UNIT F6 Solving Inequalities: Tet F6 Solving Inequalities F6. Inequalities on a Number Line An inequalit involves one of the four smbols >,, < or The following statements illustrate the meaning of each of them. > : is greater than : is greater than or equal to < : is less than : is less than or equal to Inequalities can be represented on a number line, as shown in the following worked eamples. Worked Eample Represent the following inequalities on a number line. (a) (b) < (c) < (a) The inequalit,, states that must be greater than or equal to. This is represented as shown. Note that solid mark,, is used at to show that this value is included. (b) The inequalit < states that must be less than. This is represented as shown. Note that a hollow mark, o, is used at to show that this value is not included. (c) The inequalit < states that is greater than and less than or equal to. This is represented as shown. 6 Note that o is used at because this value is not included and is used at because this value is included. CIMT, Plmouth Universit
F6. UNIT F6 Solving Inequalities: Tet Worked Eample Write an inequalit to describe the region represented on each number line below. (a) (b) (a) The diagram indicates that the value of must be less than or equal to, which would be written as (b) The diagram indicates that must be greater than or equal to and less than. This is written as < CIMT, Plmouth Universit
F6. UNIT F6 Solving Inequalities: Tet F6. of Linear Inequalities (Inequations) Inequalities such as 6 7 can be simplified before solving them. The process is similar to that used to solve equations. Worked Eample Solve the inequalit and illustrate the result on a number line. 6 7 Begin with the inequalit Adding 7 to both sides gives Dividing both sides b 6 gives 6 7 6 This is represented on the number line below. Worked Eample Solve the inequalit ( ) > CIMT, Plmouth Universit
F6. UNIT F6 Solving Inequalities: Tet Begin with the inequalit ( ) > First divide both sides of the inequalit b to give > Then adding to both sides of the inequalit gives > 7. Worked Eample Solve the inequalit Begin with the inequalit 6 9 6 9 In this case, note that the inequalit contains a ' 6 ' term. The first step here is to add 6 to both sides, giving 9 + 6 Now 9 can be added to both sides to give Then dividing both sides b 6 gives 6 or Worked Eample Solve the inequalit < 6 + Begin with the inequalit < 6 + The same operation must be performed on each part of the inequalit. The first step is to subtract, which gives < 6 Then dividing b 6 gives < CIMT, Plmouth Universit
F6. UNIT F6 Solving Inequalities: Tet The result can then be represented on a number line as shown below. An alternative approach is to consider the inequalit as two separate inequalities: () < 6 + and () 6 + These can be solved as shown below. () < 6 + ( ) () 6 ( ) + < 6 ( 6) 6 ( 6) < Both inequalities can then be displaed as shown below. 6 7 8 Where the two lines overlap gives the solution as 6 CIMT, Plmouth Universit 6
F6. UNIT F6 Solving Inequalities: Tet F6. Inequalities Involving Quadratic Terms Inequalities involving rather than can still be solved. For eample, the inequalit < 9 will be satisfied b an number between and. So the solution is written as < < If the inequalit had been > 9, then it would be satisfied if was greater than or if was less than. So the solution will be > or < The end points of the intervals are defined as 9 = ±. CIMT, Plmouth Universit 8
F6. UNIT F6 Solving Inequalities: Tet Note For this tpe of inequalit it is ver eas to find the end points but care must be taken when deciding whether it is the region between the points or the region outside the points which is required. Testing a point in a region will confirm whether our answer is correct. For eample, for > 9, test =, which gives > 9. This is not true, so the region between the points is the wrong region; the region outside the points is needed. Worked Eample Show on a number line the solutions to: (a) 6 (b) < (a) The solution to 6 is which is shown below. or (b) The solution of < is which is shown below. < < Worked Eample Find the solutions of the inequalities (a) + 6 > (b) 7 (a) B subtracting 6 from both sides, the inequalit becomes Then the solution is + 6 > > 9 < or > CIMT, Plmouth Universit 9
F6. UNIT F6 Solving Inequalities: Tet (b) Begin with the inequalit Adding 7 to both sides gives Dividing both sides b gives Then the solution is 7 8 6 Worked Eample Solve the inequalit > The left-hand side of the inequalit can be factorised to give ( ) ( + ) > The inequalit will be equal to when = and =. This gives the end points of the region as = and =, as shown below. Points in each region can now be tested. = gives > or 6 > This is not true. = gives 6 > or 6 > This is true. = gives 6 > or 6 >. This is true. So the inequalit is satisfied for values of greater than, or for values of less than. This gives the solution < or > CIMT, Plmouth Universit
F6. UNIT F6 Solving Inequalities: Tet Investigation Find the number of points (, ) where and are positive integers which lie on the line + = 9. F6. Graphical Approach to Inequalities When an inequalit involves two variables, the inequalit can be represented b a region on a graph. For eample, the inequalit + is illustrated on the graph on the right. 7 The coordinates of an point in the shaded area satisf +. 6 + Note The coordinates of an point on the line satisf + =. 6 7 + = 7 If the inequalit had been + >, then a dashed line would have been used to show that points on the line do not satisf the inequalit, as in the second graph. 6 + > 6 7 + = CIMT, Plmouth Universit
F6. UNIT F6 Solving Inequalities: Tet Worked Eample Shade the region which satisfies the inequalit = 7 7 (, ) The region has the line so first of all the line = 7 as a boundar, = 7 is drawn. (, ) The coordinates of points on this line are (, 7), (, ) and (, ) (, ) These points are plotted and a solid line is drawn through them. A solid line is drawn as the inequalit contains a ' ' sign which means that points on the boundar are included. Net, select a point such as (, ). (It does not matter on which side of the line the point lies.) 6 7 (, 7) If the values, = and =, are substituted into the inequalit, we obtain ( ) 7 or This statement is clearl false and will also be false for an point on that side of the line. = 7 7 (, ) Therefore the other side of the line should be shaded, as shown. 6 7 CIMT, Plmouth Universit
F6. UNIT F6 Solving Inequalities: Tet Worked Eample Shade the region which satisfies the inequalit + < The line + = will form the boundar of the region, but will not itself be included in the region. To show this, the line should be drawn as a dashed line. Before drawing the line, it helps to rearrange the equation as = Now points on the line can be calculated, for eample (, ),, This line is shown below. ( ) and, ( ). 6 (, ) (, ) (, ) (, ) + = 6 7 Net, a point on one side of the line is selected, for eample (, ), where = and =. Substituting these values for and into the inequalit gives + < or 8 < This is clearl true and so points on this side of the line will satisf the inequalit. This side of the line can now be shaded, as below. 6 + < (, ) + = 6 7 CIMT, Plmouth Universit