QA #2 Equations, Expressions and Inequalities

Questions and Answers Sheet 2 Equations, Expressions and Inequalities Question #1 For exercise, solve the equations and inequalities. Write the solutions sets to the inequalities in interval notation. √𝑡 + 3 + 4 = 𝑡 + 1 Answer: On simplification, we get √𝑡 + 3 + 4 = 𝑡 + 1 ⇒ √𝑡 + 3 = 𝑡 − 3 Squaring on both sides, we get 2 ⇒ (√𝑡 + 3) = (𝑡 − 3)2 ⇒ 𝑡 + 3 = 𝑡 2 − 6𝑡 + 9 ⇒ 𝑡 2 − 7𝑡 + 6 = 0 ⇒ 𝑡 2 − 6𝑡 − 𝑡 + 6 = 0 ⇒ 𝑡 (𝑡 − 6 ) − (𝑡 − 6 ) = 0 ⇒ (𝑡 − 6)(𝑡 − 1) = 0 ⇒ 𝑡 − 6 = 0 𝑜𝑟 𝑡 − 1 = 0 ⇒ 𝑡 = 6 𝑜𝑟 𝑡 = 1 But the given equation cannot satisfy when 𝑡 = 1 . ∴𝑡=6 Question #2 For exercise, solve the equations and inequalities. Write the solutions sets to the inequalities in interval notation if possible. 1 3 2 5 𝑥 + 5 > 6𝑥 − 1 Answer: Step 1 𝑥 3 2 5 5 6 + > 𝑥−1 2 Subtract 5 from both side 𝑥 3 2 2 5 2 +5− 5 > 6𝑥 − 1− 5 Step 2 𝑥 3 5 7 > 6𝑥 − 5 5 Subtract 6 𝑥 from both side 𝑥 3 − 5𝑥 6 2𝑥−5𝑥 6 > 5𝑥 > −7 6 5 7 −5− 5𝑥 6 −3𝑥 6 𝑥 2 > −7 5 7 <5 𝑥< 14 5 14 𝑥 ∈ (−∞, 5 ) Question #3 For exercise, solve the equations and inequalities. Write the solutions sets to the inequalities in interval notation if possible. −5𝑥(𝑥−3)2 2+𝑥 ≤0 Answer: Step 1 To solve the following inequality: −5𝑥(𝑥−3)2 2+𝑥 ≤0 Step 2 Calculation: Since, −5𝑥(𝑥−3)2 2+𝑥 ≤0 Multiplying both sides by −1, we get 5𝑥(𝑥−3)2 2+𝑥 ≥0 Dividing both sides by 5, we get, 𝑥(𝑥−3)2 2+𝑥 ≥0 The critical points are 𝑥 = 0, 𝑥 = −2, 𝑥 = 3 Step 3 Sign of 𝑥(𝑥−3)2 2+𝑥 Range of x can be summarized in the following table. Sign of 𝑥 (𝑥 − 3 )2 2+𝑥 𝑥 < −2 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑥 = −2 𝑛𝑜𝑡 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 −2 < 𝑥 < 0 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑥=0 𝑧𝑒𝑟𝑜 0<𝑥<3 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑥=3 𝑧𝑒𝑟𝑜 𝑥>3 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑥 Hence, (𝑥−3)22 + 𝑥 ≥ 0 is satisfied when 𝑥 < −2, 𝑥 = 0,0 < 𝑥 < 3, 𝑥 > 3 Therefore, given inequality is satisfied is 𝑥 < −2 𝑎𝑛𝑑 𝑥 ≥ 0 Hence, given inequality is satisfied if 𝑥 ∈ (−2,0] ∪ [0, ∞) Question #4 For exercise, solve the equations and inequalities. Write the solutions sets to the inequalities in interval notation if possible. 5 𝑦−4 3𝑦 = 𝑦+2 − 2𝑦 2 −14𝑦 𝑦 2 −2𝑦−8 Answer: Step 1 5 2𝑦 2 −14𝑦 3𝑦 = 𝑦+2 − 𝑦 2−2𝑦−8 𝑦−4 5 3𝑦 5 3𝑦 2𝑦 2 −14𝑦 ⇒ 𝑦−4 = 𝑦+2 − 𝑦 2−4𝑦+2𝑦−8 2𝑦 2 −14𝑦 ⇒ 𝑦−4 = 𝑦+2 − 𝑦(𝑦−4)+2(𝑦−4) ⇒ 5 𝑦−4 = 5 ⇒ 𝑦−4 = ⇒5= 3𝑦 𝑦+2 2𝑦 2 −14𝑦 − (𝑦−4)(𝑦+2) 3𝑦(𝑦−4)−(2𝑦 2 −14𝑦) (𝑦−4)(𝑦+2) 3𝑦(𝑦−4)−(2𝑦 2 −14𝑦) (𝑦+2) ⇒ 5(𝑦 + 2) = 3𝑦(𝑦 − 4) − (2𝑦 2 − 14𝑦) ⇒ 5(𝑦 + 2) = 3𝑦 2 − 12𝑦 − 2𝑦 2 + 14𝑦 ⇒ 5𝑦 + 10 = 𝑦 2 + 2𝑦 Step 2 ⇒ 𝑦 2 + 2𝑦 − 5𝑦 − 10 = 0 ⇒ 𝑦 2 − 3𝑦 − 10 = 0 ⇒ 𝑦 2 − 5𝑦 + 2𝑦 − 10 = 0 ⇒ 𝑦 (𝑦 − 5) + 2(𝑦 − 5) = 0 ⇒ (𝑦 − 5)(𝑦 + 2) = 0 Either 𝑦 = 5 𝑜𝑟 𝑦 = −2 is not possible ∴𝑦=5 Question #5 Solve the following logarithmic equations and inequalities: a. log 5 (𝑥 + 1) − log 5(𝑥 − 1) = 2 1 b. 2 log(𝑥 4 ) − log(2𝑥 − 1) = log(𝑥 2 ) + log 2 c. log1 3 + log 1 𝑥 ≤ log 1 2 2 2 5 + log 1(𝑥 − 2) 2 Answer: a. Given: log 5(𝑥 + 1) − log 5 (𝑥 − 1) = 2 Apply logarithmic property: log 𝑚 𝑥+1 log 5 (𝑥−1) = 2 𝑎 − log 𝑚 𝑎 𝑏 = log 𝑚 𝑏 Again, apply logarithmic property: log 𝑚 𝑥+1 𝑥−1 𝑎 = 𝑐 ⇒ 𝑎 = 𝑚𝑐 , = 52 𝑥 + 1 = 25(𝑥 − 1) 26 = 24𝑥 13 𝑥 = 12 b. Given: 1 2 log(𝑥 4 ) − log(2𝑥 − 1) = log(𝑥 2 ) + log 2 Apply logarithmic property: 𝑥 log 𝑚 𝑎 = log 𝑚 𝑎𝑥 , 1 log(𝑥 4 )2 − log(2𝑥 − 1) = log(𝑥 2 ) + log 2 log(𝑥 2 ) − log(2𝑥 − 1) = log(𝑥 2 ) + log 2 Again, apply logarithmic property: log 𝑚 log 𝑚 𝑎 + log 𝑚 𝑏 = log 𝑚 𝑎 − log 𝑚 𝑏 = log 𝑚 𝑎 𝑏 and 𝑎𝑏 , 𝑥2 log (2𝑥−1) = log(2𝑥 2 ) Remove log to both sides of the equation: 𝑥2 = 2𝑥 2 2𝑥−1 𝑥 2 = 4𝑥 3 − 2𝑥 2 4𝑥 3 − 3𝑥 2 = 0 𝑥 2 (4𝑥 − 3) = 0 3 𝑥 = 0, 4 But for log(𝑥 4 ) 𝑍𝑆𝐾𝑎𝑛𝑑𝑃𝑆𝐾 log(𝑥 2 ) to be defined: 𝑥>0 Therefore, the only solution of given equation is: 3 𝑥=4 c. Given: log 1 3 + log1 𝑥 ≤ log 1 2 2 2 5 + log 1(𝑥 − 2) 2 Rewrite above equation as: log (2−1) 3 + log (2−1) 𝑥 ≤ log (2−1) 5 + log (2−1) (𝑥 − 2) 1 Apply logarithmic property: log 𝑚 𝑥 𝑎 = 𝑥 log 𝑚 𝑎 , − log2 3 − log2 𝑥 ≤ − log 2 −(log 2 3 + log 2 𝑥 ) ≤ −(log 2 5 − log 2(𝑥 − 2) 5 + log 2(𝑥 − 2)) Now, removing minus sign to both sides will change sign of inequality: log 2 3 + log 2 𝑥 ≥ log 2 5 + log 2(𝑥 − 2) Again, apply logarithmic property: log 𝑚 log 2 3𝑥 ≥ log2 5(𝑥 − 2) Remove log to both sides: 3𝑥 ≥ 5(𝑥 − 2) 10 ≥ 2𝑥 5 ≥ 𝑥 ... ...(1) Also, For log 1𝑥 to be defined: 2 𝑎 + log 𝑚 𝑏 = log 𝑚 𝑎𝑏 , 𝑥 > 0 ... ...(2) and, For log 1(𝑥 − 2) to be defined: 2 𝑥−2>0 𝑥 > 2 ... ...(3) From (1), (2) and (3): 2<𝑥 ≤5 Question #6 Solve for B. 9𝐵 + 𝐶 = 𝐴 𝐵 =? Answer: Step 1 We have to solve Step 2 9𝐵 + 𝐶 = 𝐴 9𝐵 = 𝐴 − 𝐶 Answer: 𝐵 = (𝐴 − 𝐶 )/9 Question #7 Solve the below equations and inequalities for the given variable. a) 8 − 6𝑛 = −64 b) 10(𝑥 − 5) ≥ −10 c) −5 = 1 − 8𝑛 + 5𝑛 Answer: Step 1 8 − 6𝑛 = −64 or, −6𝑛 = −64 − 8 or, −6𝑛 = −72 or, 𝑛 = −72/(−6) or, 𝑛 = 12 Step 2 10(𝑥 − 5) ≥ −10 10𝑥 − 50 ≥ −10 10𝑥 ≥ 50 − 10 10𝑥 ≥ 40 𝑥 ≥ 40/10 𝑥 ≥4 Step 3 −5 = 1 − 8𝑛 + 5𝑛 −5 = 1 − 3𝑛 −5 − 1 = −3𝑛 −6 = −3𝑛 𝑛 = −6/(−3) 𝑛=2 Answer: a) 𝑛 = 12 b) 𝑥 ≥ 4 c) 𝑛 = 2 Question #8 Find the domain of the function. 𝑓 ( 𝑥 ) = √𝑥 + 4 − √1−𝑥 𝑥 Answer: Step 1 The domain is a set of x-values where the function is defined. In other words, the domain is the interval of x-values in which the graph of the function is defined. We have to find the domain of 𝑓 (𝑥 ) = √𝑥 + 4 − √1−𝑥 𝑥 . Since we know that function is not defined if its denominator becomes 0. Here, x os in denominator. Therefore, x should not be 0. i. e. 𝑥 ≠ 0 Step 2 We know that in square root, the value must not be negative. i.e. the values should be greater than or equal to 0. Here, we have two square root expressions √𝑥 + 4 and √1 − 𝑥 . To defined 𝑓 (𝑥 ), 𝑥 + 4 ≥ 0 and 1 − 𝑥 ≥ 0 . Solve both the inequalities. 𝑥+4 ≥0 𝑥 ≥ −4 ...(1) 1−𝑥 ≥ 0 𝑥−1 ≤0 𝑥 ≤ 1 ...(2) From the inequalities (1) and (2), −4 ≤ 𝑥 ≤ 1 . But 𝑥 ≠ 0 . So we have to remove 𝑥 = 0 value from 4 ≤ 𝑥 ≤ 1 . Hence, the domain of f(x) is [−4,0) ∪ (0,1] . Question #9 The function 2(4x+1) and 8x+2 are either equivalent inequalities, equivalent equation, equivalent expression, or not equivalent. Answer: Given information: The provided equations, 2(4𝑥 + 1) and 8𝑥 + 2 Consider the expression, 𝑓(𝑥 ) = 2(4𝑥 + 1) .....(1) 𝑔(𝑥 ) = 8𝑥 + 2 .....(2) Apply distributive law, an expression 2(4𝑥 + 1) . 8𝑥 + 2 ......(3) So, the expression (2) and (3) are equal. Hence, the expressions 2(4𝑥 + 1) and 8𝑥 + 2 is an equivalent Expression. Question #10 A. Characterized the x-intercept and the y-intercept of the given linear equations below. Then use them to draw and characterized the graph of the linear equation. Note: characterized means to tell whether the slope of the line is positive or negative and to tell whether the line is pointing upward to the right or downward to the right. 1. −6𝑥 + 3𝑦 = −18 2. 3𝑥 + 2𝑦 = 9𝑥 + 12 Answer: Step 1 (A) Given, 1. −6𝑥 + 3𝑦 = −18 2. 3𝑥 + 2𝑦 = 9𝑥 + 12 As we know, Y-intercept form of Linear Equation, 𝑦 = 𝑚𝑥 + 𝑐 ....... (1) Where, 𝑚 = slope of Line 𝑐 = y-intercept Also, If we put 𝑥 = 0 in Linear Equation and we get the y-intercept value. and If we put 𝑦 = 0 in Linear Equation and we get x-intercept Value. If Slope is Positive then Line is pointing upward to the Right. If Slope is Negative then Line is Pointing Downward to the Right. Step 2 1. −6𝑥 + 3𝑦 = −18 Add '6x' both Sides, −6𝑥 + 3𝑦 + 6𝑥 = −18 + 6𝑥 3𝑦 = 6𝑥 − 18 Divide Both sides by '3', 3𝑦 6𝑥 − 18 = 3 3 𝑦 = 2𝑥 − 6 .........(2) Compare this equation with (1) we get, 𝑚 = 𝑠𝑙𝑜𝑝𝑒 = 2 and 𝑐 = −6 = y-intercept If we put 𝑦 = 0 in equation (2) 0 = 2𝑥 − 6 2𝑥 = 6 𝑥=3 Here x-intercept is 3 Here, The slope is Positive Hence line Is Pointing Upward to the right. x-intercept = 3 y-intercept = −6 The Line is Pointing Upwardto the Right. Step 3 2. 3𝑥 + 2𝑦 = 9𝑥 + 12 Subtract '3x' both Sides, 3𝑥 + 2𝑦 − 3𝑥 = 9𝑥 + 12 − 3𝑥 2𝑦 = 6𝑥 + 12 Divide Both sides by '2', 2𝑦 2 = 6𝑥+12 2 𝑦 = 3𝑥 + 6 ...........(3) Compare this equation with (1) we get, 𝑚 = 𝑠𝑙𝑜𝑝𝑒 = 3 and 𝑐 = 6 = y-intercept If we put 𝑦 = 0 in equation (2) 0 = 3𝑥 + 6 −3𝑥 = 6 𝑥 = −3 Here x-intercept is −3 Here, The slope is Positive Hence line Is Pointing Upward to the right. Answer: x-intercept = −3 y-intercept = 6 The Line is Pointing Upward to the Right. Question #11 Solve the compound inequality. 4𝑥 + 1 ≥ 21 and 3𝑥 − 2 ≤ −17 Write the solution in interval notation. Answer: Step 1 (A) Given 1. −6𝑥 + 3𝑦 = −18 2. 3𝑥 + 2𝑦 = 9𝑥 + 12 As we know, Y-intercept form of Linear Equation y= 𝑚𝑥 + 𝑐 ....... (1) Where, 𝑚 = slope of Line 𝑐 = y-intercept Also, If we put 𝑥 = 0 in Linear Equation and we get the y-intercept value. and If we put 𝑦 = 0 in Linear Equation and we get x-intercept Value. If Slope is Positive then Line is pointing upward to the Right. If Slope is Negative then Line is Pointing Downward to the Right. Step 2 1. −6𝑥 + 3𝑦 = −18 Add '6x' both Sides, −6𝑥 + 3𝑦 + 6𝑥 = −18 + 6𝑥 3𝑦 = 6𝑥 − 18 Divide Both sides by '3', 3𝑦 3 = 6𝑥−18 3 𝑦 = 2𝑥 − 6 .........(2) Compare this equation with (1) we get, 𝑚 = 𝑠𝑙𝑜𝑝𝑒 = 2 and 𝑐 = −6 = y-intercept If we put 𝑦 = 0 in equation (2) 0 = 2𝑥 − 6 2𝑥 = 6 𝑥=3 Here x-intercept is 3 Here, The slope is Positive Hence line Is Pointing Upward to the right. x-intercept = 3 y-intercept = −6 The Line is Pointing Upwardto the Right. Step 3 2. 3𝑥 + 2𝑦 = 9𝑥 + 12 Subtract '3x' both Sides, 3𝑥 + 2𝑦 − 3𝑥 = 9𝑥 + 12 − 3𝑥 2𝑦 = 6𝑥 + 12 Divide Both sides by '2', 2𝑦 2 = 6𝑥+12 2 𝑦 = 3𝑥 + 6 ...........(3) Compare this equation with (1) we get, 𝑚 = 𝑠𝑙𝑜𝑝𝑒 = 3 and 𝑐 = 6 = y-intercept If we put y=0 in equation (2) 0 = 3𝑥 + 6 −3𝑥 = 6 𝑥 = −3 Here x-intercept is −3 Here, The slope is Positive Hence line Is Pointing Upward to the right. Answer: x-intercept = −3 y-intercept = 6 The Line is Pointing Upward to the Right. Question #12 Solve the compound inequality 4𝑥 + 1 ≥ 21 and 3𝑥 − 2 ≤ −17 Write the solution in interval notation. Answer: Step 1 We first solve for x from 4𝑥 + 1 ≥ 21 4𝑥 + 1 ≥ 21 4𝑥 + 1 − 1 ≥ 21 − 1 4𝑥 ≥ 20 𝑥≥ 20 4 𝑥 ≥5 Step 2 Then we find x from 3𝑥 − 2 ≤ − 17 3𝑥 − 2 ≤ −17 3𝑥 − 2 + 2 ≤ −17 + 2 3𝑥 ≤ − 15 𝑥≤ −15 3 𝑥 ≤ −5 Step 3 Combining 𝑥 ≥ 5 and 𝑥 ≤ −5 we get solution region = (−∞, −5] ⋃[5, ∞) Answer: (−∞, −5] ⋃[5, ∞) Question #13 Solve the compount inequality −2𝑢 < 6 or 3𝑢 + 3 ≥ 18 Write the solution in interval notation. Answer: Step 1 The first inequality is, −2𝑢 < 6 −𝑢 < 3 𝑢≻3 Step 2 Then for this inequality u belongs to, 𝑢 ∈ (−3, ∞) Step 3 The second inequality is, 3𝑢 + 3 ≥ 18 3𝑢 ≥ 15 𝑢 ≥5 Step 4 Then for this inequality u belongs to, 𝑢 ∈ [5, ∞) Step 5 The interval which represent −2𝑢 > 6 or 3𝑢 + 3 ≥ 18 is, 𝑢 ∈ (−3, ∞) Question #14 The organizers of a talent show have budgeted $1800 to buy souvenir clothing to sell at the event. They can but shirts for $10 each and hats for $8 each. They plan to buy at least 5 times as many shirts as hats. The organizers used these inequalities to determine the number of shirts, s, and hats, h, that can be ordered. 10𝑠 + 8ℎ ≤ 1800 ℎ ≥ 5𝑠 What error did the organizers make? A. The first inequality should be 10𝑠 + 8ℎ ≥ 1800 . B. The second inequality should be 𝑠 ≥ 5ℎ . C. The inequalities should be equations. D. The first inequality should be 𝑠 + ℎ ≤ 1800 . Answer: Step 1 Given 10𝑠 + 8ℎ ≤ 1800 ℎ ≥ 5𝑠 To find the error did the organizers make? Step 2 As per the question let s be the no. g shirts and h be the no. g hats. 10𝑠 + 8ℎ = 1800 Also ℎ = 5𝑠 Answer: (c) The inequalities should be equation. Question #15 10 What is wrong with the following equation: 4 + ( 5 ) = 6 − 2 A) Equations should not involve expressions on both sides. B)The equation mixes different operations like +, / and -. C) The left side's expression is longer than the right side. D) The two sides to not evaluate to the same value. Answer: Step 1 10 Given: 4 + ( 5 ) = 6 − 2 Step 2 Simplify the left side 10 4+( ) 5 ⇒4+2 ⇒6 Now simplify the right side 6−2 ⇒4 Here left side is not equal to the right side. So, the option (D) is wrong. The two sides do not evaluate to the same value. Question #16 Match the equivalent expression. 3𝑦 + 12 𝑥 − 3𝑦 + 12 4𝑦 + 3𝑦 + 3𝑥 − 6𝑦 − 10 − 2 12 − 3𝑦 − 2𝑥 + 𝑥 + 2𝑥 3𝑥 + 2𝑦 − 2𝑥 + 𝑦 + 12&𝑥 + 3𝑦 𝑥 + 3𝑦 + 12 + 12 3𝑥 + 𝑦 − 12 5 + 2𝑦 + 7𝑥 − 4𝑥 + 3𝑦 − 17 ? ⇒? Answer: 3𝑦 + 12 ⇒ 𝑥 + 3𝑦 + 2𝑥 − 3𝑥 + 7 + 5 𝑥 − 3𝑦 + 12 ⇒ 12 − 3𝑦 − 2𝑥 + 𝑥 + 2𝑥 𝑥 + 3𝑦 + 2𝑥 − 3𝑥 + 7 + 5 3𝑥 + 2𝑦 − 2𝑥 + 𝑦 + 12 ⇒ 𝑥 + 3𝑦 + 12 3𝑥 + 𝑦 − 12 ⇒ 4𝑦 + 3𝑦 + 3𝑥 − 6𝑦 − 10 − 2 Question #17 Select all of the points that are a solution to the system of inequalities 6𝑥 ≤ 3𝑦 − 24 10𝑥 + 5𝑦 ≻ 15 Answer: Step 1 Given: Given inequalities are 6𝑥 ≤ 3𝑦 − 24 10𝑥 + 5𝑦 ≻ 15 For a solution of the above inequalities, a solution should satisfy both the equations and not only one equation. Step 2 Let us consider a random solution Choosing random points (𝑥, 𝑦) = (1,10) For equation (1) 6𝑥 = 6(1) =6 3𝑦 − 24 = 3(10) − 24 =6 Since, the solution is satisfying the equation as 6𝑥 = 3𝑦 − 24 Hence, (1,10) is a solution. But, we have to check it for the second equation. Step 3 Considering second equation At (𝑥, 𝑦) = (1,10) 10𝑥 + 5𝑦 = 10(1) + 5(10) = 60 Hence, the solution satisfies the condition as 10𝑥 + 5𝑦 ≻ 15 Thus, (x,y)=(1,10) is a solution of inequalities as it satisfies both the equations. We can also select other points and check for them in a similar way. Question #18 Use Cramer's rule to solve the system of equations, if D=0, use another method to determine the solution set. 5x+5y-10z=8 2x+y-z=7 -x-y+2z=3 Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. There is one solution set is . (Type integers or simplified fractions.) B. There are infinitely many solutions. The solution set is , where z is any real number. (Simplify your answers. Use integers or fractions for any numbers in the expressions.) C. There is no solution. The solution set is ⊘ . Answer: Step 1 Use cramer’s rule to solve the system of equation. If D = 0, use another method to determine the solution set 5𝑥 + 5𝑦 − 10𝑧 = 8 2𝑥 + 𝑦 − 𝑧 = 7 −𝑥 − 𝑦 + 2𝑧 = 3 Step 2 First, find the determinant of the coefficient matrix.  5 5 −10  8    A =  2 1 −1  and b =  7   −1 −1 2   3  𝐷 = |𝐴 | ⇒ 𝐷 = 5(2 − 1) − 5(4 − 1) − 10(−2 + 1) ⇒ 𝐷 = 5 − 15 + 10 ⇒𝐷=0 Step 3 To solve the system of equation with Gaussian elimination using augmented matrix Write as an augmented matrix  5 5 −10 8  [ A : b] =  2 1 −1 7   −1 −1 2 3  Apply row operation 1  R3  R3 + R1 5   5 5 −10 8    2  19  R2  R2 − R1 0 −1 3 5  5  23  0 0  0 5  Rank (𝐴) = 2 and Rank (𝐴: 𝑏) = 3 ⇒ (𝐴) ≠ 𝑅𝑎𝑛𝑘(𝐴: 𝑏) ⇒ System of equation has no solution Step 4 Answer: (C ) There is no solution. The solution set is empty Question #19 We measure speed with the 𝑟𝑎𝑡𝑒 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒/𝑡𝑖𝑚𝑒 . 𝑆= 𝑑 𝑡 From this equation, find an equation d, distance and for t, time. 𝐷𝐼𝑆𝑇𝐴𝑁𝐶𝐸 𝑇𝐼𝑀𝐸 Answer: Step 1 𝑠𝑝𝑒𝑒𝑑 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑖𝑚𝑒 ; i.e. 𝑠 = 𝑑/𝑡 Step 2. 𝐷𝐼𝑆𝑇𝐴𝑁𝐶𝐸 𝑠𝑝𝑒𝑒𝑑 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑖𝑚𝑒 𝑇𝐼𝑀𝐸 𝑠𝑝𝑒𝑒𝑑 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑖𝑚𝑒 𝑖. 𝑒 𝑠 = 𝑑/𝑡 𝑠 = 𝑑/𝑡 𝑑 = 𝑠 ×𝑡 𝑡 = 𝑑/𝑠 Question #20 Write three linear inequalities that are equivalent to 𝑦 < 3 . Answer: Concept used: In mathematics, an inequality is a relation which makes a non-equal comparison between two numbers or other mathematical expression. It is used most often to compare two numbers on the number line by their size. An inequality compares two values, showing if one is less than, greater than, or simply not equal to another value. 𝑎 ≠ 𝑏 says that a is not equal to b. 𝑎 < 𝑏 says that a is less than b. 𝑎 > 𝑏 says that a is greater than b. There are four different types of inequalities: Greater than -(>); Less than -(<); Greater than or equal to −(≥) ; Less than or equal to −(≤) For inequality equation: If 𝑏 > 𝑐 ⇒ 𝑐 < 𝑏 𝑜𝑟 𝑏 < 𝑐 ⇒ 𝑐 > 𝑏 Rules for solving inequality equations: These things do not affect the direction of the inequality: -Add (or subtract) a number from both sides -Multiply (or divide) both sides by a positive number -Simplify a side But these things do change the direction of the inequality (''<''becomes''>'' for example): -Multiply (or divide) both sides by a negative number -Swapping left and right hand sides Calculation: The three inequalities that are equivalent to 𝑦 < 3 . A. 3𝑦 + 1 < −8 B. 5𝑦 + 3 < −12 C. 8(𝑦 − 6) < −72 Checking each solution: 𝑆𝑡𝑒𝑝𝑠 A 𝐸𝑥𝑝𝑙𝑎𝑛𝑎𝑡𝑖𝑜𝑛 3𝑦 + 1 < −8 𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡 1 𝑡𝑜 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠. 3𝑦 + 1 − 1 < −8 − 1 𝐷𝑖𝑣𝑖𝑣𝑑𝑒 𝑒𝑎𝑐ℎ 𝑠𝑖𝑑𝑒 𝑏𝑦 3 3𝑦 < −9 3𝑦 −9 = 3 3 B 5𝑦 + 3 < −12 𝑆𝑢𝑏𝑡𝑟𝑎𝑐𝑡 3 𝑡𝑜 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠. 5𝑦 + 3 − 3 < −12 − 3 𝐷𝑖𝑣𝑖𝑑𝑒 𝑒𝑎𝑐ℎ 𝑠𝑖𝑑𝑒 𝑏𝑦 5 5𝑦 < −15 5𝑦 −15 = 5 5 𝑦 < −3 C 8(𝑦 − 6) < −72 𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒 𝑡𝑜 𝑜𝑝𝑒𝑛 𝑡ℎ𝑒 𝑝𝑎𝑟𝑒𝑛𝑡ℎ𝑒𝑠𝑖𝑠 8𝑦 − 48 < −72 𝐴𝑑𝑑 48 𝑡𝑜 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠. 8𝑦 − 48 + 48 < −72 + 48 𝐷𝑖𝑣𝑖𝑑𝑒 𝑒𝑎𝑐ℎ 𝑠𝑖𝑑𝑒 𝑏𝑦 5 8𝑦 < −24 8𝑦 −24 = 8 8 𝑦 < −3 Thus, the three inequalities that are equivalent to 𝑦 < 3 A. 3𝑦 + 1 < −8 B. 5𝑦 + 3 < −12 C. 8(𝑦 − 6) < −72