- Chapter A Chapter A - Solving Equations Introduction and Review of Linear Equations An equation is a statement which relates two or more numbers or algebraic expressions. For example, the equation 6 is a statement relating the equality of the expression with the number 6. This statement is always true. The equation x 3 1 is a statement relating the equality of the expression x 3 with the number 1. This statement may or may not be true depending on what numerical value the variable x is assigned. A solution is a value of the variable which makes the equation true. It can be seen by inspection that x 9 is a solution of the above equation since 9 3 1 To verify that a number is a solution to an equation, replace the variable with that value. If the resulting statement is true, the value is a solution, otherwise, it is not a solution. We will explore methods for solving many different kinds of equations in this section. The simplest kind is the linear equation, which is an equation that can be written in the form ax b 0 This kind of equation is solved by isolating the variable, namely, undoing what has already been done to it. Example: Solve the equation 6 3x x forx. Solution: Begin by moving all the variable expressions to one side of the equation and all the numbers to the other side: 6 3x x subtract 6 from both sides 3x 6 6 x 6 Note the commutative property 3x x 10 now subtract x from both sides 3x x x x 10 x x cancels to 0 x 10 divide both sides by x 5 Final solution We can check our solution by replacing x with 5 in the original equation: 6 3 5 5? 6 15 9? 9 9 Therefore, our solution is correct.
- Chapter A Question: Why does the strategy in the previous example work? Answer: Itiseasytoverifythestrategyforanequationoftheformax b 0usingthe properties of real numbers: ax b b 0 b ax 0 0 b Additive Inverse Property ax b Identity for Addition 1 a ax 1 a b Multiplying by 1 a is the same as dividing by a 1 x b a Multiplicative Inverse x b a Identity for Multiplication
- Chapter A Solving Quadratic Equations A quadratic equation is an equation that can be written in the form ax bx c 0 This is referred to as the standard form of a quadratic equation. We will discuss three methods for solving quadratic equations: factoring, completing the square, and using the quadratic formula. I. Solving by Factoring The following theorem is the basis for solving equations by factoring: Theorem: Ifab 0, then a 0orb 0 or both. Therefore, the following strategy can be applied to solve equations by factoring: 1. Write the equation in standard form (if it is not already in standard form). Factor 3. Set each factor equal to zero and solve. Example 1: Solution: Solve x 5x 6forx We first write the equation in standard form: x 5x 6 x 5x 5x 5x 6 x 5x 6 x 5x 6 6 6 x 5x 6 0 x x 3 0 x 0orx 3 0 x orx 3 subtract 5x from both sides add6tobothsides now factor the left set each factor equal to 0 and solve We can check both of our solutions: 5 6? 3 5 3 6? 10 6? 9 15 6? 9 9 Bothand3aresolutions. Example : Solution: Solve a 3a 1 0fora The equation is already in standard form, so factor and solve:
- Chapter A a 3a 1 0 a 7 a 0 a 7 0ora 0 a 7 ora a 7 or a We can check both of our solutions: 7 3 7 1 0? 3 1 0? 9 3 7 1 0? 3 1 0? 9 1 1 0? 8 6 1 0? 9 1 8 0? 0 0 0 0 Therefore, both 7 and are solutions to the equation. II. Solving by Completing the Square The equation x 9 can easily be solved by knowing that 3 9(sox must be 3). However, there is another number whose square is 9. Recall that the product of two negative numbers is a positive number. Thus x 3 is also a solution to the equation since 3 3 3 9. In general, the solution to the equation x c (where c is positive) is x c or x c. This can be written more concisely as x c. This method is often referred to as the square root extration method or simply the square root method. Example 3: Solution: Solve each of the following equations. a) x 0 b) x 17 a) The solution is given by (remember to simplify your radicals!) x 0 b) Note that if then x 10 x 17 x 17 or (adding to both sides) x 17 Given a quadratic equation ax bx c 0 (especially one which is not easily factored), we can rewrite the equation so that the left-hand side is a perfect square (like the second example above). The process, which works on every quadratic expression, is described below:
- Chapter A Example : Solution 1: Solve y 7y 3 0 by completing the square We begin by arranging the left-hand side to look like y ky. y 7y 3 0 y 7y 3 subtract 3 from both sides Now we half and square the linear coefficient (the number in front of y) and add this number to BOTH sides of the equation: y 7y 3 add 7 or 9 to both sides y 7y 9 y 7y 9 1 3 9 9 y 7y 9 37 get a common denominator on right add the numerators on right Why did we do this? Because the left-hand side is now a perfect square. Recall the special product x a x ax a Notice that if we take one-half the coefficient of x and square it, we obtain a a This will work no matter what a is. Since the left hand side is a perfect square we can undo the square by taking the square root of both sides. y 7y 9 37 factor the left y 7 37 now solve as example above y 7 37 simplify the radical y 7 37 subtract 7 from both sides y 7 37 7 37 final solution Solution : follows: Later it will be more useful when completing the square to organize our numbers as y 7y 3 0 y 7y 3 0 y 7y 7 3 7 0 From here, we add 37 y 7 to both sides and continue as before. 37 0
- Chapter A Example 5: Solve 9x 10x 1 0 by completing the square. Solution 1: Since the coefficient of x is not 1, we have to be more careful when writing the left-hand side as x kx: 9x 10x 1 0 subtract 1 from both sides 9x 10x 1 divide both sides by 9 x 10 9 x 1 9 x 10 9 x 5 81 1 9 5 81 x 10 9 x 5 81 9 81 5 81 16 81 x 5 16 9 81 x 5 16 9 81 9 x 5 9 9 half and square 10 9 5 or 5 9 81 common denominator on right factor left side (perfect square) solve using square roots subtract 5 9 from both sides x 5 9 9 1 9 or x 5 9 9 1 Solution : square here: Then proceed as above. Question: Why 9 5 9 You must also be extremely careful if you use the alternate method of completing the 9x 10x 1 0 9 x 10 9 x 1 0 9 x 10 9 x 5 1 9 5 0 9 9 in step 3? 9 x 5 9 9 x 5 9 16 9 0 x 5 9 16 9 16 81 Answer: We added the 5 inside the parentheses, so the 9 distributes over it: 9 9 x 10 9 x 5 9x 10x 9 5 9 9 so we have actually added 9 5 to our expression. To balance this, we must subtract 9 5 from 9 9 the same side or add it to the other side.
- Chapter A III. Solving with the Quadratic Formula We can use the technique of completing the square on the general quadratic equation ax bx c 0 to derive a general formula for the solutions to a quadratic equation: x b a x x b a x x b a x x x x ax bx c 0 ax bx c subtract c from both sides divide both sides by a x b a x c a half and square b a b a or b a b a b a b c a a b a b a b a a b x b a x b c a a a b ac a b ac a b a get a common denominator on right add numerators factor left side use square root method b ac a simplify radical ( a a) b ac a b ac a b ac a subtract b a from both sides combine common denominator fractions The final step gives us the Quadratic Formula. The solution to any quadratic equation of the form ax bx c 0 is given by x b b ac a This formula must be memorized. Remember: a is the coefficient of x, b is the coefficient of x, andc is the constant term in this formula. Example 6: Solve x 7x 3 0 using the Quadratic Formula Solution: For the quadratic formula a 1, b 7, and c 3.
- Chapter A x 7 x x x x 7 1 3 1 7 9 1 7 37.Thus,wehave 7 37 0. 586 or 7 37 6. 51 substitute in formula Example 7: Solve 5x x 1 using any method. Solution: First write the equation in standard form. The quadratic formula is generally the easiest method to use when the quadratic does not easily factor. 5x x 1 5x x 1 0 For the quadratic formula, a 5, b (be careful with signs), and c 1. x 5 1 5 x 0 simplify the complex radical (*See Note Below) 10 x i 5 now factor and cancel the common 10 x i 5 10 x i 5 5 Note that the solutions above are complex numbers. Be sure to simplify any radicals in your final solutions! *Note: In simplifying the complex radical, recall that i 1 So 0 5 1 5 1 5i i 5
- Chapter A Equations in Quadratic Form An equation is in quadratic form when one power of x is twice as large as the (only) other power of x. More formally, an equation in quadratic form can be written as ax n bx n c 0. To solve an equation of this type, we first change the variable by renaming things. For example, if we define u x n in the equation above, then u x n x n. Therefore, our equation above becomes au bu c 0 which is a quadratic equation we can solve for u. Once we solve for u, wefindx by raising our solution(s) to the 1 n th power. Remember the 1/n th power of a number a, a1/n, is a number b such that b n a. Thatis, a 1/n b b n a. Example 1: Solve b 9b 11 0 Solution: The equation above is in quadratic form, so let u be the smaller power of b; thatis, u b. Then u b b, so our equation becomes u 9u 11 0. Thus, u 9 9 1 11 1 Since u b, we have u 9 59 u 9 3 u 16 or u 7 b 16 or b 7 b orb i 7 Example : Solve x 3 x 1 3 3 Solution: Write in standard form, and then let u x 1 3. Then u x 3 : x 3 x 1 3 3 x 3 x 1 3 3 0 u u 3 0 u 3 u 1 0 u 3oru 1 Since u x 1 3 x 1 3 3orx 1 3 1 x 1 3 3 33 or x 1 3 x 7 or x 1 3 1 3 Question: In the equation x 10 7x 5 8 0, what should u equal? Answer: Set u x 5. The equation is then transformied into u 7u 8.
- Chapter A Rational Equations A rational equation is an equation which involves rational expressions, or fractions. The easiest way to solve a rational equation is to eliminate the fractions by multiplying both sides of the equation by the least common denominator. Example 1: Solve x 3 x 1 Solution: We begin as if we were going to add the rational expression on the left-hand side. The LCD is x x. However, instead of getting a common denominator, we multiply both sides of the equation by this LCD: x 3 x x 1 x x x x x x 3 x x x x x x 3x x x x 3x x x x x x x 3x 0 x x 1 0 x orx 1 Let s check both solutions: 3 1? 3 1 1 1? 1 3 1? 3 3 1? 1 1 1 1? 1 1 This next example is very important as it demonstrates how easy it is to get a wrong answer when solving an equation involving rational expressions in x. Example : Solve 5 x x x 1 Solution: The LCD is x (remember the most it occurs in any one fraction is once). Multiply both sides of the equation by x :
- Chapter A 5 x x 5 x x x 1 x x x x 1 x 8 5 x 1 x 3 x 1 x But watch what happens when we check our solution: 5 1 5 0 5 0 UNDEFINED!!! Therefore, our solution is not really a solution at all (this is called an extraneous solution). Question: How did we get x as a solution? Answer: The reason is based on restrictions on our variable. In the first equation 5 x x x 1 x cannot be, since division by 0 is not allowed. However, when we eliminate the fractions, this restriction is eliminated: x 5 x 1 5 1 5 5 Not only is a valid number for this equation, it is a solution! This is why you must check solutions to rational equations. We always need to be on the lookout for the introduction of false solutions to the original equation. So if x is not a solution, what is a solution to this equation? It turns out that this particular equation does not have any solutions. The easiest way to see that there are no real solutions is to look at the graphs of 5 x and x x 1. There is a real solution if and only if their graphs intersect. 5 + in red x - x + 1 x - in blue Notice that for any x the graph of 5 always lies above the graph of x 1. Thus, the equation x x has no real solution. You will be asked in one of the exercises to show algebraically that this equation has no solution.
- Chapter A Example 3: Find all solutions of the equation 1 x 1 8 x. Solution: As in the previous examples we multiply by the least common denominator, which is x. Remember, if we get x as a solution, it is extraneous and not valid. 1 x 1 8 x x 1 x x 1 x 8 x x 1 x 8 x 3 x 6 x 3 6 x 9 x Let s check that x 9 is indeed a solution. 1 9 1 8 9 1 7 1 8 7 1 1 7 7 7 7 8 7 15 15 7 7 multiply by x cancel denominators subtract x from both sides add3tobothsides
- Chapter A Radical Equations When an equation contains one or more radicals, the following strategy can be used: 1. Isolate one radical. Raise both sides of the equation to the appropriate power to remove the radical (for example, if the radical is a square root, square both sides of the equation) 3. If necessary, repeat the process until all radicals have been removed, then solve the resulting equation. Example 1: Solve 3 x 7 8 Solution: Isolate the radical x 7, then square both sides. 3 x 7 8 addtobothsides 3 x 7 1 divide both sides by 3 x 7 square both sides x 7 square and square root cancel x 7 16 subtract 7 from both sides x 9 Let s check our answer: 3 9 7 8? 3 16 8? 3 8? 1 8? 8 8 As with rational equations, checking our solution will be very important in finding any extraneous solutions. Example : Solve x 5 1 x Solution: x 5 1 x x 5 x 1 subtract 1 from both sides square both sides x 5 x 1 cancel on left, FOIL on right x 5 x x 1 subtract x from both sides 5 x 3x 1 subtract 5 from both sides 0 x 3x factor right side 0 x x 1 set each factor equal to 0 and solve x orx 1 But watch what happens when we check our solutions:
- Chapter A 5 1? 1 5 1 1? 9 1? 1 1? 3 1? 1 1? 3 1 X Therefore, our only solution is x. The second solution, x 1, is an extraneous solution which arose when we squared both sides of the equation. As with rational equations, it is very important that we check solutions to make sure they satisfy the original equation. Sometimes it is necessary to isolate and eliminate more than one radical in an equation. In many cases, the only way to do this is one at a time! Example 3: Solve x x 3 3 Solution: Although it does not matter which radical is eliminated first, it is often best to save the simplest one for last. Therefore, we will eliminate x 3 first. x x 3 3 add x 3 to both sides x 3 x 3 subtract 3 from both sides x 3 x 3 square both sides x 3 x 3 FOIL left, cancel on right x 1 x 9 x 3 subtract x from both sides 3x 1 x 9 3 add 1 x and3tobothsides 3x 1 1 x divide by 1 3 x 1 x factor and cancel numerator x x square both sides again x x FOIL numerator on left, cancel on right x 8x 16 16 x multiply by 16 to eliminate fraction x 8x 16 16x subtract 16x from both sides x 8x 16 0 factor left side x 0 set factor equal to 0 and solve x Let s check our answer: Thus, our solution is valid. 3 3? 1 3? 1 3? 3 3
- Chapter A Absolute Value Equations The method for solving equations which contain the absolute value function comes from the definition of absolute value. Recall that x x if x 0 x if x 0 Therefore, for the absolute value of an expression to equal a number, either the expression or the negative of the expression must equal that number. To solve such equations we rewrite every absolute value equation as two separate equations. Example 1: Solve x 5. Solution: There are two cases to consider the first case is x 0 and the second is x 0. If x 0, then x x 5 So x 5 is a solution, but we need to be careful here. Is 5 0? Remember we got the solution by assuming that x 0 so we need to check that our solution satisfies this condition. For the other case, if x 0 x x 5 x 5. Notice that in this case the solution x 5 satisfies the assumption that x 0. Thus, we have two solutions x 5. Example : Solve y 7 3 Solution: We again consider the cases where the argument of the absolute value function is greater or equal than zero, or less than zero. That is y 7 0ory 7 0.We rewrite the equation into two equations: y 7 3and y 7 3 (Remember that we must change all or none of the signs of the expression! y 7 is NOT the same as y 7!!!) case 1: y 7 0 y 7 y 7 3 case : y 7 0 y 7 y 7 3 y 10 y 7 3 y y Notice: in both cases the solution minus 7 satisfies the appropriate assumption. That is, 10 7 0and 7 0. Let s check our solutions: 10 7 3? 7 3? 3 3? 3 3? 3 3 3 3 Both solutions are valid.
- Chapter A A geometrical interpretation of absolute value inequalities. Recall that a b refers to the distance between the points a and b on the number line. Therefore, the equation x 7 3 can be thought of as, the distance between x and 7 is 3?, or more appropriately, what number or numbers are 3 units from 7 on the number line? See the illustration below: 3 units each 7 10 Example 3: Solve the equation x x 1. Solution: As in the previous examples we break the problem into two cases. Case 1: x x 1 0 x x 1 x x 1 x x 0 x 1 x 0 Thus, we have two solutions x 1 andx. Case : x x 1 0 x x 1 x x 1 x x 0 x 1 1 8 x 1 7 i In this case there are no real solutions. Thus, the final answer is x 1 andx.
- Chapter A Equations in Several Variables Many equations or formulas involve several variables. It is often useful to solve these types of equations for one particular variable in terms of all the others. All of the techniques discussed in this section can be used to solve such equations; the key is to determine what kind of equation you have in a given variable. Example 1: The formula for the surface area of a cylinder is given by S r rh where r is the radius of the cylinder and h is the height. Solve this equation for h. Solution: Although there is a square in the equation, h is not squared, so this is a linear equation in h. We solve by isolating h. As we do, we treat every other variable as if it were just a number S r rh S r r r rh rh S r rh r S r r h S r r subtract r from both sides divide both sides by r note we cannot cancel the r!!! Example : Solve the equation in Example 1 for r. Solution: This time the desired variable r is squared, so the equation is a quadratic equation. We use the quadratic formula to solve for r. S r rh 0 r rh S Here a, b h, andc S: r h h S substitute into formula h h 8 S h h S simplify radical h h S factor in numerator h h S reduce r h h S It appears that we have two solutions. However, for this particular equation the variable r represents a length so we must have r 0. Thus, we can discard the solution with the minus sign in front of the radical, and the radius of a cylinder in terms of its height and surface area equals r h h S As you can see, solving an equation for one variable can be very different from solving the same equation for a different variable.
- Chapter A Exercises for Chapter A - Solving Equations For problems 1-3, show the following are solutions to the given equation. 1. x 1 x 1 x x 1 1 ; x 0. 1 1y 10y; y 3 3. x x 0; x 5 For problems -6, solve the equations for the given variable.. 1 1y 10y 5. x 1 x 1 6. 1 y 3 y y 5y 6 For problems 7-11, solve by factoring 7. x 7x 8 0 8. x 3x 9. 7x 16x 9 10. 16x 18 x 11. x 3 x 30x 0 For problems 1-15, solve by completing the square 1. x 6x 10 0 13. 3y 1y 8 0 1. a 3a 9 15. x 5 x 16. Solve equation 1 by using the quadratic formula 17. Solve equation 13 by using the quadratic formula 18. Solve equation 15 by using the quadratic formula. For problems 19-3, solve the equation by the method of your choice. 19. x 13x 30 0 0. 3c 1 0 1. 6t 5t 3. y y 1 0 3. 3x x 15 0. Brady throws a ball into the air. The equation s 16t 30t gives the height of the ball s (in feet) t seconds after throwing it. When will the ball hit the ground after he throws it? For problems 5-9, solve the equations for the given variable. 5. x 10x 9 0 6. x 3 x 1 3 3 7. b 3b 8 0 8. y 1 y 1 1 0 9. 16x 6x 1 1 0 For problems 30-36, solve for the given variable. 30. 1 5 x 3
- Chapter A 31. y 1 1 y 3. x 1 x 1 x x 1 1 33. a 1 3. a 11 a 1 y 3 y y 35. x 36. x x x y 16 x 3x 5 0 x 5x 6 x 0 37. We saw in an example that the equation 5 x x x 1 algebraically that this equation has no solutions. has no real solutions. Show, For problems 38-9, solve for the given variable. 38. 3 x 1 7 39. 3 x 7 8 0. y y 1. x 5 x 1. z z 0 3. y 5 y 1. x 9 3 5. 6b 3 1 0 6. 5 x 5 3 7. y 6 y 8. 3 x 1 6 9. 5 x 1 6 9 For problems 50-56, solve the equation for the indicated variable. 50. V r h for h 51. V r h for r V 1 5. V P P 1 for V 1 53. V 1 V P P 1 for P 1 5. A 1 h b 1 b for b 1 55. P F F U for F 56. v as for s
- Chapter A Answers to Exercises for Chapter A - Solving Equations 1. Substitute the value into the equation:. 0 1 0 1 0 0 1 1? 1 1 1 1? 1 1 1 1 3 10 3? 1 1 9 10 3? 3.. 5. 1 7 15? 15 15 5 5 0? 5 5 0? 0 0 1 1y 10y 1 1y 1y 10y 1y 1 y 1 y y 6 x 1 x 1 x 1 1 x 1 1 x x x x x x x 0 x 0 x 0 6. Although this does not look like a linear equation at first, it will become one when we multiply both sides by the lowest common denominator: 1 1 y 3 y 1 y 5y 6 1 y 3y 1y 10y y 1 y y 1 y y y 1 y
- Chapter A 7. x 7x 8 0 x 8 x 1 0 x 8 0orx 1 0; x 8orx 1 8. Remember to put all terms on one side of the equation: x 3x x 3x 0 x x 3 0 x 0orx 3 0; x 3 NOTICE that if we had divided both sides by x, we would have lost the solution x 0: x x 3x x x 3 ONLY 9. 7x 16x 9 7x 16x 9 0 7x 9 x 1 0 7x 9 0orx 1 0 7x 9 orx 1 x 9 or x 1 7 10. 16x 18 x 16x x 18 0 8x x 9 0 8x 9 x 1 0 8x 9 0orx 1 0 8x 9 orx 1; x 9 8 or x 1 11. Although this equation is not quadratic, it can still be factored: x 3 x 30x 0 x x x 30 0 x x 6 x 5 0 x 0orx 6 0orx 5 0 x 0orx 6orx 5 1. x 6x 10 0 x 6x 9 10 9 0 x 6x 9 19 x 3 19 x 3 19 x 3 19
- Chapter A 13. 3y 1y 8 0 3 y y 8 0 3 y y 8 1 0 (don t forget to distribute the 3!) 3 y y 3 y 3 y 3 (now rationalize) 1. y 3 3 y 3 3 y 6 3 3 a 3a 9 a 3a 3 9 3 (common denominator) a 3 9 9 15. a 3 a 3 5 5 a 3 5 3 3 5 x 5 x x x 5 0 x 1 x 5 0 x 1 x 1 5 1 0 (simplify radical) x 1 x 1 x 1 39 8 39 16 39 16 x 1 39 i x 1 39 i x 1 39 i
- Chapter A 16. Let a 1, b 6, and c 10 in the quadratic formula: x 6 6 1 10 1 6 36 0 6 76 6 19 3 19 3 19 17. Let a 3, b 1, and c 8 in the quadratic formula: y 1 1 3 8 3 1 1 96 6 1 8 6 1 3 6 6 3 6 6 3 3 18. x x 5 0 Let a, b 1, and c 5 in the quadratic formula: x 1 1 5 1 1 0 1 39 1 39 i 19. This equation factors easily, although other methods could be used: x 13x 30 0 x 3 x 10 0 x 3 0orx 10 0 x 3orx 10 0. We can easily solve this equation by isolating the variable, although other methods also work: 3c 1 0 3c 1 c 7 c 7
- Chapter A 1. The quadratic formula is probably best for this (here a 6, b 5, and c 3): 6t 5t 3 0 t 5 5 6 3 6 5 7 1 5 7 i 1. Again, we use quadratic formula with a 1, b, and c 1: y 1 1 1 1 3 3 3 3. We will use the quadratic formula to make an observation later. Here a 3, b, and c 15: x 3 15 3 196 6 1 6 x 18 6 or x 10 6 x 3orx 5 3. Set s 0andsolve: 16t 30t 0 t 30 30 16 16 30 1156 3 30 3 3 t 6 3 or t 3 t ort 1 8 (not valid) seconds after releasing the ball. 5. Let u x. Then u x, so our equation becomes
- Chapter A Therefore, u 10u 9 0 u 9 u 1 0 u 9oru 1 x 9orx 1 x 3orx 1 6. Let u x 1 3. Then u x 3, so our equation becomes u u 3 u u 3 0 u 3 u 1 0 u 3oru 1 x 1 3 3orx 1 3 1 x 3 3 or x 1 3 x 7 or x 1 7. Let u b. Then u b, so our equation becomes u 3u 8 0 u 7 u 0 u 7oru b 7orb b 7 or b i 8. Let u y 1. Then u y 1, so our equation becomes u u 1 0 u 6 u 0 u 6 oru y 1 6(no solution) or y 1 y 16 9. First, rewrite the equation in fraction form: 16 x 6 x 1 0 Now eliminate the fractions by multiplying by x : 16 6x x 0 x 6x 16 0 x 8 x 0 x 8orx 30. LCD is x:
- Chapter A Check: x 1 x 5 x x 3 x 0 3x x 0 0 x 0 1 5 0 3? 1 1 3 x 0 is the solution to the equation. 31. LCD is y 1 y (or simply cross-multiply since the fractions are isolated on both sides): Check: y 1 y y 1 y 1 y 1 y y 1 y 1 y y 1 y 5 5 1 1 5? 6 1 3 Therefore, x 5 is the solution. 3. LCD is x 1 x 1 : x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 remember to FOIL x x 1 x x 1 x 0 x 0 Check: Therefore, x 0 is the solution 33. LCD is a 1 : a 1 a 1 Check: 0 1 0 1 0 0 1 1? 1 1 a 1 a 11 a 1 a 1 a 1 a 11 a a 1 a quadratic equation a a 1 0 a 7 a 3 0 a 7ora 3
- Chapter A 7 1 7 11 7 1? 3 1 3 11 3 1 6 18 6 3 3? 8?? Therefore, x 7andx 3 are the solutions 3. Factor first: y 3 y y y 16 y 3 y y y y The LCD is y y y : Check: y y y y 3 y y y y y y y y y 3 y y y 1 y y y 1 0 y y 3 0 y ory 3 3 3 3 3 3 16 6 9 1 9 16 3 16 1 16 16 16 16 6 1 7 1 0 0 Undefined Therefore, the only solution is y 3. 35. The LCD is x x 5 : x x 5 x x 5 3x 0 x x 5 x x 5 x 5 3x x 0 x 10 3x 1x 0 3x 10x 10 0 use quadratic formula x 10 10 0 6 10 i 5 6 5 i 5 6 5 i 5 3 10 3 10 3 Since this is a complex solution, it will not be an extraneous solution. 36. First, see if factoring helps. Note that since this is multiplication, the LCD is not necessary!
- Chapter A Check: x x x x 5x 6 0 x x x x 3 0 x x x x 3 0 x orx 3 5 6 0? 3 3 3 3 5 3 6 3 0 0 0 0 UNDEFINED 5 1 0 0? 5 0 0 0 Therefore, the only solution is x 3. 37. Suppose 5 x x x 1 has a solution. Then it must be true that 5 x x x 1 x 5 x x x x 1 x 8 5 x x x 1 The only way two fractions with the same denominator can be equal is if the numerators are equal. Thus, we must have x 8 5 x 1 x 3 1 x. Thus, if we have a solution it must equal, but is not acceptible, since we cannot divide by 0. 38. First, isolate the radical, then square both sides: 3 x 1 7 x 1 x 1 x 1 x 3 Check: 3 3 1 7? 3 7? 3 7 Therefore, x 3 is the solution. 39. Again, isolate the radical and square both sides:
- Chapter A 3 x 7 8 3 x 7 1 x 7 x 7 16 x 9 Check: 3 9 7 8? 3 16 8? 1 8 Therefore, x 9 is the solution. 0. Isolate the radical and square both sides. This time, you have a quadratic equation to solve after squaring: y y Check: y y y y 8y 16 (remember FOIL) 0 y 9y 0 0 y 5 y y 5ory 5 5?? 1 5? 0? 1 5 0 Therefore, the solutions are x 5andx. 1. As in the previous problem, isolate (already done), square and solve the resulting quadratic equation: x 5 x 1 x 5 x x 1 0 x 3x 0 x x 1 x or x 1 Check: 5 1? 1 5 1 1? 9 3 X Therefore, the only solution is x.. Isolate the radical, square, and solve the resulting quadratic equation:
- Chapter A Check: z z 0 z 0 z z 0z 00 z z 1z 00 0 z 1 1 1 00 1 9 z 1 9 5 or z 1 9 5 5 0? 16 16 0? 5 5 0 16 0 X 16 Therefore, the only solution is x 5. 3. This time we must isolate one radical at a time: y 5 y 1 y 5 y y 1 (FOIL!!!) y 5 y 5 y 10 y 5 y 5 y y 1 Check: 1 5 1 1? 16 5 1? 5 1 Therefore, the solution is x 1.. The absolute value leads to equations: x 9 3 x 9 3 x 9 3 x 1 x 9 3 x 6 x 6 Therefore, the solutions are x 1 and x 6. It is advisable to check that both solutions are valid. 5. Again write equations (note the 1 does not change sign since it is not in the absolute value):
- Chapter A 6b 3 1 0 6b 3 1 0 6b 3 1 0 6b 15 0 6b 3 1 0 6b 15 6b 9 0 b 5 6b 9 b 3 Therefore, the solutions are b 5 and b 3. It is advisable to check that both solutions are valid. 6. Write the equations: 5 x 5 3 5 x 5 3 5 x 5 3 5 x 8 5 x 5 3 x 0 5 x x 5 Therefore, the solutions are x 0 and x 5. It is advisable to check that both solutions are valid. 7. Although both absolute values change sign, you only have to change one side to solve (the other leads to the same equations): y 6 y y 6 y y 6 y y 8 y 6 y y 8 3y y 3 Therefore, the solutions are y 8andy. It is advisable to check that both solutions are valid. 3 8. It is recommended to isolate the absolute value before writing the equations to avoid sign errors: Therefore the solutions are x 7 6 valid. 9. Again isolate the absolute value: 3 x 1 6 3 x 1 10 x 1 10 3 x 1 10 3 x 7 3 x 7 6 and x 13 6 x 1 10 3 x 1 10 3 x 13 3 x 13 6. It is advisable to check that both solutions are
- Chapter A 5 x 1 6 9 5 x 1 3 x 1 3 5 Recall that absolute value represents the distance from zero on the number line. Therefore, it cannot be negative, so this equation has no solutions. 50. The equation is linear in h, so we solve by isolating the variable: V r h divide by r V r h 51. The equation is now quadratic in r. This quadratic can be solved by isolating r and taking the square root: V r h divide by h V h r r V h (In general, r represents the radius of a cylinder, so only the positive square root would make practical sense). 5. The LCD is V P 1, so multiply both sides: V P 1 V 1 P V V P P 1 1 P 1 V 1 P V V 1 P V P 1 53. As in the previous problem, multiply by the LCD V P 1 : V P 1 V 1 P V V P P 1 1 P 1 V 1 P V P 1 P V V 1 5. The equation is linear in b, so solve by isolating the variable: A 1 h b 1 b A 1 hb 1 1 hb A 1 hb 1 hb 1 A hb hb 1 A hb h b 1
- Chapter A Other methods of solution are possible. 55. This equation is rational in F, so multiply by the LCD F U: F U P F F U F U PF PU F isolate F terms PF F PU F P 1 PU F 1 PU P 56. Square both sides to eliminate the radical: v as v as s v a
- Chapter B Chapter B - Solving Inequalities Introduction An inequality is a statement comparing two quantities which may not be the same. The symbols,,, or are used in place of to indicate which quantity is larger. The symbols are read as follows: a b a b a b a b a is less than b a is less than or equal to b a is greater than b a is greater than or equal to b Question: Ifa 5andb 8, which of the following statements are true? a b (Answer: False) a b (Answer: False) a b (Answer: False) a b (Answer: True) a b (Answer: True) Example 1: We learned earlier that 3 5 3 5 By evaluating each side, we find that 3 5 3 5 8 Therefore, 3 5 3 5 Example : Find values of x which satisfy the inequality x Solution: Note that, if x, then the inequality is true since (This may not make sense at first, but remember that the symbol means greater than or equal to ). Since 6, then 6 also satisfies the inequality. So does 1. In fact, any value or larger will make this inequality true. Each of these numbers (like, 6, and 1 ) is a solution to the inequality. To solve an inequality means to find all possible solutions of the inequality. Notice that this inequality has an infinite number of solutions. We will find that this is true for most inequalities. Therefore, it is not feasible to list all the solutions to an inequality. In general, there are three ways to describe the solutions of an inequality: In a statement or algebraic form x In graphical form on a number line Using interval notation,
- Chapter B Interval notation is used to describe a set of real numbers between given numbers. The set a,b represents the set of all real numbers x such that a x b. (All numbers between a and b). The set a,b represents the set of all real numbers x such that a x b. In this latter case we include the two endpoints of the interval. The symbols and are used to describe all numbers less than a given and greater than a given number, respectively. The set, is used to represent all real numbers. In summary open parentheses mean exclude the endpoint and square brackets mean include the endpoint. Question: Which of the following numbers ( 7, 5, 3, 1, 0) are in the interval 5, 1? Answer: 7 is not in the interval, 5 isintheinterval, 3 isintheinterval, 1 is not in the interval, 0 is not in the interval
- Chapter B Linear Inequalities To solve a linear inequality, isolate the variable in the same fashion as solving a linear equation. However, there is one major difference: Whenever you multiply or divide an inequality by a negative number, you must switch the inequality symbol (from to or vice versa). Why? Consider the following example: 5 7 If we multiply both sides of the inequality by 1, the result is 5 7 This is due to the fact that, on the number line, the negative numbers are written backwards, namely, the smaller the absolute value, the larger the number. Example 1: Solve for x: 7 x 11 Solution: Solve for x the same way you would solve the equation 7 x 11: 7 x 11 subtract 7 from both sides x divide by x remember to switch inequality! x The solution can be written algebraically (x ), using interval notation, or a number line which is shown below. Example y 1 Example : Solve for y: 3y 3 5 1 Solution: It is best to multiply by the LCD (least common denominator) first, which is 15: 15 y 1 15 3y 15 1 3 5 5 y 1 3 3y 15 distribute 5y 5 9y 15 y 5 15 add 5 y 10 divide by y 10 5, 5 5 Notice how we indicate that 5/ is included in the solution set by using a solid dot. If we want to indicate that 5/ is not included an unfilled circle will be used. Sometimes two inequalities are combined in the same statement, such as a x b. As stated earlier, this means the number x is between a and b. More formally, we say that
- Chapter B a x and x b Note:inorderforthistobetrue,a must be less than b to start with. A statement such as x cannot be true; there are no numbers greater than and simutaneously less than. Linear inequalities of this type may be solved by simply splitting them into two inequalities, solving each separately, and finding all values common to both solutions. Question: Is there anything wrong with the statement 5 x 5? Answer: Yes, there is no number which is both less than 5 and greater than 5. Example 3: Solve for x: 10 3x 5 17 Solution 1: The combined inequality can be rewritten as 3x 5 10 and 3x 5 17 subtract 5 from both sides 3x 15 3x 1 divide both sides by 3 x 5 x The solution is all real numbers greater than 5 and less than or equal to. This can be written as 5 x, or 5,, or we can use a number line as shown below. 5 As usual the red line is used to indicate the solution set, and filled and unfilled circles indicated whether a point is in the solution set or not. Solution : Since the variable appears only in the middle, we can isolate it as we do for a single inequality. Remember to do the same operations to all 3 parts and switch the inequality when necessary! 10 3x 5 17 15 3x 1 5 x As found earlier. Example : Solution: Solve for x: x x 7 3 3x The combined inequality can be rewritten as x 7 x x 7 3 3x add7tobothsides x x 11 subtract x from both sides x 10 3x add 3x to both sides x 11 5x 10 divide both sides by 5 x Note that the solution is all real numbers greater than 11 and less than. Since no real numbers satisfy both statements, there is no solution to the inequality. Note also that these two inequalities cannot be solved together since there is no way to isolate x in the center.
- Chapter B Absolute Value Inequalities Inequalities involving absolute value can be rewritten using the concept of distance. For example, the inequality x 3 can be thought of as what numbers are less than 3 units from 0 on the number line? 3 units left 3 units right 3 0 3 From the diagram above it can be seen that any number between 3 and 3 satisfies the inequality x 3. Similarly, the inequality x 3 can be thought of as what numbers are greater than 3 units from 0 on the number line? In this case, x can be greater than (to the right of) 3 or less than (to the left of) 3. The following generalizes this concept: x a is equivalent to a x a x a is equivalent to x a OR x a The number a in the above is assumed to be positive. We have a similar statement for and. x a if and only if a x a x a if and only if x a OR x a Example 1: Solve for x: x 1. Solution: From the statement above, x 1 can be rewritten as 1 x 1 (addtoallsides) 1 x 3 which can also be written in interval notation as 1,3. Notice that the numbers 1 and 3 are both 1 unit away from on the number line. Example : Solve for x: 3x 6 3. Solution: We first isolate the absolute value term before trying to solve the inequality. 3x 6 6 3 6 3x 3 Recall that absolute value represents distance; therefore, it cannot be negative; hence, it cannot be less than 3. Therefore, this inequality has no solution. Question: Does x 5 satisfy the inequality x 1? Answer: No.
- Chapter B Example 3: Solve for x: x 7 Solution: Before applying our rule for rewriting absolute value inequalities, we must first isolate the absolute value term on one side of the inequality (Be careful not to write the left-hand side as x 6) x 7 subtract from each side x 7 x 5 This must be rewritten as two inequalities: x 5 or x 5 subtract from both sides x 1 or x 9 divide both sides by x 1 or x 9 Therefore, our solution is x 9 or x 1. This can be written in interval notation one of two ways. Either use the word or between the intervals:, 9 or 1,, or we can use the union symbol, which means either/or:, 9 1,. A number line description of the solution set is shown below. 9 Notice that the points 9 and 1 are each 5 units away from the point. The following gives an algebraic explanation of this. x 7 factor a from the absolute value term x 7 divide by x 1 7 subtract 1 x 5 x 5 1 Thus, the solution set is the set of points which are 5 or further from the point.
- Chapter B Nonlinear Inequalities If an inequality is not linear, we can use the following method: 1) Move every term of the inequality to one side. ) Find the critical values- That is those values for which the expression is equal to zero or is undefined. 3) The critical values will divide the real number line into intervals. Do not include the critical points in the intervals.. Each interval is entirely in the solution set or entirely out of the solution set. To determine which, test a number in the interval by substituting it into the inequality. If the chosen number satisfies the inequality, then so does every other number in the interval, and if the number does not satisfy the inequality, then no other number from that interval will satisfy the inequality. ) Do this for each interval of the number line. Example 1: Solve for x: x 3 x 1 0 Solution: Since every term of the inequality is on one side, we find the critical values (Where x 3 x 1 0 ) Recall that this equation is solved by setting each factor equal to 0. Therefore, the critical values are x 3 orx 1 These critical values divide the number line into 3 intervals., 3, 3, 1, and 1,. We choose a number in each interval, and test it by substituting into the original inequality as follows: Therefore, the only interval which is in the solution set is between 3 and 1. Since the inequality is greater than or equal to 0, the critical values (where the expression is equal to zero) are also part of the solution. Therefore, our solution set is 3 x 1, or the interval 3, 1 Question: What are the critical values of the expression x 3x? Answer: x 3x x 1 x Thus, the critical values are 1 and. Here we present a more detailed explanation of why the method described above works. Example : Solve for x: x 3 x 1 0 Solution: Before we begin, let s recall the rules for signed numbers. When mulitplying or dividing numbers, if the signs are the same the result is positive. If the signs are different, the result is negative. Here we can view the left-hand side as a product of factors, x 3andx 1. The inequality states that the product is less than 0 (negative) or equal to 0. Based on the rules for signed numbers, this will occur when one factor is positive and the other is negative. Rather than deal with the different possibilities, we can test numbers between the critical values to determine whether the expression is positive or negative. Recall that the critical values occur at those values of x in which the expression is zero or undefined. We look at the critical values because those are the only places in which a
- Chapter B polynomial expression could possibly change from positive to negative or negative to positive. This can be illustrated on the number line (as the previous solution showed) or using a sign chart to analyze each interval (NOTE: remember that the critical values are part of the solution since the expression is equal to 0. You may wish to indicate this in your chart to remind you): Interval x 3 x 1 x 3 x 1 x 3 or, 3 negative negative positive 3 x 1or 3, 1 positive negative negative x 1or 1, positive positive positive Since the expression is to be less than zero (negative) or equal to zero, the solution set contains only the interval 3, 1 or 3 x 1. Since 0 is an allowed value, the finite endpoints of the intervals are also members of the solution set. Either the sign chart or the number line can be used to test each interval. Example 3: Solve for x: x x 1 Solution: Remember, for nonlinear inequalities we first move all terms to one side of the inequality so that we have 0 on the other side. x x 1 0 To find the critical values, we solve the equation x x 1 0 x 6 x 0 x 6orx These critical values divide the number line into 3 intervals: x, x 6, and x 6. Notice that since the inequality is strictly greater than 0, the critical values will not be part of the solution. The sign chart is given below: Interval x 6 x x 6 x x or, negative negative positive x 6or, 6 negative positive negative x 6or 6, positive positive positive Since the inequality is greater than zero (positive), the solution is x orx 6 In interval notation we have, 6, You may have noticed in both of these examples that the overall sign alternates from positive to negative to positive. In the next example, we show that this is not always the case. Example : Solve for x: x 3x 3 x 0 Solution: Since all terms are on the same side, we find the critical values by solving the equation x 3x 3 x 0 x x 3x 0 remember common factors first! x x x 1 0 x 0orx orx 1 These critical values divide the number line into intervals:,,, 0, 0, 1, and 1,. Notice that we are including the critical values since the inequality is greater than or equal to 0. The sign chart is given below (Note that if you are choosing a specific number in each interval you must choose a fraction for the interval 0, 1 )!
- Chapter B Interval x x x 1 x x x 1, or x positive negative negative positive, 0 or x 0 positive positive negative negative 0, 1 or 0 x 1 positive positive negative negative 1, or x 1 positive positive positive positive Since the inequality is greater than (positive) or equal to zero, the solution set is, or 1,. Question: Answer: Notice in this example that there is no sign change at x 0. Why isn t there? Notice that when we factored our expression, we obtained x x x 1 The critical value x 0 came from the factor x, which is squared. Whenever you square a real number, the result is always positive (notice the second column of the sign chart). This is why the sign of the expression does not change at x 0. Here we present examples of inequalities involving rational expressions. The important difference between rational equations and inequalities is this: You cannot get rid of fractions by mulitplying the Least Common Denominator in an inequality. The reason for this is that the expression we multiply by will be positive for some values of x and negative for other values. This doesn t matter in equations, but it does in inequalities as the direction of an inequality sign changes when we multiply by a negative number. We must use the same strategy as before. That is, locate the critical values, and then determine what happens in the intervals defined by the critical values. Example 5: Solve for x: x 1 x 3 Solution: We begin by moving all terms to one side, in this case by subtracting : x 1 0 get a common denominator x 3 x 1 x 3 0 x 3 x 3 x 1 x 6 0 distribute the negative sign x 3 x 1 x 6 0 x 3 x 5 x 3 0 To find the critical values, we must find where this expression is zero or undefined. The expression is zero when the numerator is zero: x 5 0 x 5 x 5 and the expression is undefined when the denominator is zero: x 3 0 x 3 The critical values 3 and 5 separate the number line into 3 intervals:,3, 3, 5, and 5,. Note that we are including the critical value 5 since the expression is equal to zero, but we are not including the critical value 3 since this makes the expression undefined. In general, a critical value where the expression is undefined will not be included in the solution. The sign chart is given below:
- Chapter B Interval x 5 x 3 x 5 x 3,3 positive negative negative 3, 5 positive positive positive 5, negative positive negative Since the expression must be greater than (positive) or equal to 0, the solution is 3, 5 x :3 x 5 Question: Why can t you multiply by the LCD? Answer: Recall that when you multiply or divide an inequality by a negative number, you must switch the inequality Example: x 16 x However, to remove the fractions in our example, we would have to multiply our inequality by x 3, which can be positive or negative depending on x. Therefore, we do not know whether to switch the inequality or not. You can analyze the different possibilities, but the method we used does so automatically. Example 6: Solve for x: x x 1 x 6 Solution: Remember that you cannot get rid of the fractions in the inequality-you must move everything to one side, and get a common denominator: x x 1 x 6 x x 1 0 get a common denominator x 6 x x 6 x 1 x 6 x 1 0 FOIL/distribute in the numerators x 6 x 1 x x x 8 0 distribute the negative sign x 6 x 1 x x x 8 0 combine like terms x 6 x 1 x x 0 factor the numerator x 6 x 1 x x 6 x 1 0 The critical values are x (where the expression is 0), x 6, and x 1 (where the expression is undefined). Since the inequality is strictly less than 0, none of the critical values are in the solution. The critical values divide the number line into intervals:,,, 6, 6, 1, and 1,. The sign chart is given below: Interval x x 6 x 1 x x 6 x 1, positive negative negative positive, 6 positive negative negative positive 6, 1 positive positive negative negative 1, positive positive positive positive Since the expression must be less than 0 (negative), the solution is 6, 1 x : 6 x 1
- Chapter B Exercises for Chapter B - Solving Inequalities For problems 1-3, determine whether the given values are solutions to the given inequality. 1. 3x 0; x 10, x 3, x 1. 6x 13 13x ; x 1; x 5; x 8 3. x 10x 1 0; x 1; x 5; x 1 For problems -9, write the following in interval notation, if possible. If not possible, state why.. 1 x 7 5. 9 x 10 6. x 1 7. 8 x 5 8. x 3orx 11 9. 5 x 7 For problems 10-37, solve each of the following, if possible. 10. x 1 0 35. x 1 3 x 1 11. 5x 7 36. 1 x 1 x 1 1. 5x 9 6 37. x 1 x 5 3 x 13. 7 x 3 10 1 1. x 11 1x 13 15. 1 x 9 11 16. 9 x 8 17. 3x 10 x 5x 1 18. x 11 6x 10 9x 7 19. x 1 8 0. 9 x 6 1. 5x 1 8. x 8 3. x 5 9 1. 3x 7 1 9 5. 11 x 9 10 6. x x 6 0 7. x x 5 8. x x 3 0 9. x 1 x 30. x 3 x x 0 31. x x 3 x 0 3. 1 x 0 33. x 7 x 5 3 3. 3x 6 x 1
- Chapter B Answers to Exercises for Chapter B - Solving Inequalities 1. Test each value by substituting into the inequality: 3 10 0? 6 0X x 10 is not a solution to the inequality. 3 3 0? 13 0 x 3 is a solution to the inequality 3 1 0? 1 0 x 1 is a solution to the inequality.. 6 1 13 13 1? 7 15 x 1 is a solution to the inequality 6 5 13 13 5? 7 67 x 5 is a solution to the inequality 6 8 13 13 8? 61 10 X x 8 is not a solution to the inequality 3. 1 10 1 1 0? 1 10 1 0? 3 0 x 1 is a solution to the inequality 5 10 5 1 0? 5 50 1 0? 0X x 5 is not a solution to the inequality 1 10 1 1 0? 1 10 1 0? 5 0 x 1 is a solution to the inequality. 1,7 5. 9,10 6.,1 7. 5,8 (The inequality can be rewritten as 5 x 8)
- Chapter B 8.,3 or 11, Another way of writing this would be,3 11, 9. This is not possible since we are looking for values of x whicharebothlessthan5and greater than 7 (at the same time). The combined inequality implies that 5 7. 10. x 1 0 x 1 x 3or 3, 11. 5x 7 5x 5 x 1 or, 1 1. 5x 9 6 5x 15 x 3 or, 3 Remember to switch the inequality when you divide by 5 13. We begin by multiplying by 0 (the LCD) to eliminate the fractions: 0 10 7 x 0 3 1 0 1x 15 0 1x 35 x 35 1 x 5 or, 5 1. x 11 1x 13 x 1x 13 10x 35 x 35 10 x 7 or 7, 15. We can solve both inequalities simultaneously: 1 x 9 11 8 x 0 x 10 or,10 16. We can solve the inequalities simultaneously: 9 x 8 11 x 10 11 17. We must solve these inequalities separately: x 5 or 11, 5
- Chapter B 3x 10 x and x 5x 1 x 10 and 6x 1 x 10 and x x 5 and x Therefore, the solution is x 5 or, 5 18. Solving the inequalities separately: x 11 6x 10 and 6x 10 9x 7 10x 1 and 3x 3 x 1 10 and x 1 The solution set is what is common to both intervals. If a number is larger than 1 10 will already be larger than 1. Therefore, our solution is x 1 10 or 1 10, 19. The inequality can be rewritten as 8 x 1 8 7 x 9 (.1, it 7 x 9 or 7, 9 0. The inequality can be rewritten as 6 9 x 6 15 x 3 15 x 3 or 3,15 1. Isolate the absolute value term first before rewriting: 5x 1 8 5x 0 0 5x 0 x or,. The inequality must be split into two inequalities: x 8 or x 8 x 10 or x 6 3. Since these are or rather than and, the solution is valid and can be written this way or, 6 10, using interval notation. x 5 9 1 x 5 1 x 5 1 or x 5 1 x 6 or x 16 x 13 or x 8
- Chapter B. 5. 6. In interval notation,, 8 13,. 3x 7 1 9 3x 7 3 Since the absolute value of an expression cannot be less than 3, there is no solution 11 x 9 10 x 9 1 What happens here? Since the absolute value of an expression cannot be negative, it will always be greater than 1. Therefore, the solution is all real numbers or, x x 6 0 x 3 x 0 The critical values are where the expression is equal to 0, namely, where x 3orx. Since the inequality is strictly less than 0, we do not include the critical values. These divide the number line into 3 intervals. The sign chart is given below: Interval x 3 x x 3 x, negative negative positive,3 negative positive negative 3, positive positive positive Since the inequality is less than 0 (negative), the solution is,3 ( x 3) 7. Remember to first move all terms to one side: x x 5 0 x 9 x 5 0 The critical values are x 9andx 5. Since the inequality is less than or equal to, we include both critical values in the solution. The critical values divide the number line into 3 intervals. The sign chart is given below: Interval x 9 x 5 x 9 x 5, 5 negative negative positive 5,9 negative positive negative 9, positive positive positive 8. Since the inequality is less than (negative) or equal to, the solution is 5,9 ( 5 x 9) x x 3 0 x 3 x 1 0 The critical values are x 3 or x 1. Since the inequality is greater than or equal to, these are included in the solution. The critical values divide the number line into 3 intervals. The sign chart is given below:
- Chapter B Interval x 3 x 1 x 3 x 1, 3 negative negative positive 3,1 positive negative negative 1, positive positive positive 9. 30. 31. Since the inequality is greater than or equal to, the solution is, 3 1, (x 3 or x 1) x 1 x x x 1 0 x x 1 0 x 6 x 0 At this point, you can divide by the negative 1, but be sure to switch the inequality! If you do not divide, you must take the negative 1 into account when you do the sign chart (critical values 6 and not included in the solution): Interval x 6 x x 6 x, negative negative negative b/c of -1,6 negative positive positive 6, positive positive negative Since the inequality is less than 0, the solution is, 6, (x orx 6) x 3 x x 0 x x x 0 x x 6 x 0 The critical values are x 0, x 6, and x, which are included in the solution. The critical values divide the number line into intervals. The sign chart is given below: Interval x x 6 x x x 6 x, negative negative negative negative,0 negative negative positive positive 0,6 positive negative positive negative 6, positive positive positive positive Since the inequality is less than or equal to 0, the solution is, 0,6 (x or 0 x 6) x x 3 x 0 x x x 0 x x 6 x 0 The critical values are x 0, x 6, and x, which are included in the solution. The critical values divide the number line into intervals. The sign chart is given below: