# HSC Extension 1 and 2 Mathematics/2-Unit/Preliminary/Basic arithmetic and algebra

The start of the 2-unit mathematics course is really just revision of year 10 work.

## Converting between fractions, decimals and percentages

### Converting recurring decimals to fractions

So, you're in an exam and the test paper gives you a recurring decimal, say 2.35555..., and asks you to convert it into a fraction. Well, that's easy, it's just 2.3 plus 5 9ths divided by 10. simple enough. But then you get another question asking you to convert 5.676767... into a fraction. What fraction makes the repeating 0.676767...? Ahh, now you're stuck.

That's what would have happened if you hadn't learnt how to convert recurring decimals into fractions algebraically before sitting for your exam. What use does converting recurring decimals into fractions have? Can't calculators convert between decimal and fraction easily? Yes they can. This topic is probably most useful in getting you to think like a mathematician, applying maths (in this case algebra) to problems. There isn't much more use for this, other than providing an endless source of questions to test your basic algebra.

So let's start off with our recurring decimal number 5.676767... We are going to assign that value to x, so 'x = 5.676767.... Now we have something to work with. To convert it into a fraction, we need to get rid of all those annoying decimal places. To get rid of them we are going to need to multiply x by 100 to give us 100x = 567.676767..., and then we can subtract x, or 5.676767..., from that.

${\displaystyle x=5.{\dot {6}}{\dot {7}}}$

${\displaystyle 100x=567.{\dot {6}}{\dot {7}}}$

${\displaystyle 100x-x=567.{\dot {6}}{\dot {7}}-5.{\dot {6}}{\dot {7}}}$

${\displaystyle 99x=562\;}$

${\displaystyle x={\frac {562}{99}}}$

${\displaystyle x=5{\frac {67}{99}}}$

And there you go. This is the method of removing the first period

When looking at problems with recurring decimals, you need to see how many numbers are in the repeating part. In this example, the start of the repeating part was 0.67, and there were 2 numbers in each repeat. Because there were two numbers in the repeating part being repeated, we had to move the number 2 places to the left, by multiplying the number by 100, or 102. If you have a number with a non-repeating part a and the start of the repeating part r, and a repeating length, or period, P, then we can write a general list of equations for converting the recurring decimal number into a fraction.

${\displaystyle x=a+{\dot {r}}}$

where the dotted r is the repeating part of the equation that we are going to get rid of.

${\displaystyle 10^{P}x=10^{P}a+10^{P}r+{\dot {r}}}$

where 10Pr is the start of the repeating part, shifted P places to the left.

${\displaystyle 10^{P}x-x=10^{P}a+10^{P}r+{\dot {r}}-a-{\dot {r}}}$

Now we subtract one equation from the other to get rid of the repeating part by making it subtract from itself, leaving just the first period of the repeating part which we shifted left P places.

${\displaystyle (10^{P}-1)x=(10^{P}-1)a+10^{P}r\;}$

Factorize both sides

${\displaystyle x={\frac {(10^{P}-1)a+10^{P}r}{10^{P}-1}}}$

make x the subject

If the non-repeating number a had a decimal part, like in the number 2.35555..., where a was 2.3, then the numerator would be a decimal (but not a recurring decimal) and you would have to multiply the numerator and denominator by a power of ten to make both numbers integers, and then you would have to simplify it. Calculators make this easy with their fraction buttons which can convert divisions into their simplest fractions. You could make an even more general formula if you let D, for decimal, equal the amount of decimal places that the non-repeating part has. Then you can just multiply the numerator and denominator in your final answer by 10D.

${\displaystyle x={\frac {[(10^{P}-1)a+10^{P}r]\times 10^{D}}{(10^{P}-1)x\times 10^{D}}}}$

It is really more important that you remember the basic method and idea behind it, cancelling out the repeating part by subtracting it from itself. This formula is just for fun.

An interesting thing here is that the numerator will be equal to the non-repeating number with its decimal place removed with the repeating part of the number on the end, minus the non-repeating number with its decimal place removed without the repeating part on the end. So for 2.3555..., a = 2.3, r = 0.05, P = 1, D = 1. This makes the numerator (10×2.3 - 2.3 + 0.5)×10, so 235 - 23. 235 is a with no decimals and with r on the end, and then you are just subtracting a without decimals from that. Another thing to notice is that for every recurring number in the recurring part of the decimal, there will be a 9 in the denominator. 10P - 1 is a 1 followed by zeros when P is greater than 1. So when you take one away, all you are left with is nines. 100 - 1 = 99, 1000 - 1 = 999, etc. Also, for every decimal place in the non-repeating part, there will be a zero on the end of the denominator, caused by the multiplication of 10D.

This is where the general algorithm of subtracting the non-repeating plus start of repeating as an integer from the non-repeating as integer to get the numerator, and then adding nines for every repeating decimal number and zeros for every non-repeating decimal number comes from.

## Scientific notation and approximation

significant figures

Rules:

1. All non-zero digits are significant. 2. In a number without a decimal point, only zeros BETWEEN non-zero digits are significant. 3. In a number with a decimal point, all zeros to the right of the first non-zero digit are significant.

For example: 1. 1500 has two significant figures because the two zeros after 15 do not count. If the question asks to round to 1 significant figure then the answer is 2000.

2. 1050 has three significant figures since the 0 between 1 and 5 counts. If the question asks to round to 2 significant figures then the answer is 1100.

3. 0.001500 has four significant figures because the 3 zeros on the left hand side of the 1 do not count. However, the 2 zeros on the right hand side of 5 DO count, thus there are four significant figures. If the question asks to round to 1 significant figure then the answer is 0.002.

## Evaluation of expressions involving all this stuff

If The N+1 Number Of S.f More Than 5 We Add 1+ To N Number , If It Is Less Than 5 nothing is change

## Surds

A surd is any irrational expression, i.e. it cannot be expressed rationally (as a fraction of two integers). Common examples include √2 and ${\displaystyle \pi }$. These numbers have no known patterns in their digits, and so cannot be expressed as the fraction of two integers, they can only be expressed as a decimal with the decimal places continuing on forever. Because of this, surds can only ever be evaluated as approximations to a number of decimal places. The word 'surd' actually comes from the Arabic surdus which means root. The Arab mathematicians saw numbers like plants, growing out of their roots.

### Like and Unlike surds

Basically, two surds are like if they have the same number under the radical, which is called the radicand. For example, ${\displaystyle 3{\sqrt {3}}}$ and ${\displaystyle 8{\sqrt {3}}}$ are like surds. Unlike surds do not have the same radicand, and hence, for example, ${\displaystyle 3{\sqrt {37}}}$ and ${\displaystyle 3{\sqrt {43}}}$ are unlike.

### Simplifying surds

Some surds can be simplified as a product of a rational number and a surd. For example,

${\displaystyle {\sqrt {45}}}$ can be simplified to ${\displaystyle 3{\sqrt {5}}}$

This happens when the number within the surd is expressed as a product of two numbers, one of them being a perfect square:

${\displaystyle {\sqrt {45}}}$ ${\displaystyle ={\sqrt {9\times 5}}}$

Using the law of surds, this surd can be expressed as a product of two different, separate surds:

${\displaystyle {\sqrt {9}}\times {\sqrt {5}}}$

We can then evaluate ${\displaystyle {\sqrt {9}}}$ as ${\displaystyle 3}$:

${\displaystyle 3\times {\sqrt {5}}}$ ${\displaystyle ={\sqrt {9\times 5}}}$

This method of simplification can also be applied to roots of any power.

### Addition and subtraction of surds

${\displaystyle a{\sqrt {b}}\pm c{\sqrt {b}}=\left({a\pm c}\right){\sqrt {b}}}$

### Multiplication of surds

${\displaystyle {\sqrt {a}}\times {\sqrt {b}}={\sqrt {ab}}}$

### Division of surds

One surd (root) divided by another surd (root) is equal to the root of the fraction of the two radicands.

### Rationalizing the denominator

The correct form of writing a fraction requires that the denominator be rational, and hence, not a surd. To rationalise the denominator, we multiply both numerator and denominator by the denominator, resulting in a rational denominator.

## Inequalities and absolute values

### Inequations

Inequations are mathematical statements that don't use the equals sign to express relationships between the two sides of the statement, but instead use signs such as greater than, greater than or equal to, less than, and less than or equal to.

 greater than ${\displaystyle >\ }$ greater than or equal to ${\displaystyle \geq }$ less than ${\displaystyle <\ }$ less than or equal to ${\displaystyle \leq }$

While equations give exact values for pronumerals, such as x = 2, where x is exactly 2, inequations give a range of values which the pronumeral can have, such as x > 2, where x can posses any value greater than 2.

Just like when dealing with equations, adding or subtracting an amount to both sides doesn't change the inequation. If y + 2 > 1 then by subtracting 2 from both sides y > -1.

When multiplying or dividing both sides by positive number it also works the same as an equation, for example if ${\displaystyle 8x\geq 4}$ then by dividing both sides by 8 you get ${\displaystyle x\geq {\begin{matrix}{\frac {1}{2}}\end{matrix}}}$.

When multiplying or dividing both sides of an inequation by a negative number the sign reverses. for example if you have ${\displaystyle -2x\leq 3}$ and you divide both sides by -2, the less than or equal sign changes to a greater than or equal sign as ${\displaystyle x\geq -1{\begin{matrix}{\frac {1}{2}}\end{matrix}}}$. This is because if one number ${\displaystyle a\ }$ is greater than another number ${\displaystyle b\ }$, i.e. ${\displaystyle a>b\ }$, then on a number line ${\displaystyle a\ }$ is further away from 0 in the positive direction (right) than ${\displaystyle b}$. But if you multiply the two numbers by -1, they both move to the left of the number line. Both numbers are still the same distance from 0, but because they are on the negative side of 0 ${\displaystyle a\ }$ is now more negative than ${\displaystyle b\ }$, and so ${\displaystyle a. This applies to any two numbers, positive or negative.

### Absolute Values

What are absolute values? Lets say you have a number x, and you put it on a number line. The value of x is the value that it has on the number line, lets say -3. The absolute value of x, expressed in mathematics as ${\displaystyle |x|}$ is the distance of x from 0. Because distance cannot be negative, ${\displaystyle |x|}$ is always greater than or equal to 0. This means that if x = -3 then ${\displaystyle |x|}$ would be 3, because the distance of -3 from 0 is 3. So if x is greater than or equal to 0, ${\displaystyle |x|}$ will just be its normal value x. But if x is less than 0, then ${\displaystyle |x|}$will be its negative value (the negative of a negative number), and so will be positive. This can be expressed as

${\displaystyle |x\vert ={\begin{cases}x,&x\geq 0\\-x,&x<0\end{cases}}}$

where |x| is the absolute value of x.

${\displaystyle |ab|=|a|.|b|}$

${\displaystyle |a+b|\leq |a|+|b|}$

## Algebraic manipulation

### Simplification

Removing grouping symbols and collecting like terms...

### Substitution

Evaluating expressions using giving values and formulas. Usually requires you to make something the subject first.

### Factorization

distributive law?

#### Difference of two squares

x2 - y2. This can be factorized. Can you guess how? It's probably a good idea to add some maths in here instead of just giving you the formula, so lets look at the geometrical interpretation of this. Both pronumerals are squared. The terms squared and cubed come from the formula for area of a square and volume of a cube. The area of a square is equal to the length of one side multiplied by the length of another, and because the sides of a square are all equal, the area is the length of one side multiplied by itself, so it is raised to the power of 2, or squared =).

So if you have one squared number subtracted from another squared number, you could interpret this as one area of a square being subtracted from another area of another square. One square has side length x, while the other square has side length y. Does it matter which square is larger? Not really, if the area being subtracted is larger, you will end up with a negative area, and even though area, like distance, cannot be negative, we are not talking about real area, and x2 - y2 is allowed to be negative because it is an algebraic expression. If you don't like that, then think of it this way x2 - y2 = -(y2 - x2), so if y is larger than x, then just subtract the area of the x square from the area of the y square to get a positive area, then make that negative.

So, think about this in terms of a larger square with sides x having a smaller square with sides y subtracted from it. Subtracting a square from a square is going to leave a square shaped hole in the larger square, lets say in the corner. Now what used to be the larger square is going to have 6 edges. The two edges that were not cut will still have a length x. The other two sides of the original square are going to have a length of x-y, and the two inner edges created by the hole are going to have a length y. From this we can calculate the remaining area. There are two rectangles, each with area (x-y)×y, and one square with area (x-y)(x-y). If we expand and simplify these we get 2xy - 2y2 + x2 - 2xy + y2, which can be written as y2 + 2xy + x2 - 2xy, which factorizes to y(y+2x) + x(x-2y), then y(x+y) - xy - x(x-y) + xy, which can be further factorizes to (x-y)(x+y) simplifies to x2-y2. So

${\displaystyle x^{2}-y^{2}=(x+y)(x-y)\;}$

You could try the same thing with x2+y2, but it doesn't get you anything really useful, just a(a-b) + b(a+b).

Another way to looks at this kind of factorisation problem, especially if there aren't any geometrical methods to factorize it, is to look for an expression that looks similar, but gives slightly different results, and then figure out what needs to be done in order to get from that expression to the first.e.g.

${\displaystyle (x+y)^{2}=x^{2}+2xy+y^{2}}$

${\displaystyle x^{2}-y^{2}=x^{2}+2xy+y^{2}+a}$

${\displaystyle a=x^{2}-y^{2}-(x^{2}+2xy+y^{2})}$

${\displaystyle a=-2xy-2y^{2}}$

${\displaystyle x^{2}-y^{2}=x^{2}+2xy+y^{2}-2xy-2y^{2}}$

${\displaystyle x^{2}-y^{2}=x^{2}+xy+y^{2}-xy-2y^{2}+xy-xy}$

${\displaystyle x^{2}-y^{2}=x(x+y)-y(y+x)}$

${\displaystyle x^{2}-y^{2}=(x-y)(x+y)}$

A third way to look at it is to add something to both sides.

${\displaystyle xy-xy=0}$

${\displaystyle x^{2}-y^{2}=x^{2}-y^{2}}$

${\displaystyle x^{2}-y^{2}+0=x^{2}+xy-xy-y^{2}}$

${\displaystyle x^{2}-y^{2}=x(x+y)-y(x+y)}$

${\displaystyle x^{2}-y^{2}=(x-y)(x+y)}$

probably went into too much detail on this. Awell.

#### The sum and difference of two cubes

${\displaystyle a^{3}-b^{3}=(a-b)(a^{2}+ab-b^{2})}$

${\displaystyle a^{3}+b^{3}=(a+b)(a^{2}-ab-b^{2})}$

### Algebraic fractions

#### Reduction

factorizing numerator and denominator and canceling any common factors

LCM

## Linear equations

Syllabus Point: 1.4 (i)

${\displaystyle ax+b=0,\,a\neq 0}$

## Linear inequalities

Syllabus Point: 1.4 (ii)

if both sides of an inequality are multiplied by a negative number, the direction of the inequality is reversed.

Syllabus Point: 1.4 (iii)

${\displaystyle ax^{2}+bx+c=0,a\neq 0}$

### Factorisation

Example:

${\displaystyle 2x^{2}+11x+12=0}$

${\displaystyle (2x+3)(x+4)=0}$

${\displaystyle (2x+3)=0}$ OR

${\displaystyle (x+4)=0}$

${\displaystyle x={\frac {-3}{2}}}$ OR

${\displaystyle x=-4}$

## Simultaneous equations

Syllabus Point: 1.4 (iv)

Simultaneous equations are also referred to as systems of liner equations. The solution of simultaneous equations is any ordered pairs of the variables that can satisfy the constraints of the system.

Example:

${\displaystyle 3x+4y=25}$

${\displaystyle 5x-3y=3}$

In this example, the ordered pair (3,4) can satisfy all the equations, thus making it a solution of this system. When ${\displaystyle x=3}$ and ${\displaystyle y=4}$, both equations( ${\displaystyle 3*3+4*4=25}$ and ${\displaystyle 5*3-3*4=3}$) are satisfied.

The content below represent the methods to solve simultaneous equations.

### Elimination

The elimination method is sometimes also known as the linear combination method. This method attempts to add or subtract equations so that the equation set can be reduced to one simple linear equation.

Example:

${\displaystyle x+3y=13}$

${\displaystyle 3x-2y=-5}$

From this stage,  we can try to eliminate x and solve for y. (It is possible to eliminate y and solve for x.) Start by multiplying equation 1 by 3 to match the coefficient of x for equation 1 and equation 2.

${\displaystyle 3*(x+3y)=3*(13)}$

${\displaystyle 3x+9y=39}$

Please note that when operating on one side of the equation, you must also operate the other side to keep the Left hand side equal to the Right hand side.

With the coefficient of X equal on both equations, it is now possible to eliminate x by subtracting equation 2 from equation 1.

${\displaystyle 3x+9y=39}$

- ${\displaystyle (3x-2y=-5)}$

= ${\displaystyle (3x-3x)+(9y--2y)=39--5}$

= ${\displaystyle 11y=44}$

From here we can solve this equation like a simple linear equation.

${\displaystyle 11y=44}$

${\displaystyle 11y/11=44/11}$

${\displaystyle y=4}$

We can now substitute this back to either equation 1 or 2 to get the full answer.

${\displaystyle x+3*4=13}$

${\displaystyle x+12=13}$

${\displaystyle x=13-12}$

${\displaystyle x=1}$

${\displaystyle x=1}$ and ${\displaystyle y=4}$