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# What does BIDMAS stand for?

By Filsan on the 10th of January, 2013

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Bidmas is an order of operation used when doing equations, which include both addition and subtraction, or Multiplication and Division.

This is the order it goes in.

Brackets
Indicies
Division
Multiplication
Subtraction

Another anagram for this is "bodmas" identical, but the "o" stands for "power of" i.e.

Brackets
power Of
Division
Multiplication
Subtraction

e.g. When given a simple example:
(56 - X)(Y + 7) where X=6 and Y=3.
You must use bidmas. First solve the brackets, which include addition and subtraction but are exempt from the rule as they are inside the brackets and then multiply the two values.

(56 - 6) x (3 + 7) =
50 x 10 = 500

By henry warren on the 14th of January, 2013

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BIDMAS is an acronym reminding students of the order of operation used when evaluating expressions involving a number of different operations. The letters of BIDMAS stand for:

• Brackets,

• Indicies,

• Division,

• Multiplication,

• Subtraction.

When only addition and subtraction (or only multiplication and division) are left in an expression work them out in the order you find them, starting from the left and working towards the right.

By AMRITA CHAKRAVARTY on the 21st of January, 2013

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By jagadish on the 14th of January, 2013

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What does BIDMAS stand for?

In Order of Operation:

B: Brackets

I: Indicies

D: Division

M: Muliplication

S: Subtraction

You may also see this referred to as BODMAS

THE PART OF BODMAS MOST OFTEN FORGOTTEN

It is quite common for people to forget that brackets always come first - which means you can calculate the answer of whatever is in the brackets before you attempt to calculate the rest of the problem.

Once you've worked out everything in the brackets, normally these sort of problems become very easy!

By Bridget Gaskin on the 17th of January, 2013

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The BIDMAS stands for the line of operation of mathematical functions it represents Brackets,Indicies,Division,Multiplication,Addition and Subtraction. For example for a mathematical operation  (2+3-4*6)-4*9, we need to do the operation inside the bracket and then we have tocomplete the operations out side.

By Adersh Antony on the 19th of January, 2013

## Solve a pair of simultaneous equations algebraically

If we have 2 equations with 2 unknown variables, there is only one solution set that satisfies both equations.

Algebraically, there are two methods for finding this solution:

The Method of Elimination to solve the simualtaneous equations– this is widely regarded as the simpler of the two methods, and the most commonly taught, both equations must be linear to use this method.

Elimination involves combining the two equations together to make one equation in terms of only one variable, which we can solve easily.

How?

1. Prepare the two equations such that if you add or subtract the two equations together one variable cancels out.

In other words, you want to multiply or divide an equation (or both) so that the number (not necessarily the ± sign) next to the variable in both equations is the same.

2. You add/subtract the equations to form one equation with a one variable to solve for this variable.

3. You plug the answer you got in 2, back in to one of your original equations, to find the other variable.

4. Check these are correct by plugging both your answers into the other equation.

For example,

a) 2y = x + 2
b) y = 16 - 2x

1. We have two easy options here:

We can either multiply equation a) by 2, which gives you

a*) 4y = 2x + 4

which means the number multipled by x, 2, is the same as in b) (remember the ± sign doesn't matter here)

or we can multiply equation b) by 2, which gives you

b*) 2y = 32 - 4x

which means the number multipled by y, 2, is now the same as in a).

Lets carry on with our first option and a*.

2.

The equations we are now working with are

a*) 4y = 2x + 4
b) y = 16 - 2x

If we add these 2 equations together, the x variable gets eliminated (hence the name):

a*+b : 5y = 20

Remember to keep each term on the correct side, add up terms on the left together, add up terms on the right together.

We can no solve 5y = 20 easily! Just divide both sides by 5.

y = 4

3. Plug this y value into one of the original equations, lets take a).

a) 2y = x + 2

Using y = 4

2(4) = 8 = x + 2
x = 6

4. Lets check this combination first with our second original equation b).

b) y = 16 - 2x

16 - 2(6) = 16 -12 = 4 = y

It works.

We are done.

So x = 6, y = 4.

Nothing in this section yet. Why not help us get started?

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