> [!NOTE] Content on this page is adapted from [lecture notes](https://spakula.github.io/fyA/index.html) provided by [Ján Špakula](https://spakula.github.io) ## Simultaneous Linear Equations **[Ján's notes](https://spakula.github.io/fyA/simeq.html)** A linear equation in one variable (or unknown) is an equation in the form $ax=b$ where $a$ and $b$ are constant numbers and $a\ne 0$, for example, $10x=15$. A linear equation in two unknowns is in the form $ax+by=c$ where $a,b \ne 0$. If $c=0$ it is called *homogeneous*. For example, $3x-2y=4$. When you have a pair of linear equations, you can solve them by substitution or elimination. $ 4x+y=9 \qquad -x+y=1 $ ### Substitution Arrange one of the equations to make an unknown the subject, and substitute it into the other equation to eliminate one of the unknowns and solve the other unknown. $ y=x+1 \qquad 4x+x+1=9 \qquad 5x=8 \qquad \boxed{x=1.6} $ Then solve the first equation. $ y=\frac{8}{5}+1 \qquad y=\boxed{2.6} $ ### Elimination Subtract one equation from the other to eliminate at least one of the unknowns $ 4x+y-(-x+y)=9-(1) \qquad 5x=8 \qquad \boxed{x=1.6} $ Then solve for the other term. $ y=9-(4\times 1.6) \qquad \boxed{y=2.6} $ ## Cartesian Coordinates **[Ján's notes](https://spakula.github.io/fyA/rectangular-cartesian-coordinates.html)** The graph of a function in the form $f(x)=mx+c$ is a straight line, where $m$ is called the *slope* and $c$ is called the *intercept*. Note that $c=f(0)$. Given two points $P_{1}$ and $P_{2}$, you can calculate $m$ as the ratio of **rise over run** (vertical change over horizontal change, or gradient) $ P_{1}=(x_{1},\ y_{1}) \qquad P_{2}=(x_{2},\ y_{2}) $ $ m=\frac{y_{2}\ -\ y_{1}}{x_{2}\ -\ x_{1}} $ ## Factorisation **[Ján's notes](https://spakula.github.io/fyA/factorising.html)** Factors are the individual constituents of a product expression. For example: $ 1\cdot 2 \cdot 2 \cdot 3 \qquad x(y+z) \qquad (a+b)(b+3) $ Factorising is writing something as a product. For example: $ 12=1 \cdot 2 \cdot 2 \cdot 3 \qquad xy+xz=x(y+z) \qquad b^2+ab+3a+3b=(a+b)(b+3) $ Factors of a whole number $a$ always include $1$ and $a$. A prime number has no other factors, other than $1$ and itself. It is not always possible to sensibly factorise a given algebraic expression. ### Difference of Squares It is useful to remember the difference of squares formula, which is a general solution to factorising two square numbers. $ a^2-b^2=(a-b)(a+b) $ For example: $ x^2-25=(x+5)(x-5) \qquad m^2-49n^2=(m+7n)(m-7n) $ $ 4y^2-1=(2y+1)(2y-1) \qquad 81a^2-9b^2=(9a+3b)(9a-3b) $ ### Magic X Method The magic X method is a technique for factorising quadratic expressions in the form $ax^2+bx+c$. It works by finding two numbers that multiply to give $ac$ (the product of the coefficient of $x^2$ and the constant term) and add to give $b$ (the coefficient of $x$). 1. Draw an X diagram 2. Write $ac$ at the bottom and $b$ at the top 3. Find two numbers that multiply to $ac$ and add to $b$, and write them either side of the X 4. Divide each number by $a$ (the coefficient of $x^2$) and simplify 5. Write the factors as $(dx+e)(fx+g)$ where the denominators become the coefficients of $x$ and the numerators become the constants #### Examples **Example 1** $ 2x^2+7x+3 $ $ a=2 \qquad b=7 \qquad c=3 \qquad ac=6 $ We need two numbers that multiply to $6$ and add to $7$. $ \begin{array}{c} & 7 & \\ 6 & \times & 1 \\ & 6 & \end{array} $ The numbers are $6$ and $1$. Divide by $a=2$ and simplify: $ \frac{6}{2}=\frac{3}{1} \qquad \frac{1}{2} $ $ \boxed{2x^2+7x+3=(x+3)(2x+1)} $ **Example 2** $ 6x^2-7x-3 $ $ a=6 \qquad b=-7 \qquad c=-3 \qquad ac=-18 $ We need two numbers that multiply to $-18$ and add to $-7$. $ \begin{array}{c} & -7 & \\ -9 & \times & 2 \\ & -18 & \end{array} $ The numbers are $-9$ and $2$. Divide by $a=6$ and simplify: $ \frac{-9}{6}=\frac{-3}{2} \qquad \frac{2}{6}=\frac{1}{3} $ $ \boxed{6x^2-7x-3=(2x-3)(3x+1)} $ **Example 3** $ x^2+5x+6 $ $ a=1 \qquad b=5 \qquad c=6 \qquad ac=6 $ We need two numbers that multiply to $6$ and add to $5$. $ \begin{array}{c} & 5 & \\ 3 & \times & 2 \\ & 6 & \end{array} $ The numbers are $3$ and $2$. Divide by $a=1$ and simplify: $ \frac{3}{1} \qquad \frac{2}{1} $ $ \boxed{x^2+5x+6=(x+3)(x+2)} $ When $a=1$, the magic X method simplifies to finding two numbers that multiply to $c$ and add to $b$, which directly become the constants in the factors.