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This expression contains brackets and . The brackets should be worked out first, and then the indices should be calculated before the multiplication. The addition should be worked out last.

\[d + (3e + f)^2 = 2 + (3x - 3 + 1)^2\]

\[2 + (3x - 3 + 1)^2 = 2 + (-9 + 1)^2 = 2 + (-8)^2\]

The bracket has been worked out first.

\(2 + (-8)^2 = 2 + 64\) . \(-8^2\) means \(-8 \times -8\) , which is +64 using sign rules (two signs that are the same make a positive).

\(2 + 64 = 66\) . The addition part of the expression has to be worked out last following BIDMAS.

This is the final answer.

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Substitution - Algebraic formulae - Edexcel - GCSE Maths Revision - Edexcel - BBC Bitesize

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This expression contains brackets and . The brackets should be worked out first, and then the indices should be calculated before the multiplication. The addition should be worked out last. \[d + (3e + f)^2 = 2 + (3x - 3 + 1)^2\] \[2 + (3x - 3 + 1)^2 = 2 + (-9 + 1)^2 = 2 + (-8)^2\] The bracket has been worked out first.

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We use the term simultaneous equations when we have multiple equations that share a bunch of variables and are all true at the same time. We've seen how to solve one equation with one variable, but if there's an equation with two variables,

we can't really "solve" it in the same way. However, if you have 2 equations both with 2 variables, like

2x+y=10\,\,\,\text{ and }\,\,\,x+y=4

then there is a solution for us to find that works for both equations. These are equations are called simultaneous for this reason. In order to be able to solve simultaneous equations you will need to be confident in the following topics.

How to Solve Simultaneous Equations

There are two ways to solve equations,

For higher tier simultaneous equation questions you will need to be able to solve quadratics by factorising and through use of the quadratic formula.

Example 1: Solving Simultaneous Equations

Find the solution to the following simultaneous equations.

To do this, we'll use a process called elimination - we're going to eliminate one of the variables by subtracting one equation from the other. We will write one equation on top of the other and draw a line underneath, as with normal subtraction.

So, we eliminated the y term and now have 2x=6.

Dividing both sides of this equation by 2, we get

Now that we've found x, finding y is easier - we use substitution, which means that we pick one of the equations, we'll pick x+y=4, and replace x with 3, as that's what we now know it to be. The equation x+y=4 becomes

Then, subtracting 3 from both sides of the equation, we get

Thus, the solution to these simultaneous equations is

Example 2: Applied Simultaneous Equation Question

A store sells milkshakes and ice creams. If you buy 4 milkshakes and 3 ice creams, it will cost £12. If you buy 2 milkshakes and 2 ice creams, it will cost £7. Work out the cost of an individual milkshake and an individual ice cream.

In these sentences, there is a hidden pair of simultaneous equations, solving these will require application of knowledge.

Let's say that the price of a milkshake is a, and the price of an ice cream is b. This creates the equation

If we do the same thing with "2 milkshakes and 2 ice creams cost £7", we get

Thus, we have our simultaneous equations.

In order to use the elimination method, you need to have the a or b variables the same in both equations. We can multiply both sides of the second equation by 2, to make this work:

So the a values of the first and third equation are the same. Subtracting the first equation from the third, we get

We have immediately that the price of an ice cream, b, is £2. Then, substituting this back into the first equation, we get

Subtract 6 from both sides to get

and then divide both sides by 4 to get

Therefore the price of a milkshake is £1.50.

Example 3: Solving Quadratic Simultaneous Equations

Solve the following simultaneous equations.

Because one of these equations is quadratic, we can't use elimination like before. Instead, we have to use substitution.

The 2nd equation tells us y is equal to x+3, so we will replace the y in the first equation with x+3:

If we expand the bracket and then subtract 9 from both sides, this becomes

Observing that -1\times3=-3 and -1+3=2, we can factorise this quadratic:

The solutions for this are

Then, to find the corresponding y solutions, we must put these values back into the 2nd equation. Doing so, we get

\text{When }x=1, y=1+3=4,\,\,\,\text{ and when }x=-3, y=-3+3=0.

Thus, we have two solution pairs:

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Simultaneous Equations Worksheets | Questions and Revision | MME

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We use the term simultaneous equations when we have multiple equations that share a bunch of variables and are all true at the same time. We've seen how to solve one equation with one variable, but if there's an equation with two variables, we can't really "solve" it in the same way.

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1) remove x from each side
2) remove 1 from each side
3) repeat these steps until you have a simple equation with a single side of x

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Equations that have more than one unknown can have an infinite number of solutions. For example, \(2x + y = 10\) could be solved by:

To be able to solve an equation like this, another equation needs to be used alongside it. That way it is possible to find the only pair of values that solve both equations at the same time. These are known as simultaneous equations .

An example of this is:

The unknowns of \(x\) and \(y\) have the same value in both equations. This fact can be used to help solve the two simultaneous equations at the same time and find the values of \(x\) and \(y\) .

Solving simultaneous equations by elimination

The most common method for solving simultaneous equations is the elimination method which means one of the unknowns will be removed from each equation. The remaining unknown can then be calculated. This can be done if the of one of the letters is the same, regardless of sign.

Solve the following simultaneous equations:

First, identify which unknown has the same coefficient. In this example this is the letter \(y\) , which has a coefficient of 1 in each equation.

Either add or subtract the two equations from each other to eliminate the letter \(y\) . In this example the equations will need to be subtracted from each other as \(y - y = 0\) .

If the equations were added together, then \(y + y = 2y\) , and so the letter \(y\) would not be eliminated.

\[\begin{array}{ccccc} 3x & + & y & = & 11 \\ - && - && - \\ 2x & + & y & = & 8 \\ = && = && = \\ x &&& = & 3 \end{array}\]

\[3~\mathbf{\times~3} + y = 11\]

Check the answers by substituting both values into the other original equation. If the equation balances, then the answers are correct:

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Simultaneous equations - Solving simultaneous equations - Edexcel - GCSE Maths Revision - Edexcel - BBC Bitesize

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Equations that have more than one unknown can have an infinite number of solutions. For example, \(2x + y = 10\) could be solved by: To be able to solve an equation like this, another equation needs to be used alongside it.

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Algebra - Simultaneous Equations Linear (Video 1)

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Direct proportion = y/x = multiplier
Inverse proportion = k = yx

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Identifying Proportional and Non-Proportional Relationships in Tables

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1) K = xy
2) K is always constant (the same)
3) xy has to equal k for it to be inversely proportional
4) k/y = x
5) k/x = y

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Maths - Solving inverse proportion word problems - English

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1) y/x = multiplier
2) the goal is to identify the multiplier and complete the grid, or state a relationship between x and y

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8 : 16
/8  /8
1 : 2

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This video explains how to answer questions on Ratio - Expressing as 1:n

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9 : 18
/18 /18
0.5 : 1

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1) add the two (or more) digits in the ratio together
2) divide the total number by this
3) times the number you get by each side of the ratio
e.g.
share 300 into 7 : 8
7 + 8 = 15
300/15 = 20
20 x 7 = 140
20 x 8 = 160
Ans = 140 : 160

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1) Find a common factor
2) Divide both sides of the ratio by this until it cannot be divided as a whole number any more

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How to Simplify Ratios | Simplifying Ratios

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A surd is the square root of a number that cannot be reduced as whole

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Surds / Radicals - the basics - multiplication made easy with radical math

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Method 1:
1) Use knowledge of decimal places to make the powers equal on both sides
2) Add/subtract whole numbers
3) Leave powers the same

Method 2:
1) Translate standard form into whole numbers
2) Add/subtract these as whole numbers
3) Convert whole number back into standard form (be aware that only numbers 1-9 are used)

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How to Add or Subtract Numbers in Standard Form (Part 4/4)

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One to one maths interventions built for GCSE success

Weekly online one to one GCSE maths revision lessons available in the spring term

Find out more

Here we will learn about multiplying and dividing numbers in standard form including how to multiply, divide and square numbers in standard form.

There are also multiplying and dividing numbers in standard formworksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you're still stuck.

What is multiplying and dividing in standard form?

Multiplying and dividing numbers in standard form is where we multiply and divide integers and decimals and apply the laws of indices to simplify the powers of ten. You also need to check that your final answer is in standard form.

E.g.

\[(4\times10^{3})\times(2\times10^{6})\]

\[10^{3}\times10^{6}\quad=\quad10^{9}\]

\[(4\times10^{3})\times(2\times10^{6})\quad=\quad8\times10^{9}\]

E.g.

\[(8\times10^{9})\quad \div \quad(2\times10^{6})\]

\[10^{9}\quad \div \quad10^{6}\quad=\quad10^{3}\]

\[(8\times10^{9})\quad \div \quad(2\times10^{6})\quad=\quad4\times10^{3}\]

There are many applications of multiplying and dividing standard form numbers which includes applying it to area, volume, surface area, equations and Pythagoras' theorem.

How to multiply and divide in standard form

In order to multiply and divide numbers in standard form:

1. Multiply or divide the integers or decimal numbers.

2. Multiply or divide the powers of ten by adding or subtracting the indices.

3. Write the solution in standard form, checking that the first part of the number is \pmb{1\leq{x}\lt10}.

Multiplying and dividing standard form worksheet

Get your free multiplying standard form and dividing standard form worksheet of 20+ questions and answers. Includes reasoning and applied questions.

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Multiplying and dividing standard form worksheet

Get your free multiplying standard form and dividing standard form worksheet of 20+ questions and answers. Includes reasoning and applied questions.

COMING SOON

Multiplying and dividing standard form examples

Example 1: multiplying numbers in standard form

Work out:

\[ (3 \times 10^{4}) \times (2 \times 10^{5})\]

Multiply or divide the integers or decimal numbers.
2Multiply or divide the powers of ten by adding or subtracting the indices.

\[10^{4} \times 10^{5} = 10^{9}\]

3Write the solution in standard form, checking that the first part of the number is \pmb{1\leq{x}\lt10}.
This number is already in standard form.

Example 2: multiplying numbers in standard form

Work out:

\[(3 \times 10^{5}) \times (6 \times 10^{-2})\]

To compensate, you need to multiply the power of ten by 10 which has the effect of adding one to the power, which gives the result 10^{4} .

The final result is 1.8 \times 10^{4}.

Example 3: dividing numbers in standard form

Work out:

\[(8 \times 10^{7}) \div (2 \times 10^{5})\]

This number is already in standard form.

Example 4: dividing numbers in standard form

Calculate:

\[(3 \times 10^{4}) \div (6 \times 10^{-3})\]

Write your answer in standard form.

To compensate, you need to divide the power of ten by 10 which has the effect of subtracting one from the power, which gives the result 10^{6}.

The final result is 5\times10^{6}.

Example 5: squaring numbers in standard form

Calculate:

\[(8 \times 10^{4})^{2}\]

Write your answer in standard form.

We know that

\[\left(8 \times 10^{4}\right)^{2}=\left(8 \times 10^{4}\right) \times\left(8 \times 10^{4}\right)\]

Let's first work out 8\times8 = 64

Simplify 10^{4} \times 10^{4} = 10^{8}

To compensate, you need to multiply the power of ten by 10 which has the effect of adding one to the power, which gives the result 10^{9}.

The final result is 6.4 \times 10^{9}.

Example 6: cubing numbers in standard form

Calculate:

\[(6 \times 10^{4})^{3}\]

Write your answer in standard form.

We know that (6\times10^{4})^{3}= (6\times10^{4})^{3}\times(6\times10^{4})^{3} \times (6\times10^{4})^{3}

Let's first work out 6\times6\times6 = 216

Simplify 10^{4} \times 10^{4} \times 10^{4} = 10^{12}

To compensate, you need to multiply the power of ten by 100 which has the effect of adding 2 to the power, which gives the result 10^{14}.

The final result is 2.16 \times 10^{14}.

Common misconceptions

When multiplying numbers in standard form, a common mistake is multiplying the powers rather than adding the powers.

E.g.

(4\times10^{5})\times(2\times10^{3}) = 8\times10^{8} .
The power of ten is 10^{8} , NOT 10^{15}.

When dividing numbers in standard form, a common mistake is dividing the powers rather than subtracting the powers.

E.g.

(8\times10^{9})\div(2\times10^{3}) = 4\times10^{6} .
The power of ten is 10^{6} , NOT 10^{3}.

Not converting solutions to standard form

After calculating with standard form, a common mistake is not checking that the first part of the number is 1 ≤ n < 10.

E.g.

45 x 10^{6} is not in standard form.

Not adding or subtracting negative powers correctly, especially when dividing by a negative power as here you would be subtracting a negative number which has the same effect as adding.

Practice multiplying and dividing standard form questions

Multiplying and dividing standard form GCSE questions

1. A cube has side lengths of 2\times10^{3} metres. Calculate the surface area of the cube. Write your answer in standard form.
(3 marks)

Finding the area of one face: 4\times10^{6}
(1)
Multiply by 6 or correct answer not in standard form.
(1)

2.4\times10^{7}m^{2}
(1)

2. Work out \frac{9.3\times10^{8}}{3.1\times10^{2}} . Give your answer in standard form.
(2 marks)

9.3\div 3.1 = 3 or 10^{8} \div 10^{2} = 10^{6}
(1)
3\times10^{6}
(1)

3.

a) Write 75000 in standard form.

b) Write 0.03 in standard form.

c) Using your answers to a) and b), and also the formula F=ma, calculate the value of F when m=75000 and a=0.03. Write your answer in standard form.
(5 marks)

a) 75000 in standard form is 7.5 \times 10^{4}
(1)
b) 0.03 in standard form is 3 \times 10^{-2}
(1)
c) F = ma

Therefore,

F=7.5 \times 10^{4} \times 3 \times 10^{-2}
(1)
7.5 \times 3 =22.5

10^{4} \times 10^{-2} = 10^{2}
(1)
22.5 \times 10^{2} = 2.25 \times 10^{3}

F=2.25 \times 10^{3}
(1)

Learning checklist

You have now learned how to:

Multiply numbers in standard form

Divide numbers in standard form

Square and cube numbers in standard form

Still stuck?

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Multiplying and dividing in standard form

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One to one maths interventions built for GCSE success Weekly online one to one GCSE maths revision lessons available in the spring term Find out more Here we will learn about multiplying and dividing numbers in standard form including how to multiply, divide and square numbers in standard form.

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Rules:
1) Only numbers between 1 and 9 can be used for standard form
2) these numbers can have multiple decimal figures
3) powers affect the number if they have decimal digits
e.g.
8.927 x 10 (3) = 8927
8 x 10 (4) = 80000
2.926 x 10 (-4) = 0.0002926

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Standard Form: Introduction [Video Lesson]

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1) simplify the indices
2) find a fraction or whole number suitable for addition
3) (powers cannot be added or subtracted like multiplication)
4) add or subtract digits relativley

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√ The Addition and Substraction of Indices by Index Laws Explained. Watch this Video to Find out!

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1) Find a square number factor in the surd
2) Change the format to (square root x square root)
3) Square root the square number
4) leave the multiplier number as a square root
e.g.
,/162 = 81 x 2
,/81 x ,/2
Ans = 9 ,/2

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Simplifying surds - Surds - AQA - GCSE Maths Revision - AQA - BBC Bitesize

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Learn about and revise surds, including how to add, subtract, multiply and divide them with GCSE Bitesize AQA Maths.

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1) Take the denominator of the power and root the integer by the denominator (if numerator is 1)
OR
2) Take the denominator and root the integer by the denominator, put this in brackets, then apply the numerator as an external power (if the numerator and denominator are both N)

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a (0) = 1
a (-1) = 1/a
a (-2) = 1/a2
a (-3) = 1/a3
a (-4) = 1/a4

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√ Negative Indices | Negative Exponents | Index laws | Mathematics

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Rules:
- add powers when multiplying
- subtract powers when dividing
- multiply powers when using brackets to square/cube
- these rules only apply when the integers are equal

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Indices - Simplifying expressions using the laws of indices - National 5 Maths Revision - BBC Bitesize

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Revise how to simplify expressions involving indices at part of National 5 maths.