# A-level Mathematics/OCR/C1/Equations

An equation consists of two expressions joined by the equals sign ($=$). Everything on the left-hand side is equal to everything on the right-hand side, for example $2+3=4+1$. Some equations contain a variable, usually denoted by $x$ or $y$ , though any symbol can be used.

## Manipulating expressions

Sometimes, expressions will be messier than they need to be, and they can be represented in an easier to understand form. The skills here are essential to the rest of the A-level course, although it is very likely that you have already covered them at GCSE.

### Collecting like terms

When collecting like terms, you simply add all the terms in $x$ together, all the terms in $y$ together, and all the terms in $z$ together. The same applies for any other letter that represents a variable.
For example, $2x+4y+8z-3x-7y-2z+4x$ becomes:
$2x-3x+4x=3x$
$4y-7y=-3y$
$8z-2z=6z$
So, by adding all the answers, $2x+4y+8z-3x-7y-2z+4x$ simplified is $3x-3y+6z$.

### Multiplication

Multiplication of different variables such as $a\times b$ becomes $ab$. Single variables become indices, so $x\times x$ is $x^{2}$.
Like addition and subtraction, you keep like terms together. So, for example:
$2x^{2}z\times 3yz^{2}\times 4xy^{3}$ becomes: $24{x^{3}}{y^{4}}{z^{3}}$

### Fractions

It is quite often that fractions are encountered. Therefore we need to learn how to handle them properly. When working with fractions, the rule is to make all of the denominators equal, and then write the expression as one fraction. You need to multiply both the top and bottom by the same amount to keep the meaning of the fraction the same.
For example, for ${\frac {3x}{2}}+{\frac {2y}{5}}-{\frac {z}{10}}$, the common denomimator is $10$.
Multiply both parts by $5$: ${\frac {15x}{10}}$
Multiply both parts by $2$: ${\frac {4y}{10}}$
Leave this as it is: ${\frac {z}{10}}$
You now have ${\frac {15x}{10}}+{\frac {4y}{10}}-{\frac {z}{10}}$, which becomes ${\frac {15x+4y-z}{10}}$.

## Solving equations

Often, to solve an equation you must rearrange it so that the unknown term is on its own side of the equals sign. By rearranging $2+x=5$ to $x=5-2$, $x$ has been made the subject of the equation. Now by simplifying the equation, you can find that the solution is $x=3$.
An equation with a variable will only hold true for certain values of that variable. For example $2+x=5$ is only true for $x=3$. The values that the variables have when the equation is true are called the solutions of the equation. Therefore $x=3$ is the solution of the equation $2+x=5$.

### Changing the subject of an equation

You will usually be given equations that are more complex than the example above. To move a term from one side of the equals sign to the other, you have to do the same thing on both sides of the equals sign. For example, to make $x$ the subject of $y={\frac {4a(x^{2}+b)}{3}}$:
| Multiply both sides by $3$ | $3y=4a(x^{2}+b)$ |
| Divide both sides by $4a$ | ${\frac {3y}{4a}}=x^{2}+b$ |
| Subtract $b$ from both sides | ${\frac {3y}{4a}}-b=x^{2}$ |
| Square root both sides | ${\sqrt {{\frac {3y}{4a}}-b}}=x$ or $-{\sqrt {{\frac {3y}{4a}}-b}}=x$ |

### Solving quadratic equations

Quadratic equations are equations where the variable is raised to the power of 2 and, unlike linear equations, there are a maximum of two roots. A root is one value of the variable where the equation is true, and to fully solve an equation, you must find all of the roots. For a quadratic equation you can factorise it and then easily find which values make the equation valid. The example above is quite a simple case. You will usually be given a more complicated equation such as $2x^{2}+5x+3=0$. If the equation isn't already in the form $ax^{2}+bx+c=0$, rearrange it so that it is. The steps needed to factorise $2x^{2}+5x+3=0$ are:
| Multiply $2$ by $3$ (coefficient of $x^{2}$ multiplied by the constant term) | $2\times 3=6$ |
| Find two numbers that add to give $5$ (coefficient of $x$) and multiply to give $6$ (answer from previous step) | $2\times 3=6$, $2+3=5$ |
| Split $5x$ to $2x+3x$ (from the results of the previous step) | $2x^{2}+2x+3x+3=0$ |
| Simplify | $2x(x+1)+3(x+1)=0$ |
| Simplify further | $(2x+3)(x+1)=0$ |
So $(2x+3)(x+1)=0$ is $2x^{2}+5x+3=0$ in factorised form. You can now use the fact that any number multiplied by $0$ is $0$ to find the roots of the equation. The numbers that make one bracket equal to $0$ are the roots of the equation. In the example, the roots are $-1.5$ and $-1$.
It is also possible to solve a quadratic equation using the quadratic formula or by completing the square.

## Simultaneous equations

Simultaneous equations are useful in solving two or more variables at once. Basic simultaneous equations consist of two linear expressions and can be solved by three different methods: elimination, substitution or by plotting the graph.

### Elimination method

The basic principle of the elimination method is to manipulate one or more of the expressions in order to cancel out one of the variables, and then solve for the correct solution.
An example of this:
| $2x+3y=10$ | (1) | (Assigning the number (1) to this equation) |
| $2x+6y=6$ | (2) | (Assigning the number (2) to this equation) |
From this, we can see that by multiplying equation (1) by a factor of 2 and then subtracting this new equation from (2), the $y$-variable will be eliminated.
(1) $\times 2\rightarrow 4x+6y=20$ (1a) (Assigning the number (1a) to this equation)
Now subtracting (2) from (1a):
| | $4x+6y=20$ | (1a) |
| $-$ | $2x+6y=6$ | (2) |
| $=$ | $2x+0y=14$ | |
Now that we have $2x=14$, we can solve for $x$, which in this case is $7$.
$x=7$.
Substitute the newly found $x$ into (1):
$2\times 7+3y=10$
$14+3y=10$
And we find that $y=-4/3$
So, the solution to the two equations (1) and (2) are:
$x=7$
$y=-4/3$

### Substitution method

The substitution method relies on being able to rearrange the expressions to isolate a single variable, in the form variable = expression. From this result this new expression can then be substituted for the variable itself, and the solutions evaluated.
An example of this:
| $2x+3y=12$ | (1) | (Assigning the number (1) to this equation) |
| $x+y=6$ | (2) | (Assigning the number (2) to this equation) |
From this expression, it is possible to see that (2) is the most simplistic expression, and thus will be the better choice to rearrange.
Taking (2), and rearranging this into $x=6-y$. (2a)
Subbing (2a) into (1) we get
$2(6-y)+3y=12$
Solving this, we get that $y=0$
Again we can sub this result into one of the original equations to solve for $x$. In this case $x=6$.
Note that for situations in which one of the equations is non-linear, you must isolate one variable in the linear equation and substitute it into the non-linear one. Then you can solve the quadratic equation with one of the methods above.
Another form of substitution is if you've got a similar expression in both equations, like in this case:
| $2x+3y=10$ | (1) | (Assigning the number (1) to this equation) |
| $2x+6y=6$ | (2) | (Assigning the number (2) to this equation) |
Here, $2x$ is found in both equations, so:
| $2x=10-3y$ | (1) |
| $2x=6-6y$ | (2) |
And since $2x=2x$, you could do:
$10-3y=6-6y$
$6y-3y=6-10$
$3y=-4$
$y=-4/3$
Now you've got $y$, and finding $x$ will be the same as above.
Graph by first solving for y. 2x+3y=12

### Solving problems with simultaneous equations

Often, you will be given problems which you must be able to write out as a pair of simultaneous equations. You will need to recognise such problems, and write them out correctly before solving them. Most problems will be similar to these examples with some differences.
Example
At a record store, 2 albums and 1 single costs £10. 1 album and 2 singles cost £8. Find the cost of an album and the cost of a single.
Taking an album as $a$ and a single as $s$, the two equations would be:
$2a+s=10$
$a+2s=8$
This is part of the C1 (Core Mathematics 1) module of the A-level Mathematics text.