Posts Tagged ‘solve’


November 1, 2009 Leave a comment

An inequality(or inequation) is similar to an equation accept for instead of saying both side of the inequality are equal we say one side is greater than (or equal to depending upon the type of inequality) the other, this is done using the greater than (> ), greater than or equal to (\geq ), less than (< ) and less than or equal to (\leq ).


Some simple examples which contain only one variable are:

5x > 3
2x-7 < x+5
x^2 - 4 \leq x

Solving and Manipulating Inequalities

Inequalities can be solved by rearranging them and isolating the variable you want to find in a similar way to normal equation (see the post quadratic inequalities to see how to solve quadratics). However, rather than getting an exact value such as x=3 we get a range (open or closed) of values such as x<2 or -3<-1.

Much of the manipulation is the same though there are slight variations when dividing or multiplying by negative numbers or taking the reciprocal. The important thing to remember is that like normal equations we must do the same to both sides.

Addition and Subtraction

Addition and subtraction are exactly the same to equalities. We can add or subtract whatever we like as long as we do the same to both the sides. This enables us to take expressions “to the other side” by reversing their sign. For example all the following manipulations are valid.

x + 3 < 4  \Leftrightarrow x < 4-3 = 1
x - 3 < 4  \Leftrightarrow x < 4+3 = 7
x < 4  \Leftrightarrow x + 3 < 4+3 = 7

Multiplication and Division

Again we can perform multiplication and division in a similar way to the way we perform it with equalities by doing the same to both sides. However, if we are multiplying or dividing by a negative number we must reverse the direction of the inequality since
-x < y \Leftrightarrow x>-y
This means we must be careful when diving by an unknown since by definition we don’t know whether or not it is positive or negative. If this has to be done you should consider both the cases it is positive and negative separately and if it is only positive or negative then the other inequality should lead to a contradiction which can easily be spotted such as x<0 and x>3.

Examples of valid manipulation are below:
2x < 6 \Leftrightarrow x<3
-2x < 6 \Leftrightarrow x>3
\frac{4}{x} < 3 \Leftrightarrow \frac{4}{3} < x for x \geq 0 and/or \frac{4}{3} > x for x<0


When taking the reciprocal or “one over” of an expression you must reverse the inequality so
x < y \Leftrightarrow \frac{1}{x} > \frac{1}{y}


Simultaneous Equations

September 4, 2009 Leave a comment

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

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

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

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.


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