# Daily Maths Revision – Week 10 Walkthrough

If you’re struggling with this week’s questions, this walkthrough should help you get the skills you need.

This week, we’re exploring:

### Question 1: Adding and Subtracting Algebraic Fractions

The key to adding or subtracting fractions is having a common denominator. This is the case whether or not the fractions are algebraic – we need to find a common denominator for the fractions before adding or subtracting.

Make sure you are confident with expanding double brackets. Take a look at week 9 if you need a reminder.

Simplify the expression below:

\begin{aligned} \boldsymbol{\frac{x}{x-7}-\frac{x-1}{x+3}}\end{aligned}

We need to get both terms in this expression to have the same denominator. We find this by multiplying the denominators of the two fractions together: $x-7$ and $x+3$. As with numerical fractions, whatever we do to the denominator we have to also do to the numerator.

\begin{aligned} \frac{x}{x-7}-\frac{x-1}{x+3} &= \frac{x(x+3)}{(x-7)(x+3)} - \frac{(x-1)(x-7)}{(x+3)(x-7)}\end{aligned}

Now, we can expand the brackets on the numerators and the denominators.

\begin{aligned} \frac{x(x+3)}{(x-7)(x+3)} - \frac{(x-1)(x-7)}{(x+3)(x-7)} &= \frac{x^2+3x}{x^2-4x-21} - \frac{x^2-8x+7}{x^2-4x-21} \end{aligned}

Next, we simply subtract the numerators and simplify.

\begin{aligned} \frac{x^2+3x}{x^2-4x-21} - \frac{x^2-8x+7}{x^2-4x-21} &= \frac{x^2+3x-(x^2-8x+7)}{x^2-4x-21} \\ &= \boldsymbol{\frac{11x-7}{x^2-4x-21}} \end{aligned}

### Question 2: Bearings

Bearings are a particular way of describing direction. There are 3 rules you need to remember when working with bearings:

1. Bearings must be measured from north.
2. Bearings must be measured in a clockwise direction.
3. Bearings must be written as three figures.

You may be asked for the bearing of A from B. You will always measure the angle at the point the bearing is from.

Calculate the bearing of X from Y.

First, we use co-interior angles to find the acute angle at Y. Remember, co-interior angles add up to 180°.

180 – 107 = 73°

This isn’t our final answer. The angle needs to be measured clockwise from north. To find the bearing, we need to subtract 73° from 360°.

360 – 73 = 287°

Now, quickly run through the checklist. The angle has been measured clockwise from north and has been written as three figures. If our answer was 2-digits long, we would add a 0 to the front of the angle. This won’t change the angle but will mean the bearing will be written as three figures.

### Question 3: Converting Recurring Decimals to Fractions

A recurring decimal is a decimal that repeats the same number, or group of numbers, forever. Any recurring decimal can be written as a fraction – you may be able to recognise some straight away but others are more complicated.

Write $\boldsymbol{0.\dot{2}\dot{3}}$ as a fraction.

We start by assigning the number a letter – this is simply to communicate the steps we do.

$x = 0.\dot{2}\dot{3}$

Next, we multiply this number to get a second number with the same recurring digits. In this case, we multiply by 100.

$100x = 23.\dot{2}\dot{3}$

Now, we find $100x - x$. As the numbers have the same recurring digits, these will cancel out when we subtract and leave us with an integer value:

$99x = 23$

Finally, we divide by 99 to get the fractional value of $x$:

$x = \frac{23}{99}$

The key to any question like this is getting two values which have the same recurring digits. If you are not sure what to multiply to get this, try 10, 100 and so on until you have two that match.

Write $\boldsymbol{3.0\dot{1}\dot{9}}$ as a mixed number in its simplest form.

First, we assign this number a letter.

$x = 3.0\dot{1}\dot{9}$

This time, it’s not as clear what we should multiply by so let’s find $10x$, $100x$ and so on until we find a pair that will work. It can help to write out the first few numbers in the recurring decimal.

\begin{aligned} x &= 3.019191...\\10x&= 30.191919...\\100x&=301.919191...\\1000x&=3019.191919...\end{aligned}

We can see that $1000x$ and $10x$ have the same recurring digits, so we find the value of $1000x - 10x$

\begin{aligned} 990x&=2989\\ x&=\frac{2989}{990} \end{aligned}

The question asks for a mixed number in its simplest form, so we need to write our fraction in that form. From the decimal, we know that the fraction is a little more than 3 so we can use this to help us.

\begin{aligned} x&=\frac{2989}{990}\\&=\boldsymbol{3\frac{19}{990}}\end{aligned}

### Question 4: Expanding Triple Brackets

Expanding triple brackets is very similar to expanding double brackets – you just need to do it twice!

We start by focusing on two of the brackets and expanding. Then, we write the result in a bracket and multiply by the remaining expression.

Expand $\boldsymbol{(2x+5)(3x-2)(x+3)}$.

Let’s start by expanding the first two brackets. We will do this using the grid method. Remember, you write one of the expressions across the top and the other down the side and multiply each row by each column. If you want a reminder of this, take a look as last week’s walkthrough.

Now, we write this expression across the top of a new grid and write the final bracket down the side.

Just as before, we multiply each column by each row in turn.

\begin{aligned} (6x^2+11x-10)(x+3)&=6x^3+11x^2-10x+18x^2+33x-30\\ &= \boldsymbol{6x^3+29x^2+23x-30} \end{aligned}

### Question 5: Substituting into Functions

Substituting into functions is no different to any other substitution – it’s just a case of understanding the notation.

The function f($x$) is read as f of $x$. It is a function (which is similar to an equation) with the variable $x$. Sometimes, it will have a number instead of $x$ – such as f(3). In this case, you substitute that number in for $x$.

Given that f($\boldsymbol{x}$) = $\boldsymbol{3x^2-5x+\frac{2}{x}}$ , find the value of f(4).

In this case, we need to substitute in $x =4$.

f($4$) = $3\times4^2 - 5\times4+\frac{2}{4}$
= $48-20+\frac{1}{2}$
= $\boldsymbol{28.5}$

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