Chapter 6 Rational Epressions and Equations Section 6. Rational Epressions Section 6. Page 7 Question a) (6) 8 (7 ) 4, 0 5 5(6) 0 5 5(7 ) 5 c) 4 7 44 d) 77 e) (6) f) 8(6) 8 + 4( + ) 4 + 8 4( ) 4 y ( y )( y+ ) y y + y Section 6. Page 7 Question a) Divide the numerator and the denominator by pq. pq pq p pq q pq Multiply the numerator and the denominator by ( 4). ( 4) + 4 ( + 4)( 4) 8 6 c) Divide the numerator and the denominator by (m ). 4( m ) 4( m ) ( m ) m 9 ( m )( m+ ) ( m ) 4 m + MHR Pre-Calculus Solutions Chapter 6 Page of 7
d) Multiply the numerator and the denominator by (y + y). ( y + y) y ( y )( y + y) y y + y y Section 6. Page 7 Question a) The denominator of 4 is zero when 0. The denominator of c c is zero when c 0. That is, when c. c) The denominator of d) The denominator of denominator zero. y y + 5 m + 5 is zero when y + 5 0. That is, when y 5. is never zero. There is no value of m that makes the e) The denominator of d is zero when d 0. That is, when d, or d ±. f) The denominator of + is zero if + 0. This never occurs, because + for all values of. There is no value of that makes the denominator zero. Section 6. Page 7 Question 4 In each the denominator cannot be zero, as division by zero is not defined. a a) For, 4 a 0. So, the non-permissible value is a 4. 4 a For e + 8, e 0. So, the non-permissible value is e 0. e c) For y. ( y + 7), (y 4)(y + ) 0. So, the non-permissible values are y 4 and ( y 4)( y+ ) MHR Pre-Calculus Solutions Chapter 6 Page of 7
7( r ) d) For, (r )(r + ) 0. So, the non-permissible values are r and ( r )( r+ ) r. k + 8 e) For, k 0. So, the non-permissible value is k 0. k 6 8 f) For, ( 4)( + 5) 0. So, the non-permissible values are ( 4)(+ 5) 5 and. Section 6. Page 8 Question 5 4 a) For πr 8πr, 8πr 0. So, the ecluded value is r 0. For c) For t+ t t, t 0. So, the ecluded values are t ±., 0 5 0. So, the ecluded value is. 0 5 g d) For, g 9g 0. Then, g 9g g(g 9) g(g )(g + ). So, the ecluded g 9g values are g 0 and g ±. Section 6. Page 8 Question 6 a) c c ( c 5) ( c 5), c 0, 5 w (w + ) (w + ), w,0 w (w + ) (w + ) c) ( 7) ( + 7) ( ) ( 7) + 7,,7 MHR Pre-Calculus Solutions Chapter 6 Page of 7
d) 5 0 + ( a ) ( a ) ( a) ( a + ),, a Section 6. Page 8 Question 7 a) is not a factor, it is a term. Dividing out the would be like reducing 5 5 to 0 0 : you cannot just strike out part of each number. Determine what values of the variable make the denominator zero. Factor the denominator, then set each factor equal to zero and solve. + 0 ( + )( ) 0 + 0 or 0 The non-permissible values are and. c) If possible, factor the numerator and the denominator. Then, divide both the numerator and the denominator by any common factors. ( ) ( + ) + ( + ) ( ) +,, + Section 6. Page 8 Question 8 a) 6 4 r r r p p p 4 r, p 0, r 0 p 6 ( ) 0 5 5( ), 5 c) b + b b+ b b 7 ( b 6) 4 ( 6)( 4) ( b + 6) ( b 4) ( b 6) ( b+ 6) b 4, b ± 6 ( b 6) MHR Pre-Calculus Solutions Chapter 6 Page 4 of 7
d) e) 0k + 55k+ 75 5(k + k+ 5) 0k 0k 50 0(k k 5) 4 ( + 4) 4 4, 4 5 0 (k + 5) (k + 5) ( k + ) ( k ) k + 5, k, ( k ) f) 5( ) y y+ y 5( y) ( + y) ( y) ( y) 5( + y), y y Section 6. Page 8 Question 9 + 5 ( )( + 5) + 5, The statement is sometimes true. It is true for all values of ecept. Section 6. Page 8 Question 0 The original epression may have had other factors that have been divided out. yy ( + ) Eample:, y 6,. ( y 6)( y+ ) Section 6. Page 8 Question Yes, Mike is correct, because (5 ) 5 +, or 5. The only thing to remember is that the factor in the denominator leads to a non-permissible value that should be stated in the reduced epression. MHR Pre-Calculus Solutions Chapter 6 Page 5 of 7
Section 6. Page 8 Question Eamples: Start with a factored epression and epand to create the question for your friend. ( + )( ) + ( + )( + ) + 4+ 4 Other similar eamples are + 6 6, + + 5+ 6 +. Section 6. Page 8 Question In the third step, Shali cancelled + from both the numerator and the denominator. This is not correct because is not a factor of the numerator. The correct solution is as follows. g 4 ( g ) ( g + ) g 4 ( g ) g +, g Section 6. Page 8 Question 4 4p 4p Eample: ( p )( p+ ) p p Section 6. Page 9 Question 5 d a) t, so an epression for the time is: r n + n+ n n + n+ (n+ )( n+ 4) n ( n 6) (n+ )( n+ 4) ( n 4)( n+ 4) n +, n ± 4 ( n 4) MHR Pre-Calculus Solutions Chapter 6 Page 6 of 7
Section 6. Page 9 Question 6 a) c) The denominator, 4, cannot be zero, so 0. d) π 4 Area of circle π Area of square 4 e) Percent π 4 00% The percent of paper in the circle is 79%, to the nearest percent. Section 6. Page 9 Question 7 a) The non-permissible value is p, but this negative value has no meaning since p is a mass. The least etra yield occurs when the epression 900 p + p etra yield. is least. If p 0, there is no 900 p 900( 450) c) Try p 450, 896 + p + 450 900 p 900( 900) Try p 900, 898 + p + 900 900 p 900( 000) Try p 000, 898 + p + 000 900 p 900( 6000) Try p 6000, 900 + p + 6000 The greatest etra yield seems to be 900 kg. A graph of the related function, confirms this. 900 p y, + p MHR Pre-Calculus Solutions Chapter 6 Page 7 of 7
Section 6. Page 9 Question 8 a) Substitute d 00 and r q into t 00 q 50 t, q 0 q d t. r Substitute d 00 and r p 4 into 00 t p 4 00 t, p 4 p 4 d t. r Section 6. Page 9 Question 9 a) Cost for 0 students 50 + 0(9) 60 The total cost for 0 students is $60. Cost for n students 50 + 9n So, cost per student 50 + 9 n, n 0 n 50 + 9n c) Substitute n 0 into. n 50 + 9(0) Cost per student 0 0.67 The cost per student, if 0 students go, is $0.67. Section 6. Page 0 Question 0 a) No. Terri divided out the term 5 but it is not a factor of the denominator. Eample: when m 5 5 5 or but or m+ 5 + 5 8 m+ + 4 MHR Pre-Calculus Solutions Chapter 6 Page 8 of 7
Section 6. Page 0 Question a) To change into 5, 4 0 5( ) 5 5(4) 0 multiply numerator and denominator by 5. 6 To change into, 4 4 8 ( ) 6 4( ) 4 8 multiply numerator and denominator by ( ). Section 6. Page 0 Question a) c) 4( ) 4() 4 8 ( ) () 6 9 ( )(+ 5) ( + 5) + 0 5, 6 + 5 Section 6. Page 0 Question a) 5(5 5 5b 5 b, b 0 5b + ( + ) (4 a (4 a 4ab+ 4ab, 0, a b 0 a b MHR Pre-Calculus Solutions Chapter 6 Page 9 of 7
c) a b ( a 7 (7 ) b a, 0 4 Section 6. Page 0 Question 4 a) Use the formula for area of a triangle, QR is the base. So, QR (Area) height ( 6) QR ( )( + ) QR QR ( + ) c) The non-permissible value is. Section 6. Page 0 Question 5. bh A, where b is the base and h is the height. a) + 6 ( )( ) 9 ( )(+ ) +, ± + n + n n+ n 5) 5 ( )( 5n n n(5 n) ( n+ )(n 5) n(n 5) n + n n 5, n 0, n MHR Pre-Calculus Solutions Chapter 6 Page 0 of 7
Section 6. Page 0 Question 6 a) + + + + 9 9 + 8 9 ( + 6)( ) ( + )( ) ( ) ( ) 0 4 4 0 + 6, ± + 4( 9) ( + ) [( 9) + ( + )][( 9) ( + ) + 6+ 9 ( + )( + ) + + ( + )( + ) [ 8 ][ 8 ] + ( + )( + ) [ 5][ ] ( 5) ( + ) ( 7) ( + ) ( + ) ( + ) ( 5)( 7), c) d) ( ) 8( ) + [( ) 6][( ) ] ( 4) ( ) [( 4) + ( )][( 4) ( )] ( 6)( ) ( + 6)( ) ( )( + ),,, ( + )( ) ( + 4+ 4) 0( + 4+ 4) + 9 [( + 4+ 4) ][( + 4+ 4) 9] (+ ) ( + ) [(+ ) + ( + )][(+ ) ( + )] + + + (+ )( ) ( 4 )( 4 5) ( + ) ( + )( + 5)( ) ( + ) ( ) ( + )( + 5), ± MHR Pre-Calculus Solutions Chapter 6 Page of 7
Section 6. Page 0 Question 7 Use the formula for area of a parallelogram, A bh, where b is the base and h is the height, to find the base, AB, and height, BC, of ABC. For parallelogram ABFG, 6 AB 4 (4 )(4+ ) (4 ) 4+, 4 For parallelogram BCDE, 6 BC (+ 4)( ) ( ) + 4, (4+ )(+ 4) Then, the area of ABC + 9+ 4 9 6 + + 9 The area of ABC is ( 6 + +) square units,,. 4 Section 6. Page Question 8 a) Visualize the carpet laid flat: the eposed edge is a rectangle with length is L and thickness is t so its area is Lt. A πr πr A π(r r ) A π(r r)(r + r) MHR Pre-Calculus Solutions Chapter 6 Page of 7
A c) L t π( R r)( R+ r) L, t > 0, R> r t Also, r, R, and t should be epressed in the same units. Section 6. Page Question 9 Eamples: a),, 5 ( + )( 5) ( + ) + ( )( + ) +,, The given epression has non-permissible value, so I multiplied numerator and denominator by ( + ) to introduce the other non-permissible value,. Section 6. Page Question 0 a) Eample: Use y. y 4 4 4 y 5y ( ) 5( ) 8y + 4 8() + 4 6 5 ( )( y y y+ y 8y+ 4 4(y+ ) y, y 4 ) c) The algebraic approach, in part, proves that the two epressions are equal. Substituting a particular value does not prove the result for all permissible values. MHR Pre-Calculus Solutions Chapter 6 Page of 7
Section 6. Page Question vertical change a) Use slope horizontal change. p 5 slope p+ p p 8, p p + Eample: If p, the slope is 8 7. + c) When p the slope is undefined and the line is vertical. Section 6. Page Question Eample: To epress a fraction in lowest terms you divide the numerator and the denominator by any common factors. The same principle applies when simplifying a rational epression. 0 (0) 0 + 0 0( + ) 80 8(0) 50 80 0(5 8 ) 8 +, 5 5 8 8 Section 6. Multiplying and Dividing Rational Epressions Section 6. Page 7 Question a) 5 m m c f f 5 c 4 m 9 m, c 0, f 0, m 0 ( a ( a ) ( a+ 5) ( a 5) ( a+ 5) 5 ( a 5 a 5, a, 5, a b 5( a ) c) ( y 7) ( y+ ) (y ) (y+ ) 4(y + ) 4( y 7), y ±,, ( y + ) ( y ) (y )( y ) MHR Pre-Calculus Solutions Chapter 6 Page 4 of 7
Section 6. Page 7 Question a) d 00 6 ( d 0) ( d + 0) 44 d + 0 44 4 d 0, d 0 4 a 9 a+ a a+ a+ a + ( a + ) ( a ) ( a ) ( a+ ) a, a, ± a 6 d +0 c) 4z 5 z 4 z z+ 0 4z+ 0 (z 5) (z 5) (z + 5) ( z 4) 5, z ±,4 z 4 (z + 5) d) (p ) p + 5p p p p 6p p + p p ( p + ) p +, p,,, ( p ) ( p + ) (p ) ( p + ) p ( p ) Section 6. Page 7 Question a) The reciprocal of t is t. The reciprocal of is. c) The reciprocal of 8 y is y y or. 8 8 d) The reciprocal of p p is p p. MHR Pre-Calculus Solutions Chapter 6 Page 5 of 7
Section 6. Page 7 Question 4 4 t t 4 t s s s s t a) The non-permissible values are s 0, t 0. r 7r r r 7r r+ 7 r 49 r+ 7 r 49 r r 7r r+ 7 ( r 7)( r+ 7) r The non-permissible values are r 0, ±7. 5 0 5 n ( n ) c) n+ n n+ 0 n The non-permissible values are n ±. Section 6. Page 7 Question 5 6 + ( ) + + +, Section 6. Page 7 Question 6 y y y y y 9 y ( y ) ( y + ) y, y 0, ± y + y y Section 6. Page 7 Question 7 a) p ( + p) p p, p MHR Pre-Calculus Solutions Chapter 6 Page 6 of 7
7k ( 7k + ) k 7k k 7k or, k 0, k k 7 Section 6. Page 7 Question 8 a) c) w w 6 w+ (w + ) ( w ) w+ 6 w+ ( w + ) 5 5 5 v v v v v v v w, w, v v ( v 5) ( v + ) v, v 0,5, v + 9 5 ( ) (+ ) + 5 + 5 w + w + ( + ) ( ) ( ) or,, 5, + 5 + 5 d) 8 4 y y y+ y y y y y + (4y ) (y ) ( y ) ( y + ) (y + ) ( y ) ( y ) or, y ±,,, y y 4 y + 4y Section 6. Page 7 Question 9 5 + 5 + + + The non-permissible values in the original epression are and. Then, when the division is converted to multiply by the reciprocal, is also non-permissible. MHR Pre-Calculus Solutions Chapter 6 Page 7 of 7
Section 6. Page 7 Question 0 Height of stack number of sheets thickness of each sheet n 4 ( n ) ( n + ) ( n ) n + n + n n +, n, n + An epression for the thickness of one sheet of plywood is Section 6. Page 8 Question n +, n,. n + a) metres per minute; 60 6 metres in 60 min 5 5 The outer edge of the windmill blade travel ( 6) metres in h. distance speed time 600 speed 900 n + n + 900 600 n +, n The average speed of the plane is n + kilometres per hour, n. c) volume height area of base + + height ( )( + ) ( + ) ( + ) ( ) ( + ) +,, An epression for the height of the bo, in metres, is +,,. MHR Pre-Calculus Solutions Chapter 6 Page 8 of 7
Section 6. Page 8 Question m+ m+ m+ ( m )( m+ ) m m m m+ m + m+ m+ m+ m m m ( m )( m+ ) m+ m + The answers are reciprocals of each other. This is always true. Compare the following general rational epressions: a a y ay b y b b a b b y b y a ay Section 6. Page 8 Question Eample: yd ft in..54 cm yd 9.44 cm ft in. Section 6. Page 8 Question 4 a) Tessa s mistake is in the first step: she took the reciprocal of the dividend, not the divisor. c 6 c+ 6 ( c 6) ( c+ 6) c 8c c 4c 8c c + 4 cc ( 6) or 4c 4 c, c 0, 6 6 c) The correct answer is the reciprocal of Tessa s answer. Taking reciprocals of either factor produces reciprocal answers. See Question above. MHR Pre-Calculus Solutions Chapter 6 Page 9 of 7
Section 6. Page 8 Question 5 Area length width 9 length + ( ) ( + ( + ) ) ( ) ( + ) +,, An epression for the length is +,,. Section 6. Page 8 Question 6 Area base height 7 8 + Area PQR 4 8 ( 8) ( + ) + ( ) ( + ) 8 +, ±,8 ( ) An epression for the area of the triangle is +, ±,8. ( ) Section 6. Page 9 Question 7 P a) In K, substitute m h m w. P h K w Pw K, m 0, w 0, h 0 h In y π d, substitute d. r π y r r y π π r y, 0, r 0, d 0 MHR Pre-Calculus Solutions Chapter 6 Page 0 of 7
v c) Substitute r into a w r. w v a w w a wv, w 0 Section 6. Page 9 Question 8 V If V T VT, then V. Substitute V T T n n n 6 + 4 V n 6 ( n 4) ( n+ 4) n V n V ( n 4), n, ± 4 6 n + 4 n 6, T n n and T n + 4. 6 Section 6. Page 9 Question 9 a) Yes. The product has the difference of squares pattern. When you epand ( 5)( + 5) 5. + 7 ( + ) 7 ( + ) ( ) 7 ( 7) ( + 7) 7 7 7 + 7 c) 7 7 + 7 + 7 ( 7)( 7 + 7 This is the same epression as obtained by factoring ) 7 7 in part. MHR Pre-Calculus Solutions Chapter 6 Page of 7
Section 6. Page 9 Question 0 V sin a) In, substitute V 85, 5, and g 9.8. g V sin ( 85) sin 5 g (9.8) 90 The canister reaches a height of approimately 90 m. V sin + In, substitute V and 0. g 5 + sin 0 V sin 5 g g ( + ) ( 5) ( + ) 4 g( 5) In this case, the canister reaches a height of g ( + ) 4 g ( 5) metres. Section 6. Page 0 Question I agree. In both, you first convert any division to multiplication by multiplying by the reciprocal. Then, you divide numerators and denominators by any common factors. ()() 5 Eample: and 5 ()(5) 5 0 5 ( + ) ( + ) ( + )( + ) ( + ) ( + ) ( + )( + ) + +, + 6+ 9 ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ( + ) ( + ),, ( + ) MHR Pre-Calculus Solutions Chapter 6 Page of 7
Section 6. Page 0 Question vertical change a) Use slope horizontal change. p+ ( p+ ) slope p (p 5) p+ p p p+ 5 p + 4 p The slopes of perpendicular lines are negative reciprocals. So, slope of line (4 p) p 4 perpendicular to MN is p+ p+. Section 6. Page 0 Question a) tan B b a sin B cos B b c b a a c c) They are the same, sin B tan B. cos B 6. Adding and Subtracting Rational Epressions Section 6. Page 6 Question a) 4 7 6 6 6 7 + 0, 0 MHR Pre-Calculus Solutions Chapter 6 Page of 7
c) 5 t+ t+ 5 5 t+ + t+ + 5 0 0 0 8t + 8 0 (4t + 4) 0 5 4t + 4 5 d) e) m m m m + + m+ m+ m+ m ( m+ ) m + m, m a a a a a 4 a 4 a 4 a 4 ( a+ ) ( a 4) a 4 a+, a 4 Section 6. Page 6 Question 7 6+ 7 7+ 6+ 7 + 9 9 9 9 9 Section 6. Page 6 Question a) 4 4( ) ( )( + ) ( + ) ( )( + ) 4 + ( )( + ) 4,, ( )( + ) MHR Pre-Calculus Solutions Chapter 6 Page 4 of 7
5 + 5 + + + + 8 0 4 ( + 0)( ) ( )( + ) ( 5)( + ) + (+ )( + 0) ( + 0)( )( + ) 0 + + + 0 ( + 0)( )( + ) + 8 ( + 0)( )( + ) ( + 6), 0, ± ( + 0)( )( + ) Section 6. Page 6 Question 4 a) Eamples for common denominators: 4, 6 LCD Eamples for common denominators: 0a y, 50a y LCD 0a y c) Eamples for common denominators: (9 )( + ), 9 LCD 9 Section 6. Page 6 Question 5 a) 5 () + + a 5a 5( a) (5a) 5 6 + 5a 5a, a 0 5a () ( ) + + 6 ( ) 6( ) 9 + 6 6 9 +, 0 6 MHR Pre-Calculus Solutions Chapter 6 Page 5 of 7
c) d) 6 4(5 ) 6 4 5 5 5 0 6 5 5 0 6 5 (0 ), 0 5 4z 9 4 z( z) 9 ( ) y yz y( z) yz( ) 4z 9 yz yz 4z 9 yz (z )(z+ ), 0, y 0, z 0 yz e) s 6 s(6 t) ( t ) 6() + + 5 0 5 5 (6 ) 0 ( ) 5 ( t t t t t t t t st t + 0t 0t 0t st + t 0t (4st + t 4) 0 t 0 st + t 0t 4 4, t 0 ) f) 6y 6 y( by) ( a) a b y + + ab aby abby ( ) ( abya ) aby 6 + by a a b y aby aby aby 6 +, a 0, b 0, y 0 by a a b y aby MHR Pre-Calculus Solutions Chapter 6 Page 6 of 7
Section 6. Page 6 Question 6 8 5 8 5( ) a) 4 + ( )( + ) ( + )( ) 8 5 + 0 ( )( + ) 8 5, ± ( )( + ) ( 4) + + + ( 4)( + ) ( + )( 4) + ( 4)( + ), 4, ( 4)( + ) c) ( ) ( + ) + ( + )( ) ( + )( ) 6 ( + )( ) 8 ( + )( ) ( 4), ± ( + )( ) d) 5 y 4 5( y) ( y+ ) ( y 4) y+ y y+ y ( y+ ) y yy ( + ) yy ( + ) 5y y y+ 4 yy ( + ) y + yy ( + ) ( y + ) y ( y+ ), y 0, y MHR Pre-Calculus Solutions Chapter 6 Page 7 of 7
e) h h h h + + h 9 h + 6h+ 9 h ( h )( h+ ) ( h+ )( h+ ) h f) hh ( + ) hh ( ) ( h+ )( h+ ) + ( h )( h+ )( h+ ) ( h )( h+ )( h+ ) ( h )( h+ )( h+ ) 6 8 7 h + h+ h h h h ( h )( h+ )( h+ ) 5h 7 ( h )( h+ )( h+ ) (5h + 9), h ± ( h )( h+ )( h + ) + + + 6 + ( + )( ) ( + )( ) ( )( ) ( ) + ( + )( )( ) ( + )( )( ) 6 + ( + )( )( ) 6 + ( + )( )( ) Section 6. Page 6 Question 7 a) ( + 5) + + 5 + 9 5 ( 5)( + 5) + 5 4 ( )( + ), 0,,, ( + )( )( ) ( ) 5 9 5 6 0 ( + ) ( ) ( + 5) ( + 5) (+ )( 5) + ( 5)( + 5) ( + 5)( 5) + + ( 5)( + 5) + ( 5)( + 5) ( + 5), ± 5, ( 5)( + 5) MHR Pre-Calculus Solutions Chapter 6 Page 8 of 7
8 + 6 5 4 ( + )( ) c) 8 ( 8) ( + ) ( ) ( + )( ) ( + )( ) + ( + )( ) 4, 0,,,8 ( + )( ) n+ 6 n+ 6 + + 5 + 6 7 + ( )( ) ( )( 4) n n n n n n n n d) ( n+ )( n 4) 6( n ) + ( n )( n )( n 4) ( n )( n )( n 4) n n + 6n ( n )( n )( n 4) n + 5n 4 ( n )( n )( n 4) ( n+ 8)( ) ( n ) ( n )( n 4) n + 8, n,,4 ( n )( n 4) w w 6 w w 6 + 5 + 6 + 6 + 8 ( + )( + ) ( + 4)( + ) w w w w w w w w ww ( + 4) ( w 6)( w+ ) ( w+ )( w+ )( w+ 4) ( w+ )( w+ )( w+ 4) w w w w ( w+ )( w+ )( w+ 4) + 8 + + 8 w + w+ 8 ( w+ )( w+ )( w+ 4) ( w+ 9)( w+ ) ( w+ )( w+ ) ( w + 4) w + 9, w,, 4 ( w+ )( w+ 4) MHR Pre-Calculus Solutions Chapter 6 Page 9 of 7
Section 6. Page 6 Question 8 In the third line, multiplying by 7 should give 7 + 4. Also, Linda has forgotten to list the non-permissible values. 6 4 7 6( + ) + 4 7( ) + 4 + ( )( + ) 6+ + 4 7+ 4 ( )( + ) + 0, ± ( )( + ) Section 6. Page 6 Question 9 Yes, the epression can be simplified further. Factor from the numerator then simplify. + 5 ( 5) ( 5)( + 5) ( 5) ( + 5), ± 5 + 5 Section 6. Page 7 Question 0 a) 6 9 6 9 ( ) ( ) ( + ), 0, ± + + 6 6 t t+ t t t t t( t+ 6) t+ 6 t ( t + ) t ( t + 6) t ( t )( t+ ) ( t + 6), t 0,,, 6 ( t ) MHR Pre-Calculus Solutions Chapter 6 Page 0 of 7
c) m m+ (m+ ) ( m) (m+ ) + m + m(m+ ) m (m+ ) m m+ m 6m+ 9 m m (m + ) m (m + ) 6m+ 9+ m ( m + ) m ( m + ) ( m + ) m, m 0,, m + d) + + 4 4 ( 4) + ( + 4) + 4 + ( 4)( + 4) ( 4)( + 4) 6 + 4 ( 4) ( + 4), ± 4, ( 4) ( + 4) ( ) Section 6. Page 7 Question a) AD + C B AD + CB D D B AD + CB B D AD + CB BD AD CB + BD BD A C + B D MHR Pre-Calculus Solutions Chapter 6 Page of 7
AB + C D D AB + E F C D EF + + F D AB + CD + EF Section 6. Page 7 Question Let h represent the length of the hypotenuse. h h h + 4 + + 4 6 4 + + 6 h h + 6 5 + 4 5 Section 6. Page 7 Question a) 00 m tells the epected number of weeks to gain 00 kg; tells the number of m 004 + weeks to gain 00 kg when the calf is on the healthy growth program. 00 00 m m+ 4 00 00 00( m+ 4) 00m c) m m+ 4 m( m+ 4) 800, m 0, 4 mm ( + 4) Yes, the simplified epression still represents the difference between the epected and the actual times the calf took to gain 00 kg because the epressions are equivalent. MHR Pre-Calculus Solutions Chapter 6 Page of 7
Section 6. Page 7 Question 4 a) At a typing speed of n words per minute, the time to type 00 words is 00 n minutes. The time to type the three assignments is 00 500 000 + + n n n minutes 00 500 000 700 c) + +. The epression tells the number of minutes to type all n n n n three assignments at a typing speed of n words per minute. d) If the speed decreases by 5 words per minute with each new assignment, then the etra time to type them is given by 00 500 000 700 500 500 000 + + + + n n 5 n 0 n n n 5 n 0 500( n 5)( n 0) + 500 n( n 0) + 000 n( n 5) nn ( 5)( n 0) 500 500 75 000 500 5000 000 5000 n + n + n n+ n n nn ( 5)( n 0) 500n 75 000 nn ( 5)( n 0) If the typing speed deceases by 5 words per minute for each new assignment then it will 500n 75 000 take minutes longer to type the three assignments. nn ( 5)( n 0) Section 6. Page 8 Question 5 a) + ( ) ( + ) ( + ) + + + 5 6 4 + 5 ( ) ( + ) ( 4) ( )( 4) ( + )( + 5) + ( + 5)( 4) ( + 5)( 4) 6+ 8+ + 6+ 5 ( + 5)( 4) +, 0,, 4,, 5 ( + 5)( 4) MHR Pre-Calculus Solutions Chapter 6 Page of 7
+ + + ( ) ( + ) ( 4)( + ) ( ) ( ) + ( 4)( + ) ( )( ) ( )( + ) ( )( + ) 8 ( + ) ( )( + ) 9, 0,,,, ( )( + ) c) + ( ) ( + ) ( + ) + 5 6 4 + 5 ( ) ( + ) ( 4) ( )( 4) ( + )( + 5) ( + 5)( 4) ( + 5)( 4) 6+ 8 6 5 ( + 5)( 4) ( + 5)( 4) ( 4 ), 0,, 4, 5, ( + 5)( 4) d) + 4 + 7+ + ( )( + ) ( + ) + 6 + + + 6 ( + ) ( + ) ( + ) ( + )( + ) ( )( + 6) ( + 6)( + ) ( + 6)( + ) + 4+ 4+ ( + 6)( + ) 5, 0,, 6, ( + 6)( + ) MHR Pre-Calculus Solutions Chapter 6 Page 4 of 7
Section 6. Page 8 Question 6 Let represent her speed, in kilometres per hour, during the first 0 km. Then, represents her speed for the remaining 6 km. distance Use the formula time to write an epression for the total time. speed 0 6 Total time + 0 6 An epression for the total time of her bike ride is + hours. Section 6. Page 8 Question 7 Eample: Jo runs km/h faster than Sam. Write an epression for how much longer it takes Sam to run 0 km. Let represent Sam s running speed. Then, + represents Jo s running speed. 0 0 An epression for how much longer it takes Sam to run 0 km is + hours. Section 6. Page 8 Question 8 a) Incorrect: a b a b. Find the LCD first, do not just combine pieces. b a ab ca cb a + b Incorrect:. Factor c from the numerator and from the denominator, c+ cd + d remembering that c() c. 6 6 c) Incorrect: a b a + b. Distribute the subtraction to both terms in the 4 4 4 numerator of the second rational epression by first putting the numerator in a bracket. b d) Incorrect:. Simplify the denominator first, then divide. a b a b e) Incorrect:. Multiplying both numerator and denominator by, which is a b b a the same as multiplying the whole epression by, changes every term to its opposite. MHR Pre-Calculus Solutions Chapter 6 Page 5 of 7
Section 6. Page 8 Question 9 a) Agree. Each term in the numerator is divided by the denominator, and then can be simplified. Disagree. Eamples: + y + y +, but this may not be the original. y y y y y + + simplifies to the same epression. y Section 6. Page 9 Question 0 a) In R + + R R R R, R, and R 4. R + + 4 R 6 + 4 + R substitute The total resistance is ohms. R + + R R R R R + RR + RR R RRR RRR R RR + RR + RR c) In R RRR substitute R R R RR RR, R, and R 4. + + ()() 4 R 4 ( ) + 4 ( ) + () 4 R 6 R The total resistance is ohms. d) Eample: I prefer the simplified version from part because it is not a double fraction. MHR Pre-Calculus Solutions Chapter 6 Page 6 of 7
Section 6. Page 9 Question Eample: Arithmetic: 6 If, then 9 6 9 4 6 Algebra: If, then 6 6 4 Section 6. Page 9 Question vertical change a) Use slope horizontal change. p p slope 4 AB p p 4p (p ) ( p ) p 6 9 p 6 p 9 p, p ( p ) When p the slope is undefined, so the line is vertical. c) When p < and p is an integer, the slope is negative. Eample: When p, slope AB.5. d) When p 4, the slope is positive, from p 5 to p 0 the slope is always negative. MHR Pre-Calculus Solutions Chapter 6 Page 7 of 7
Section 6. Page 9 Question p q r + + + + p q r p q r p q r + + p p q q r r p q r + + p q r + + Section 6. Page 9 Question 4 Eamples: + + and 5 5 5 5 () + (5) + 5 5 5 + + and ( y) + ( ) y+ + y y y Section 6. Page 40 Question 5 a) The student s suggestion is correct. Eample: Find the average of and. 4 + + 4 4 5 8 Halfway between and, or 4 and 6 5, is. 4 8 8 8 7 6+ 7 + a a a, a 0 4a MHR Pre-Calculus Solutions Chapter 6 Page 8 of 7
Section 6. Page 40 Question 6 Yes. Eample: + 5 and + 5 6 6 6 5 + y + y + and + y y y y y + y Section 6. Page 40 Question 7 a) + f u v u+ v uv Substitute u 80 and v 6.4. 80 + 6.4 f 80( 6.4) 86.4 f 5 5 f 5.9 8.4 6 When u 80 and v 6.4, f 5.9 cm. c) Given that u + v, then f f uv uv u+ v. Section 6. Page 40 Question 8 Step Yes Step 4a) A, B Step 4 A, B ( ) + ( 4) Step 5 Always: + 4 ( 4)( ) + 5 ( 4)( ) MHR Pre-Calculus Solutions Chapter 6 Page 9 of 7
Section 6.4 Rational Equations Section 6.4 Page 48 Question a) 4 5 5 + 4 6 5 5 + 4 6 ( ) ( 5) 4 (5) 4( ) ( 5) 5 + ( ) + 6 + 7 + + 5 + 0 + 7 + + 5 ( + 5) + 7 ( + 5) + ( + 5) + 5 ( + 5) + ( + 5) + ( + 5 + 5) (+ ) + ( + 5) 7 ( + 5) 7 ( + 5) c) ( ) ( + ) 4 5 9 + 4 5 ( )( + ) + 4 5 ( )( + ) ( + )( ) ( )( + ) + 4 5 ( )( + ) ( )( + ) ( ) ( + ) + 4 5( ) ( )( + ) MHR Pre-Calculus Solutions Chapter 6 Page 40 of 7
Section 6.4 Page 48 Question a) f + f 6( f + ) 6( f ) 6() ( f + ) ( f ) f + 9 f + 4 f y + y 4 y y y + y y y 4 y 4( y) + y 6 4y+ y 6 6 y, y 0 c) 9 4 8 w w 6 w 9w+ 8 9 4 8 w w 6 ( w )( w 6) 9 4 8 ( w )( w 6) ( w )( w 6) w w 6 ( w )( w 6) 9( w 6) 4( w ) 8 9w 54 4w+ 8 5w 60 w, w, 6 Section 6.4 Page 48 Question a) 6 t + 4 t 6 t t + t(4) t t t 8 + 8t+ 0 ( t 6)( t ) 0 t 6 or t, t 0 t 6 c + 5 c c 9 6 c + c ( c )( c+ ) 5 6 ( c ) ( c ) 5 c c 6 5( c ) 6 5c + 5 5c 0 c, c ± MHR Pre-Calculus Solutions Chapter 6 Page 4 of 7
c) d d + d + 4 d + d 4 d d d + d + 4 ( d )( d + 4) d d d ( d )( d + 4) ( d )( d + 4) d 4 + ( d )( d 4) d + + ( d ) d d + d + 4 d d 6 0 ( d )( d + ) 0 d or d, d, 4 d) + + 5 + + + 5 + ( )( + ) + + 5 ( )( + ) ( )( + ) + ( )( + ) + + + ( )( ) ( )( ) 5 + + + 5 0 0 ( )( + ) Then,, ±. Section 6.4 Page 48 Question 4 y 6y 9 The solution for + 6 cannot be y because this is a non-permissible y y value for the equation. MHR Pre-Calculus Solutions Chapter 6 Page 4 of 7
Section 6.4 Page 48 Question 5 a) Length Width ( ), 0 For the width to eist, > 0. Area length width ( ), 0, > 0 c) Perimeter (length + width) 4 4 0 8 + + 4 + (7+ )( ) 0 Since cannot be negative, 0.5 cm. Section 6.4 Page 48 Question 6 6 a) + b+ 5 b 6 ( b+ 5)( b ) ( b+ 5)( b ) + b+ 5 b 6( b ) ( b+ 5)( b ) + ( b+ 5) b b + b b + b+ 6 5 5 0 5 0 b 0b+ 57 Substitute a, b 0, and c 57 into the quadratic formula. b ( 0) ± ( 0) 4( )( 7) ( ) 0 ± 7 b b 6.56 or b.44 5 MHR Pre-Calculus Solutions Chapter 6 Page 4 of 7
c 6 c+ c 4 c 6 c+ ( c )( c+ ) c 6 ( c )( c+ ) ( c )( c+ ) c ( c )( c ) + + cc ( ) ( c )( c+ ) 6 c c c + 6 0 c + c 8 c + c 9 0 Substitute a, b, and c 9 into the quadratic formula. ± 4( )( 9) c ( ) ± 7 c c.54 or c. 54 Section 6.4 Page 48 Question 7 Substitute w 0 into l l + w. w l l l+ 0 0 l l 0l+ 900 l 0l 900 0 Substitute a, b 0, and c 900 into the quadratic formula. ( 0) ± ( 0) 4( )( 900) l ( ) 0 ± 4500 l 0 ± 0 5 l l 5 ± 5 5 l 5( ± 5) The length of the frame is eactly 5( + 5) cm or 48.5 cm, to the nearest tenth of a centimetre. MHR Pre-Calculus Solutions Chapter 6 Page 44 of 7
Section 6.4 Page 48 Question 8 Let represent the first number. Then, the second number will be 5. + 5 4 4(5 ) + 4 (5 ) 00 4 + 4 5 5+ 00 0 ( 0)( 5) 0 So, 0 or 5. The two numbers are 5 and 0. Section 6.4 Page 49 Question 9 + 6 9 + ( + 6) 9( ) + 9 9 7 The numbers are and 4. Section 6.4 Page 49 Question 0 Let represent the number of students epected to go on the trip. 540 540 + 6 540( 6) + ( 6) 540 540 40 + 8 540 8 40 0 6 080 0 ( 6)( + 0) 0 So, 6. Thirty students actually went on the trip. MHR Pre-Calculus Solutions Chapter 6 Page 45 of 7
Section 6.4 Page 49 Question Let n represent the first integer. Then, the net consecutive integer is n +. + n n+ 0 0( n+ ) + 0n n( n+ ) n+ n + n 60 0 n n 49 0 0 (n+ 6)( n 5) 0 6 n or n 5 Then, since n represents an integer, the two numbers are 5 and 6. Section 6.4 Page 49 Question a) With both taps running it should take less than min because more water is going in at the same time. Time to Fill Tub (min) Cold Tap Hot Tap Both Taps Fraction Filled Fraction Filled in in min minutes c) + d) + 6 + 6() + 6 5 6. The time to fill the tub with both taps running is. min. MHR Pre-Calculus Solutions Chapter 6 Page 46 of 7
Section 6.4 Page 49 Question Time to Fill Pool (h) Hose A Hose B Both Hoses 6 6() 6 Fraction Filled Fraction Filled in in h hours 6 It would take 6 h to fill the pool using only hose B. Section 6.4 Page 49 Question 4 a) Distance (km) Rate (km/h) Downstream 8 + Upstream 8 Time (h) 8 + 8 8 8 + 8 8 c) + 8( ) 8( + ) 8 54 8+ 4 0 78 7.8 The rate of the kayakers in still water is 7.8 km/h. d) ± MHR Pre-Calculus Solutions Chapter 6 Page 47 of 7
Section 6.4 Page 50 Question 5 Time to harvest all (h) Fraction of harvest done per hour Fraction of harvest done in t hours t Nikita alone 7 7 7 Neighbour t 48 alone 48 48 Together t t t t + 7 48 48t+ 7 t (7)(48) 0t 456 t 8.8 If they work together, the harvest will take 8.8 h. Section 6.4 Page 50 Question 6 Distance (km) Speed (km/h) To Forde Lake 70 5 Beyond Forde Lake 60 70 60 4(4) 5 70 96 ( 5) 60( 5) 70 96 + 480 60 00 0 96 490 00 0 48 45 50 Time (h) 70 5 60 ( 45) ± ( 45) 4(48)( 50) (48) 45 ± 88 85 96 5.6566... The average speed of the herd beyond Forde Lake was 5.7 km/h, to the nearest tenth of a kilometre per hour. MHR Pre-Calculus Solutions Chapter 6 Page 48 of 7
Section 6.4 Page 50 Question 7 Let kilometres per hour represent Ted s speed east of Swift Current. Distance Speed Time (km) (km/h) (h) West of Swift 75 75 0 Current 0 East of Swift 00 00 Current 75 00 + 0 ( )75 00()( 0) + ( 0) 550 600 6000 + 0 0 + 40 6000 0 ( + 00)( 60) 00 or 60 Ted s average speed east of Swift Current was 60 km/h and west of Swift Current it was 50 km/h. Section 6.4 Page 50 Question 8 Let kilometres per hour represent the speed of the current. Distance (km) Speed (km/h) Time (h) Up river 6 6 Down river 6 + 6 + Use the fact that the total time to paddle up river and back is h to write an equation. + 6 6+ (6 + ) + (6 ) (6 + )(6 ) + + 6.5 The speed of the current is approimately.5 km/h. MHR Pre-Calculus Solutions Chapter 6 Page 49 of 7
Section 6.4 Page 50 Question 9 Let pages per day represent the reading rate for the first half. Reading Rate in Number of Number Pages per Day Pages Read of Days First 59 59 Half Second 59 + 59 Half + 59 59 + + 59( + ) + 59 ( + ) 59+ 08 + 59 + 5 0 66 08 0 8 444 b b ac ± 4 a ( 8) ± ( 8) 4( )( 444) () 8 ± 677 6 0 The reading rate for the first half of the book is about 0 pages per day. Section 6.4 Page 50 Question 0 a) For the 0% solution A 0.. Substitute A 0., s, and C 0.. A C s+ w 0. 0. + w 0.( + w) 0. 0. + w 0. w w To get a 0% solution, L must be added to the -L bottle of 0% solution. MHR Pre-Calculus Solutions Chapter 6 Page 50 of 7
For a 0% solution, A 0.. Substitute A 0., s 0.5, and C 0.0. A C s+ w 0. 0.0 0.5 + w 0.0(0.5 + w) 0. 0. 0. 5 + w 0.0 w 5 0.5 w 4. 5 To get a % solution, 4.5 L must be added to the half-litre bottle of 0% solution. Section 6.4 Page 50 Question Since b a, a. Substitute to eliminate b. b a b 4 + 5 a b a a 4 + a 5 a a 4 + a 5 a + a 5 5a 4+ 4a 9a 5( ) 4( ) a a ± 9 MHR Pre-Calculus Solutions Chapter 6 Page 5 of 7
Section 6.4 Page 5 Question a) + a b b+ a ab b+ a ab ab a + b aa ( + 8) 6 a+ a+ 8 a a a a+ 48 a + 6a 6( + 8) + 6 0 a + 4a 48 0 a + a 4 0 ( a+ 6)( a 4) So, a 6 or a 4. The two numbers are 6 and, or 4 and. Section 6.4 Page 5 Question a) a or a y y y ay y a y ay + y y ( ay+ ) y ay y y ay+ ay + y ( ay+ ) y ay + In both, 0, y 0, ay. d v0t+ gt d gt v0t d gt v0, t 0 t MHR Pre-Calculus Solutions Chapter 6 Page 5 of 7
c) E I r R + n r E R + n I r E R n I r E IR n I ri r n, R, n 0, I 0, E IR E IR n Section 6.4 Page 5 Question 4 a) Rational epressions combine operations and variables in one or more terms. Rational equations involve rational epressions and an equal sign. Eample: + is a rational epression, which can be simplified but not solved. y + 5 is a rational equation that can be solved. 5 5 ( ) ( ) ( ) Multiplyeach term by the LCD. ( ) ( 5) ( ) ( ) Divide common factors. 5 5 Multiply. 5 Collect like terms. 5 Divide. c) Eample: Add the second term on the left to both sides, to give 5 MHR Pre-Calculus Solutions Chapter 6 Page 5 of 7
Section 6.4 Page 5 Question 5 a) Let represent the number of pages that the ink-jet printer prints per minute. Then, the laser printer prints + 4 pages per minute. Write an equation for the printing done by both in 4 min. 4 + 4( + 4) 490 8 + 6 490 8 54 5.5 The ink-jet printer prints 5.5 pages per minute. Eample: I try to use new paper rarely. However, if I consider all the paper entering my life, such as the new phone book each year, flyers, newspapers, and so on, the total is more than I initially thought. I doubt if it is 0 000 pages per year, though. c) Eamples: Re-use paper so both sides are used. Read newspapers and magazines on line rather than getting a hard copy. Section 6.4 Page 5 Question 6 a) Let n represent your average score for the net 4 quizzes. Total score on first 6 quizzes 6(6) Total score on net 4 quizzes 4n Total score if average is to be 40 on 0 quizzes 40(0) 6(6) + 4n 40(0) 6 + 4n 400 4n 84 n 46 Your average mark on the net 4 quizzes needs to be 46. 45 50 is 90%, so 0(40) + 5( ) 45. For this equation to be true, you would need 55 on 5 each of the remaining quizzes, which is not possible. MHR Pre-Calculus Solutions Chapter 6 Page 54 of 7
Section 6.4 Page 5 Question 7 a) Tyler made an error in the last term on the left of the third step. 5 + ( + ) ( )( + ) 5 ( ) 5 5 + + 0 8 7 5 Use the quadratic formula with a 8, b 7, c 5. ± b b 4ac a ( 8)( ( 8) 7) ± ( 7 ) 4( 5) 7 ± 09 6 c) To the nearest hundredth,.4 or 0.47. Chapter 6 Review Chapter 6 Review Page 5 Question a) b 0, because division by zero is not defined. Eample: Some rational epressions have non-permissible values. For, may not take on the value. Chapter 6 Review Page 5 Question Agree. Eample: There are an unlimited number of ways of epressing and unlimited equivalent epressions can be created by multiplying, or dividing, by. Chapter 6 Review Page 5 Question a) y 0, so y 0 + 0, so c) The denominator is always, so no non-permissible values. MHR Pre-Calculus Solutions Chapter 6 Page 55 of 7
d) (a )(a + ) 0, so a, e) m m 0 (m + )(m ) 0 So m,. f) t 8 0 (t )(t +) 0 So t ±. Chapter 6 Review Page 5 Question 4 a) s 8 s 6 s s s 6, s 0 c) b b 4b 8 4( b ) 5 ( 5 ) 5 5, 5, b 4 Chapter 6 Review Page 5 Question 5 a) (5) 0 ( ) 6, 0 9 ( ) ( + ), ± + c) c d ( c d) f ( f) c 6d, f 0 9 f d) m+ m+ m 4 m+ 4 m+ 4 m 4 m m 4, m 6 m ± 4 MHR Pre-Calculus Solutions Chapter 6 Page 56 of 7
Chapter 6 Review Page 5 Question 6 a) Factor the denominator(s), set each factor equal to zero and solve. m 4 m 4 Eample: since, the non-permissible values are ±. m 9 ( m+ )( m ) i) ii) iii) 0 (+ )( 5 ) + + 5, ( ) 9 ( + )( ) a a a a a a a a, a ± a + y ( y ) 4 4y 4( y) iv), y 4 8 6 4 (9 )(9 ) + 8 4 (9 ) 9, 9 Chapter 6 Review Page 5 Question 7 a) Αrea L ength Width Length ( + )( ) +, A simplified epression for the length is +. The non-permissible values are, as this value would make the width zero, and, as this value would make the length zero. MHR Pre-Calculus Solutions Chapter 6 Page 57 of 7
Chapter 6 Review Page 5 Question 8 Eample: The same processes are used for rational epressions as for fractions. Multiplying involves finding the product of the numerators and then the product of the denominators. To divide, you multiply by the reciprocal of the divisor. The differences are that rational epressions involve variables and may have non-permissible values. ()() + + ( + )( + ) 5 ()(5) 5 ()(5) + 5+ 6 5 0 + 4 4 4 + + 4 + +, ( + ) Chapter 6 Review Page 5 Question 9 a) p 0q r 8p 6mt r p 4mt 4 4 5 0 q 8 p 4 5 q, p 0, r 0 r m m t 6 4 m t 4 t m, m 0, t 0 4t c) a+ b 4 ( a+ b ) 4 8 a+ b 8 a+ b, a b MHR Pre-Calculus Solutions Chapter 6 Page 58 of 7
d) ( ) ( + ) ( + 5) + 5 ( + ) 4 + 0 + 5 + ( )( + 5), 0, + 5 e) ( d + ) d + d + d + 6 d + d + 5d + 6 ( d + ) ( d + ), d,, ( d + ) ( d + ) ( d + ) f) y 8y 9 y 9y+ 8 y 5 0 + 9 5 y y y y ( y 9) ( y 9) ( y + ) ( y 8)( y ) ( y ) ( y + ) ( y ) ( y 5) ( y + 5) 5 y ( y 8)( y+ 5), y ±,5,9 y Chapter 6 Review Page 5 Question 0 a) 4 t t 4 8t a a a b 4 4 b b b a, a 0, b 0 b 7 5 7 y y y ( y) ( + y) 5 c), ± y 5( + y) 5 d) a+ 9 a + 6a+ 9 ( a + ) a a a a ( a + ) ( a + ), a ± a + MHR Pre-Calculus Solutions Chapter 6 Page 59 of 7
e) ( + ) ( + ) + + + + ( + + ) ( + ) ( ) 9 4 8 4 + ( + )( + ), 0, ±,, + f) 4 6 4 6 ( + ) ( ) 6 ( + ), ( ) Chapter 6 Review Page 5 Question a) 9 m 9 m m m m m m, m 0 ( ) ( ) 4 + + ( ) ( + ) + +, 0, ±, + ( + ) + a 0 5a 0 a c) a 4 a+ a 9 a 4 a + 0 6 5 ( a ) ( a 4) ( a + ), a ±,4 6 MHR Pre-Calculus Solutions Chapter 6 Page 60 of 7
+ 5 d) 5 + 4 + 4 ( + 4) ( + 4) ( ) 4,, 4, 5 + 4 + 4 5 5 Chapter 6 Review Page 5 Question Height Volume length width + 5 Height ( )( + 4) + ( )( + 4) ( 5 ) ( )( + 4) ( )( + 4) An epression for the height of the prism is centimetres. Chapter 6 Review Page 5 Question Eample: The advantage of using the LCD is that less simplifying needs to be done. a) LCD is 0 LCD is ( )( + ) Chapter 6 Review Page 5 Question 4 a) m m+ + m m m m 5 5 5 m, 0 c) y + y + + y + y + y, y d) + ( + ) MHR Pre-Calculus Solutions Chapter 6 Page 6 of 7
y y y y ( y)( + y) ( y )( y+ ) e) ( ) y ( y)( + y) + y ( y)( + y), ± y y Chapter 6 Review Page 5 Question 5 a) 4 (4 ) ( ) 6 4 8 6 + 6 5 y y y 8 (y ) ( y ) y 8 + + y y 6y 6y 6y 6y 4y + y 6 y+ 8 6y 6y 6y, y 0 9 7 9( + ) 7 c) + + 9 ( )( + ) ( )( + ) 9 + 7+ 7 ( )( + ) 9 + 4, ± ( )( + ) MHR Pre-Calculus Solutions Chapter 6 Page 6 of 7
a a a a a( a ) d) a+ a + a 6 a+ ( a+ )( a ) aa ( ) aa ( ) ( a+ )( a ) + ( a+ )( a ) a a a a a, a, ( a+ )( a ) e) a ab b a ab b + + a b a b a+ b a b ( a ( a+ a+ b aa ( + ab+ ba ( ( a ( a+ a + ab ab+ ab b ( a ( a+ a b a b, a ± b f) + + + + + + + 4 9 5 ( )( ) ( )( ) ( + ) + ( ) (+ )( + ) ( )(+ )( + ) 5 + + ( )(+ )( + ) 6, ±, ( )(+ )( + ) Chapter 6 Review Page 5 Question 6 a) a + + b a b ab Left Side + a b b a + ab ab a+ b ab Right Side MHR Pre-Calculus Solutions Chapter 6 Page 6 of 7
Chapter 6 Review Page 54 Question 7 a+ b+ c Eam mark, d ; a+ b+ c Final mark + d a+ b+ c+ d 6 Eample: 60 + 70 + 80 d 60 + 70 + 80 + (70) 70 6 Chapter 6 Review Page 54 Question 8 a) i) c + 0 represents the amount that Beth spends per chair, $0 more per chair than she planned ii) c 0 represents the amount that Helen spends per chair, $0 less per chair than planned 00 iii) represents the number of chairs Helen bought c 0 50 iv) represents the number of chairs Beth bought c + 0 00 50 v) + represents the total number of chairs purchased by the two sisters c 0 c+ 0 450c 500 50(9c 0) or, c ±0 c 00 ( c 0)( c+ 0) Chapter 6 Review Page 54 Question 9 Eample: When solving a rational equation, you multiply all terms by the LCD to eliminate the denominators. In addition and subtraction of rational epressions, you use a LCD to simplify by grouping terms over one denominator. Add or subtract. Solve. + + 5 + 0 + 6 6 5 0 5 6 6 MHR Pre-Calculus Solutions Chapter 6 Page 64 of 7
Chapter 6 Review Page 54 Question 0 a) s s + s ( s+ ) s s+ 6 s 9, s + + + ( + )( ) ( + )(+ ) + + + 6 + + 0 0 8 + + 0 ( 5 4) 0 ( + 4)( + ) 4 or,, z 4 c) + z 5 5z 5( z ) + z 4 5z 5z 5z 0+ z 4 6z 6 z, z 0 d) m m + m m+ mm ( + ) + ( m )( m+ ) (m )( m ) m + m+ m m m+ 9 8 0 m + m 9 0 (m+ )( m ) 0 m or m, m ± MHR Pre-Calculus Solutions Chapter 6 Page 65 of 7
e) ( ) + 9 4, There is no solution, since is non-permissible. f) g) + + ( )( ) (+ ) + ( )(+ ) 4 + + 5 0 6 ± or ±, 0, 5 + + 6 5 + + ( )( + ) ( ) + 5( + ) ( )( + ) 9+ 5+ 0 + + 6 + 4 5 0 ( + 5)( ) 0 5 or,, Chapter 6 Review Page 54 Question Let n represent the first number. Then, n represents the other number. + n n 8 8( n) + 8n n( n) 96 8n+ 8n 6n n n 6n+ 96 0 n n+ 0 ( n 8)( n 4) 0 So, n 8 or n 4. The two numbers are 4 and 8. MHR Pre-Calculus Solutions Chapter 6 Page 66 of 7
Chapter 6 Review Page 54 Question Let hours represent the time it would take Elaine to paint the room by herself. Time to Paint Fraction Painted Fraction Painted Room (h) in h in hours Matt alone 5 5 5 Elaine alone Working together 5 5 5 5 7.5 It would take Elaine 7.5 h to paint the room by herself. Chapter 6 Review Page 54 Question a) Let metres per second represent the speed of the elevator on the way up. Then, the speed as the elevator descends is + 0.7 metres per second. Distance (m) Speed (m/s) Time (s) Going Up 60 60 Descending 60 + 0.7 60 + 0.7 60 60 6.5(60) + 0.7 60 60 + 6 +.5(60) + 0.7 60 60 + 50 6 + 0.7 60( + 0.7) + 60 ( + 0.7)(4) + + + 60 60 4 79.8 0 4 40. 0 570 0 560 MHR Pre-Calculus Solutions Chapter 6 Page 67 of 7
Use the quadratic formula with a 570, b 0, and c 560. ± 4 a b b ac ( 0) ± ( ( 570) 0± 79 0 40. 5 The rate of ascent is.5 m/s. 0) 4( 570) 560 ( ) c).5 m/s.5(60)(60) km/h. The rate of ascent is 9 km/h. 000 Chapter 6 Practice Test Chapter 6 Practice Test Page 55 Question ( + ) For,,. ( )( + ) Answer D is correct. Chapter 6 Practice Test Page 55 Question 7 + 6 ( )( 6) 4 ( 6)( + 4) + 4 Answer B is correct. Chapter 6 Practice Test Page 55 Question 8 5y 5 8(8) + 5 y(6 y) 5( y) + y 4 8 4y Answer A is correct. 64 0y 5 + 4y y MHR Pre-Calculus Solutions Chapter 6 Page 68 of 7
Chapter 6 Practice Test Page 55 Question 4 4 9 ( 4) 9 Answer A is correct. 4 Chapter 6 Practice Test Page 55 Question 5 6 4 t t+ 4 6( t+ 4) 4( t ) 6t+ 4 4t t 6 t 8 Answer D is correct. Chapter 6 Practice Test Page 55 Question 6 5 6 5 ( ) + 9 5 + ( )( + ) ( 5)( + ) The non-permissible values are ±,, 5. Chapter 6 Practice Test Page 55 Question 7 + k 0 5 + 7+ 6 + + k 0 5 (+ )( + ) + + k + ) 0 ( 5)( + Therefore, k. k 0 0 MHR Pre-Calculus Solutions Chapter 6 Page 69 of 7
Chapter 6 Practice Test Page 55 Question 8 5y y+ + 6 y y 6 5y y+ + 6 y ( y ) 5 yy ( ) + 6 ( y+ ) 6( y ) 5 0 6 y y+ y 6( y ) y y+ 6( y ) 5 4 (5y )( y ) 6( y ) 5y, y 6 Chapter 6 Practice Test Page 55 Question 9 Let represent the time for the smaller auger to fill the bin. 6 6 + 5 Chapter 6 Practice Test Page 55 Question 0 Eample: For both you use a LCD. When solving, you multiply by the LCD to eliminate the denominators, while in addition and subtraction of rational epressions, you use the LCD to group terms over a single denominator. Add or subtract. Solve. + 6 4 7 5 7 4 5 + 5 5(6) 8 8 5 + 5 40 8 8 40 0 MHR Pre-Calculus Solutions Chapter 6 Page 70 of 7
Chapter 6 Practice Test Page 55 Question 5 + 6 + 5 + ( )( + ) + ( )( + ) 5 ( + )( ) 5 9 8 0 ( 4)( + ) 0,, Therefore the solution is 4. Chapter 6 Practice Test Page 55 Question Use the fact that the difference between successive terms of an arithmetic sequence is constant. 5+ 5 (5 + ) 5( ) 5( ) 0( ) 0 + 6 0+ 5 0 5 0 + 0 46 0. MHR Pre-Calculus Solutions Chapter 6 Page 7 of 7
Chapter 6 Practice Test Page 55 Question Let kilometres per hour represent the speed of the plane in calm air. Winnipeg to Calgary Distance (km) Speed (km/h) Time (h) Against Strong 00 00 50 Headwind 50 In Calm Air 00 00 00 00 + 50 (00) (00)( 50) + ( 50) 400 400 0 000 + 50 0 50 0 000 ± 4 a 50 ( ) b b ac ( 50) ± ( ) 4( )( 0 000) 50 ± 48 500 7 The speed of the plane in calm air is 7 km/h, to the nearest kilometre per hour. MHR Pre-Calculus Solutions Chapter 6 Page 7 of 7