Name: Questions and Answers #5 Equations, Expressions and Inequalities
Brand: EduBirdie
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University:
Stanford University
Course:
MATH 20 | Calculus
Academic year:

2021
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25

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11
Author:

MathPaladin

Questions and Answers Sheet 5 Equations, Expressions and Inequalities Question #1 Is 2(6-3x)x equivalent to 2(3𝑥) + 𝑥 + 12 Answer: Step 1 The expressions are given by 2(6 − 3𝑥)𝑥 𝑎𝑛𝑑 2(3𝑥) + 𝑥 + 12 Step 2 We can simplify the expression 2(6-3x) as following: 2(6 − 𝑥)𝑥 = 12𝑥 − 6𝑥 2 .........(1) We can simplify the expression 2(3x)+x+12 as following: 2(3𝑥) + 𝑥 + 12 = 7𝑥 + 12 ...........(2) From equations (1) and (2) we get, 2(6 − 3𝑥)𝑥 ≠ 2(3𝑥) + 𝑥 + 12 Hence, 2(6 − 3𝑥)𝑥 is not equivalent to 2(3𝑥) + 𝑥 + 12 Question #2 To solve: |x+6|>0 General strategy to solve the inequalities that involve absolute value: Absolute value inequalities deal with the inequalities (<,>, ≤, and ≥) on the expressions with absolute sign. General properties for solving inequalities that involve absolute value are 1) |x|>k is equivalent to x<-k or x>k , where k is positive number. 2) |x|-k and x-k and x0 The given equation is of the form of |x|>k. From the property we know that |x+6|>0 is equivalent to x+6<0 or x+6>0 First we have to isolate the absolute value expression on one side of the inequality before solving the inequality, so we have to subtract 6 from both sides. x+6<0, x+6>0 x+6-6<0-6 or x+6-6>0-6 x<-6, x>-6 The solution is x<-6 or x>-6 The solution set is (−∞, −6) ∪ (−6, ∞) Conclusion: The solution set is (−∞, −6) ∪ (−6, ∞) Question #3 To solve: |3x-1|>-2 General strategy to solve the inequalities that involve absolute value: Absolute value inequalities deal with the inequalities (<,>, ≤, and ≥) on the expressions with absolute sign. General properties for solving inequalities that involve absolute value are 1) |x|>k is equivalent to x<-k or x>k , where k is positive number. 2) |x|-k and x-k and x-2 The given equation is of the form of |x|>k. However, the standard method for solving inequalities that involve absolute value cannot be applied to |3x-1|>-2 because k is negative. We will solve the given equation |3x-1|>-2 by inspection. |3x-1|>-2 is satisfied by all real numbers because the absolute value of (3x-1) , regardless of what number is substituted for x, will always be greater than -1. The solution set is the set of all real numbers, which we can express in interval notation as (−∞, ∞) . Conclusion: Absolute values are always greater or equal to zero. So |3x-1|>-2 is true for all x and the interval notation is (−∞, ∞). Question #4 Tell whether the expression is a monomial. If it is, name the variable(s) and the coefficient, and give the degree of the monomial. If it is not a monomial, state why not. 2𝑥 3 Answer: Step 1 A monomial is a polynomial, which has only one term. As we know, the polynomials are the equations or algebraic expressions which consists of variables and coefficients and has one or more terms in it. Each term of the polynomial is a monomial. 2𝑥 3 is a monomial name the variable(s): x the coefficient: 2 the degree of the monomial: 3 Step 2 2𝑥 3 is a monomial name the variable(s): x the coefficient: 2 the degree of the monomial: 3 Question #5 ∑8𝑛=1(3𝑛 − 4) 1) Campfire Mocha 2) -1 3) 76 4) 20 Answer: Step 1 In algebra, an equation can be defined as a mathematical statement consisting of an equal symbol between two algebraic expression that have the same value. The most basic and common algebraic equations in math consist of one or more variables. For instance, x-9=7 is an equation, in which x-9 and 7 are two expressions separated by an ‘equal’ sign. Step 2 Now substitute values of n from 1 to 8 in the equation and add them to obtain the summation of the series. 8 ∑(3𝑛 − 4) = (3 × 1 − 4) + (3 × 2 − 4) + (3 × 3 − 4) + (3 × 4 − 4) + (3 × 5 − 4) + (3 × 6 − 4) 𝑛=1 +(3 × 7 − 4) + (3 × 8 − 4) = −1 + 2 + 5 + 8 + 11 + 14 + 17 + 20 =76 Question #6 To calculate: The solution of 6𝑏𝑦 − 4𝑦 ≤ 7𝑏𝑦 + 10 Answer: Formula used: The addition principle for inequalities: For any real numbers a, b and c: ab is equivalent to a+c>b+c Similar statement holds for ≤ and ≥ The multiplication principle for inequalities: For any real numbers a and b, and for any negative number c, ab is equivalent to ac>bc . Similar statement holds for ≤ and ≥ The distributive property tells us how to solve expressions in the form of a(b+c)=ab+ac . The distributive property is sometimes called the distributive law of multiplication. Calculation: Consider, the equation 6𝑏𝑦 − 4𝑦 ≤ 7𝑏𝑦 + 10 . Subtract -7by on both the sides of the equation 6𝑏𝑦 − 4𝑦 ≤ 7𝑏𝑦 + 10 . 6𝑏𝑦 − 7𝑏𝑦 − 4𝑦 ≤ 7𝑏𝑦 − 7𝑏𝑦 + 10 −1𝑏𝑦 − 4𝑦 ≤ 10 𝑦(−𝑏 − 4) ≤ 10 Divide (-b-4) into both the sides of the equation 𝑦(−𝑏 − 4) ≤ 10. (−𝑏−4) 10 𝑦 (−𝑏−4) ≤ (−𝑏−4) 10 𝑦 ≥ (−𝑏−4) or − 10 𝑏+4 We reversed the inequality symbol when we divided (-b-4) to the inequality because b>0 then -b-4<0 . Hence, the solution to inequality {x | x  − 10 }. (b + 4) Question #7 Whether the point (4,0) is solution set of the system of equations 𝑥 + 𝑦 ≤ 4, 𝑥 − 𝑦 < 4 or not, and also give reasons. Answer: Consider the provide system are, 𝑥 + 𝑦 ≤ 4 and x-y<4 . For check whether an ordered pair (4,0) is a solution of a system of linear inequalities 𝑥 + 𝑦 ≤ 4 and 𝑥 − 𝑦 < 4 , put the ordered pair (4,0) in the system of linear inequalities 𝑥 + 𝑦 ≤ 4 and 𝑥 − 𝑦 < 4 , Check whether the inequalities are satisfied or not. Putting (4,0) in the first linear inequality 𝑥 + 𝑦 ≤ 4 gives, 4+𝑥 ≤4 4 ≤4 Therefore, point (4,0) is satisfied for the linear inequality 𝑥 + 𝑦 ≤ 4 . Putting (4,0) in the second linear equation x-y<4 gives, 4-0<4 4<4 Therefore, point (4,0) does not satisfied for the linear equation x-y<4 . Here the point (4,0) is true for first inequality, but false for second inequality, so it is not part of the solution set of provided inequalities. Therefore, point (4,0) is not a solution of the provided system of inequalities. Question #8 Solve each equation. 𝑛2 − 13𝑛 + 36 = 0 Answer: Step 1 An equation can be defined as a mathematical statement consisting of an equal symbol between two algebraic expressions that have the same value. The most basic and common algebraic equations in math consist of one or more variables. Step 2 The given equation is 𝑛2 − 13𝑛 + 36 = 0 . Simplify the equation and write it in factored form by taking the common terms out and then find solution for the variable n as follows; 𝑛2 − 13𝑛 + 36 = 0 𝑛2 − 4𝑛 − 9𝑛 + 36 = 0 𝑛(𝑛 − 9) − 9(𝑛 − 4) = 0 ................... Take the common terms out. (𝑛 − 9)(𝑛 − 4) = 0 .................... Factored the factors. (n-9)=0 or (n-4)=0 ............ Seperate the factors. n=9 or n=4 Hence, the solution for the given equation 𝑛2 − 13𝑛 + 36 = 0 is n=4 or 9. Question #9 a. To find: The logical first step in solving equation |2𝑥 − 1| = 5 b. To find The logical first step in solving the inequality |3𝑥 + 2| ≤ 8 Answer: a) Approach: Given below are the steps to solve an absolute value equation: 1. If fractions are present, multiply both sides of the equation by the LCD of the fractions. 2. Use the addition property of equality to rewrite the equation so that the absolute value is on one side of the equation and constant terms are on the other side. 3. Use the following property of absolute value: |x|=c is equivalent to |𝑥| = ±𝑐 This property says that to solve an absolute value equation, we must solve two separate equations. For example, |x|=6 is equivalent to two equations x=6 and x=-6 Calculation: The given equation is |2x-1|=5 . In this equation, only absolute value term is one side of the equation. Hence, we can use the property of absolute value to obtain two equations. Thus, |2x-1|=5 is equivalent to 2x-1=5 and 2x-1=-5 Final solution: The logical first step in solving the equation |2x-1|=5 will be use the property of absolute value which says |x|=c is equivalent to |𝑥| = ±𝑐. Hence, the given equation |2x-1|=5 is equivalent to 2x-1=5 and 2x-1=-5 b) Approach: Absolute value inequality: The table given below shows various inequalities and their solution sets: Inequality Solution Set: |x|c x<-c or x>c |𝑥| ≥ 𝑐𝑥 ≤ −𝑐 or 𝑥 ≥ 𝑐 The above properties hold true even if x is replaced by an algebraic expression. Calculation: The given inequality is |3𝑥 + 2| ≤ 8 As it is clear from the above table, solution set of inequality can be written as: −8 ≤ 3𝑥 + 2 ≤ 8 Once we have written the inequality in this form, we can split write it in terms of two inequalities: 3𝑥 + 2 ≥ −8 and 3𝑥 + 2 ≤ 8 Final solution: The first logical step in solving the inequality |3𝑥 + 2| ≤ 8 would be to use the properties of absolute value inequalities and obtain the following inequalities: 3𝑥 + 2 ≥ −8 and 3𝑥 + 2 ≤ 8 Question #10 To solve: |x-2|>6 General strategy to solve the inequalities that involve absolute value: Absolute value inequalities deal with the inequalities (<,>, ≤, and ≥) on the expressions with absolute sign. We can use the property |x|>k is equivalent to x<-k or x>k , where k is a positive number When solving an absolute value inequality it's necessary to first isolate the absolure value expression on one side of the inequality before solving the inequality. |ax+b|0 =-cc , where c>0 =ax+b<-c or ax+b>c We can replace > above with ≥ and < with ≤. Answer: Calculation: To solve: |x-2|>6 The number, x-2 , must be more than 6 units away from zero. Thus, |x-2|>6 is equivalent to x-2<-6 or x-2>6 Now we have to solve the above disjunction. First we have to isolate the absolute value expression on one side of the inequality before solving the inequality, so we have to add 2 to both sides. x-2<-6, x-2>6 x-2+2<-6+2 or x-2+2>6+2 x<-4, x>8 The solution set is (−∞, −4) ∪ (8, ∞) Conclusion: The solution set is (−∞, −4) ∪ (8, ∞) Question #11 Whether the given statement is true or false. If false, then to provide the correct statement. 𝑥+3 Given: The statement ''The inequalities (𝑥 + 3)(𝑥 − 1) ≥ 0 and ≥ 0 have the same solution set.'' 𝑥−1 Answer: Consider the inequality (𝑥 + 3)(𝑥 − 1) ≥ 0 and the function 𝑓(𝑥) ≥ 0. It can be written as f(x)=(x+3)(x-1) It is clear that the function has x-intercept at -3, 1. 𝑥+3 In the expression ≥ 0, multiply both sides by (x-1)^{2} : 𝑥+3 𝑥−1 𝑥+3 𝑥−1 𝑥−1 ≥0 × (𝑥 − 1)2 ≥ 0 × (𝑥 − 1)2 (𝑥 + 3)(𝑥 − 1) ≥ 0 Since both expressions are same, both have the same solution sets. Conclusion: The provided inequalities have the same solution sets. Question #12 To calculate: The solution of 𝑎(𝑏𝑦 − 2) ≥ 𝑏(2𝑦 + 5). Answer: Formula used: The addition principle for inequalities: For any real numbers a, b and c: ab is equivalent to a+c>b+c Similar statement holds for ≤ and ≥. The multiplication principle for inequalities: For any real numbers a and b, and for any negative number c, ab is equivalent to ac>bc . Similar statement holds for ≤ and ≥. The distributive property is sometimes called the distributive law of multiplication and division. Calculation: Consider, the equation 𝑎(𝑏𝑦 − 2) ≥ 𝑏(2𝑦 + 5). Apply distributive law on inequality, 𝑎(𝑏𝑦 − 2) ≥ 𝑏(2𝑦 + 5). 𝑎𝑏𝑦 − 2𝑎 ≥ 2𝑏𝑦 + 5𝑏 Subtract 2by on both the sides of the inequality 𝑎𝑏𝑦 − 2𝑎 ≥ 2𝑏𝑦 + 5𝑏 . 𝑎𝑏𝑦 − 2𝑏𝑦 − 2𝑎 ≥ 2𝑏𝑦 − 2𝑏𝑦 + 5𝑏 𝑎𝑏𝑦 − 2𝑏𝑦 − 2𝑎 ≥ 5𝑏 Add 2a on both the sides of the inequality 𝑎𝑏𝑦 − 2𝑏𝑦 − 2𝑎 ≥ 5𝑏 . 𝑎𝑏𝑦 − 2𝑏𝑦 − 2𝑎 + 2𝑎 ≥ 2𝑎 + 5𝑏 . 𝑎𝑏𝑦 − 2𝑏𝑦 ≥ 2𝑎 + 5𝑏 𝑦(𝑎𝑏 − 2𝑏) ≥ 2𝑎 + 5𝑏 Divide (ab-2b) into both the sides of the inequality 𝑦(𝑎𝑏 − 2𝑏) ≥ 2𝑎 + 5𝑏. (𝑎𝑏−2𝑏) 2𝑎+5𝑏 𝑦 (𝑎𝑏−2𝑏) ≥ (𝑎𝑏−2𝑏) 2𝑎+5𝑏 𝑦 ≥ (𝑎𝑏−2𝑏) The inequality symbol will not change when we divided (ab-2b) to the inequality because a>2 and b>0 then (ab-2b)<0. Hence, the solution to inequality {y | y  2a + 5b }. (ab − 2b) Question #13 Whether the statement ''The equations 5x-4=2x and 3x=4 are equivalent equations'' is true or false. Answer: Formula used: Addition property for inequalities: For any real number a, b and c, ab is equivalent to a+c>b+c . Similar statement hold for ≤ and ≥. Multiplication property for inequalities: For any real number a, and b and for any positive number c, ab is equivalent to ac>bc . Similar statement hold for ≤ and ≥. Distributive law: For any real numbers a, b and c, 𝑎(𝑏 + 𝑐) = 𝑎𝑏 + 𝑎𝑐. Calculation: Consider the equation 5x-4=2x . Simplify the equation 5x-4=2x as follows. 5x-4=2x 5x-4+4=2x+4 [Add 4 to both sides] 5x=2x+4 5x-2x=2x+4-2x [Add -2x to both sides] On further simplification, 3x=4 1 1 1 3 3 3 ( ) 3𝑥 = 4 ( ) [Multiply by ( ) on both sides] 𝑥= 4 3 Now, consider the equation 3x=4 . Solve the above equation as follows. 3x=4 1 1 1 3 3 3 ( ) 3𝑥 = 4 ( ) [Multiply by ( ) on both sides] 𝑥= 4 3 The solution of both the equation is same and it is equivalent equation. Thus, the statement ''The equations 5x-4=2x and 3x=4 are equivalent equations'' is true. Question #14 The given set of equations has no solution or infinitely many solutions. Given: 6𝑥𝑦 ≤ 24 and 6𝑥 − 𝑦 > 24 are the given set of inequalities. Answer: For the first equation: 𝑦 ≥ 6𝑥 − 24 and for second equation: 𝑦 < 6𝑥 − 24 The given set of equations has solutions in common, as can be seen from the inequalities, as in one equation one side will be shaded and in another opposite side will be shaded. Conclusion: The given set of equations has no solutions. Question #15 Solve each equation. 𝑠 2 − 4𝑠 = 21 Answer: Step 1 An equation can be defined as a mathematical statement consisting of an equal symbol between two algebraic expressions that have the same value. The most basic and common algebraic equations in math consist of one or more variables. Step 2 The given equation is 𝑠 2 − 4𝑠 = 21. Solve the equation to find the solution for s by taking common terms out after rearranging the equation as follows; 𝑠 2 − 4𝑠 = 21 𝑠 2 − 4𝑠 − 21 = 21 − 21 ........................ Subtract 21 from both sides. 𝑠 2 − 4𝑠 − 21 = 0 𝑠 2 + 3𝑠 − 7𝑠 − 21 = 0 𝑠(𝑠 + 3) − 7(𝑠 + 3) = 0 ............................ Take the common terms out. (𝑠 + 3)(𝑠 − 7) = 0 ............................. Factored form. (s+3)=0 or (s-7)=0 ......... Seperate the factors. s=-3 or s=7 Hence, the solution for the given equation 𝑠 2 − 4𝑠 = 21 is 𝑠 = −3 or 7. Question #16 To solve: |3𝑥 + 1| ≤ 13. General strategy to solve the inequalities that involve absolute value: Absolute value inequalities deal with the inequalities (<, >, ≤, and ≥) on the expressions with absolute sign. We can use the property |x|-k and x-k and x2 . |x|<2 , represents the distance between x and 0 that is less than 2. Whereas |x|>2 , represents the distance between x and 0 that is greater than 2. We can write an absolute value inequality as a compound inequality (i.e.)-20 =-cc , where c>0 =ax+b<-c or ax+b>c We can replace > above with ≥ and < with ≤ . Answer: Calculation: To solve: |3𝑥 + 1| ≤ 13 Let's continue to think in terms of distance on a number line. The number, 3x+1 , must be less than or equal to 13 units away from zero. |3𝑥 + 1| ≤ 13 is equivalent to −13 ≤ 3𝑥 + 1 ≤ 13 By using the property |x|-k and x2 or not, Answer: Calculation: The system of equation given is x+2y<3 and -4x+y>2 . Substitute the ordered pair (-5, 2) into each inequality to determine whether this point lies in the region. If the ordered pair satisfies both the equations, then this ordered pair lies in the region. Substitute the value of x and y from the ordered pair (-5, 2), in x+2y<3 . -5+2(2)<3 -5+4<3 -1<3 This is true. Now, check for the equation -4x+y>2 , -4(-5)+2>2 20+2>2 22>2 This is also true. Since the ordered pair satisfies both the inequalities, therefore, the ordered pair lies in the region. Question #18 The solution of the equations or inequalities and express it in interval notation where approptiate. a. 4<2y or -3(y+2)>-y+1 b. 4<2y and -3(y+2)>-2y+1 Answer: Calculation: a. Inequality is 4<2y or -3(y+2)>-2y+1 Solution is given as: 4<2y or -3y-6>-2y+1 y>2 or -3y+2y>6+1 y>2 or -y>7 y>2 or y<-7 Solution in interval notation form is 8, -7?(2,8) b. Inequality is 4<2y and -3(y+2)>-2y+1 Solution is given as: 4<2y and -3y-6>-2y+1 y>2 and -3y+2y>6+1 y>2 and -y>7 y>2 and y<-7 Solution in interval notation form is 8, -7n(2,8) Conclusion: Hence, the solutions for the below equations and inequalities are as follows: a. 4<2y or -3(y+2)>-2y+1 : Solution is y<-7 or y>2 b. 4<2y and -3(y+2)>-2y+1 : Solution is y>-7 and y>2 Question #19 Explain why x+2=9 and x+2=-6 are not equivalent equations. Answer: Equivalent expressions are expressions that are the same, even through they may look a little different. If we plug in the same variable value into equivalent expressions, they will each give us the same value when we simplify. Calculation: Considering the first equation x+2=9 . Subtracting 2 from both sides we get, x+2=9 ⇒ 𝑥+2−2 =9−2 ∴𝑥=7 Considering the second equation x+2=-6 . x+2=-6 ⇒ 𝑥 + 2 − 2 = −6 − 2 \therefore x=-8 Therefore, the solutions of both equations are not same so they are not equivalent equations. Conclusion: The equations x+2=9 and x+2=-6 are not equivalent equations. Question #20 To calculate: Thesolution of compound inequality 4−𝑥 −2 ≤ ≤ 6, 3 if possible then write the answer in interval notation. Answer: Consider the compound inequality, −2 ≤ 4−𝑥 3 ≤6 Simplify the left and middle side of the compound inequality, −2 ≤ −2 ≤ 4−𝑥 3 4−𝑥 4−𝑥 3 ≤6 3 ≥ −2 Multiply both side by 3, 3(4−𝑥) 3 ≥ 3 ⋅ (−2) 4 − 𝑥 ≥ −6 −𝑥 ≥ −10 𝑥 ≤ 10 4−𝑥 Simplify the right and middle side of the compound inequality, −2 ≤ ≤6 4−𝑥 3 3 ≤6 Multiply both side by 3, 3(4−𝑥) 3 ≤6⋅3 4 − 𝑥 ≤ 18 −𝑥 ≤ 14 x\geq-14 Thus, 𝑥 ≤ 10 and 𝑥 ≥ −14 , in interval notation, it can be denoted as [-14, 10] Hence, the solution of compound inequality −2 ≤ 4−𝑥 3 ≤ 6 is [-14, 10]

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