Forming and solving equations key points?

Solving equations.

Linear:

get everything that is not relevant to the subject to the opposite side of the equation, and everything relevant on the other side of the equation, via + or -

e.g. 5x + 8 = 3x + 2; 2x = -6

either multiply or divide through to single out one unit of the subject

x = -3

Equations with higher power:

simplify as much as possible, using the methods for linear equations, then factorise.

Simplify

3x^2 + 15x + 18 = 0; x^2 + 5x + 6 = 0

Factorise (it's always easier if you have 0 on one side)

(x+2)(x+3) = 0; x = -2 or -3

if it's difficult to factorise, use the formula. if the equation contains a power higher than 2 and is difficult to solve, try dividing the equation with simple factors to see if you can get it down to a nice quadratic. Simple factors include x+1, x+2, x+3, x+4, x+5, x-1, x-2, x-3, x-4, x-5, x. If the equation contains factors that are either not obvious or simple, then it's not something you should be solving; i.e. there's software that solves equations like that for you; you're just learning how things are done.

Simultaneous equation:

like linear equations, single out 1 of the factors and find its value. With simultaneous equations, you try to single a subject in terms of the other(s), then substitute your answer back into an equation to solve the other subject. This is true, irrespective of which method you use.

e.g. Equation 1: 3x + y = 54;

Equation 2: 2x + 8y = 63

Eqn1: y = 54 - 3x; sub in Eqn 2.

Eqn2: 2x + 8(54 - 3x) = 63; x = 369/22; sub back into either equations (I choose Eqn 1)

3(369/22) + y = 54; y = 81/22

If you have more than 2 unknowns, single out 1 unknown in relation to the other 2, and then you try to single out another unknown like they're simultaneous equations, but with 3 equations as oppose to 2. These answers should be simple, if not obvious; if they aren't, then you should not either be tested with them or your answers may be wrong. For 4 unknowns or above, use software

linear programming:

it's simultaneous equations with graph work. The answer is usually one of the corners where the lines intercept each other.

Common sense:

to give the right answer, you sometimes need to use common sense. e.g. areas and lengths don't have negative values (e.g. -3cm or -8m^2), so always go for answers that give you the positive result. e.g. if you're asked what is the optimum quantity to produce in economic questions, and you get extreme results (e.g. 50 vs 95843), but you know you are restricted to a low range (e.g. your factory can only produce a max of 800), go for the result that makes sense and fits the criteria of the question.

Forming equations:

Worded problems can always be hard to structure in mathematical equations. This is true at all levels from GCSE to PhD.

You should always try to keep things as simple as they can be without oversimplifying the situation. e.g. modeling London Bridge only as a straight line is oversimplifying; if you added in the Tower and the major support, then it's simple enough. Make sure your model and equation is adequate for your use.

Always go for the simplest approach. We can all do things the long way round and complicate things, but it is often not necessary. You would not want to do extra unneeded work when you are short on time or you have other things to do.

Always try to have the simplest answer. Anyone with an ounce of intelligence can make any answer complicated, but to make a complicated answer beautiful and eloquent and it is simple and useful is the challenge. The questions you usually come across tend to involve simple equations, so try not to over-complicate.

When you get a worded problem that involves graphs or geometry of any sort, always sketch it out. It always give you a clearer picture of what you are doing and it serves as a reminder of context of the question.

Other tips:

if there is an unknown value, we tend to label it a letter e.g. x, y, z. It's easier to envision than using words. These sort of questions would involve relative values. e.g. you have a right angled triangle with lengths that are 1cm and 2cm longer than one side with an unknown value (say x). These questions would involve using the material you have learnt in lessons and you occasionally get things that involve more than one thing. e.g. to solve the triangle problem, we would use algebra and Pythagoras theorem: x^2 + (x+1)^2 = (x+2)^2

Turning a word problem into a mathematical problem is usually the hardest part of answering the question. Once you have done that right, everything else is usually routine and simple. If you don't get the answer you are looking for, or if the answer looks complicated, it is usually because you have the wrong equation to start off with or there is a more suitable equation you have not found yet.

Hope this helps