of the above equations to get a value of x. Substituting into 1 gives

x = 4 + 1/4

x=5/4

So that we now have our values of x and y that satisfy both of the equations

**x=5/4 and y=1/4**

If you have any questions or comments please them in the form below

## Implicit Differentiation

Implicit differentiation involves differentiating an equation that hasn’t been arranged such that all of one variable, eg y, is on one side and all of the other variable, eg x, is on the other side. For example differentiating the equation with respect to x (ie find dy/dx):

3x^{2} +2y^{3} + 6 = 3x^{2}y

TO do this you need to remember that the derivative of y with respect to x is dy/dx, hence when you differentiate a function such as

y=3x

you get

dy/dx = 3

Where the right has gone to the derivative of 3x and the left has gone to the derivative of y, dy/dx.

If you want to differentiate more complex terms involving x you can use the chain rule, since y can be written as a function of x.

so if f(x) = g(y) then

f'(x) = dy/dx g'(y)

(or a simple way of doing it is treat any y’s like x’s and stick a dy/dx on the end).

So for example y^{2} differentiated becomes 2ydy/dx

All the other rules like the product rule and quotient rule still apply.

So to finish lets consider our original equation

3x^{2} +2y^{3} + 6 = 3x^{2}y

This becomes

6x + 6y^{2}dy/dx = 6xy +3x^{2}dy/dx

Which we can re-arrange to get

dy/dx = (6xy-6x)/(6y^{2}-3x^{2})=(2xy-2x)/(2y^{2}-1x^{2})

By David Woodford

## First Order Differential Equations

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. To do this you treat the derivative like a fraction and then multiply through by the bottom of that one derivative is on one side (eg dx) and the derivative is on the other side (eg dy), you then want all the x’s to be multiplied by the dx and all the y’s to multiplied by the dy such that there are no terms not being multiplied by either. You can then integrate both sides, the one by dx and the other by dy – remembering to add a constant to one side.

Then simply re arranging will give the general solution. If you need the particular solution substitute the given values 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

## Quadratic Formula

The quadratic formula is a quick(unless you can factorise) way of solving quadratic equations. You basically take the coefficient’s of x, x^{2} 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

ax^{2}+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.

**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

### Other posts relating to the quadratic formula

**Derive the Quadratic formula**: an explanation of how to derive and prove the quadratic formula by completing the square**Factorise Quadratics**: how to solve a quadratic without using the formula

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