Questions and Answers #2 Equations and Inequalities

Questions and Answers Sheet 2 Equations and Inequalities Question #1 Determine the solution 𝑆𝑥 of the equation and inequalities over Z. 1) |4𝑥 + 5| = 11 2) |7𝑥 + 6| = 25 3) |2𝑥 − 11| = 20 4) |3𝑥 − 1| = −2 5) |2𝑥 + 3| < 8 Answer: 1) |4𝑥 + 5| = 11 4 x + 5 if 4 x + 5  0 (| 4 x + 5 |=  ) −(4 x + 5) if 4 x + 5  0 4x+5=11(if 4x+5>0) ⇒ 4𝑥 = 11 − 5 = 6 6 3 ⇒𝑥= = 4 2 -(4x+5)=11(if 4x+5<0) ⇒ 4𝑥 + 5 = −11 ⇒ 4𝑥 = −16 ⇒ 𝑥 = −4 ∴ |4𝑥 + 5| = 11 ⇒ 𝑆𝑥 = [−4] 2) |7x+6|=25 ⇒ 7𝑥 + 6 = 25(𝑖𝑓 7𝑥 + 6 > 0) ⇒ 7𝑥 = 25 − 6 = 19 19 ⇒𝑥= 7 -(7x+6)=25(if 7x+6<0) ⇒ 7𝑥 + 6 = −25 ⇒ 7𝑥 = −31 31 ⇒𝑥=− 7 There doesn't exist any solution over Z Question #2 Solve the the system equations 8x-4y=4 2x+3y=61 Answer: 8x-4y=4 (1) 2x+3y=61 (2) Multiply equation (2) by 4 8x+12y=244 8x-4y=4 16y=240 240 𝑦= 16 y=15 Put y=15 in equation (1) 8x-4(15)=4 8x=4+60 8x=64 x=8 Thus, the solution is (x,y)=(8,15) Question #3 Solve the equations and inequalities. a) 32𝑥 − 4 × 3𝑥 + 3 = 0 b) sin 𝜃 = sin 2 𝜃, 0 ≤ 𝜃 ≤ 2𝜋 c) sin2 𝜃 − cos 2 𝜃 = 1 + cos 𝜃 , 0 ≤ 𝜃 ≤ 2 𝜋 Answer: a) 32𝑥 − 4 × 3𝑥 + 3 = 0 ⇒ (3𝑥 )2 − 4(3𝑥 ) + 3 = 0 4 ± √16 − 12 ⇒ 3𝑥 = 2 4 ± √4 = 2 4±2 = 2 ⇒ 3𝑥 = 4 + 2 or 3𝑥 = 4 − 2 ⇒ 3𝑥 = 3 or 3𝑥 = 1 = 30 ⇒ 𝑥 = 1 or x=0 b) sin 𝜃 = sin 2 𝜃, 0 ≤ 𝜃 ≤ 2 𝜋 ∵ sin(2𝜃) = 2 sin 𝜃 cos 𝜃 ⇒ sin 𝜃 = 2 sin 𝜃 cos 𝜃. ⇒ sin 𝜃 (1 − 2 cos 𝜃) = 0 ⇒ sin 𝜃 = 0 or 1 − 2 cos 𝜃 = 0 1 𝜋 5𝜋 2 3 ⇒ 𝜃 = 0, 𝜋, 2𝜋 or cos 𝜃 = ⇒ 𝜃 = , 3 when 0 ≤ 𝜃 ≤ 2𝜋 Question #4 Solve the following rational equations and inequalities, please. 1) 2) 4 2𝑥−1 2𝑥+4 𝑥+1 ≤ 1 𝑥+1 ≥0 Answer: Find the domain 4 1 1) ≤ 4 2𝑥−1 1 − 𝑥+1 ≤0 2𝑥−1 𝑥+1 4𝑥+4−2𝑥+1 ≤0 (2𝑥−1)(𝑥+1) 2𝑥+5 ≤0 (2𝑥−1)(𝑥+1) 1 ⇒ 𝑥 ≠ , −1𝑥 ≤ 2 𝑥 ∈ (−∞, 2) 2𝑥+4 𝑥+1 −5 2 −5 2 ] ≥0 ⇒ 𝑥 ≠ −1 and 2𝑥 + 4 ≥ 0 (or) 𝑥 + 2 ≥ 0 𝑥 ≥ −2 𝑥 ∈ [−2, ∞) Question #5 Solve the inequalities and equations a) b) c) d) 𝑥−1 𝑥+2 5 ≥0 − 12 1 2𝑥 3𝑥−15 𝑥 3𝑥+4 𝑥2 = 1 3𝑥 ≤ 5−𝑥 = 2 𝑥 Answer: I did the first one 𝑥−1 a) ≥0 𝑥+2 𝑥 − 1 ≥ 0 and x+2>0 𝑥 ≥ 1 and x>-2 then 𝑥 ∈ [1, ∞) or 𝑥 − 1 ≤ 0 and x+2<0 𝑥 ≤ 1 and x<-2 𝑥 ∈ (−∞, −2) Combining, we get 𝑥 ∈ (−∞, −2) ∪ [1, ∞) Question #6 Solves for the value of x for each of the exponential equations and inequalities a) 82𝑥−2 = 2 1𝑥 1 2 8 b) ( ) < c) 5𝑥 = 25𝑥−2 Answer: 1) 82𝑥−2 = 2 We know that 8 = 23 so, 23(2𝑥−2) = 21 Since, bases are the same so, their exponents are also the same, 3(2x-2)=1 6x-6=1 Add 6 both sides, 6x=7 Divide both sides by 6, 7 𝑥= 6 So, 𝑥 = 2) 1𝑥 2 1𝑥 2 < < 13 7 6 1 8 2 So, x<3 3) 5𝑥 = 25𝑥−2 5𝑥 = 52(𝑥−2) Thus, bases are same so, their power should also same means equal. So, x=2(x-2) x=2x-4 -x=-4 Multiply both sides by -1, x=4 Hence, x=4 Question #7 Solve the inequality 5 ≤ 3 + |2𝑥 − 7| Answer: 5 ≤ 3 + |2𝑥 − 7| Simplify the expression 3 + |2𝑥 − 7| ≥ 5 3 + |2𝑥 − 7| − 3 ≥ 5 − 3 (subtract 3 from both sides) |2𝑥 − 7| ≥ 2 By the absolute rule, if |𝑢| ≥ 𝑎, 𝑎 > 0 then |𝑢| ≤ −𝑎 and |𝑢| ≥ 𝑎 2𝑥 − 7 ≤ −2 and 2𝑥 − 7 ≥ 2 (by the absolute rule) 2𝑥 − 7 + 7 ≤ −2 + 7 and 2𝑥 − 7 + 7 ≥ 2 + 7 (add 7 on both sides) 2𝑥 ≤ 5 and 2𝑥 ≥ 9 2𝑥 2 ≤ 𝑥≤ 5 2 5 2 and 2𝑥 2 9 ≥ (divide by 2) and 𝑥 ≥ 9 2 2 5 9 2 2 Thus, the solution is (−∞, ] ∪ [ , ∞) Question #8 Describe the sets of points in space whose coordinates satisfy the given inequalities or combinations of equations and inequalities. x=y,z=0 Answer: We have x=y,z=0 Let the points with coordinates (𝛼, 𝛽, 𝛾) for 𝛼, 𝛽, 𝛾 ∈ 𝑅 satisfies the given equation, by equation x=y , we get 𝛼 = 𝛽(1) And from equation z=0m we get 𝛾 = 0(2) Therefore from (1), (2) the coordinates of the required point are (𝛼, 𝛼, 0) for 𝛼 ∈ 𝑅 , Hence the solution is 𝑆 = (𝛼, 𝛼, 0): 𝛼 ∈ 𝑅 Question #9 Solve the equation 3𝑥(𝑥 − 1) = 𝑥 + 6 Answer: We have: 3𝑥(𝑥 − 1) = 𝑥 + 6 Quadratic formula: 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 𝑥= −𝑏±√𝑏 2 −4𝑎𝑐 2𝑎 Thus, 3𝑥(𝑥 − 1) = 𝑥 + 6 3𝑥 2 − 3𝑥 = 𝑥 + 6 3𝑥 2 − 3𝑥 − 𝑥 − 6 = 0 3𝑥 2 − 4𝑥 − 6 = 0 Now, 𝑥= 𝑥= 𝑥= 𝑥= 𝑥= −(−4)±√(−4)2 −4×3×(−6) 2×3 4±√16+72 6 4±√88 6 4±2√22 6 2±√22 3 Thus, the values are 𝑥 = 2+√22 3 and 𝑥 = 2−√22 3 Question #10 Describe the sets of points in space whose coordinates satisfy the given inequalities. 𝑦 ≥ 𝑥 2, 𝑧 ≥ 0 Answer: Draw a parabola in two dimensional space. Imagine an infinite number of the same parabolas are stacked upon each other. Since the first equation 𝑦 ≥ 𝑥 2 allows z to equal anything, the second equation essentially cuts the stack in half as it rids the bottom of the stack. The two inequalities together represent the region inside and on the parabolic cylinder 𝑦 = 𝑥 2 that is above and on the xy-plane. Question #11 Please, describe the sets of points in space whose coordinates satisfy the given inequalities and equation. 𝑥 ≥ 0 , 𝑦 ≥ 0 , 𝑧 = 0 Answer: Since, it is given: z=0 , we will consider x-y plane because in x-y plane, z-value is always zero. Since, 𝑥 ≥ 0 , 𝑦 ≥ 0 , the set of points lies within first quadrant with positive x-axis and positive y- axis and origin as well. Question #12 Solve the following equations and inequalities, please a) −𝑥 2 + 2𝑥 > 1 b) [1 − 3𝑥] = 5 c) [4𝑥 − 3] ≤ 2 Answer: a) −𝑥 2 + 2𝑥 > 1 −𝑥 2 + 2𝑥 > 1 Multiply the equation by -1 𝑥 2 − 2𝑥 < −1 Add 1 to both sides 𝑥 2 − 2𝑥 + 1 < −1 + 1 𝑥 2 − 2𝑥 + 1 < 0 (𝑥 − 1)2 < 0 x-1<0 x<1 b) [1-3x]=5 3x=1-5 3x=-4 Divide by 3 4 𝑥=− 3 c) [4𝑥 − 3] ≤ 2 Add 3 to both sides 4𝑥 − 3 + 3 ≤ 2 + 3 4𝑥 ≤ 5 5 𝑥≤ 4 Question #13 Describe the sets of points in space whose coordinates satisfy the given inequality and equation 𝑥 2 + 𝑦 2 ≤ 1, 𝑧 = 0 Answer: We are given: 𝑥 2 + 𝑦 2 ≤ 1, 𝑧 = 0 Here, the given inequality of circle represents it is an interior circle. Thus, the boundary of a circle is centered at (0,0,0) and has radius 1 in the xy-plane. Question #14 Solve the equation log 2 (3𝑥 − 1) = log 2 (𝑥 + 1) + 3 Answer: Given: log 2 (3𝑥 − 1) = log 2 (𝑥 + 1) + 3 𝑚 Used property: log 2 (𝑚) − log 2 (𝑛) = log 2 ( ) 𝑛 Simplify the equation log 2 (3𝑥 − 1) = log 2 (𝑥 + 1) + 3 log 2 (3𝑥 − 1) − log 2 (𝑥 + 1) = 3 log 2 ( ( 3𝑥−1 𝑥+1 3𝑥−1 𝑥+1 )=3 ) = 23 (3x-1)=8(x+1) (3x-1)=8x+8 3x-8x=8+1 -5x=9 9 𝑥=− 5 Question #15 Solve the equation (𝑥 2 − 9)2 − 2(𝑥 2 − 9) − 35 = 0 Answer: Given (𝑥 2 − 9)2 − 2(𝑥 2 − 9) − 35 = 0 𝑥 4 + 81 − 18𝑥 2 − 2𝑥 2 + 18 − 35 = 0 𝑥 4 − 20𝑥 2 + 64 = 0 Let t=x , thus, 𝑡 2 − 20𝑡 + 64 = 0 𝑡 2 − (16 + 4)𝑡 + 64 = 0 𝑡 2 − 16𝑡 − 4𝑡 + 64 = 0 𝑡(𝑡 − 16) − 4(𝑡 − 16) = 0 (𝑡 − 16)(𝑡 − 4) = 0 t=16 and t=4 Substitute 𝑡 = 𝑥 2 𝑥 2 = 16 and 𝑥 2 = 4 Thus, 𝑥 = ±4 and 𝑥 = ±2 Question #16 Solve the following equations and inequalities: 1) −𝑥 2 + 2𝑥 > 1 2) |1 − 3𝑥| = 5 3) |4𝑥 − 3| ≤ 2 Answer: a) −𝑥 2 + 2𝑥 > 1 −𝑥 2 + 2𝑥 − 1 > 0 −(𝑥 2 − 2𝑥 + 1) > 0 𝑥 2 − 2𝑥 + 1 < 0 (𝑥 − 1)2 < 0 Square of any number cannot be negative. Thus, there is no solution. Question #17 B and C are sets of real numbers defined as follows. 𝐵 = {𝑧|𝑧 > 4} 𝐶 = {𝑧|𝑧 ≤ 7} Write 𝐵 ∩ 𝐶 and 𝐵 ∪ 𝐶 using interval notation. If the set is empty, write ∅ . Answer: 𝐶 = {𝑢|𝑢 > 4} 𝐷 = {𝑢|𝑢 ≤ 7} 𝐶 ∪ 𝐷 = (−∞, ∞) 𝐶 ∩ 𝐷 = (4,7] Question #18 B and C are sets of real numbers defined as follows. Write 𝐵 ∩ 𝐶 and 𝐵 ∪ 𝐶 using interval notation. If the set is empty, write ∅. 𝐵 = {𝑧|𝑧 > 3} 𝐶 = {𝑧|𝑧 ≤ 9} Answer: 𝐵 = (3, ∞) 𝐶 = (−∞, 9] 𝐵 ∩ 𝐶 = (3,9] 𝐵 ∪ 𝐶 = (−∞, ∞) Question #19 C and D are set of real numbers 𝐶 = [𝑥|𝑥 < 4|] = (−∞, 4) 𝐷 = [𝑥|𝑥 ≤ 7] = (−∞, 7] Write 𝐶 ∩ 𝐷 and 𝐶 ∪ 𝐷 using interval notation Answer: Given 𝐶 = [𝑥|𝑥 < 4|] = (−∞, 4) 𝐷 = [𝑥|𝑥 ≤ 7] = (−∞, 7] Thus, 𝐶 ∩ 𝐷 = 𝑥 < 9 = (−∞, 4) 𝐶 ∪ 𝐷 = 𝑥 ≤ 7 = (−∞, 7] Question #20 1) 2𝑥 + 4.5(25 − 𝑥) = 25 6 2) − 𝑦 − 4 = −22 5 Solve, please Answer: If x=a is a zero of an equation f(x)=0 , then f(a)=0 1) 2x+4.5(25-x)=25 2x+112.5-4.5x=25 -2.5x=25-112.5 -2.5x=-107.5 𝑥= −107.5 −2.5 x=43 6 2) − 𝑦 − 4 = −22 6 5 − 𝑦 = −22 + 4 5 6 − 𝑦 = −18 6 5 5 𝑦 = 18 𝑦 = 18 × y=15 5 6