Chapter 2 Further algebraic skills
Transcript of Chapter 2 Further algebraic skills
Further algebraic skills
The main mathematical ideas investigated are:
▶ adding and subtracting like algebraic terms
▶ multiplying and dividing algebraic terms
▶ adding and subtracting algebraic fractions
▶ establishing and applying the index laws in
algebra
▶ solving linear equations.
ALGEBRA AND MODELLING
Syllabus references: AM3CEC
Outcomes: MG1H-3, MG1H-9, MG1H-10
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2A Simplifying algebraic expressions Simplifying algebraic expressionsThis section begins with a review of some of the simplifi cations used in the Preliminary General Mathematics course.
WORKED EXAMPLE 1Simplify, where possible, by collecting like terms.
a 9a − a b 3xy − 5xy c 4x − x2 + 7x
Solve Think Apply
a 9a − a = 8a 9a and a are like terms. Collect the like terms by carrying the
sign in front of each term. Like terms
have exactly the same pronumerals,
hence x and x2 are not like terms.
b 3xy − 5xy = −2xy 3xy and 5xy are like terms.
c 4x − x2 + 7x = 11x − x2 4x and 7x are like terms.
EXERCISE 2A1 Simplify, where possible, by collecting like terms.
a 3x + 7x b 6x − 3x c x + x d 3p − 2p
e x2 − 7x2 f x2 + 3x g 7ab + 5ba h x + 7 + 2x
i 8a2 + 14a2 − 6a2 j p + 3p − 8 k p2 + 2p + 4p2 l 7k + k − 8
WORKED EXAMPLE 2Simplify the following by collecting like terms.
a 4x − 9x + 3 − x b 4 − 2a + 3a + 7 c a + 8 − 3a + 7
Solve/Think Apply
a 4x − 9x + 3 − x = 4x − 9x − x + 3
= −6x + 3
Collect the like terms by carrying the
sign in front of each term.
b 4 − 2a + 3a + 7 = −2a + 3a + 4 + 7
= a + 11
c a + 8 − 3a + 7 = a − 3a + 8 + 7
= −2a + 15
2 Complete the following.
a 5x + 7 + 2x − 3 = 5x + 2x + 7 − 3 = ___
b 4a + 3b − 2a − 5b = 4a − 2a + 3b − 5b = ___
c −3m − 4 − 2m + 3 = −3m − 2m − 4 + 3 = ___
d x2 + 3x − 5x + 4x2 = x2 + 4x2 + 3x − 5x = ___
3 Simplify the following by collecting like terms.
a x + 2 + 3x + 6 b 4 + 5a + 7 − 2a c 2a + 4b + 4a − 2b
d n2 − 2n + 3n + 2n2 e 8 − 7x − 6 + 2 f x2 + 2x + 3 − 5x
g a2 + a + a + a2 h 5t − 4t + 7 + t i −5a − 3a + 3 − 5
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25Chapter 2 Further algebraic skills
4 Simplify, where possible, by collecting like terms.
a −8l − 4 + 2l − 7 b x2 + 2x − 7x − x2 c 4x − 2y − (−x) + y
d ab + b − 2ab − 5b e −x − 6 − 2x − 3 f 5t − (−t) + 6 − 3t
g 3a − 2 − a + 2 − a h a2b + a2b − 3a2b + 8a i 5d − 2c + d − 2c + 4
j 4p5 + 5p4 − 2p5 − 6p4 k 8m2 − 5n2 + 7n2 − 4m2 l 6s2t + 5s2 − 8s2t − 6s2
WORKED EXAMPLE 3Simplify:
a −2 × 5x b 5a × 3b
Solve Think Apply
a −10x −2 × 5x = −2 × 5 × x
= (−2 × 5) × x
= −10x
Multiply the coeffi cients and
multiply the pronumerals. The
coeffi cient of an algebraic terms is
the number by which the pronumeral
is being multiplied.b 15ab 5a × 3b = 5 × a × 3 × b
= (5 × 3) × (a × b)
= 15ab
5 Complete the following to simplify the expressions.
a 4x × 5y = 4 × x × 5 × y b −3a × 2b = −3 × a × 2 × b
= (4 × 5) × (x × y) = ___ = (−3 × ___) × (a × ___) = ___
c 5 × 3x2 = 5 × 3 × x2 d −4ab × −3c = −4 × a × b × −3 × c
= ___ = (−4 × ___) × (a × ___ × ___)
= ___
6 Simplify:
a 5 × 3x b 4y × 7 c −3 × 8x d −5 × 7a
e 4p × 5q f −5a × 3b g −2k × −3m h −6w × −z
WORKED EXAMPLE 4Simplify:
a 10x ÷ 2 b 20pq ÷ −5p c 8ab ÷ 10b
Solve Think Apply
a 5x 10x ÷ 2 = 10x
____ 2
= 510 × x
_______ 12 = 5x
Expand each term. Divide the
coeffi cients and the pronumerals.
b −4q 20pq ÷ −5p = 20pq
_____ −5p
= 420 × 1p × q
___________ 1−5 × 1p
= 4q
___ −1 = −4q
c 4a
___ 5 8ab ÷ 10b =
8ab ____
10b
= 48 × a × 1b
__________ 510 × 1b =
4a ___
5
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7 Complete the following to simplify.
a 12x ÷ 6 = □12 × x
_______ □6 b −12y ÷ 8 =
−□12 × y _________ □8
c 6pq ÷ 8p = □6 × 1p × q
___________ □8 × 1p
= ___ = ___ = ___
8 Simplify the following.
a 8a ÷ 4 b 25y ÷ 5 c 35x ÷ (−7)
d 2a ÷ 4 e 8a ÷ 2a f −6y
____ 3
g 12z
____ −3z h 20xy
____ 10x
i 16ab
_____ 12b
j −4pq
_____ 6p k
−20mn _______ −4m l
−16xy ______
20xy
WORKED EXAMPLE 5Simplify the following.
a 2 __
7 ×
5 __
9 b
4x ___
7 ×
3 ___
5y
Solve Think Apply
a 10
___ 63
2 __
7 ×
5 __
9 =
2 × 5 _____
7 × 9 Multiply the numerators
and multiply the
denominators.b 12x
____ 35y
4x
___ 7 ×
3 ___
5y = 4x × 3
______ 7 × 5y
9 Complete the following to simplify.
a 3 __
7 ×
k __
5 =
3 × □ ______ 7 × □ = ___ b
x __
5 ×
3 __ y =
x × □ ______ 5 × □ = ___
Simplify the following.
a 5 __
6 ×
m __
3 b
x __
2 ×
y __
5 c
m __
4 ×
3 __ n d
2a ___
7 ×
3 ___
4b
WORKED EXAMPLE 6Simplify these algebraic fractions.
a 16
___ 9 ×
27 ___
28 b
16a ____
9 ×
27 ____
28b c
6ab ____
5 ×
10 ___
9a
Solve Think Apply
a 416
___ 19 ×
327 ___ 728 =
12 ___
7
The HCF of 16 and 28 is 4. Divide the
numerator and the denominator by 4. The HCF
of 9 and 27 is 9. Divide the numerator and the
denominator by 9.
Divide the numerator and
denominator by the highest
common factor (HCF).b 416a
____ 19 ×
327 ____ 728b =
12a ____
7b
c 26 1a b
_____ 15 ×
210 ____ 39 1a =
4b ___
3
The HCF of 6 and 9 is 3. Divide the numerator
and the denominator by 3. The HCF of 5 and 10
is 5. Divide the numerator and the denominator
by 5. a is a common factor so divide the
numerator and the denominator by a.
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27Chapter 2 Further algebraic skills
Complete the following.
a 10x
____ 9 ×
3 ____
16y =
510x ____ □9
× □3
____ 816y b
12ab _____
5 ×
10 ___
3a =
□121ab ______ 15
× 210
____ □31a
= ___ = ___
Simplify the following.
a 3x
___ 5 ×
2n ___
7 b
2a ___
9 ×
6b ___
5 c
4z __
3 ×
9w ___
8
d 8m
___ 5 ×
15n ____
16 e
3k ___
5 ×
10 ___
9m f
5a ___
3 ×
9 ___
2a
g 7 ___
2p × 3p
___ 14
h 2xy
___ 3 ×
6 ___
2y i
8ab ____
9 ×
15 ___
3a
j 6xy
___ 5 ×
25 ___
3y k
9c ____
10d × 25de _____
12c2 l a2
___ bc2 × bc
___ a
WORKED EXAMPLE 7Simplify these algebraic fractions.
a 3x
___ 7 ÷
8 ___
5y b 4a
___ 3b ÷
16 ___
9b
Solve Think Apply
a 3x
___ 7 ÷
8 ___
5y = 3x
___ 7 ×
5y ___
8
= 15xy
____ 56
To divide by 8 ___
5y invert it and
multiply, or multiply by its
reciprocal.
To divide by a fraction invert it
and multiply or multiply by its
reciprocal.
b 4a
___ 3b
÷ 16
___ 9b
= 14a
____ 131b ×
391b ____ 416
= 3a
___ 4
To divide by 16
___ 9b
invert it and
multiply.
Complete the following to simplify these algebraic fractions.
a 5x
___ 3 ÷
4y ___
7 =
5x ___
3 × ___ b
8m ___
5n ÷ 4m
___ 15
= 8m
___ 5n × ___
= ___ = ___
Simplify these algebraic fractions..
a 4x
___ 3 ÷
7x ___
2 b
11a ____
5 ÷
6 ____
10b c
15p ____
4q ÷ 3p
___ 8q
d 18xy
____ 7 ÷
6x ___
14 e
12a ____
9 ÷
4b ___
3c f
8 ___
5x ÷ 12y
____ 15
2B Expanding using the distributive lawExpanding using the distributive lawThe expression a × (b + c) can be written in expanded form as a × b + a × c. This is known as the
distributive law.
To expand a(b + c), each term inside the grouping symbols is multiplied by the term outside the grouping symbols.
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12
13
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WORKED EXAMPLE 1Expand and simplify:
a 4(2a + 7) b 5(3m − 4)
Solve Think Apply
a 8a + 28 4(2a + 7) = 4 × 2a + 4 × 7 a(b + c) = ab + ac
a(b − c) = a(b + (−c))
= a × b + a × (−c)
= ab + (−ac)
= ab − ac
b 15m − 20 5(3m − 4) = 5(3m + (−4))
= 5 × 3m + 5 × (−4)
= 15m + (−20)
= 15m − 20
EXERCISE 2B1 Complete to expand the following.
a 5(x + 3) = 5 × ___ + 5 × ___ = ____
b 7(2y − 5) = ___ × 2y − ___ × 5 = ____
c 3a(4a − 7) = 3a × ___ − 3a × ___ = ____
2 Expand and simplify the following.
a 2(x + 3) b 2(a − 2) c 3(b − 5) d 9(xy + z)
e 3(2m + 1) f 6(3a − 7) g 7(1 − 3d) h a(a + 5)
i x(3 − x) j 3x(4x − 1) k w(w − 1) l 6a(4a − 7b)
WORKED EXAMPLE 2Expand and simplify the following.
a −3(x + 7) b −4(5 − 2y)
Solve Think Apply
a −3x − 21 −3(x + 7) = (−3) × x + (−3) × 7
= −3x − 21
Multiply each term inside the
parentheses by the negative
number at the front:
−a(b + c) = −ab − ac
−a(b − c) = −ab + ac
b −20 + 8y −4(5 − 2y) = (−4) × 5 − (−4)(2y)
= −20 − (−8y)
= −20 + 8y
3 Complete to expand the following.
a −2(x + 5) = (−2) × x + (−2) × 5 b −6(3y − 5) = (−6) × ___ − (−6) × ___
= ___ + ___ = ___ − ___
= ___ − ___ = ___ + ___
4 Expand and simplify the following.
a −4(x + 3) b −5(4 − x) c −2(4x + 1)
d −4(2 − x) e −2(3p + 4) f −3(m + 2n)
g −x(x + y) h −a(2b + 3d) i −6(x + 2y + z)
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29Chapter 2 Further algebraic skills
WORKED EXAMPLE 3Expand the following.
a −(7 + y) b −(y − 2)
Solve Think Apply
a −7 − y −(7 + y) = −1 × (7 + y)
= −7 − yA negative sign at the front of the
parentheses is an abbreviation for −1;
that is, −(a + b) = −1 × (a + b).
The eff ect of multiplying by −1 is to
change the sign of each term inside the
parentheses.
b −y + 2 −(y − 2) = −1 × (y − 2)
= −y + 2
5 Expand the following.
a −(x + 2) b −(x − 4) c −(5 + x)
d −(7p + 6) e −(4x + 2) f −(3 − 5x)
g −(4y + 8x) h −(8p − 7q) i −(4x2 − 3x)
6 Expand the following.
a −2a(3 − b) b −3y(x + 4) c −4a(a − 4)
d −4m(p + m) e −x(x + 6) f −3c(4c − d)
g −3a(2a + b) h −5x(x − 2y) i −7p(2p + 5q)
WORKED EXAMPLE 4Expand and simplify the following.
a 3 + 2(2 − 5x) b 4 − 3(x − 2)
Solve/Think Apply
a 3 + 2(2 − 5x) = 3 + 4 − 10x
= 7 − 10x
When expanding brackets, multiply every term inside
the brackets by the term outside the brackets.
Expand and collect like terms.b 4 − 3(x − 2) = 4 − 3x + 6
= 10 − 3x
7 Complete the following.
a 4 + 3(5x + 2) = 4 + 15x + ___ b 7 + 2(3a − 5) = 7 + 6a − ___
= ___ + ___ = ___+ ___
= ___ − ___
c 4 − 3(a + 2) = 4 − 3a − ___ d 10 − 4(p − 3) = 10 − 4p + ___
= ___ − ___ = ___ − ___
8 Expand and simplify the following.
a 7 + 2(3x + 5) b 4 − 3(1 + 2x) c 18 + 2(5x + 8)
d 7x − (3 − 4x) e 5x + 3(2x + 1) f 7 − 5(1 − 3x)
g 2x − (3x + 11) h 7 − 2(x − 3) i 5 − 6(3x − 1)
j 2(3x − 5) + 7x k 3(x − 2) + 4 l 5(x − 3) − 3x
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WORKED EXAMPLE 5Expand and simplify these expressions.
a 5(x + 2) + 2(5 − x) b 3(2 − x) − 4(x − 3)
Solve/Think Apply
a 5(x + 2) + 2(5 − x) = 5x + 10 + 10 − 2x
= 3x + 20
Expand and collect like terms.
b 3(2 − x) − 4(x − 3) = 6 − 3x − 4x + 12
= −7x + 18
9 Complete the following.
a 3(a + 4) + 2(6 − a) = 3a + 12 + ___ − ___
= ___ + ___
b 2(y − 3) − 4(y − 2) = 2y − ___ − 4y + ___
= ___ + ___
Expand and simplify these expressions.
a 3(x + 4) + 2(x − 1) b 4(y + 1) + 2(y + 3)
c 3(p + 1) + 4(p − 5) d 2(1 − x) + 4(x − 2)
e d(d + 2) + 3(d − 3) f 2x(x + 1) + 3(x + 2)
g 2x(x + 2) − 4(4 − x) h 3x(x + 4) + 5(x + 2)
i 6(x − 3) − (5x + 4) j −(x + 6) − 7(x + 3)
k −(3 − x) + 2(4 − x) l 3(5 − 2x) − (4 − x)
2C Adding and subtracting algebraic fractions Adding and subtracting algebraic fractionsWORKED EXAMPLE 1Simplify the following.
a 9t
___ 20
+ 7t
___ 15
b 5k
___ 6 − 2k
Solve Think Apply
a 9t
___ 20
+ 7t
___ 15
= 27t
___ 60
+ 28t
___ 60
= 55t
___ 60
= 11t
___ 12
Lowest common denominator = 60:
9t
___ 20
= 27t
___ 60
and 7t
___ 15
= 28t
___ 60
55t
___ 60
= 55t ÷ 5
_______ 60 ÷ 5
= 11t
___ 12
Change the fractions to
equivalent fractions with a
common denominator.
Add or subtract the
numerators and simplify if
possible.b
5k ___
6 − 2k =
5k ___
6 −
12k ____
6
= − 7k
___ 6
Lowest common denominator = 6:
2k = 12k
____ 6
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31Chapter 2 Further algebraic skills
EXERCISE 2C1 Complete the following to simplify the expression.
7y
___ 8 −
3y ___
4 =
7y ___
8 −
□ __ 8 = ___
2 Simplify the following expressions.
a 11x
____ 20
+ 8x
___ 20
b 2y
___ 3 +
3y ___
4 c
7a ___
8 −
2a ___
3 d
14t ___
9 −
3t __
5
e 4x
___ 5 +
3x ___
2 f
11p ____
10 −
p __
4 g
5a ___
7 −
3a ___
14 h
5z __
6 +
z __
3
i 3y
___ 7 + y j 2p +
2p ___
5 k m −
2m ___
3 l 2k −
3k ___
8
2D Indices IndicesINVESTIGATION 2.1
From Investigation 2.1 we see that the laws for the use of indices are:
am × an = am + n
am ÷ an = am − n
(am)n = amn
(ab)n = anbn
WORKED EXAMPLE 1Simplify the following.
a p7 × p8 b t 16 ÷ t 4 c (m5)4
Solve Think Apply
a p7 × p8 = p15 p7 × p8 = p7 + 8 Use the index laws to simplify.
b t 16 ÷ t 4 = t 12 t 16 ÷ t 4 = t 16 − 4
c (m5)4 = m20 (m5)4 = m5 × 4
EXERCISE 2D1 Complete the following.
a a7 × a5 = a7 + □ b m15 ÷ m7 = m15 − □ c (k3)5 = k3 × □
= ___ = ___ = ___
2 Simplify the following.
a k10 × k6 b p12 ÷ p7 c (t3)5
d m11 × m10 e x25 ÷ x5 f (y7)5
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WORKED EXAMPLE 2Simplify the following.
a 3x7 × 4x8 b x2y3 × x5y c 7x4y3 × 3x2y8
Solve Think Apply
a 12x15 3x7 × 4x8 = (3 × 4) × (x7 × x8)
= 12 × x15
Group the coeffi cients
and like pronumerals.
Multiply the coeffi cients
and use the index law for
multiplication to simplify
the pronumerals.
b x7y4 x2y3 × x5y = (x2 × x5) × (y3 × y)
= x7 × y4
c 21x6y11 7x4y3 × 3x2y8 = (7 × 3) × (x4 × x2) × (y3 × y8)
= 21 × x6 × y11
3 Complete the following.
a 4p5 × 3p7 = (4 × 3) × (p5 × p7) = ___
b k3m4 × km7 = (k3 × k) × (m4 × m7) = ___
c 4x2y5 × 5x3y2 = (4 × 5) × (x2 × ___) × (y5 × ___) = ___
4 Simplify the following expressions.
a 5a6 × 3a8 b 4y9 × 3y c 6m11 × 2m4
d p8q3 × p4q2 e a7b8 × ab4 f xy3 × x7y
g 3m3n2 × 4m7n3 h 5a4b × 3a6b5 i 4p7q10 × 10p4q11
j 3k7 × 1
_ 4 k3 k
2
_ 5 x7y4 × 4xy l 5a3b4c2 × 2a2bc3
WORKED EXAMPLE 3Simplify the following.
a (2x7)4 b (m2n3)5 c (5x6y5)3
Solve Think Apply
a (2x7)4 = 24 × (x7)4
= 16x28
(2x7)4 = 2x7 × 2x7 × 2x7 × 2x7
= (2 × 2 × 2 × 2) × (x7 × x7 × x7 × x7)
= 24 × (x7)4
(ab)n = anbn
b (m2n3)5 = (m2)5 × (n3)5
= m10n15
(m2n3)5 = m2n3 × m2n3 × m2n3 × m2n3 × m2n3
= (m2 × m2 × m2 × m2 × m2)
× (n3 × n3 × n3 × n3 × n3)
= (m2)5 × (n2)5
c (5x6y5)3 = 53 × (x6)3 × (y5)3
= 125x18y15
(5x6y5)3 = 5x6y5 × 5x6y5 × 5x6y5
= (5 × 5 × 5) × (x6 × x6 × x6)
× (y5 × y5 × y5)
= 53 × (x6)3 × (y5)3
5 Complete the following.
a (2x3)5 = 2□ × (x3)□ b (a7b4)8 = (a7)□ × (b4)□
= ___ = ___
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33Chapter 2 Further algebraic skills
6 Simplify:
a (7x12)2 b (2z10)4 c (5m6)3
d (a7b4)5 e (x4y5)10 f (pq5)3
g (10x3y5)4 h (3a7b9)2 i (2k5m4)3
WORKED EXAMPLE 4Simplify the following.
a 6m7
____ 3m2 b
p7q9
____ p3q2 c
8a6b5
______ 10a3b3
Solve Think Apply
a 2m5
6m7
____ 3m2 =
6 __
3 ×
m7
___ m2
= 2 × m5
Group the coeffi cients and like
pronumerals.
Divide the coeffi cients and use
the index law for division to
simplify the pronumerals.b p4q7
p7q9
____ p3q2 =
p7
___ p3 ×
q9
__ q2
= p4 × q7
c 4a3b2
_____ 5
8a6b5
______ 10a3b3 =
8 ___
10 ×
a6
__ a3 ×
b5
__ b3
= 4 __
5 × a3 × b2
7 Complete the following.
a 15k6
____ 3k2 =
15 ___
3 ×
k6
__ k2 = ___ b
x11y8
____ x4y2 =
x11
___ x4 ×
y8
__ y2 = ___
c 9m6n7
______ 12m4n6 =
9 ___
12 ×
m6
___ m4 ×
n7
__ n6 = ___
8 Simplify:
a 10k8
____ 5k3 b
12w9
_____ 4w2 c
15z6
____ 20z3
d p6q10
____ p5q4 e
a12b9
____ a10b4 f
m8n8
____ m4n5
g 6k3m7
_____ 2km2 h
6w4v6
_____ 8w2v2 i
18b10c8
______ 27b3c8
WORKED EXAMPLE 5
Simplify 9x4y3
_____ 5x2y
× x7y2
____ 3xy2 .
Solve Think Apply
9x4y3
_____ 5x2y
× x7y2
____ 3xy2 =
3x8y2
_____ 5
9x4y3
_____ 5x2y
× x7y2
____ 3xy2 =
9x11y5
______ 15x3y3
= 9 ___
15 ×
x11
___ x3 ×
y5
__ y3
= 3 __
5 × x8 × y2
Multiply and divide the
coeffi cients, then multiply and
divide the pronumerals using
the index laws.
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9 Complete the following.
a a4b6
____ a2b3 ×
a10b4
____ a8b5 =
a14b10
_____ a□b□ b x5y7 ×
x6y3
____ x8y9 =
x□y□
____ x8y9 c
(p7q4)5
______ p10q15 =
p35q□
_____ p10q15
= ___ = ___ = ___
Simplify the following.
a a3b4 × a7b2
__________ a5b3 b
m7n8
___________ m2n4 × m3n3 c
x7y3
____ x2y2 ×
x5y4
____ x4y2 d
(a2b3)4
______ a5b2
e 4x8y10
___________ 5x3y × 2x2y2 f
a5b4 × a2b5
__________ a3b9 g
x5y4
____ x2y2 ×
x6y5
____ x7y4 h
3m5n2 × 2m7n4
_____________ 4m3n × 5m6n3
WORKED EXAMPLE 6Expand and simplify the following.
a 2x3(x2 + 4) b 3a2(4a7 − 5a)
Solve Think Apply
a 2x5 + 8x3 2x3(x2 + 4) = 2x3 × x2 + 2x3 × 4 Use a(b + c) = ab + ac
and the index laws.
b 12a9 − 15a3 3a2(4a7 − 5a) = 3a2 × 4a7 − 3a2 × 5a
Complete the following.
a 4k3(3k + 2) = 4k3 × 3k + 4k3 × 2 b m6(4m3 − m2) = m6 × 4m3 − m6 × m2
= ___ = ___
Expand and simplify the following.
a y4(y2 + 3) b n7(5 + n2) c p3(2p2 − 4)
d 2b4(b2 − 7) e 3k(2k2 + 4m) f 4w3(2w2 − 3z)
g a6(a3 + a2) h 2t7(3t4 − t2) i 3w4(4w3 − 2w6)
WORKED EXAMPLE 7Expand and simplify 2a4(a + 3) + 3a4(a − 5).
Solve/Think Apply
2a4(a + 3) + 3a4(a − 5) = 2a4 × a + 2a4 × 3 + 3a4 × a − 3a4 × 5
= 2a5 + 6a4 + 3a5 − 15a4
= 2a5 + 3a5 + 6a4 − 15a4
= 5a5 − 9a4
Expand using the distributive law
and index law. Collect like terms.
Complete the following.
3a2(a + 5) + a2(4a − 1) = 3a2 × a + 3a2 × 5 + a2 × 4a − a2 × 1
= 3a□ + 15a□ + ___ − a2 = ___
Simplify the following.
a 3x2(6x − 3) + 2x2(7x + 5) b 4x3(2x3 + 7) − x3(3x3 − 4)
c 2p4(5p3 − p) + 3p4(2p3 − 4p) d 3x5(2x4 − 5x2) − x5(3x4 − 4x2)
e 5y2(2y5 + 3y3) − 4y2(3y5 + 2y3) f 7x3(2x2 − 3x) + 5x2(x3 − 3x2)
10
11
12
13
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35Chapter 2 Further algebraic skills
2E Equations EquationsWORKED EXAMPLE 1Solve these equations.
a 6x = 41 b x __
3 = −4 c x + 7 = 3 d x −4 = −1
Solve Think Apply
a x = 6 5
_ 6
6x ___
6 =
41 ___
6
x = 6 5
_ 6
Add or subtract the same number to
or from both sides of the equation.
Multiply or divide both sides by the
same number.b x = −12
x __
3 × 3 = −4 × 3
x = −12
c x = −4 x + 7 − 7 = 3 − 7
x = −4
d x = 3 x − 4 + 4 = −1 + 4
x = 3
EXERCISE 2E1 Solve the following equations.
a 5p = −30 b 7k = 31 c x __
7 = 4 d
p __
6 = −5
e p + 7 = 16 f x + 8 = 16 g x − 3 = 9 h y − 4 = 26
i k − 24 = −25 j x + 11 = 0 k x − 17 = 0 l x __
5 = 0
WORKED EXAMPLE 2Solve the following equations.
a 8x − 7 = −31 b 19 − 3x = 7
Solve Think Apply
a 8x − 7 = −31
8x = −24
x = −3
8x − 7 + 7 = −31 + 7
8x = −24
8x
___ 8 = −
24 ___
8
x = −3
Add or subtract the same number to
or from both sides of the equation,
then divide both sides by the same
number.
b 19 − 3x = 7
−3x = −12
x = 4
19 − 3x − 19 = 7 − 19
−3x = −12
−3x
____ −3 =
−12 ____ −3
x = 4
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2 Solve these equations.
a 3x − 10 = 2 b 4p + 7 = 3 c 6x − 25 = −5
d 9a + 10 = −8 e 4x − 5 = 9 f 17 − 2x = −3
g 5 − 4x = 21 h 12 − 7x = 15 i 18 = 17 + 5m
j 5t + 19 = 4 k −3p − 11 = −2 l −2k + 7 = −13
WORKED EXAMPLE 3Solve the following equations.
a x __
5 + 7 = 3 b 4 −
x __
3 = 2
Solve Think Apply
a x __
5 + 7 = 3
x __
5 = −4
x = −20
x __
5 + 7 − 7 = 3 − 7
x __
5 = −4
x __
5 × 5 = −4 × 5
x = −20
Add or subtract the same number to
or from both sides of the equation,
then multiply both sides by the same
number.
b 4 − x __
3 = 2
− x __
3 = −2
x = 6
4 − x __
3 − 4 = 2 − 4
− x __
3 = −2
− x __
3 × −3 = −2 × −3
x = 6
3 Solve these equations.
a m
__ 3 + 1 = 6 b
k __
2 − 3 = 7 c
p __
5 + 6 = 2
d y ___
10 − 3 = −7 e 5 −
x __
2 = 2 f 9 −
m __
4 = −1
g 1 − t __
6 = 5 h 2 −
q __
7 = −1 i
w __
5 − 9 = 6
j k __
3 + 5 = 1 k
m __
4 − 3 = −5 l 5 −
z __
3 = 1
WORKED EXAMPLE 4
Solve 2x
___ 3 = 7.
Solve Think Apply
2x
___ 3 = 7
2x = 21
x = 10 1
_ 2
2x
___ 3 × 3 = 7 × 3
2x = 21
2x
___ 2 =
21 ___
2
x = 10 1
_ 2
Multiply both sides by the same number
then divide both sides by the same
number.
4 Solve the following equations.
a 2x
___ 3 = 5 b
3x ___
4 = 2 c
2x ___
5 = −4
d 5x
___ 4 = 3 e
3x ___
2 = −5 f
5x ___
6 = 14
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37Chapter 2 Further algebraic skills
WORKED EXAMPLE 5
Solve 2x
___ 3 =
5 __
4 .
Solve Think Apply
2x
___ 3 =
5 __
4
2x = 15
___ 4
x = 15
__
8 or 1 7
_ 8
2x
___ 3 × 3 =
5 __
4 × 3
2x = 15
___ 4
2x
___ 2 =
15 ___
4 ×
1 __
2
x = 15
__
8 or 1 7
_ 8
Multiply both sides of the
equation by the same number.
then divide both sides by the same
number.
(Dividing by 2 is the same as
multiplying by 1
_ 2 .)
5 Solve the following equations.
a 2x
___ 5 =
3 __
4 b
3m ___
4 =
1 __
2 c
2p ___
3 =
4 __
7
d 3k
___ 5 = −
2 __
3 e
5t __
9 = −
5 __
4 f
7y ___
5 = −
3 __
5
WORKED EXAMPLE 6Solve 3(p − 2) = 21.
Solve Think Apply
3(p − 2) = 21
3p − 6 = 21
3p = 27
p = 9
Expanded 3(p − 2) = 3p − 6
3p − 6 + 6 = 11 + 6
3p = 27
p = 9
Expand and solve the resulting
equation.
6 Solve these equations.
a 5(n + 1) = 65 b 3(p − 7) = 21 c 2(x + 11) = 25
d 4(y − 3) = 7 e 3(2m − 5) = 24 f 2(4x + 3) = 19
WORKED EXAMPLE 7Solve the following equations.
a 5x + 3 = 3x − 1 b 14 − 3x = 2 − x
Solve Think Apply
a 5x + 3 = 3x − 1
2x + 3 = −1
2x = −4
x = −2
5x + 3 − 3x = 3x − 1 − 3x
2x + 3 = −1
2x + 3 − 3 = −1 − 3
2x = −4
Eliminate the variable on the
right-hand (or left-hand) side
by adding it to or subtracting it
from both sides of the equation.
Complete the solution as for
previous examples.b 14 − 3x = 2 − x
14 − 2x = 2
−2x = −12
x = 6
14 − 3x + x = 2 − x + x
14 − 2x = 2
14 − 2x − 14 = 2 − 14
−2x = −12
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7 Solve the following equations.
a 6x + 2 = 2x + 14 b 3x + 8 = 13 − 2x
c 4x − 5 = 3x − 7 d 10 − 2x = 4 − 5x
e 2x − 10 = x + 4 f 5x − 7 = 13 + 7x
g 3x − 6 = 9 − 2x h 4x − 5 = 5x − 2
i 3a + 2 = a − 5 j 2k − 3 = 12 − 3k
k −4m + 1 = 3 − 5m l −2t − 3 = 7 − 4t
WORKED EXAMPLE 8Solve
4 __ n = 22.
Solve Think Apply
4 __ n = 22
4 = 22n
4 ___
22 = n
n = 2
__ 11
4 __ n × n = 22 × n
4 = 22n
4 ___
22 =
22n ____
22
∴ n = 2
__ 11
Multiply both sides of the
equation by the same number,
then divide both sides by the
same number.
8 Solve these equations.
a 75
___ n = 3 b 48
___ m = 6 c 6 __ k = −2 d
39 ___ t = −3
e 3 __ p = 4 f
7 __ a = 2 g
12 ___ w = 7 h
5 __ z = −6
WORKED EXAMPLE 9Solve this equation:
w __
2 +
w __
5 = −14
Solve Think Apply
w
__ 2 +
w __
5 = −14
5w + 2w = −140
7w = −140
w = −20
Lowest common denominator = 10
( w __ 2 +
w __
5 ) × 10 = −14 × 10
w
__ 2 × 10 +
w __
5 × 10 = −140
5w + 2w = −140
7w = −140
7w
___ 7 =
−140 _____
7
w = −20
Multiply both sides of the
equation by the lowest common
denominator of the fractions.
9 Solve the following equations.
a k __
3 +
k __
2 = 10 b
z __
3 +
z __
5 = 4 c
2x ___
3 +
3x ___
4 = 1
d y __
2 −
y __
6 = 3 e
3x ___
5 −
x __
2 = −4 f
4k ___
3 −
3k ___
4 = 2
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39Chapter 2 Further algebraic skills
WORKED EXAMPLE 10Solve
x __
3 + 4 = 2x − 11.
Solve Think Apply
x __
3 + 4 = 2x − 11
x + 12 = 6x − 33
−5x + 12 = −33
−5x = −45
x = 9
Lowest common denominator = 3
( x __
3 + 4) × 3 = (2x − 11) × 3
x + 12 = 6x − 33
x + 12 − 6x = 6x − 33 − 6x
−5x + 12 = −33
−5x + 12 − 12 = 33 − 12
−5x = −45
Multiply both sides of the
equation by the lowest common
denominator of the fractions.
Solve the following equations.
a x __
3 + 2 = x − 4 b x + 3 =
x __
2 + 1 c 2x − 5 =
3x ___
4 + 1
d 2x
___ 3 − 1 = 3x − 15 e
x __
5 − 1 =
x __
3 + 2 f 11 −
4x ___
3 =
x __
2 − 1
WORKED EXAMPLE 11Solve these equations.
a x2 = 9 b y3 = 7
Solve Think Apply
a x = 3 or −3 x = √ __
9 or x = − √ __
9 Take the square root of both sides.
Note: 3 or −3 is often written as ±3.
b y = 3 √ __
7
= 1.9 (1 decimal place)
3 √ __
y3 = 3 √ __
7 Take the cube root of both sides.
Solve the following equations.
a y3 = 8 b x2 = 100 c y3 = 35
d x2 = 36 e y3 = 10 f x2 = 7
WORKED EXAMPLE 12Solve the following equations.
a √ __
x = 5 b 3 √ __
x = 4
Solve Think Apply
a x = 25 ( √ __
x )2 = 52 Square both sides.
b x = 64 ( 3 √ __
x )3 = 43 Cube both sides.
Solve these equations.
a √ __
x = 3 b 3 √ __
x = 2 c √ __
m = 7
d 3 √ __
k = 1 e √ ___
2p = 10 f 3 √ ___
2w = 5
10
11
12
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2F Practical equations Practical equationsA formula links two or more pronumerals according to a rule. The pronumeral that is expressed in terms of the
others is called the subject of the formula. If the pronumeral to be found is the subject of the formula then its
value is found directly by substitution. If the pronumeral to be found is not the subject then, after substitution, the
resulting equation is solved to fi nd its value.
WORKED EXAMPLE 1a Find the value of t when v = 117, u = 5, a = 8 and v = u + at.
b Find the value of N when R = 23, I = 4 and R = I __ N + 1.
Solve Think Apply
a v = u + at
117 = 5 + 8t
112 = 8t
t = 14
In the formula, replace v by 117,
u by 5 and a by 8. Solve the
resulting equation.
Substitute the given values
into the formula and solve the
resulting equation.
b R = I __ N + 1
23 = 4 __ N + 1
22 = 4 __ N
22N = 4
N = 4 ___
22
= 2 ___
11
Replace R by 23 and I by 4. Solve
the resulting equation.
EXERCISE 2F1 If v = u + at, fi nd the value of:
a t when v = 102, u = 18, a = 7 b a when v = 54, u = 12, t = 14
2 Use the formula d = 1
_ 2 ct to fi nd the value of:
a c when d = 100, t = 8 b t when d = 320, c = 16
3 If M = 5k
___ 18
fi nd k when M = 15.
4 If V 2 = gR fi nd R when V = 12, g = 10.
5 If B = m
__ h2 fi nd m when B = 23, h = 1.63.
6 Use the formula I = Prn
____ 100
to fi nd the value of:
a P when I = 19 500, r = 3, n = 25 b n when I = 2100, P = 5000, r = 6
c r when I = 2560, P = 8000, n = 8.
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41Chapter 1 Further algebraic skills
7 Given s = ut + 1
_ 2 at2, fi nd u when s = 360, t = 8 and a = 10.
8 If S = n __
2 (a + l) fi nd a when S = 560, n = 20 and l = 53.
9 If S = n __
2 {2a + (n − 1)d} fi nd the value of a given that S = 610, d = 3 and n = 20.
Use the formula A = 180 − 360
____ n to fi nd n when A = 120.
Use the formula a = 2Rn
_____ n + 1
to fi nd the value of:
a R when a = 12, n = 24 b R when a = 19.5, n = 39
Use the formula I = E _____ R + r to fi nd the value of:
a E when I = 8, R = 15, r = 3 b R when I = 2, E = 24, r = 3
Given R = I __ N + 1, fi nd the value of N when:
a R = 3, I = 7 b R = 11, I = 5
Use t = √ __
l __ g to fi nd l when t = 1.3, g = 10.
Use r = 3 √ __
V
__ k to fi nd V when r = 3, k = 4.2.
Use the formula E = 1
_ 2 mv2 to fi nd the value of v when E = 135 and m = 2.8.
INVESTIGATION 2.1Complete the following.
1 y3 × y4 = y × y × y × ___ × ___ × ___ × ___ = y□
2 k8 ÷ k5 = k × k × k × k × k × k × k × k
__________________________ k × □ × □ × □ × □ = k□
3 (p2)3 = p2 × p2 × p2 = p□
4 (km)3 = km × km × km = k × k × k × ___ × ___ × ___= k□m□
10
11
12
13
14
15
16
02006 Photo of students workinh on algebraic
equations (indices or substituting into
formula?)
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REVIEW 2 FURTHER ALGEBRAIC SKILLS
Language and terminologyHere is a list of terms used in this chapter. Explain each term in a sentence.
algebraic fraction, expand, expression, equation, formula, index, index law, indices, like terms, pronumeral,
substitute, value, variable
Having completed this chapter you should be able to:• add and subtract like algebraic terms
• multiply and divide algebraic terms
• add and subtract algebraic fractions
• establish and apply index laws in algebra
• solve linear equations.
2 REVIEW TEST1 8x − x =
A 8 B 8x C 7x D 7
2 5x − 7 + 4x − 3 =A x − 10 B 9x − 10 C 9x − 4 D x − 4
3 −5d − 3c + 2d − 4c + 6 =A 3d − 7c + 6 B 6 C −3d + 7c + 6 D −3d − 7c + 6
4 5x × −3y =A 2xy B −2xy C −15xy D 15xy
5 8xy
___ 6x
=
A 8y B 3 ___
4y C 4xy
___ 3 D
4y ___
3
6 9ab
____ 10
× 20
___ 6a2 =
A 3b
___ a B 3b
___ 2a C
2b ___
3a D 2b
___ a
7 5x
___ 4 ÷
3y ___
7 =
A 15xy
____ 28
B 35x
____ 12y C
12y ____
35x D 28
____ 15xy
8 6(3m − 4) =A 18m − 4 B 9m − 4 C 9m − 10 D 18m − 24
9 −3(2a − 5) =A −6a − 15 B −6a + 15 C −5a − 5 D −5a − 8
3 + 4(5 − 2x) =A 23 − 8x B 35 − 14x C 23 −2x D 35 − 2x
10
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Chapter 2 Further algebraic skills 43
5 − 3(2a − 1) =A 4a − 1 B 4a − 2 C 8 − 6a D 2 − 6a
2x(x − 3) + x(x − 7) =A x2 − 13x B x2 + x C 3x2 + 13x D 3x2 − 13x
4t
__ 5 +
t __
2 =
A 5t
__ 7 B
13t ___
20 C
13t ___
10 D
4t2
___ 7
4p6 × 3p7 =A 12p42 B 7p42 C 12p67 D 12p13
(2m7n4)3 =A 2m21n12 B 8m21n12 C 8m10n7 D 2m10n7
x4y3 × x3y7
_________ x7y5 =
A y5 B xy5 C y2 D xy2
2x5(4x2 − 3x) =A 8x7 − 6x6 B 8x10 − 6x6 C 8x7 − 6x5 D 8x10 − 6x5
2a2(3a − 1) − 3a2(1 − 2a) =A −5a2 B −a2 C 12a3 − 5a2 D 12a3 − a2
The solution of 5p + 3 = 7 is:
A p = 2 B p = − 4
_ 5 C p =
4
_ 5 D p = 1
1
_ 4
The solution of 7x − 4 = 2x + 6 is:
A x = −2 B x = 2 C x = 1 1
_ 9 D x =
9
__ 10
The solution of 3x
___ 4 = 7 is:
A x = 9 1
_ 3 B x =
3
__ 28 C x = 5
1
_ 4 D x =
4
__ 21
The solution of 5 __ n = 3 is:
A n = 15 B n = 2 C n = 3
_ 5 D n = 1
2
_ 3
The solution of t ___
15 − 50 = 175 is:
A t = 15 B t = 125 C t = 1875 D t = 3375
The solution of w
__ 3 +
w __
2 = −10 is:
A w = −25 B w = −8 1
_ 3 C w = −12 D w = −24
Given the formula V = IR − E, the value of I when V = 13, R = 4 and E = 7 is:
A 1 1
_ 2 B
1
_ 4 C 5 D
2
_ 3
If you have any diffi culty with these questions, refer to the examples and questions in the sections listed in the table.
Question 1–7 8–12 13 14–18 19–24 25
Section A B C D E F
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
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2A REVIEW SET1 Simplify:
a 7ab − 5ab b 7k + k c 5p − 3 + 7p − 1 d x2 + 2x + 2x2 + x
2 Simplify:
a 5 × −3b b 15pq
_____ 10p c
8ab ____
15 ×
5 ___
4a d
2x ___
3 ÷
3y ___
5
3 Expand and simplify:
a 5m(2m − 3) b −(x + 7) c 2(3x − 5) + 8x d 2(4x − 1) + 3(2x + 1)
4 Simplify:
a 2x
___ 3 +
x __
4 b
k __
6 + 2k
5 Simplify:
a 5m3n2 × 7m4n b 2k7m10
______ 8k4m7
c (a2b3)5
d 4p6q5 × 6p2q8
____________ pq × 3p3q3
e 5a3(2a2 + 7) + 2a3(3a2 − 4)
6 Solve:
a 8x = 57 b x __
5 = −2 c x − 5 = −3
d x + 8 = 5 e 4x − 5 = 3 f 5 − 3x = −4
g 3(m + 5) = 17 h 15 + 4x = 3 + x i x __
2 + 1 = 3x − 9
j z __
5 +
z __
2 = 1 k x2 = 16 l 3 √
__ y = 5
7 Use the formula E = 1
_ 2 mv2 to fi nd the value of m when E = 390 and v = 20.
2B REVIEW SET1 Simplify:
a 4x − x b 12x2 − 8x2 c 4x − 5y − 7x + y d 4y + x2 − 2x2 + y
2 Simplify:
a −7m × −4n b 12pq ÷ 18q c 6km
____ 5 ×
4 ___
3m d
5t __
3 ÷
10t ___
9
3 Expand and simplify:
a −6(5p − 2) b −(m − 7) c 7 − 4(2x + 3) d 2(2 − 3x) + 4(x − 1)
4 Simplify:
a 2m
___ 5 +
3m ___
4 b x −
3x ___
4
5 Simplify:
a 4k2m5 × 1
__ 4 km2 b
4w13
____ 8w8 c (2p3)4
d 15p7q8
____________ 3p4q2 × 5p3q4 e 4p2(3p − 5q2) + 2p2(7p + 3q2)
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Chapter 2 Further algebraic skills 45
6 Solve:
a 7p = 64 b 3x
___ 2 = 5 c
x __
3 + 12 = 7 d 8 − x = 15
e 3x − 7 = 5 f 4 − 7x = −2 g 2(p − 3) = 7 h 12 + 7x = 3x + 4
i y3 = 8 j √ __
w = 6 k 4 __ x = 3 l 3 √
__ x = 5
7 Use the formula A = 1
_ 2 bh to fi nd b given that A = 72 and h = 8.
2C REVIEW SET1 Simplify:
a 8p − p b 6x + 7x c 2y − 5x − 5y − 2x d 5 − x2 + 3x2 + 2
2 Simplify:
a 4xy × 3yz b 10m ÷ 15mn c 4 ____
5ab ×
15a ____
16 d
4t __
3 ÷
8t __
9
3 Expand and simplify:
a 4z(3 − z) b −(5 − 2x) c 4 + 2(x − 3) d 4(x + 1) − 5(2x + 1)
4 Simplify:
a 5m
___ 6 −
2m ___
3 b 3k +
3k ___
4
5 Simplify:
a v6w5 × 2
_________ 5v3w2
b m7n10
_____ m3n4
c (3yz3)2
d 6t 7v3
_____ 10t4v
× 5t3v4
____ tv5
e 3x2(4x3 − 2) + x2(3x3 + 9)
6 Solve:
a 5x = 16 b 4x
___ 3 = −2 c 4 − x = −9 d
x __
2 + 5 = 3
e 4x − 11 = 8 f 3 − 5x = −9 g 5(a + 1) = 35 h y __
3 − 1 = y + 5
i 3z
__ 2 −
5z __
6 = 8 j
4 __ x = 2 k y2 = 3 l √
_ t = 14
7 Use the formula T = a + (n − 1)d to fi nd n given that T = 41, a = 3 and d = 2.
2D REVIEW SET1 Simplify:
a 3x − 2x b 4x2 + 8x + x2
c 3x − 4y + 4x − y d a2 − 5a2b2 + 2a2b2 − a2
2 Simplify:
a −7pq × 3qr b 16a × 20ab c 6 ___
5x × 20xy
____ 9 d
8k ___
3 ÷
4 ___
6k
3 Expand and simplify:
a −5w(w − 4) b −(2x − 11) c 8 + 3(3x − 2) d 2(p + 1) − 3(p − 2)
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Insight Mathematics General 12 HSC course 146
4 Simplify:
a 14p
____ 5 −
3p ___
2 b
2x ___
7 +
5x ___
9
5 Simplify:
a 2x4y5
_____ 5 ×
5x2y3
_____ 2 b
15a7b10
______ 20a6b
c (10u5v4)4
d 4a6b2 × 3a7b9
____________ 6a3b6 × 4a10b5 e 5y2(2y4 − 5) + 3y2(3y4 − 4)
6 Solve:
a 2x = 29 b y __
5 =
4 __
3 c m − 7 = −7
d 5t + 2 = −3 e w
__ 7 − 1 = 5 f 5v + 3 = 9 − 7v
g 4(2a + 1) = 11 h 5p
___ 3 −
p __
6 =
3 __
4 i
4 __ x = 3
j t 2 = 1 k 3 √ __
y = 3 l x + 7
_____ 4 = 3
7 Use the formula B = m
__ h2 to fi nd m if B = 31.25 and h = 1.6.
2 EXAMINAT ION QUEST ION (15 MARKS)a Simplify:
i 3(5x − 2y) − 2(4x − y) (2 marks)
ii 5y3 × 7y5 (1 mark)
iii (p2q3)4 (1 mark)
iv 8p6q7
______ 12p5q2 (1 mark)
b Solve:
i 3(m + 6) = 22 (2 marks)
ii w
___ 10
− 11 = 13 (2 marks)
iii 18 = 54
___ n (2 marks)
iv 5y + 3 = 3y − 7 (2 marks)
c Use the formula a = 2Rn
_____ n + 1
to fi nd the value of R when a = 95 and n = 19. (2 marks)
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