Chapter 10. Lesson is special because any non-zero number divided by itself is 1, and anything multiplied by 1 remains the same.

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1 Chapter 10 Lesson is special because any non-zero number divided by itself is 1, and anything multiplied by 1 remains the same a. yes You cannot divide by zero. c. Yes; x 3 d. Answers vary; sample solutions. x x, x+5 n2, and x+5. n 2 e. Yes, because z = 1. The fact that anything multiplied by 1 stays the same is called z the Identity Property of Multiplication a. 1, x 0 x x+5, x 0 c. 3 x 1, x 1or2 e. hk, h 0 f. 2m 5 3m+1, x 6or 1 g. 2(n 2), x h. 4x 1, x 1 4 or 3 2 d. 1, x a. Yes; you can tell by substituting any number (other than zero). No; you can tell by substituting a number (other than 1). c. They can be simplified like this when the numerator and denominator are single terms and are products of factors. d. (i) is not simplified correctly; (ii) is simplified correctly a. x+3 x 3 2x 5 3x+1 c. 1 d. x a. 1 none c. 2 d Yes, he can. a. x =2 Divide both sides by a. x <0 x a. x 4or2, x+4 x 2 2(x+2) x 2or3, (x 3) a B Answer Key 69

2 Lesson x 1 2x+3 ; x cannot be 4 or , a. 4x+3 x+2, x 5or 3 x+3 2x+1, x 1 9 or 1 2 d. (y 2)2 (y+5) 3y(y+2) c. 7+4m 3m 2, m 2 3 or 3 2, y 0, 2,or2 e. 2x, x 0 and y 0 f. 3x+1 2x 3, x 3 2, 4,or a c. 5(x 2) (x+4) 2 d. 2(2x 3) (x 5)(x+1) e. 5x 1 2(x 3) = 5x 1 2x 6 f. 1 2x a. x 2=4 For each, x =6. c. x+3=8, x = a. m = 6 5, b = (0, 7) m = 3, b = (0, 5) c. m =2, b = (0, 12) a. ( 1 3, 2) (4, 9) a. x 4 3x+2 5 x 3 c a. m = 6 x =5.5 c. k =4 d. x = after 44 minutes Lesson a. 2x 3(2x 1) = 2x 6x 3 x 4 x a. She can multiply each term in the equation by a number to eliminate the decimals. 2x 2 +3x 5 =0, (2x+5)(x 1)=0, so x = 2.5 or a. 3x 5=7, x = 4 3x 2 2x 5 =0, x = 1 or 5 3 c. 1+x=10, x = 9 d. 4x 2 +8x 5=0, x = 1 2 or x = a. x 15 =24 ; One denominator was eliminated, but there is still a fraction left over. 4 Multiply by 4; 4x 15 =96, x = = c. 24 or any multiple of 24; 24 is the least common multiple of 6 and 8; 4x 15 =96. d. x = Equations: 4x 2 =6x, 1 2x =5, 5 3x = 7, 2x+1=5, x 6 =3+2x, 4 x= 1; solutions: a: x = 1, b: x = 2, c: x =4, d: x =2, e: x = 9, f: x = a. 5(3x 1) 2(4 x+1) 1 c. p+9 3p 2 d. 4 x y= 1 3 x y< 2 x a. 2d 3 2d 3=19, 11 candies 3 70 Algebra Connections

3 Lesson a. possible equation. 10x+45=15, x = 3 Six is the smallest number that can eliminate all fractions. Any multiple of 6 works, as does multiplying first by 3 and then again by a. 3x 2x =8, x = 8 5 2x 2 =3x,x =1 or 2.5 c. 14x+7 3x 9 =168, x = 10 d. x+3+2x 4 = x+5,x = a. p 2 or 4 because they would cause one (or both) denominators to be zero. p=3 or 4. However, p= 4 is an excluded value, and therefore is extraneous. Hence, the solution is p= a. x 2 +4x+3=0, x = 1 or 3 a 2 6a+9 =0, a =3 c. 12m 2=38+10m, m =20 d. 2(x 3)+3x =4(x+5), x = a. x =4 x = 5 or 2 c. x = 16 3 d. x = a. (x 2)(x+6) (x+4)(2x+3) 2x+1 x a. t =5 seconds 100 feet t +s=27, 1 2 t s =111, t =19 Times papers See diagram at right C Lesson x = x+12 =20, x = a. Divide by 4; dividing undoes multiplication, because they are inverse operations. after dividing, x+3=5, x = 2; yes c. Using the opposite operation eliminates (or undoes ) the multiplication a. If you treat (x+3) as a group, then 4 (group) = 20. Therefore, x+3=5 x+3=5, x = 2; yes iii, 2 i, 3 ii a c. 11 d. 2 e. 25 f. 27 s a. x =10 or a. Both Hank and Frank are correct. two a. 4 or or 100 c. no solution d. 3 or (a), (b), and (c) all are equal to a. x > 2 x 12 c. x >1 d. x a. 3+4x =14, x = 11 4 rewriting Let x represent the amount of money the youngest child receives. Then x+2x+x+35=775 ; $185, $370, and $220. Answer Key 71

4 Lesson a. two solutions x =3± 12 c or Reminder. the scope of this course is limited to real numbers. a. two solutions. 4 ± 20 no solution c. two solutions. 5 or 2 d. one solution. 1 2 e. no solution f. one solution a. no solution one solution c. two solutions a. Two solutions because 2x 5 can equal 9 or 9 Looking inside offers a quick solution. c. Since 2x 5 =9 or 9, then x =7 or x = a. Answers vary Answers vary, but it should contain an absolute-value expression equal to zero a. x =3 or 11 x =14 c. x =2 d. x = No, because 1 is not greater than a. x+4 m+5 4x 3 m a. (3x 1)(3x 1) c. m m m d. w w w w w w w w w w a. 4x(x 3) 3(y+1) 2 c. (2m+1)(m+3) d. (3x 2)(x+2) t = number of toppings, 1.19(3)+0.49t =4.55, and t = 2 Lesson a. x =4 or 6 x =15 c. x =53 d. x = 3 or 7 e. x = 15 or 9 f. x =5 or If x = the length of a side of the hot tub, then (x+3) and 16 x 10. However, since the minimum side length is 4 feet, the possible measurements that Ernie can order are 4 x 10 feet See graph at right When any real number is squared, the result is either positive or zero a. 7 or 1 4 or 8 c. 3 d. no solution a. x <2 x 6 c. x >4 d. x a. x 3 3x 14 2x 1 x xxxyy 72 Algebra Connections

5 Lesson x <2.5 ; a. 2.5; No, because the inequality does not include x equality. Find the boundary point by solving the corresponding equality. Then test a point on either side of the boundary point a. x < 1 or x >5 ; ; After finding dividing points, test a point x in each possible solution region. two boundary points a. The solution points are those x-values for which y= x 2 is greater than y=3. 1 x 5 ; x ; The solution points are the x-values for which y= x 2 is below (less than) y=3, including the points of intersection. c. All numbers; y= x 2 is always above the line y= a. two. x =1 and x = 3 three. x < 3, 3<x<1, and x >1 c. 3<x<1 ; d. Yes, see graph at right Solutions vary; in general, any inequality that states that x 2 is less than a negative number will work (x+2)(2x+3)=2x 2 +7x a. 2 1 c. 0 d y= 3 x See graph at right a. 2(x+5) or 2x+10 7x x a. x =13 x =3 c. 5 x 5 d. x < 2 3 or x >2 Lesson x 2 and x 4, a. x < 3 or x > 1 4 <x<3 c. all numbers d. x 1 e. no solution f. x <2 7 or x > t 2 +96t 140, 2.5 t 3.5 seconds x = 105, 9, 1, or (y 2)(y 2)(y 2) It has no solution, because an absolute value cannot equal a negative number a. x =1 x >3 or x < 3 c. 0 x 4 3 d. x = 3 e. 2 x 3 f a. y= 2 3 x 1 3 Yes; the slope of 6x 4y=8 is a. x 4 7m 1 3m+2 c. (4z 1)2 z+2 d. x 3 x C Answer Key 73

6 Lesson Both have the same solutions. 2± or a. They have the same answer. This tells us that the equations are equivalent, which can be shown by expanding (x+2) 2. Answers vary a. Add 3 to both sides of the equation and then write the square as a product. You must add tiles to complete a square a. (x+3) 2 =4, x = 1 or 5 (x+1) 2 =6, x = 1± or a. (x 3) 2 =10, x 0.16 or 6.16; (x+2) 2 =9, x = 5 or 1; c. (x 1) 2 =5, x 3.24 or (2x+3) 2 =16, x = 1 2 or a. 4x 3 5x+1 5x+2 5x 2 4(2x+1) c. x 2 d. 2y 3x Both result in no solution a. 2 <x<2 x 2.5 c. x = 1 4 f. 5 x 3 d. no solution e. x = See graph below. The portion for x between 5 and 3 (inclusive) should be highlighted m+22+2m =80, m = 8; It will be full after 8 months. 74 Algebra Connections

7 Lesson a. 2 0 c. 1 d The constant in the parentheses will be half the coefficient of the x-term in the original quadratic equation. c. If the quadratic is in the form x 2 +bx+c, then ( b 2 )2 c unit tiles need to be added to both sides of the equation a. (w+14) 2 =144, w= 2 or 26 (x+2.5) 2 =2.25, x = 1 or 4 c. (k 8) 2 =81, k = 1 or 17 d. (z 500) 2 =189225, z =65 or a. (x+2) 2 =1, x = 3 or 1 (x 4) 2 =9, x =1 or 7 c. (x 12) 2 =15, x =12± 15, x or a. x+2 3x 2 (x 2) 2 (x 11)(x+4) a. x = <x<0 c. x 9 or x 7 d. no solution e. x = 4 or 4 f. all numbers See graph at right They will be the same after 20 years, when both are $1800. Lesson a. i. y y y y y y y, ii. 5(2m)(2m)(2m), iii. (x x x)(x x x), iv. 4x x x x x y y i. x 2, ii. x 3, iii. x 5, iv. k 8, v. 2k, vi. m 7, vii. x 12, viii. 3x 3, challenge. x a. y 7 w 3 c. x 8 d. x 22 e. 13p 2 q 3 f. x6 y 3 g. 10h 25 h. 5m 22 i. 81k 80 j. 8 g 7 =8g 7 k. m12 n 40 l. pw Answers vary a. incorrect, x 100 correct c. incorrect, 27m 6 n a. 1 h 2 x7 c. 9k 10 d. n 8 e. 8y 3 f. 28x 3 y a. Haley is correct. You cannot add unlike terms. Haley is incorrect. The bases differ a. 1 unit tile (x+1) 2 = While there are multiple ways to write the rule, one possible way is y= x 2 +5x a. x =2 k 3.76 or 1.24 c. 2 <x<10 d. x =5 Answer Key 75

8 Lesson a. x 11 x 9 c. m 12 d. x 8 y 8 e. x 3 y 8 or y8 x 3 f. 1 4 x Expression Generalization Why is this true? a. x 25 x 40 = x 65 x m x n =? x m+n x 36 x13 = x23 x m x n =? xm n If you have m x s and n x s multiplied together, you then have a product of (m+n) x s. Many of the x s on top and bottom make ones, leaving (m n) x s. c. (x 5 ) 12 = x 60 (x m ) n =? x mn If you have n sets of m x s, then you have mn x s Its value is 1, and it can be rewritten as x m m = x 0. Thus x 0 = a. 1 x = x 1 1 x 2 = x 2 c. 1 x 3 = x a. 1 k 5 1 c. x3 d. p 2 e. y f. 1 x 6 g. 1 a 2 b h. x b, c, d, f y>2x (15, 2) y= x(x 20)= x 2 +20x ; Its maximum height is 100 feet when x = Perfect square form. (x 5) 2 =0 ; x = 5; answers vary a. x 4 3x+2 5 x 3 c Algebra Connections

9 Lesson a. The result is the square root of the base. 2, 7 c. It is inductive reasoning (based on patterns) a. 4 2, 5; Rewrite each as 3 repeated factors and then take one of them. c. 27 =3 3 3, so 27 2/3 is 3 3=9 ; 32 = , so 32 3/2 = 2 2 2=8 ; 25 =5 5, so 25 3/2 =5 5 5=125 d. 5 x ; fifth root a. (3 1/2 ) 4 =3 2 =9 3 7 =2187 c. (2 5 ) 1/3 =2 5/ a. 3 8 c. 7 d. 1 e. 6 f. 4 g. 2 h a. 4,600,000,000 years meters c seconds a. 100,000,000,000,000, c. While opinions may vary, since most calculators cannot handle the calculations in standard form, scientific notation is easier a. 15x 64p 6 q 3 c. 3m ,261,392,000 or feet See solution graph at right; ( 2, 6) and (1, 6) x 20, x 3.2 hours; Vinita can rent a skateboard for up to 3.2 hours (3 hours and 12 minutes) a. The person starts walking away from the detector and then turns around and walks slightly faster toward the machine The person does not move c. This is not possible because the motion detector cannot simultaneously record two different distances. Answer Key 77