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2.1 Adding and subtracting like termsA pronumeral (letter) represents a number. It may stand for an unknown value or series of values that change. For example, in the equation x + 5 = 8, x is a pronumeral that represents a value. Its value can be determined because we know 3 + 5 = 8, so x = 3.
When a term has a pronumeral and a number, the number is written before the pronumeral and is called the coeffi cient. For example, the term 3xy has a coeffi cient of 3 and its pronumerals are written after the coeffi cient in alphabetical order.
Like termsTerms that have exactly the same pronumerals such as 2a and 5a are called like terms. Only like terms can be added and subtracted. It involves adding and subtracting the coeffi cients. Adding and subtracting like terms simplifi es the algebraic expression. It is often called collecting the like terms.
2.1
C H A P T E R
2Algebraic manipulation
Syllabus topic — AM1 Algebraic manipulation Add and subtract like terms
Multiply and divide algebraic terms
Expand and factorise algebraic expressions
Evaluate the subject of the formula through substitution
Solve linear equations involving up to 3 steps
Solve equations following substitution
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1 Find the like terms or the terms that have exactly the same pronumerals. 2 Only like terms can be added or subtracted; unlike terms cannot.3 Add or subtract the coeffi cients or the numbers before the pronumeral of the like terms.
Example 1 Adding and subtracting like terms
Simplify 2ab + 3 + 5ab − 7.
Solution1 Rewrite the expression by grouping
the like terms.2 Add and subtract the coefficients.
2ab + 3 + 5ab − 7 = (2ab + 5ab) + 3 − 7 = 7ab − 4
Example 2 Adding and subtracting like terms
Simplify 4y + 6y2 − 3y − 5y2.
Solution1 Rewrite the expression by
grouping the like terms.2 Add and subtract the coefficients.
Adding and subtracting algebraic fractions To add and subtract algebraic fractions rewrite each fraction as an equivalent fraction with a common denominator, then add or subtract the numerators. A common denominator can always be found by multiplying the denominators of both fractions together.
Example 3 Adding and subtracting algebraic fractions
Simplify x x
6 4+ .
Solution
1 Find a common denominator for 6 and 4. Both 6 and 4 divide into 12. Alternatively, multiply 6 by 4 and use 24.
2 Multiply the first fraction by 2 (6 × 2 = 12) and the second fraction by 3 (4 × 3 = 12).
3 Write the equivalent fractions.4 Add the numerators of the equivalent fractions.
+ =+ = × + ×
= += +
=
x x x x×x x×
x x
x
6 4
2x x2x x
6 2×6 2×3
4 3×4 3×2
12
3x x3x x
125
12
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d 3 8 42 23 82 23 8 42 24m m3 8m m3 83 82 23 8m m3 82 23 8 m m2 2m m2 2+ −2 2+ −2 23 82 23 8+ −3 82 23 8m m+ −m m3 8m m3 8+ −3 8m m3 82 2m m2 2+ −2 2m m2 23 82 23 8m m3 82 23 8+ −3 82 23 8m m3 82 23 8 m m−m m e e e e e2 2e e2 2e e e e2 2e e2 222 2+ +e e+ +e e2 2+ +2 2e e2 2e e+ +e e2 2e ee e+ +e e2 2+ +2 2e e2 2e e+ +e e2 2e ee e2e e+ +e e2e e2 222 2+ +2 222 2e e2 2e e2e e2 2e e+ +e e2 2e e2e e2 2e e e e−e e f d d d d+ −d d+ −d d d d+d d2 2d d2 2d d+ −2 2+ − d d+d d2 2d d+d d52 252 2
6 Which of the following is equivalent to m m m n n+ + +m m m n+ + +m m m n + ?
a 2 2m m2 2m m2 2n2 2+ +2 22 2m m2 2+ +2 2m m2 2 b m n3 2m n3 2m nm n+m n3 2+3 2m n3 2m n+m n3 2m n
c 5 4 3m m5 4m m5 4 n n− +5 4− +5 4m m− +m m5 4m m5 4− +5 4m m5 4 n n−n n d 7 4 2m m n7 4m m n7 4 2m m n2 n− +7 4− +7 4m m n− +m m n7 4m m n7 4− +7 4m m n7 4 −
e 3 23 2n n m3 2m3 2− +3 23 2n n m3 2− +3 2n n m3 23 2− −3 2 f − + − +2 5− +2 5− + 2 3− +2 3− +m m2 5m m2 5− +2 5− +m m− +2 5− + n n2 3n n2 3− +2 3− +n n− +2 3− +
7 Copy and add like terms where possible to complete the table.
+ x 3x x + y
3x
7y
x - y
2y
8 Matteo has $y for shopping. He spent $x for a pair of jeans, $3x for a shirt and $2x for a belt. Write an expression in simplified form for how many dollars he has left.
9 The perimeter of a plane shape is the distance around the boundary of the shape. The plane shape opposite is a rectangle with a length l and a breadth b. Write an expression for the perimeter of this rectangle by collecting like terms.
10 The isosceles triangle opposite has three sides whose lengths are 3x + y, 3x + y and x + 2y. Write an expression in simplified form for the perimeter of this triangle by collecting like terms.
11 Add or subtract the algebraic fractions.
a w w
4 3+ b
a a
4 5− c
x x
7
2x x2x x
3+
d z z
3 5− e
3
8 6
h h+ f 5
12 8
r r−
g u u
10
4u u4u u
15+ h
3
4 10
e e− i w w w
2 4 6+ ++ +
j 3
5 4 2
a a a− +− + k 7
10 6 3
x x x− +− + l dd d+ −+ −2 10
l
b
3x + y
3x + y
x + 2y
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2.2 Multiplication and division of algebraic termsAlgebraic terms are multiplied and divided to form a single algebraic expression. Terms usually contain a coeffi cient before a pronumeral. The multiplication sign between the coeffi cient and the pronumeral is omitted. For example, the algebraic term 4x can be written in expanded form as 4 × x. After an algebraic expression is written in expanded form, the coeffi cients can be multiplied or divided and the pronumerals can be multiplied or divided. Index notation should be used to write expressions in a shorter way such as a × a = a2. If the algebraic terms contain fractions it is easier to cancel any common factors in the numerator and denominator. This makes the calculations easier.
Multiplication and division of algebraic terms
1 Write in expanded form. 2 If the algebraic term is a fraction cancel any common factors.3 Multiply and divide the coeffi cients. 4 Multiply and divide the pronumerals.5 Write the coeffi cient before the pronumerals.6 Write the pronumerals in alphabetical order and express in index notation.
Example 4 Multiplying algebraic terms
Simplify the following.a 2cd de× (−3 ) b x x x2x x2x x× 3x x× 3x x × 4
Solution
1 Write in expanded form.2 Multiply the coefficients (2 × −3 = −6).3 Write the pronumerals in alphabetical
order.4 Express answer using index notation
(d × d = d 2).1 Write in expanded form.2 Multiply the coefficients (1 × 3 × 4 = 12).3 Write the coefficient before the
pronumerals.4 Write the pronumerals using index
notation.5 Express the answer using index notation
(x2 × x × x = x4).
a 2 2 36
6 2
cd2 2cd2 2de2 2de2 2 c d d ec d d e
cd e
× (2 2× (2 2−32 2−32 2)2 2)2 2= ×2 2= ×2 2 × ×c d× ×c d − ×3− ×3 d e×d e= − × × × ×d d e× × ×d d e
2.3 Expanding algebraic expressionsGrouping symbols in algebraic expressions indicate the order of operations. The two most commonly used grouping symbols are parentheses ( ) and brackets [ ]. They are removed by using the distributive law or a × (b + c) = ab + ac. This is illustrated below.
To expand an algebraic expression using the distributive law, multiply the number or terms inside the grouping symbols by the number or term outside the grouping symbols. The resulting algebraic expression is simplifi ed by collecting the like terms.
Make sure you remember to multiply all the terms inside the grouping symbol by the number or term outside the grouping symbols.
Expanding algebraic expressions
1 Multiply the number or term outside the grouping symbol by the a fi rst term inside the grouping symbol.b second term inside the grouping symbol.
2 Simplify and collect like terms if required. a b a b a c
ab ac( )a b( )a b c( )c+ =( )+ =( )c( )c+ =c( )c a b a× +a b a ×
= +ab= +ab a b a b a c
ab ac( )a b( )a b c( )c− =( )− =( )c( )c− =c( )c a b a× −a b a ×
= −ab= −ab
Example 7 Expanding algebraic expressions
Expand 5 2( )5 2( )5 2 3( )3 .( )y( )( )−( )
Solution1 Multiply the first term inside the parenthesis (2y) by
the number outside the parenthesis (5).2 Multiply the second term inside the parenthesis (-3)
by the number outside the parenthesis (5).3 Write in simplest form.
2.4 Factorising algebraic expressionsFactorising is the reverse process to expanding. For example, expanding the expression 5(2y − 3) produces 10y − 15, whereas factorising the expression 10y − 15 produces 5(2y − 3). The fi rst step in factorising an expression is to fi nd the largest factor of both terms or the highest common factor (HCF). In this case the HCF of 10y and 15 is 5. The HCF is written outside the grouping symbol and terms inside are found by dividing the HCF into each term.
Factorising algebraic expressions
1 Find the largest factor of each term or the HCF.2 Write the HCF outside the grouping symbol.3 Divide the HCF into each term to fi nd the terms inside the grouping symbols.4 Check the factorisation by expanding the expression.
Example 11 Factorising algebraic expressions
Factorise 3p − 6.
Solution1 Find the largest factor of each term (HCF is 3).2 Write the HCF or 3 outside the grouping symbol.3 Divide the HCF or 3 into each term to find the terms
inside the grouping symbols.4 Check by expanding the expression.
Solution1 Find the largest factor of each term (HCF is 2x).2 Write the HCF or 2x outside the grouping symbol.3 Divide the HCF or 2x into each term to find the
terms inside the grouping symbols.4 Check by expanding the expression.
11 Factorise by using the HCF for each of the following expressions.
a 2ab ac ag+ +ac+ +ac b 4x xy xx xy xx x zy xzy x+ +x x+ +x xy x+ +y xx xy xx x+ +x xy xx x c 7 1 2g g7 1g g7 1h g7 1h g7 14h g47 1− +7 17 1g g7 1− +7 1g g7 17 1h g7 1− +7 1h g7 1
d d d de2d d2d dd d3d d3d d3d d− +d d− +d dd d3d d− +d d3d d e b bc b2b b2b b 5c b5c b+ −b b+ −b bc b+ −c b f k k h khh khh kk k− +k k4 2h k4 2h k4 2k k4 2k k− +4 2− +k k− +k k4 2k k− +k k h k− +h k4 2h k− +h k24 224 2− +4 2− +2− +4 2− +
2.5 SubstitutionSubstitution involves replacing the pronumeral in an algebraic expression with one or more numbers. The resulting numerical expression is evaluated and expressed to the specifi ed level of accuracy.
Substitution of values
1 Write the algebraic expression.2 Replace the variables in the expression with the numbers given in the question.3 Evaluate using the calculator.4 Write the answer to the specifi ed level of accuracy and correct units if necessary.
Example 13 Substituting values
Evaluate 3a − 4b + c given a = 2, b = 5 and c = −10.
Solution1 Write the algebraic expression.2 Substitute the values for a, b and c into
6 Evaluate these expressions given x = 7, y = -5 and z = 21.
a x z y2 2x z2 2x z+ +x z+ +x z2 2+ +2 2x z2 2x z+ +x z2 2x z b y x3y x3y x4y x4y xy x−y x
c 4 1x z4 1x z4 14 1+ −4 14 1x z4 1+ −4 1x z4 1 d z y+z y+z y4z y4z y2
e 3 2xy
z f
6
5
2y
zx
7 The cross-sectional area of a solid is an annulus. It is evaluated using π ( )( )R r( )2 2( )2 2( )( )R r( )2 2( )R r( )( )R r( )−( )R r( ) where R is the radius of the outer circle and r the radius of the inner circle. Find the area of an annulus if R is 8 cm and r is 4 cm. Answer correct to one decimal place.
8 Determine the value of 23
2p q2p q2 given that p = 4 and q = 6.
9 Evaluate 2 33 y2 3y2 32 3+2 3 if y = 12.
10 Find the value of 2π lg
when l = 2.6 and g = 9.8. Give your answer correct to two decimal places.
11 Find the value of u as2u a2u a2u a2u a+u a+u a if u = 6, a = 7 and s = 2.
12 Find the value of 1
2π fcπ fcπ if f = 10 and c = 2. Give your answer correct to three decimal
places.
13 Find the value of 3Rr
R rR r+R r when R = 8.2 and r = 4.9. Give your answer correct to two
decimal places.
14 What is the value of yA
( )y( )y( )+( )( )12( ) when y = 9 and A = 15. Give your answer correct to the nearest whole number.
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2.6 Linear equationsAn equation is a mathematical statement that says that two things are equal. It has an equal sign. For example, these are all equations:
Linear equations have all their variables raised to the power of 1. The above three equations are linear equations. An equation such as x2 = 9 is not a linear equation as the variable is raised to the power of 2.
Solving an equationThe process of fi nding the unknown value for the variable is called solving the equation.
When solving an equation look to perform the opposite operation:• + is opposite to −• × is opposite to ÷• x2 is opposite to x
Make sure the equation remains balanced like a set of scales. The same operation needs to be done on both sides of the equal sign to keep the balance.
Solving an equation
1 Look to perform the opposite operation (+ is opposite to −, × is opposite to ÷).2 Add or subtract the same number to both sides of the equation OR3 Multiply or divide both sides of the equation by the same number.4 To solve two- or three-step equations, repeat the above steps as required. It
is often easier to fi rstly add or subtract the same number to both sides of the equation.
When a solution has been reached it can be checked. The solution of the equation must satisfy the equation. Always check your solution by substituting your answer into the original equation. The left-hand side of the equation must equal the right-hand side.
2.6
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Solving an equation using a graphics calculatorGraphics calculators have an equation mode that may be used to solve any type of linear equation. The red letters on the keypad are used as the variable. After the equation is entered, set the required variable to zero and choose the ‘solver’ key. The calculator will show the answer and the values of the left-hand side and right-hand side of the equation.
Example 20 Solving a linear equation using a graphics calculator
Solve the equation 80 - 10y = 100.
Solution1 Select the EQUA menu.2 Select SOLVER (F3).3 Enter the equation. To place the variable (y) use
the ALPHA key to access the letters in red above the keys.
4 Highlight the required variable or Y = 0.5 Press SOLV and the calculator will show the
answer. 6 The values of the left (Lft) and right (Rgt) sides of
the formula are shown and should be equal.
Example 21 Solving a linear equation using a graphics calculator
Solve the equation 6x + 5 = 7 + 5x.
Solution1 Select the EQUA menu.2 Select SOLVER (F3).3 Enter the equation. To place the variable (x) use the
ALPHA key to access the letters in red above the keys.
4 Highlight the required variable or X = 0.5 Press SOLV and the calculator will show the answer. The values of the left (Lft) and right (Rgt) sides of
the formula are shown and should be equal.
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Exercise 2F 1 Solve the following linear equations.
a y + =8 1+ =8 1+ = 7 b x + =13+ =13+ = 28 c a + =7 1+ =7 1+ = 2d c + =7 4+ =7 4+ = −7 4− e m + =9 4+ =9 4+ = f 4 54 5+ =4 54 5−4 5d4 5d4 54 5+ =4 5d4 5+ =4 5g 8 118 1+ =8 1h8 1h8 18 1+ =8 1h8 1+ =8 1 h 9 49 4+ =9 49 4+ =9 4r9 4+ =9 4 i 10 1+ = −q+ =q+ =j 5 = x + 2 k 12 = m + 7 l -3 = g + 5
2 Solve the following linear equations.a a - 7 = 3 b k - 5 = 5 c d - 9 = 14d s - 5 = - 4 e z - 12 = -7 f k - 7 = -9g 11 - v = 4 h 7 - x = 3 i 6 - j = 7j 9 = h - 5 k 11 = f - 4 l -4 = c - (-1)
3 Solve the following linear equations.a 4x = 12 b 5w = 45 c 7v = 28d 2t = -12 e 6h = -30 f -4a = 40g 2w = 13 h 3c = -23 i 7e = -8j 17 = 8k k -75 = 7d l -14 = -3e
4 Solve the following linear equations.
a y
24= b d
78= c w
64=
d f
−=
35 e
a
−=
75 f
g
92=
g d
129= − h s
−= −
113 i x
−= −
54
j 62
=−y
k 102
= m l 9
9=
−w
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2.7 Equations with fractionsEquations with fractions are solved in exactly the same way as other equations. Look to perform the opposite operation (+ is opposite to -, × is opposite to ÷) to both sides of the equation. Check your solution by substituting your answer into the original equation.
Example 22 Solving an equation with a fraction
Solve the equation e
34 10+ =4 1+ =4 1 .
Solution1 Write the equation.2 The opposite operation to adding 4 is subtracting 4.
Subtract 4 from both sides of the equation.3 The opposite operation to dividing by 3 is multiplying
by 3. Multiply both sides by 3.4 Check that the solution is correct by substituting
e = 18 into the original equation.
e
e
e
e
34 10
36
33
6 3
18
4 4
+ =4 1+ =4 1
=
× =× = 6 3×6 3
=
− −4 4− −4 4
Example 23 Solving an equation with a fraction
Solve the equation 3 5
26
3 5x3 53 5−3 5 = .
Solution1 Write the equation.2 The opposite operation to dividing by 2 is multiplying
by 2. Multiply both sides of the equation by 2.3 The opposite operation to subtracting 5 is adding 5.
Add 5 to both sides of the equation.4 The opposite operation to multiplying by 3 is dividing
by 3. Divide both sides of the equation by 3.5 Check that the solution is correct by substituting x = 5 2
3
into the original equation.
3 5
26
3 5 123 173
3
17
317
3
52
3
5 5
3 5x3 5
3 5x3 53 1x3 1x
x
3 5−3 5 =
− =3 5− =3 53 1=3 1
=
=
=
+ +5 5+ +5 5
2.7
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6 There are thirty-six times as many cars in Australia as trucks. Let C stand for the number of cars and T for the number of trucks.a Write an equation with C as the subject of the equation that correctly describes the
relationship between the number of cars and trucks.b A local community has 120 trucks. How many cars are in the community?
2.8 Using formulasA formula is a mathematical relationship between two or more variables. For example:• S D
T= is a formula for relating the speed, distance and time. S, D and T are the variables.
• P = 4L is a formula for fi nding the perimeter of a square, where P is the perimeter and L is the side length of the square. P and L are the variables.
By substituting all the known variables into a formula, we are able to fi nd the value of an unknown variable.
Using a formula
1 Write the formula.2 Replace the variables in the formula with the numbers given in the question.3 Evaluate using the calculator.4 Write the answer to the specifi ed level of accuracy and correct units if
necessary.
Example 24 Using a formula
The cost of hiring a windsurfer is given by the formula
C = 4t + 7
where C is the cost in dollars and t is the time in hours. Kayla wants to sail for 3 hours. How much will it cost her?
Solution1 Write the formula.2 Substitute the value for t into the formula.3 Evaluate. 4 Write your answer in words.
Using a formula and a graphics calculator Graphics calculators have an equation mode that may be used to enter a formula. The coloured letters on the keypad are used as the variable.
After the formula is entered, set the unknown variable to zero and enter the values for the known variables. It is not necessary for the known variable to be the subject of the formula.
Select the ‘solver’ key to obtain the answer. The calculator will show the answer and the values of the left-hand side and right-hand side of the equation.
Example 27 Using a formula and a graphics calculator
The circumference, C, of a circle with radius, r, is given by the formula C = 2π r. Find the circumference of a circle with a radius of 5 cm using a graphics calculator.
Solution1 Select the EQUA menu.2 Select SOLVER (F3).3 Enter the formula. To place the variables (C and r)
use the ALPHA key to access the letters in red above the keys.
4 Enter the known variable. The radius is 5 so R = 5.5 Highlight the required variable or C = 0.
6 Press SOLV and the calculator will show the answer. 7 The values of the left (Lft) and right (Rgt) sides of
the formula are shown and should be equal.
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2 If A = s2 find the value of A when:a s = 9b s = 6c s = 32
3 If A bA bhA b=A b1
A b1
A b2
find the value of A when:
a b = 10, h = 4
b b = 15, h = 2
c b = 2, h = 3
4 If P = 2l + 2b find the value of P when:a l = 8, b = 10b l = 2, b = 11c l = 3, b = 9
5 Find the value of T (correct to one decimal place) in the formula TMv
r=
2
if:
a M = 1.7, v = 4 and r = 3.8
b M = 2.1, v = 1 and r = 2.2
6 Use the formula ab= + −
9= +9= + 18
2 to find the value of a when:
a b = 8 b b = 1 c b = -2
7 If px
x=
+12
4 find the value of p when:
a x = 7 b x = 5 c x = -3
8 The cost of hiring a hall is given by the rule C = 30t + 1000 where C is the total cost in dollars and t is the number of hours for which the hall is hired. Find the cost of hiring the hall for:a 2 hours b 5.5 hours
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9 The distance, d km, travelled by a truck in t hours at an average speed of s km/h is given by the formula d = st. Find the distance travelled by a truck travelling at a speed of 70 km/h for 5 hours.
10 Given that HE
T= find the value of H when:
a E = 2.6 × 10−11 and T = 100 b E = 7.8 × 10−6 and T = 20
11 The formula used to convert temperature from degrees Fahrenheit to degrees Celsius is C FC F= −C FC F= −C F5C F5C F
9C F
9C F( )C F( )C F= −( )= −C F= −C F( )C F= −C F 32( )32 . Use this formula to convert the following temperatures to degrees
Celsius. Answer correct to the nearest whole number.a 40°F b 110°F
12 The formula v = u + at relates the velocity, acceleration and time. a Make u the subject of the formula.b Make a the subject of the formula.c Make t the subject of the formula.
13 The circumference, C, of a circle with radius, r, is given by the formula C = 2π r. a Make r the subject of the formula.b Find the radii of circles with the following circumferences. (Answer correct to two
decimal places.) i 3 cm ii 6.9 mm
14 The body mass index is Bm
h=
2 where m is the mass in kg and h is the height in m.
a Make m the subject of the formula.b Find m to the nearest whole number when: i B = 22.78 and h = 1.79 m ii B = 31.8 and h = 1.86 m
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15 Find the value of v (correct to one decimal place) in the formula v uv u as= += +v u= +v uv u= +v u2= +2= + 2 if: a u = 4, a = 7 and s = 12b u = -2, a = 10 and s = 5
16 Use the formula RV= 3
43
π to find the value of R (correct to two decimal places) when:
a V = 12 b V = 44 c V = 100
17 Find the value of s (correct to one decimal place) in the formula s ut at at= +s u= +s ut a= +t a1t a1t a2
t a2
t a 2 if:
a u = -5, a = 4 and t = 6 b u = 2, a = 5 and t = 15
18 The formula for calculating simple interest is IPRT=100
where P is the principal, R is the
interest rate per annum and T is the time in years. Calculate the interest earned, correct to the nearest cent, on the following investments.a $10 000 at an interest rate of 9% p.a. for 3 yearsb $88 000 at an interest rate of 11.2% p.a. for 2 years
c $24 000 at an interest rate of 73
4% p.a. for 2 years
19 The volume of a cone is evaluated using V rV r hV r=V r1
V r1
V r3
2πV rπV r where h is height and r is radius.a Write the formula with h as the subject.b Calculate the height of a cone, correct to two decimal places, if the volume of the
cone is 18 cm3 and the radius is 2 cm.
20 The volume of a sphere is given by the formula V rV rV r=V r4
V r4
V r3
3πV rπV r where r is the radius.a Write the formula with r as the subject.b What is the radius in metres of a spherical balloon with a volume of 2 m3? Answer
correct to one decimal place.
14.8
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Adding and subtracting like terms 1 Find the like terms.2 Only add or subtract like terms.3 Add or subtract the coefficients of the like terms.
Multiplication and division of
algebraic terms
1 Write in expanded form and cancel any factors in a fraction.
2 Multiply and divide the coefficients and pronumerals. 3 Write the pronumerals in alphabetical order and express
in index notation.
Expanding algebraic expressions 1 Multiply the term outside the grouping symbol by thea First term inside the grouping symbol.b Second term inside the grouping symbol.
2 Simplify and collect like terms if required.
Factorising algebraic expressions 1 Find the largest factor of each term or the HCF.2 Write the HCF outside the grouping symbol.3 Divide the HCF into each term to find the terms inside
the grouping symbols.4 Check the factorisation by expanding the expression.
Linear equations 1 Perform the opposite operation (+ and -, × and ÷).2 Add/subtract the same number to both sides of the equation.3 Multiply/divide both sides of the equation by the same
number.
Equations with fractions • Use the same steps as linear equations.
Substitution 1 Write the algebraic expression.2 Substitute the values and evaluate.
Using formulas 1 Write the formula. 2 Substitute the values and evaluate.
Chapter summary – Algebraic manipulation Study guide 2
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10 The distance d in km that a person can see the horizon from h in metres above the sea level, is given by the formula
dh= ×5= ×5= ×2
.
a Find d when h is 18 metres.b Find h when d is 10 kilometres.
11 Use the formula RV= 2
33
π to fi nd the value of R (correct to two decimal places) when:
a V = 9b V = 24c V = 200
12 Einstein’s equation E = mc2 states that the energy E in joules equals the mass of m kg multiplied by the square of the speed of light c (3 × 108 m/s). Find the amount of energy produced by a:a mass of 500 kgb mass of 200 kg
13 The cost of hiring a hall is given by the formula C = 20t + 2000 where C is the total cost in dollars and t is the number of hours for which the hall is hired. a Make t the subject of the equation.b Find t when C is $2060.
Challenge questions 2
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