1. Review of Topic Equations Changing subject of formulae Inequalities

2. 1.Solving Equations An equation requires you to find the value of the variable that makes both sides of the “=” the same value. The most basic concept you need to understand is that for the equality to remain, any operation done on one side of the equals must also be done to the other side eg to solve x – 2 = 5 we need to +2 to both sides x – 2 + 2 = 5 + 2 x = 7 Note when x = 7, the LHS has a value of 7 – 2 = 5, which is the same as the RHS The second important concept is that operations on the variable need to be removed in the reverse order that they are applied, using the inverse operation. Inverse operations “undo” the operation and are paired as follows Operation  Inverse operation Add (+)  Subtract (-) Multiply (×)  Divide (÷) Square () 2  Plus or minus square root (±  ) Square root (  )  Square () 2

3. When the variable appears only once in the equation, the order of operations on the variable is established by imagining you are substituting a value for the variable. List the operations in order to side of the equation and then list the reverse operations in the reverse order as the example below shows: Solve: 3x 2 – 5 = 7Note the order of operations on x are: () 2, ×3, - 5 Inverse operations in reverse order are: +5, ÷3, ±  Now taking these Inverse operations one at a time we can find the value of x as follows: => 3x 2 = 12 (we added 5 to both sides) => x 2 = 4(we divided both sides by 3) => x = ±  4(we found ±  of both sides) => x = ±2 Note both 2 and -2 when substituted for x in the original equation give 7 as the value to the LHS

4. 2.Equations where the variable appears more than once In such cases, you need to rearrange the terms in the equation to get the terms involving “x” on one side and the numbers only on the other side of the equals sign. Rearranging is done by doing the same thing to both sides, as shown in the example below. Eg Solve : 5x – 4 = 3x + 6 => 5x - 4 -3x = 3x + 6 -3x => 2x – 4 = 6 We now have a simple equation like the one above. => 2x = 10(we have +4 to both sides) => x = 5(we have ÷2 on both sides) There are 2 terms involving x (5x and 3x). If we subtract 3x from both sides the RHS will no longer have any terms involving x Operations on x are: ×2, -4 Remove by: +4, ÷2

5. 3. Equations involving denominators / fractions There are 2 ways of solving equations with denominators: Method 1: Use mathematical operations to make the equation of the form where a, b, c and d are algebraic expressions or numbers. Once in this form we can multiply both sides by the product of the denominators (ie in this case bd) as follows: Multiply both sides by bd and cancel the denominators Note this is the same results as cross multiplying.

6. An example of doing this is: Cross multiply => 30 = 3x => x = 30 ÷ 3 => x = 10

7. A harder example: Cross multiply => 2× 4 = 3 (2x + 1) Note brackets around 2x + 1 => 8 = 6x + 3 => 6x = 8 – 3 => x = 5/6

8. Method 2: Find the product of all denominators and multiply every term by this. This method will get rid of all denominators straight away! Note the product of the denominators is 5x Multiply each term by 5x and cancel denominators => 10 = 3x – 5 => 3x = 10 + 5 => 3x = 15 => x = 5

9. A harder example Note the product of the denominators is 2x( x-1) Multiply each term by 2x(x-1) and cancel denominators => 2 × 2x = 3 × (x-1) => 4x = 3x – 3 => x = -3

10. 4.Manipulating Formulae (Changing the subject of a formula) In the formula A =  r 2, A is said to be the subject of the formula. Changing the subject requires us to rearrange the formula so that a different variable is the subject. To do this we need to establish the operations on the variable to be made the subject and removing them using the inverse process in exactly the same way as we solve equations. Example Make r the subject of A =  r 2 Note the operations on r are: () 2, ×  Remove operations : ÷ ,  A =  r 2 Note: no need to use -  because r must be positive

11. A harder example Make h the subject of the formula: SA =  r 2 + 2  rh Operations on h are: ×2  r, +  r 2 Remove operations in reverse order: -  r 2, ÷2  r => SA -  r 2 = 2  rh (-  r 2 ) (÷2  r )

12. 5.Inequalities Graphically we can represent inequalities on a number line as in the examples below. Note unless otherwise stated we assume we are working with real numbers. Examples: Show graphically a.x ≤ 3______________________________ b.x > -2________________________________ c.-2 < x ≤ 5 _____________________________ d.x < 1 or 3 < x < 5 _________________________

13. Algebraic inequalities can be solved by rearranging in the same way as we solve equations. The only difference is that we must remember to change the direction of the inequality when we multiply or divide by a negative number. Solve: 5 – 2x < 11 Operations on x are: × (-2), + 5 Remove in reverse order: -5, ÷ (-2)* => -2x < 6 (-5) => x > -3 (÷-2) * Note the inequality changes when we divide by the -2