Expressions, equations etc (F)

The Language of Algebra is how we express a practical problem in mathematical terms. We use letters to denote unknown values.

Look out for various terms which can be translated into mathematical signs Here are the signs and the words often used to replace them in a question: + 'More than' - 'Less than' x 'The product of', 'times' ÷ 'Divided by', 'split equally' = 'Is the same as'

I am 2 years older than Kathryn. Find an expression for my age.

Let Kathryn's age be x. 'Older than' is like 'more than' so we use + My age is x + 2

Simplifying Expressions means collecting similar terms together to make the expressions easier to read.

Look for like terms ie. the same letters. Be careful to keep the correct sign with each term. Also, group constant terms (ie. normal numbers) together.

Collect like terms and simplify the expression: 3 - 4x + 5 + x

Collecting like terms we get 3 + 5 - 4x + x which becomes 8 - 3x

More complex algrebraic expressions can involve combinations of letters and powers of letters. Simplifying these also involves collecting like terms.

When there's combinations of letters and powers of letters, 'like terms' must contain the same combinations and powers so, m²n is NOT a like term to mn².

Don't forget though that the order of the letters in a term isn't important so, m²n is the same as nm²

By first collecting like terms, simplify the expression: 2xy - y² + yx + 3x²y - xy

Collecting like terms we get: 2xy + yx - xy - y² + 3x²y = 2xy - y² + 3x²y (don't forget xy = yx)

Equations come in many forms. To solve those with brackets, first expand the brackets, then solve the equation in the normal way.

Make sure you multiply any term outside a bracket with every term inside.

Solve the equation: 2(x + 1) - 3(x - 3) = 0

Expand the brackets to get: 2x + 2 - (3x - 9) = 0 ie. 2x + 2 - 3x + 9 = 0 11 - x = 0 So, x = 11 Substitute back in to the original equation to check: 2(11 + 1) = 2(12) = 24 and 3(11-3) = 3(8) = 24 so our answer is correct

Multiplying Expressions involves multiplying both numbers and letters.

Take care to multiply both numbers and letters together and to make sure the write power is used if two expressions using the same letter are multiplied together.

Multiply 4p(p - q)

Be careful to multiply each term in the bracket by the one outside. We get 4p² - 4pq

Equations come in many forms. Sometimes the variable appears on both sides eg. 2x - 3 = x + 1

When an equation has the variable on both sides, collect the terms using the variable on one side and constants on the other. Then you can solve the equation.

In order to collect terms on one side, add or subtract amounts from each side. For example: 4x - 3 = 2x + 1 To remove the 2x on the right hand side, subtract 2x from both sides of the equation. 4x - 3 - 2x = 2x + 1 - 2x So, 2x - 3 = 1. Similarly, remove the -3 from the left hand side by adding 3 to each side. 2x - 3 + 3 = 1 + 3 ie. 2x = 4 so x = 2

Remember, what ever you add or subtract to/from one side of an equation, you must add or subtract to/from the other side.

Solve the equation 5 - b = 3 + b

First, add b to both sides. 5 - b + b = 3 + b + b So, 5 = 3 + 2b Now subtract 3 from both sides So, 2 = 2b so b = 1

A Formula expresses a basic rule Eg. Area = πr² is the formula for the area of a circle.

An Expression is an arrangement of algebraic terms - it doesn't equate to anything Eg. 2mn + p is an expression

An Equation is algebraic expression that equals a number or another algebraic expression Eg. 3c - 4 = 7 - c

If there's no equals sign, it's an Expression If there's an equals sign with a single letter or a word (eg. Area) on one side and an expression involving different letters on the other side, you can be pretty certain it's a Formula. If there's an equals sign and you can rearrange the terms to collect the variable on one side and a constant on the other, it's an Equation More advanced equations might be mistaken for formulas but at this level, these rules will apply.

Decide whether these are formulas, expressions or equations (i) 4d = 1 - 2d (ii) 2(t - 1) + 2s(3p + q) (iii) A = 4πr²

(i) There's an equals sign so it's not an expression. There's just one variable, d and we can arrange it to get this on one side and a constant on the other so it's an equation.

(ii) There's no equals sign so it's an expression

(iii) There's an equals sign but we can't separate out a variable to be equal to just a constant so it's a formula.

Expansion means multiplying out brackets Simplification means gathering like terms and combining them into one term Factorisation means finding factors and using brackets to simplify further.

Expansion: Remember to multiply each term inside the bracket by what’s outside it.

Simplification: Don't forget that the order of the letters doesn't matter in an algebraic term: abc = bca = cab.

Factorisation: It's safest to factorise in steps. First find any numerical common factors. Once those have been factored out, then look for common factors made up of letters.

Expand, simplify and factorise if possible the following expression: (3a - 2b)(2b - a) + b(4b - a)

First expand (multiply) out the brackets to get 6ab - 4b² - 3a² + 2ab + 4b² - ab

Simplify by gathering the like terms together: 6ab + 2ab - ab - 4b² + 4b² - 3a² = 7ab - 3a²

Now find common factors across the terms: a(7b - 3a)

A Formula is a means of working out an unknown variable by using information we have about other variables, When the variable that is unknown changes, we need to rearrange the formula to enable us to work it out.

Formulas work like equations. Whatever you do one side of the equals sign, you must to do the other.

Rearrange the formula P = 4t - s to make t the subject

We need to get t on its own on one side of the equals sign. Firstly, add s to both sides P + s = 4t Now divide by 4 and swap the sides to get: t = (P + s)/4

A linear equation is one which the variable appears only to the power 1 Eg. 3x = 2 4(b - 1) = 2(3 + b)

Solving linear equations is straightforward. Collect all terms involving the variable on one side and all constants on the other.

Don't forget, whatever you do to one side of an equation, you must do to the other.

Solve the equation: c/5 = 4 + c

First, let's get rid of the fraction. To do this, multiply both sides by 5 c = 20 + 5c Now, subtract 5c from both sides to get -4c = 20 so, c = -5 Substitute this into the original equation to check the answer: c/5 = -5/5 = -1 4 + c = 4 - 5 = -1 so the answer is correct

Equations with the variable on both sides can be quite complex, involving several expressions.

The key is to take care in expanding brackets and simplifying. Also, it's a good habit to check your answer by substituting it back into the original question.

Solve the equation 3(6 + a) = 4(1 - a)

First, expand the brackets to get 18 + 3a = 4 - 4a Now add 4a to each side 18 + 7a = 4 Now subtract 18 from each side 7a = -14 which makes a = -2 Substitute back into the original equation 3 ( 6 - 2 ) = 12 4 ( 1 - (-2) ) = 4 ( 3 ) = 12 So our answer is correct

Setting Up Equations is the conversion of a practical problem into a mathematical one

The key thing is to understand what information the question is providing and what it wants you to find. It's normally a good move to make the piece of information the question is asking for a variable such as x and take it from there.

Last year your brother was twice as old as you are now. Next year he'll be 12. How old are you now?

They're asking for your age now so let that be x If your brother is 12 next year, then this year he's 11 and last year he was 10 We're told that last year he was twice as old as you are now so, 2x = 10 so x = 5 That is, you are now 5 years old.

When we know the value of one or more variables in an expression or formula, we can Substitute these in to simplify the expression/formula or evaluate it

This is quite straightforward. Just make sure you take care with the arithmetic.

Find the value of x² + xy - y² when x = -1, y = 2

The expression becomes: (-1)² + (-1)(2) - 2² = 1 - 2 - 4 = -5

Iteration is a method for finding approximate solutions to equations that are difficult to solve algebraically. You start with an estimate and repeatedly apply a formula to get closer and closer to the true answer.

An iterative formula has the general form:

where x_n is the current value and x_n+1 is the next (improved) value.

To use iteration to solve an equation:

1. Rearrange the equation into the form x = f(x) to create an iterative formula.

2. Choose a starting value x₀ (this is often given in the question).

3. Substitute x₀ into the formula to find x₁.

4. Substitute x₁ into the formula to find x₂, and so on.

5. Keep going until the values converge (settle down to a consistent answer to the required number of decimal places).

An equation can often be rearranged into several different iterative formulas. Not all of them will converge — the question will usually tell you which rearrangement to use.

The solution is found to a given number of decimal places when two consecutive iterations round to the same value at that precision.

A staircase diagram showing iteration converging to a solution. Starting at x₀ = 2, each step applies the formula and moves closer to where y = f(x) meets y = x.

The equation can be rearranged to x³ = 9 − 2x, giving the iterative formula:

Starting with x₀ = 2, find the solution to 1 d.p.

x₁ = ∛(9 − 2 × 2) = ∛5 = 1.7100

x₂ = ∛(9 − 2 × 1.7100) = ∛5.5800 = 1.7737

x₃ = ∛(9 − 2 × 1.7737) = ∛5.4526 = 1.7601

x₂ and x₃ both round to 1.8 to 1 d.p., so the solution is x = 1.8 (1 d.p.).

Use the iterative formula with x₀ = 2 to find a solution to correct to 1 d.p.

x₁ = ∛(4 × 2 + 2) = ∛10 = 2.1544

x₂ = ∛(4 × 2.1544 + 2) = ∛10.6176 = 2.1979

Both x₁ and x₂ round to 2.2, so the solution is x = 2.2 (1 d.p.).