The word "Algebra" comes from the Arabic word Al-Jabr in the title of a book, *Hidab al-jabr wal-muqubala*, written in Baghdad about 825 A.D. by the Arab mathematician Mohammed ibn-Musa al-Khowarizmi.

The title has been translated to mean "science of restoration (or reunion) and opposition" or "science of transposition and cancellation" and "The Book of Completion and Cancellation".

Algebra is used to find the missing number in ☐ – 6 = 10. But rather than use ☐, algebra uses **letters for unknowns**, so we write x – 6 = 10

To solve the equation, the left-side, where x is lessened by 6, is "restored" or "completed" back to x. To do this we add 6. But we also must add 6 to the right for completeness, so x = 16

**Using a balance**

If we write x + 2 = 3 then we ask how do we restore the left side to x.

We subtract 2. But we must also take 2 from the right for completeness. So x = 1. *x* is the **unknown** and it equals 1.

The process to work this out is to imagine a balance with *x* + 2 on the left and 3 on the right. If we remove the 2 from the left the balance tilts. If we also take off 2 from the right, giving 1, the balance is restored with *x* on the left and 1 on the right. So *x* = 1

Q: In *y* + 6 = 10 what is the unknown and what value does it have?

The word equate comes from Latin 'aequare' meaning identical. Instead of using the phrase "is equal to" the sign = was invented in 1557 by a Welsh mathematician.

So an equation has two parts, either side of an = sign and one thing is the same as another.

*x* + 2 = 3 is an **Equation**. It means the unknown value *x* plus 2 equals 3.

Another name for *x* is **variable** which is a symbol for a number we don't know yet. The + is called an **operator** and the numbers are **constants**.

An **Identity** is an equation that is true for all values (use the ≡ identical symbol)

e.g. 3a + 2a ≡ 5a; x^{2} + x^{2} ≡ 2x^{2}

Q1: In 2a – 6 = 3 – a what is the variable, operator and constant?

Q2: Is x^{2} = 100 an identity?

If we write *x* + 2 = 3, each part *x*, 2 and 3 are called **terms**

Q. In 2a – 6 = 3 – a What are the terms?

Can you work out the value for a?

An **expression** is a group of terms

In 2a – 6 = 3 – a, the 2a – 6 and 3 – a are both expressions. Each expression has two terms.

Q. In a + b – 2 = 2 – a – b

What are the expressions?

How many terms are in each expression?

Polynomial comes from poly- ("many") and -nomial ("term") meaning "many terms"

A polynomial can have:

- constants (like 3, -20, or ½)
- variables (like x and y)
- exponents (like the 3 in x
^{3}), but only 0, 1, 2, 3, ... are allowed - but NOT division by a variable (i.e.
^{3}/_{y}is not allowed)

Q. Which of theses are polynomials:

1) 3xy^{–3} 2) 3/(x+1) 3) 3a^{2} + 9 4) 8

An equation is a statement involving an unknown number, typically called x.

In a "linear" equation, x is only to the first power. e.g. 6x = 9; 2x + y = 7;

y^{2} – 4 = 5 is not a linear equation because it has a squared term.

A linear equation can be written as a function, with f(x) instead of y,

so f(x) = 3x – 1 rather than y = 3x – 1

Solving any equation corresponds to the four operations of arithmetic.

1. If x + a = b, then x = b – a.

"If a number is added on one side of an equation, we may subtract it on the other side."

2. If x – a = b, then x = b + a.

"If a number is subtracted on one side of an equation, we may add it on the other side."

3. If ax = b, then x = b ÷ a

"If a number multiplies one side of an equation, we may divide it on the other side."

4. If x÷a = b, then x = ab.

"If a number divides one side of an equation, we may multiply it on the other side."

Simplify means collecting the **same terms together(like terms).**.

Be careful because, ab = a×b and ba = b×a so infact ab = ba.

i.e. 2ab + 3ba = 5ab (or 5ba)

a^{2}b + ba^{2}= 2a^{2}b

Look at this example:

Mark out those things that are the same, **including their + or – sign**, then add or subtract.

In the example below only the ab's can be added. The other terms are different.

Q. Simplify

1. 3x + 2y + x – 3y =
2. 3xy^{2} + 2y + 4xy^{2}=

3. 4x^{2} + 2x – 5x^{2} =

Substitute means replacing a letter with a numerical value. This can be in a formulae and expressions, including scientific formula.

Given that v = u + at,

find v when t = 1, a = 2 and u = 7;

v = 7 + 2 × 1 = 9

Q. Calculate v = √u² + 2as

with u = 2.1, s = 0.18, a = –9.8

A factor is a number that divides into another number.

The factors of 8 are 1, 2, 4 or 8. Factors work in pairs multiplying to give 8.

The common factors of 4 and 8 are 1, 2, 4.

8 is **not** a common factor.

**Factorise means find the common factor.**

i.e. Factorise the two terms: 3 + 6.

3 goes into 3 and 6 so it is the factor.

Put the common factor at the front of a bracket with two missing terms inside.

3 ( ☐ + ☐ )

A bracket means that when we multiply the factor outside the brackets with the terms inside we get the original terms.

So now we have to work out the missing terms that multiply by the factor 3 to give 3 and 6 (the originals)

3 + 6 →3(☐ + ☐) → 3(1 + 2)

because 3 × 1 = 3 and 3 × 2 = 6

It works the same for letters.

Factorise 3x + 6y

The common factor of 3x and 6y is 3 so factorising gives 3(x + 2y)

The common factor of 4x + x^{2} is x so factorising gives x(4 + x)

Q. Factorise

1. 5x + 10y
2. 2x – 4xy
3. 2x^{2} + 4x

This is the opposite of factorising and we multiply the common factor at the front of the brackets by the terms inside.

Expand

3(2x – 5y + 1) = 6x – 5x + 3

2(2x^{2} – 4y) = 4x^{2} – 8y

2x(4x^{2} – 3x + 2) = 8x^{3} – 6x^{2} + 4x

2xy^{2}( xy + 2x – 3y^{2}) = 2x^{2}y^{3} + 4x^{2}y^{2} – 6xy^{4}

Q. Expand

1. 5(x + 2y)
2. 2x(3x – 4y)
3. 2x^{2}(x + 1)

When an equation is given a value we can solve it. e.g 4y + 16 = 10.

This means that when an unknown number (y) is multiplied by 4 and then 16 added it makes 10. You have to find y.

To work this out, do the SAME operations (e.g. adding or subtracting) to both sides of the equation to get only letters on one side of the equals sign and numbers on the other.

Get rid of the 16 by taking 16 away from the left. Make sure you do the same to the right side.

So we get 4y + 16 *–16* = 10 *–16*.

This means 4y = –6.

Now divide both sides by 4 so we just get y on the left.

This means y = –^{4}/_{6}

Q. Solve

1. 2(x + 3) = 18
2. 8 + x = 6 + 2x

3. 8 + x = 6 + 6x

By turning a word question into algebra, we can solve it:

Q. There are twice as many sheep as cows in a field. How do we write this as an equation?

Let number of cows be C and the number of sheep be S.

There are twice as many sheep as cows, so S must be twice C.

We write this as S = 2C

Q. Jess is 4 years older than Matt. Their total age is 100. How old are they?

Let Jess be x years old, using a letter, because her age is unknown.

Jess is 4 years older than Matt. So Matt's age is x – 4 years old.

Total age is x(Jess) + x – 4(Matt) = 100.

or 2x – 4 = 100; so 2x = 104; so x = 52.

So Jess is 52 and Matt is 48.

Q. Cyril, Jen and Brian combined age is 100.

Cyril is twice as old as Jen and she is 4 years older than Brian.

How old are they?

Add powers when they are multiplied:

y^{2} × y^{5} = y^{7}

Subtract powers when they are divided :

y^{8} ÷ y^{3} = y^{5}

But (y^{2})^{4} = y×y × y×y × y×y × y×y = y^{8}

Q. Solve

y³ × y² × y × y³
/
y^{4} × y^{2} × y^{3}

Remember the balance - do the same thing to both sides of the equation.

Rearrange to make y the subject: x = y + 2

Subtract 2 from both sides, x * –2* = y + 2

Make y the subject: 3x = 2y + 1

Subtract 1 from both sides, 3x – 1 = 2y

Divide both sides by 2 and y is:

3x − 1 / 2

Q. Make z the subject: 4y + 5 = 3z – 5