Algebra and Graphs

Introduction to algebra

Algebra is like a maths game where we use letters to stand for numbers we don't know yet.

For example, if I say "I have $x$ candies, and I eat 3", then $x - 3$ means how many candies are left. It is just a way to represent patterns or unknown values.

Replace $x$ with 4:

$2 \times 4 + 3 = 8 + 3 = 11$

Algebraic manipulation

Simplify expressions by collecting like terms

Like terms are terms that have the same letters and powers. You can add or subtract them.

Group like terms:

$(2a^2 + 5a^2) + (3ab - 9ab) + (-1 + 4) = 7a^2 - 6ab + 3$

Expand products of algebraic expressions

To expand means to get rid of brackets.

$3x \times 2x - 3x \times 4y = 6x^2 - 12xy$

First do two brackets with the FOIL method (First, Outside, Inside, Last) so that you do not forget any term, for $(x - 2)(x + 3)$:

• First: $x \times x = x^2$
• Outside: $x \times 3 = +3x$
• Inside: $-2 \times x = -2x$
• Last: $-2 \times 3 = -6$

Now add it all up: $x^2 + 3x - 2x - 6 = x^2 + x - 6$.

So now the question becomes $(x^2 + x - 6)(2x + 1)$. Multiply every term in the first bracket by every term in the second:

$(x^2 \times 2x) + (x \times 2x) + (-6 \times 2x) + (x^2 \times 1) + (x \times 1) + (-6 \times 1)$

$= 2x^3 + 2x^2 - 12x + x^2 + x - 6 = 2x^3 + 3x^2 - 11x - 6$

Factorise by taking out common factors

Factorising is the opposite of expanding.

What do both terms have in common? They both have $x$, and they are both divisible by 3, so you take that out:

$9x^2 + 15xy = 3x(3x + 5y)$

Factorise special expressions

Type 1: $ax^2 + bx + kay + kby$ (grouping)

Group it like this: $(2x^2 + 4xy) + (3x + 6y)$. Now take out common factors:

$2x(x + 2y) + 3(x + 2y)$

Since $(x + 2y)$ is in both, the final answer is $(2x + 3)(x + 2y)$.

Type 2: $a^2x^2 - b^2y^2$ (difference of squares)

$16x^2 - 25y^2 = (4x)^2 - (5y)^2 = (4x + 5y)(4x - 5y)$

Type 3: $a^2 + 2ab + b^2$ (perfect square)

For example, $x^2 + 6x + 9 = (x + 3)(x + 3) = (x + 3)^2$, and in the same way $x^2 - 6x + 9 = (x - 3)^2$.

Type 4: completing the square

Focus on $x^2 + 6x$.

• Take half of 6, which is 3. This is because $(x + a)^2 = x^2 + 2ax + a^2$.
• So $(x + 3)^2 = x^2 + 6x + 9 = x^2 + 2(3)x + 9$.

By adding and subtracting 9, the expression remains the same:

$x^2 + 6x + 5 + 9 - 9 = (x^2 + 6x + 9) - 9 + 5 = (x + 3)^2 - 4$

A more complex question: $2x^2 + 8x + 3$. Factor out 2 from the first two terms (to make the coefficient of $x^2$ one), then complete the square inside the bracket:

$2(x^2 + 4x) + 3 = 2\left[x^2 + 4x + 4 - 4\right] + 3 = 2\left[(x + 2)^2 - 4\right] + 3$

$= 2(x + 2)^2 - 8 + 3 = 2(x + 2)^2 - 5$

Type 5: mid-term breaking

1. Find two numbers that multiply to $2 \times 5 = 10$ and add up to 11. These are 1 and 10.
2. Break the middle: $2x^2 + 10x + x + 5$.
3. Group: $(2x^2 + 10x) + (x + 5)$.
4. Factor: $2x(x + 5) + 1(x + 5)$.
5. Final factors: $(2x + 1)(x + 5)$.

Algebraic fractions

Addition and subtraction

Make the denominators the same.

$\frac{2x}{6} + \frac{3(x - 4)}{6} = \frac{2x + 3x - 12}{6} = \frac{5x - 12}{6}$

Multiplication

Just multiply the numerators and denominators.

$\frac{2x}{3} \times \frac{x - 4}{2} = \frac{2x(x - 4)}{6}$

Division

Flip the second fraction and multiply.

$\frac{2x}{3} \div \frac{x - 4}{2} = \frac{2x}{3} \times \frac{2}{x - 4} = \frac{4x}{3(x - 4)}$

Factorise and simplify rational expressions

Factorise the top: $x^2 - 2x = x(x - 2)$.

Factorise the bottom with mid-term breaking: we need two numbers that multiply to 6 and add to $-5$. They are $-3$ and $-2$.

$x^2 - 3x - 2x + 6 = x(x - 3) - 2(x - 3) = (x - 3)(x - 2)$

Now cancel the common factor $(x - 2)$:

$\frac{x(x - 2)}{(x - 3)(x - 2)} = \frac{x}{x - 3}$

Make the denominators the same:

$\frac{1\times(x - 3)}{(x - 2)(x - 3)} + \frac{(x + 1)(x - 2)}{(x - 3)(x - 2)} = \frac{(x - 3) + (x + 1)(x - 2)}{(x - 2)(x - 3)}$

Leave the answer like this, because it is simplified.

Indices (again)

A negative index means a reciprocal, so $2x^{-3} = \dfrac{2}{x^3}$. Dividing by a fraction means multiplying by its flip:

$\frac{2}{5x^2} \div \frac{2}{x^3} = \frac{2}{5x^2} \times \frac{x^3}{2} = \frac{x}{5}$

Equations

An equation is used to describe the relationship between two or more things. For example, the area of a rectangle can be written like this: Area $=$ Length $\times$ Width. If the length is 5 cm and the width is 6 cm, the area is $5 \times 6 = 30\ \text{cm}^2$.

Solving linear equations

We can modify equations without changing the actual relationships.

Subtract 4 from both sides: $3x = 10 - 4 = 6$. Divide both sides by 3: $x = 2$.

A harder question:

1. Expand the brackets: $2x - 6 = 3x + 6$.
2. Take the 6 on the right to the left ($-6$ on the left): $2x - 6 - 6 = 3x$, so $2x - 12 = 3x$.
3. Take the $2x$ to the right ($-2x$ on the right): $-12 = 3x - 2x$.

So $x = -12$.

The very important table (memorise!)

Solving quadratic equations

Previously we learnt different techniques to factorise expressions so that we can solve different types of equations. The quadratic formula solves equations that cannot be factorised or solved through mid-term breaking.

Always compare with $ax^2 + bx + c = 0$. For example, $x^2 + 3x - 10 = 0$ gives $a = 1$, $b = 3$, $c = -10$.

Plug all the values into this equation. You will always get two answers (sometimes they might be the same).

$x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-10)}}{2(1)} = \frac{-3 \pm \sqrt{49}}{2}$

$x = \frac{-3 + 7}{2} = 2 \qquad \text{or} \qquad x = \frac{-3 - 7}{2} = -5$

Now let's solve an equation using the completing the square method.

$x^2 + 6x = -5$

$x^2 + 6x + 9 = -5 + 9$

$(x + 3)^2 = 4$

$x + 3 = \pm 2$

So $x + 3 = 2$ gives $x = -1$, and $x + 3 = -2$ gives $x = -5$.

Solving fractional equations

Make the denominators the same:

$\frac{2(x - 2) + 5(x - 1)}{(x - 1)(x - 2)} = 2$

The denominator is being divided over the right side, so it will be multiplied there:

$2(x - 2) + 5(x - 1) = 2(x - 1)(x - 2)$

Expand both sides:

$2x - 4 + 5x - 5 = 2(x^2 - 3x + 2) = 2x^2 - 6x + 4$

Take everything to one side and group the same terms:

$2x^2 - 13x + 13 = 0$

Now use the quadratic formula with $a = 2$, $b = -13$, $c = 13$:

$x = \frac{13 \pm \sqrt{169 - 104}}{4} = \frac{13 \pm \sqrt{65}}{4}$

So $x = 5.27$ or $x = 1.23$ (to 3 significant figures).

Solving simultaneous equations

Now we will find the values of two unknowns using two different equations:

$2x + 3y = 12 \qquad\qquad 4x - y = 5$

Elimination method. Our goal is to cancel one of the variables by adding or subtracting the equations. Multiply the second equation by 3 to make the coefficients of $y$ match: $(4x - y = 5) \times 3 \Rightarrow 12x - 3y = 15$.

Now add the two equations:

$(2x + 3y) + (12x - 3y) = 12 + 15 \quad\Rightarrow\quad 14x = 27 \quad\Rightarrow\quad x = \frac{27}{14}$

Substitute $x$ into one of the original equations (it is easier to use the first one):

$2\left(\frac{27}{14}\right) + 3y = 12 \quad\Rightarrow\quad \frac{27}{7} + 3y = 12 \quad\Rightarrow\quad 3y = 12 - \frac{27}{7} = \frac{57}{7} \quad\Rightarrow\quad y = \frac{57}{21} = \frac{19}{7}$

Substitution method. Make one of the variables the subject (whichever is easier) and plug it into the other equation. From $4x - y = 5$ we get $y = 4x - 5$. Plug this into the other equation:

$2x + 3(4x - 5) = 12 \quad\Rightarrow\quad 2x + 12x - 15 = 12 \quad\Rightarrow\quad 14x = 27 \quad\Rightarrow\quad x = \frac{27}{14}$

$y = 4\left(\frac{27}{14}\right) - 5 = \frac{19}{7}$

Changing the subject of the equation

This means making another letter the main subject.

$2x + 3y = 2z \quad\Rightarrow\quad 3y = 2z - 2x \quad\Rightarrow\quad 3y = 2(z - x) \quad\Rightarrow\quad y = \frac{2(z - x)}{3}$

Inequalities

Inequalities are used when the value of a variable lies in a certain range. To represent them, we use the symbols $<$, $>$, $\le$, $\ge$. For example, if $x$ can be 3 or any value above 3, we write $x \ge 3$ ($x$ is greater than or equal to 3).

Representing inequalities on a number line

This means that $x$ can be any value between $-3$ and $1$, but 1 is not included.

• Closed circle if it is $\le$ or $\ge$ (including the number).
• Open circle if it is $<$ or $>$ (strict, not including the number).

On the number line: a filled dot at $-3$ (because it includes $-3$) and an open circle at 1 (because it does NOT include 1).

Solving linear inequalities

You solve inequalities like you solve normal equations, except for one thing.

$-2x + 2 < 2x - 2 \quad\Rightarrow\quad -2x - 2x < -2 - 2 \quad\Rightarrow\quad -4x < -4$

Divide both sides by $-4$. The sign flips because we divided by a negative number:

$x > 1$

Solve each part separately:

• $-3 \le 3x - 2$ gives $-3 + 2 \le 3x$, so $-1 \le 3x$, so $x \ge -\frac{1}{3}$.
• $3x - 2 < 7$ gives $3x < 9$, so $x < 3$.

Together: $-\dfrac{1}{3} \le x < 3$.

Representing inequalities on graphs

In 2D graphs (with $x$ and $y$), inequalities are shown as lines and shaded areas.

• Use a broken (dotted) line if it is $<$ or $>$ (not including the line).
• Use a solid line if it is $\le$ or $\ge$ (including the line).
• Shade the unwanted region unless the question says otherwise.

First, check for $x$: the region is between 1 and 3 and the line is solid. Hence $1 \le x \le 3$.

Similarly, $y$ is between 2 and 4, so $2 \le y \le 4$.

Sequences

• A sequence is just a list of numbers that follow a pattern.
• Each number is called a term.
• For example, $2, 4, 6, 8, 10, 12, \dots$ is a sequence where each number goes up by 2.

Continuing sequences

The pattern is $+3$ each time, so the following numbers are $17, 20, 23$.

• The difference between the first and second terms is 1.
• For the next terms, it is 2, then 3, then 4, then 5.
• So the following numbers are $(16 + 6) = 22$, $(22 + 7) = 29$, $(29 + 8) = 37$.

Finding the $n$th term

The $n$th term is the formula to find any term in the sequence.

First find the pattern: each time you add 4. Now derive a formula for the $n$th term:

$n\text{th term} = 4n - 1$

To find the formula for arithmetic sequences such as this, multiply the difference by $n$ ($+4$) and then adjust it to match the first term (the first term is 3, and $4 \times 1 = 4$, so subtract one from each term).

The 37th term is $4 \times 37 - 1 = 147$.

• For squared terms, the formula will be $n\text{th term} = n^2 + c$, where $c$ is the adjustment, like $-1$ in the arithmetic sequence formula.
• For cubed terms, it will be $n^3$ instead of $n^2$.
• For the simple combination above ($2, 4, 7, 11, 16, \dots$), it will be $n\text{th term} = \frac{1}{2}n(n + 1) + c$, where $c$ is the adjustment. Here $c = 1$, because $\frac{1}{2}(1)(2) + 1 = 2$.

Proportion

Proportion means two things change in a connected way. You can write this using the $\propto$ symbol, "is proportional to".

• Start with $y \propto x$, so $y = kx$.
• Use the values to find $k$: $12 = k \times 4$, so $k = 3$.
• Now use it to find $y$ when $x = 9$: $y = 3 \times 9 = 27$.

• Start with $y \propto \dfrac{1}{x^2}$, so $y = \dfrac{k}{x^2}$.
• Use the values to find $k$: $5 = \dfrac{k}{2^2}$, so $5 = \dfrac{k}{4}$, so $k = 20$.
• Now find $y$ when $x = 5$: $y = \dfrac{20}{5^2} = 0.8$.

Functions

A function is like a machine: you give it a number (input), and it gives you a new number (output) based on a rule. We use special notation: $f(x)$ means "the function $f$ of $x$".

If $f(x) = 3x - 5$, then $f(2) = 3(2) - 5 = 6 - 5 = 1$.

Domain and range

These are like the rules for input and output.

• Domain: what values you are allowed to put into the function (the inputs). Usually, it is $x$.
• Range: what values can come out (the outputs). Usually, it is $f(x)$.

You can plug any real number into $x$, so the domain is all real numbers. Because squaring never gives a negative result, the smallest output is when $x = 0$: $h(0) = 2(0)^2 + 3 = 3$. So the range is $y \ge 3$.

Some functions cannot have all real numbers as their domain. For example, $y = \sqrt{x}$ has the domain $x \ge 0$ (only zero and above), because negative numbers do not have a real square root.

Inverse functions

An inverse function reverses what the actual function did.

1. Replace $f(x)$ with $y$: $y = 3x - 5$.
2. Swap $x$ and $y$: $x = 3y - 5$.
3. Make $y$ the subject: $x + 5 = 3y$, so $y = \dfrac{x + 5}{3}$.

So $f^{-1}(x) = \dfrac{x + 5}{3}$. That means: if $f(2) = 1$, then $f^{-1}(1) = 2$.

Composite functions

This means putting one function inside another. Let $f(x) = x + 2$ and $g(x) = 3x$.

To find the composite function $g(f(x))$, take the function on the right, $f(x)$, and put it into the one on the left, $g(x)$:

$g(f(x)) = g(x + 2) = 3(x + 2) = 3x + 6$

Graphs in practical situations

Using and interpreting graphs

Graphs are pictures that show information. Instead of using lots of words or numbers in a table, we use lines and shapes to tell a story visually.

A basic graph has:

• $x$-axis (independent variable, horizontal): usually shows time or input.
• $y$-axis (dependent variable, vertical): usually shows distance, speed, money or output.
• Points: little dots that show values.
• Lines or curves: connect the points to show how things change.

This graph shows a person's trip to the grocery store.

• In the first 20 minutes, the person travels 10 metres towards the store.
• He then takes a break for 10 minutes (because he didn't cover any distance).
• He then travels again for 10 minutes and reaches the store.
• In the last 20 minutes, the person travelled back to his home.

Plotting graphs

First plug the values into the equation to find the values of $y$:

For example, $x = -2$ gives $y = 2(-2) - 1 = -5$. Finally, plot the points and draw the graph.

Gradient

The gradient, also called the slope, is calculated using this formula:

To calculate it, first choose 2 points from the graph, for example $(1, 1)$ and $(2, 3)$. Then plug these values into the formula:

$m = \frac{3 - 1}{2 - 1} = 2$

For speed-time and distance-time graphs, remember this:

• The gradient of a distance-time graph $=$ speed (because it shows the distance covered divided by the time taken).
• The gradient of a speed-time graph $=$ acceleration (because it shows the change in speed divided by the time taken).

Distance-time graphs

Using the trip graph above, we can calculate the gradient of each straight part to give the speed. Since the lines are straight, the person is travelling at constant speed.

• First 20 minutes: speed $= \dfrac{10}{20} = 0.5$ m/min.
• Next 10 minutes: stationary (the line is flat).
• The next 10 minutes: speed $= \dfrac{15 - 10}{40 - 30} = 0.5$ m/min.
• The last 20 minutes: speed $= \dfrac{0 - 15}{60 - 40} = -0.75$ m/min.

In the final 20 minutes, the velocity is negative because the person travels back from where he went (going in the opposite direction).

Curved lines. A curved line that gets steeper means the person is accelerating, because he is covering more distance in the same time interval as time goes on.

A curve that gets flatter means the person is decelerating: he covers less and less distance in the same time interval, so his speed is decreasing.

Speed-time graphs

Here is a speed-time graph of a car going towards a certain destination.

We can find the distance travelled by the car in a certain time by calculating the area under the graph. Divide the graph into shapes like triangles, rectangles and trapeziums.

• First 20 s (triangle): distance $= \frac{1}{2} \times 20 \times 10 = 100$ m.
• Next 10 s (rectangle): distance $= (30 - 20) \times 10 = 100$ m.
• From 30 s to 40 s (trapezium): distance $= \dfrac{(15 + 10)}{2} \times (40 - 30) = 125$ m.
• The last 20 s (triangle): distance $= \frac{1}{2} \times (60 - 40) \times 15 = 150$ m.

The total distance is $100 + 100 + 125 + 150 = 475$ m. Since the speed was positive the whole time, the car was only travelling away from its starting point.

In the first 20 s, the graph slopes upwards. This means that the car is accelerating, because the speed of the car is increasing. From 40 s to 60 s, the slope is downwards, which means that the car is decelerating.

Converting units

First write down the units in fraction form: $\dfrac{\text{m}}{\text{s}} \to \dfrac{\text{km}}{\text{h}}$.

Next, convert the units while keeping them in fraction form. There are 1000 metres in a kilometre and 3600 seconds in an hour:

$\frac{\text{m}/1000}{\text{s}/3600} \to \times 3.6$

If we multiply the number by 3.6, we get the final answer: $3000 \times 3.6 = 10\,800$ km/h.

Graphs of functions

Linear functions

A linear function always gives a straight line on a graph. The equation will always follow this format:

For example, $f(x) = 2x - 1$ has gradient 2, and the line crosses the $y$-axis at $-1$. It is the graph we plotted above.

Polynomial functions

Polynomials usually follow the format $y = ax^n$. Some examples are $y = x^2$, $y = x^3$ and $y = x^2 + 2x + 1$.

Graphs of $x^2$ and $-x^2$

To plot the graphs of polynomial functions, plot the points on the graph, then join them using a free hand (make a curve instead of straight lines). The graph of $y = x^2$ is symmetrical about the $y$-axis.

To find the turning point of the graph, convert the equation into the completed-square form.

$x^2 + 6x + 9 - 9 + 3 = (x + 3)^2 - 6$

The $x$-coordinate comes from $x + 3 = 0$, so $x = -3$. The $y$-coordinate is the number at the end, $-6$.

Turning point: $(-3, -6)$.

Graphs of cubics

Reciprocal functions

These functions have vertical and horizontal asymptotes. Asymptotes are lines to which the graph keeps getting closer but never touches.

• The graphs of $\frac{1}{x}$ and $-\frac{1}{x}$ are symmetrical about the line $y = x$ and $y = -x$ (the dotted line on the graph).
• They have the $y$-axis as the vertical asymptote and the $x$-axis as the horizontal asymptote.

The graphs of $\frac{1}{x^2}$ and $-\frac{1}{x^2}$ are symmetrical about the $y$-axis.

Exponential functions

These have the form $y = a^x$, where $a$ is a positive number (not 1).

• The gradient of the graph keeps increasing, hence its name.
• Exponential functions always have a $y$-intercept of 1, except when a number is being added or subtracted.
• For example, the red line is the graph of $y = 2^x$, and the purple line is the graph of $y = 2^x - 1$ (it moves one unit down the $y$-axis).
• The red line has a horizontal asymptote on the $x$-axis ($y = 0$), and the purple one has an asymptote at $y = -1$.

Roots, domain and range from a graph

Another important thing in graphs is the roots of the graph ($x$-intercepts). These are very helpful in finding out what type of graph it is.

• Domain (values of $x$): goes from $-3$ to $5$. Hence $-3 \le x \le 5$.
• Range (values of $y$): goes from $-2$ to $4$. Hence $-2 \le y \le 4$.

The first is a cubic: it has the S-shaped curve, and it crosses the $x$-axis 3 times (three roots). The second is a quadratic: it has the U-shape of a parabola. A quadratic has at most two roots, and a cubic has at most three.