1 Algebra (notation and equations) cyan magenta yellow 95 Algebraic notation Algebraic substitution Linear equations Rational equations Linear inequations Problem solving Money and investment problems Motion problems (Extension) 100 50 75 A B C D E F G H 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 Contents: black Y:\HAESE\SA_10-6ed\SA10-6_01\011SA10-6_01.CDR Monday, 4 September 2006 2:15:47 PM PETERDELL SA_10-6 12 ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1) Algebra is a very powerful tool which is used to make problem solving easier. Algebra involves using pronumerals (letters) to represent unknown values, or values which can vary depending on the situation. Many worded problems can be converted to algebraic symbols to make algebraic equations. We learn how to solve equations in order to find solutions to the problems. Algebra can also be used to construct formulae, which are equations that connect two or more variables. Many people use formulae as part of their jobs, so an understanding of how to substitute into and rearrange formulae is essential. Builders, nurses, pharmacists, engineers, financial planners and computer programmers all use formulae which rely on algebra. OPENING PROBLEM Holly bought XBC shares for $2:50 each and NGL shares for $4:00 each. 0 Consider the following: ² What did Holly pay, in total, for 500 XBC shares and 600 NGL shares? ² What did Holly pay in total for x XBC shares and (x + 100) NGL shares? ² Bob knows that Holly paid, in total, $5925 for her XBC and NGL shares. He also knows that she bought 100 more NGL shares than XBC shares. How can Bob use algebra to find how many of each share type Holly bought? A ALGEBRAIC NOTATION The ability to convert worded sentences and problems into algebraic symbols, and to understand algebraic notation, is essential in the problem solving process. x2 + 3x x2 + 3x = 8 x2 + 3x > 28 ² ² ² Notice that: is an algebraic expression, whereas is an equation, and is an inequality (sometimes called an inequation). Recall that when we simplify repeated sums, we use product notation: i.e., x+x = 2 ‘lots’ of x =2£x = 2x x+x+x = 3 ‘lots’ of x =3£x = 3x and Also, when we simplify repeated products, we use index notation: magenta yellow 95 100 50 75 25 0 5 95 100 50 25 0 75 x £ x £ x = x3 and 5 95 100 50 75 0 5 95 100 50 75 25 0 5 cyan 25 x £ x = x2 i.e., black Y:\HAESE\SA_10-6ed\SA10-6_01\012SA10-6_01.CDR Wednesday, 30 August 2006 11:11:19 AM DAVID3 SA_10-6 ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1) Example 1 Self Tutor Write, in words, the meaning of: a x¡5 b a+b x¡5 a+b 3x2 + 7 a b c 13 c 3x2 + 7 is “5 less than x”. is “the sum of a and b”, or “b more than a”. is “7 more than three times the square of x”. EXERCISE 1A 1 Write, in words, the meaning of: a 2a b pq c p m d a2 e a¡3 f b+c g 2x + c h (2a)2 i 2a2 j a ¡ c2 k a + b2 l (a + b)2 Example 2 Self Tutor Write the following as algebraic expressions: a the sum of p and the square of q b the square of the sum of p and q c b less than double a a 2 Write a c e g i k m o p + q2 b (p + q)2 c 2a ¡ b the following as algebraic expressions: the sum of a and c b the sum of p, q and r the product of a and b d the sum of r and the square of s the square of the sum of r and s f the sum of the squares of r and s the sum of twice a and b h the difference between p and q, if p > q a less than the square of b j half the sum of a and b the sum of a and a quarter of b l the square root of the sum of m and n the sum of x and its reciprocal n a quarter of the sum of a and b the square root of the sum of the squares of x and y Example 3 Self Tutor Write, in sentence form, the meaning of: b+c a D = ct b A= 2 cyan magenta 95 yellow Y:\HAESE\SA_10-6ed\SA10-6_01\013SA10-6_01.CDR Tuesday, 25 July 2006 3:35:10 PM PETERDELL 100 50 75 25 0 5 95 100 50 D is equal to the product of c and t. A is equal to a half of the sum of b and c, or, A is the average of b and c. 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 a b black SA_10-6 14 ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1) 3 Write, in sentence form, the meaning of: a L=a+b b d N = bc r n K= t e a+b 2 2 T = bc h c= g K= p a2 + b2 Example 4 c M = 3d f F = ma i A= a+b+c 3 Self Tutor Write ‘S is the sum of a and the product of g and t’ as an equation. The product of g and t is gt. The sum of a and gt is a + gt, ) the equation is S = a + gt. The difference between two numbers is the larger one minus the smaller one. 4 Write the following as algebraic equations: a S is the sum of p and r b D is the difference between a and b where b > a c A is the average of k and l d M is the sum of a and its reciprocal e K is the sum of t and the square of s f N is the product of g and h g y is the sum of x and the square of x h P is the square root of the sum of d and e B ALGEBRAIC SUBSTITUTION input, x Consider the number crunching machine alongside: 5x – 7 calculator output If we place any number x, into the machine, it calculates 5x ¡ 7, i.e., x is multiplied by 5 and then 7 is subtracted. For example: if x = 2, 5x ¡ 7 =5£2 ¡ 7 = 10 ¡ 7 =3 and if x = ¡2, 5x ¡ 7 = 5 £ ¡2 ¡ 7 = ¡10 ¡ 7 = ¡17 cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 To evaluate an algebraic expression, we substitute numerical values for the unknown, then calculate the result. black Y:\HAESE\SA_10-6ed\SA10-6_01\014SA10-6_01.CDR Thursday, 3 August 2006 10:14:31 AM PETERDELL SA_10-6 ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1) Example 5 15 Self Tutor If p = 4, q = ¡2 and r = 3, find the value of: a 3q ¡ 2r a b 2pq ¡ r 3q ¡ 2r = 3 £ ¡2 ¡ 2 £ 3 = ¡6 ¡ 6 = ¡12 b p ¡ 2q + 2r p+r c 2pq ¡ r = 2 £ 4 £ ¡2 ¡ 3 = ¡16 ¡ 3 = ¡19 p ¡ 2q + 2r p+r c = 4 ¡ 2 £ ¡2 + 2 £ 3 4+3 = 4+4+6 4+3 14 7 =2 = EXERCISE 1B 1 If p = 5, q = 3 and r = ¡4 find the value of: 5p 3p ¡ 2q a e 4q 5r ¡ 4q b f c g 2 If w = 3, x = 1 and y = ¡2, evaluate: y y+w a b c w x y¡x+w xy + w e f g 2(y ¡ w) y¡x Example 6 3pq 4q ¡ 2r 3x ¡ y w x ¡ wy y + w ¡ 2x b2 = (¡2)2 = ¡2 £ ¡2 =4 magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 h 5w ¡ 2x y¡x y ¡w x Notice the use of brackets! ab ¡ c3 = 3 £ ¡2 ¡ (¡1)3 = ¡6 ¡ ¡1 = ¡6 + 1 = ¡5 b 3 If a = ¡3, b = ¡4 and c = ¡1, evaluate: a c2 b b3 c 3 3 3 e b +c f (b + c) g cyan d pqr 2pr + 5q Self Tutor If a = 3, b = ¡2 and c = ¡1, evaluate: a b2 b ab ¡ c3 a d h black Y:\HAESE\SA_10-6ed\SA10-6_01\015SA10-6_01.CDR Thursday, 3 August 2006 10:03:00 AM PETERDELL a2 + b2 (2a)2 d h (a + b)2 2a2 SA_10-6 16 ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1) Example 7 Self Tutor If p = 4, q = ¡3 and r = 2, evaluate: p p a p¡q+r b p + q2 p p¡q+r p = 4 ¡ ¡3 + 2 p = 4+3+2 p = 9 =3 a b p p + q2 p = 4 + (¡3)2 p = 4+9 p = 13 + 3:61 4 If p = 4, q = ¡1 and r = 2, evaluate: p p a p+q b p+q c p p e pr ¡ q f p2 + q 2 g p r¡q p p + r + 2q INVESTIGATION d h p p ¡ pq p 2q ¡ 5r SOLVING EQUATIONS Linear equations like 5x ¡ 3 = 12 can be solved using a table of values on a graphics calculator. We try to find the value of x which makes the expression 5x ¡ 3 equal to 12 upon substitution. This is the solution to the equation. What to do: 1 Enter the function Y1 = 5X ¡ 3 into your calculator. 2 Set up a table that calculates the value of y = 5x ¡ 3 for x values from ¡5 to 5. 3 View the table and scroll down until you find the value of x that makes Y1 equal to 12. As we can see, the solution is x = 3. 4 Use your calculator and the method given above to solve the following equations: a 7x + 1 = ¡20 b 8 ¡ 3x = ¡4 x +2=1 d 13 (2x ¡ 1) = 3 c 4 5 The solutions to the following equations are not integers, so change your table to investigate x values from ¡5 to 5 in intervals of 0:5: a 2x ¡ 3 = ¡6 b 6 ¡ 4x = 8 c x ¡ 5 = ¡3:5 6 Use a calculator to solve the following equations: magenta yellow Y:\HAESE\SA_10-6ed\SA10-6_01\016SA10-6_01.CDR Monday, 31 July 2006 9:31:44 AM PETERDELL 95 100 50 c 75 25 0 5 95 100 50 75 25 0 5 95 5 ¡ 4x = 70 b 100 50 75 25 0 95 100 50 75 25 0 5 cyan 5 3x + 2 = 41 a black 2x + 5 = 2 23 3 SA_10-6 ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1) C 17 LINEAR EQUATIONS Linear equations are equations which can be written in the form ax + b = 0, where x is the unknown (variable) and a, b are constants. SOLVING EQUATIONS The following steps should be followed when solving linear equations: Step 1: Step 2: Decide how the expression containing the unknown has been ‘built up’. Isolate the unknown by performing inverse operations on both sides of the equation to ‘undo’ the ‘build up’ in reverse order. Check your solution by substitution. Step 3: Example 8 Self Tutor a 4x ¡ 1 = 7 Solve for x: a ) 4x ¡ 1 = 7 4x ¡ 1 + 1 = 7 + 1 ) 4x = 8 ) The inverse of + is ¡ and £ is ¥ b 5 ¡ 3x = 6 fadding 1 to both sidesg 8 4x = 4 4 ) x=2 fdivide both sides by 4g Check: 4 £ 2 ¡ 1 = 8 ¡ 1 = 7 X b ) 5 ¡ 3x = 6 5 ¡ 3x ¡ 5 = 6 ¡ 5 ) ¡3x = 1 fsubtracting 5 from both sidesg 1 ¡3x = ¡3 ¡3 ) fdividing both sides by ¡3g x = ¡ 13 ) Check: 5 ¡ 3 £ ¡ 13 = 5 + 1 = 6 X EXERCISE 1C cyan magenta yellow 95 c g 100 50 75 25 0 5 95 50 75 100 5x = 45 3x ¡ 2 = ¡14 b f 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 1 Solve for x: a x+9=4 e 2x + 5 = 17 black Y:\HAESE\SA_10-6ed\SA10-6_01\017SA10-6_01.CDR Thursday, 3 August 2006 10:04:26 AM PETERDELL ¡24 = ¡6x 3 ¡ 4x = ¡17 d h 3 ¡ x = 12 8 = 9 ¡ 2x SA_10-6 18 ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1) Example 9 Self Tutor x ¡ 3 = ¡1 5 Solve for x: a a x ¡ 3 = ¡1 5 x ¡ 3 + 3 = ¡1 + 3 5 x ) =2 5 x £5 = 2£5 ) 5 ) x = 10 ) 1 5 (x b Check: ¡ 3) = ¡1 fmultiplying both sides by 5g Check: 10 5 ¡ 3 = 2 ¡ 3 = ¡1 X ¡ 3) = ¡1 ¡ 3) £ 5 = ¡1 £ 5 ) x ¡ 3 = ¡5 x ¡ 3 + 3 = ¡5 + 3 ) x = ¡2 ) 1 5 (x fadding 3 to both sidesg 1 5 (x ) 1 5 (¡2 ¡ 3) = 2 Solve for x: x a = 12 4 x+3 = ¡2 5 e b 1 5 b 1 2x f 1 3 (x fmultiplying both sides by 5g fadding 3 to both sidesg £ ¡5 = ¡1 X =6 + 2) = 3 c 5= x ¡2 d x + 4 = ¡2 3 g 2x ¡ 1 =7 3 h 1 2 (5 ¡ x) = ¡2 If the unknown appears more than once, expand any brackets, collect like terms, and then solve the equation. Example 10 Self Tutor 3(2x ¡ 5) ¡ 2(x ¡ 1) = 3 Solve for x: 3(2x ¡ 5) ¡ 2(x ¡ 1) = 3 ) 6x ¡ 15 ¡ 2x + 2 = 3 ) 4x ¡ 13 = 3 ) 4x ¡ 13 + 13 = 3 + 13 ) 4x = 16 ) x=4 fexpanding bracketsg fcollecting like termsg fadding 13 to both sidesg fdividing both sides by 4g cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 Check: 3(2 £ 4 ¡ 5) ¡ 2(4 ¡ 1) = 3 £ 3 ¡ 2 £ 3 = 3 X black Y:\HAESE\SA_10-6ed\SA10-6_01\018SA10-6_01.CDR Wednesday, 26 July 2006 10:02:41 AM PETERDELL SA_10-6 ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1) 3 Solve for x: a 2(x + 8) + 5(x ¡ 1) = 60 c 3(x + 3) ¡ 2(x + 1) = 0 e 3(4x + 1) ¡ 2(3x ¡ 4) = ¡7 b d f 19 2(x ¡ 3) + 3(x + 2) = ¡5 4(2x ¡ 3) + 2(x + 2) = 32 5(x + 2) ¡ 2(3 ¡ 2x) = ¡14 If the unknown appears on both sides of the equation, we ² ² expand any brackets and collect like terms move the unknown to one side of the equation and the remaining terms to the other side simplify and solve the equation. ² Example 11 Self Tutor Solve for x: a 3x ¡ 4 = 2x + 6 b 4 ¡ 3(2 + x) = x 3x ¡ 4 = 2x + 6 ) 3x ¡ 4 ¡ 2x = 2x + 6 ¡ 2x ) x¡4 = 6 ) x¡4+4 = 6+4 ) x = 10 a fsubtracting 2x from both sidesg fadding 4 to both sidesg Check: LHS = 3 £ 10 ¡ 4 = 26, RHS = 2 £ 10 + 6 = 26. X 4 ¡ 3(2 + x) = x ) 4 ¡ 6 ¡ 3x = x ) ¡2 ¡ 3x = x ) ¡2 ¡ 3x + 3x = x + 3x ) ¡2 = 4x 4x ¡2 = ) 4 4 1 ) ¡2 = x b fexpandingg fadding 3x to both sidesg fdividing both sides by 4g i.e., x = ¡ 12 Check: LHS = 4 ¡ 3(2 + ¡ 12 ) = 4 ¡ 3 £ cyan magenta 95 yellow Y:\HAESE\SA_10-6ed\SA10-6_01\019SA10-6_01.CDR Tuesday, 25 July 2006 4:41:29 PM PETERDELL = 4 ¡ 4 12 = ¡ 12 = RHS X 3x ¡ 4 = 5 ¡ x ¡x = 2x + 4 5x ¡ 9 = 1 ¡ 3x 5 ¡ 3(1 ¡ x) = 2 ¡ 3x 3(4x + 2) ¡ x = ¡7 + x 100 50 75 25 0 b d f h j 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 4 Solve for x: a 2x ¡ 3 = 3x + 6 c 4 ¡ 5x = 3x ¡ 8 e 12 ¡ 7x = 3x + 7 g 4 ¡ x ¡ 2(2 ¡ x) = 6 + x i 5 ¡ 2x ¡ (2x + 1) = ¡6 3 2 black SA_10-6 20 ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1) Sometimes when more complicated equations are expanded a linear equation results. Example 12 Self Tutor (x ¡ 3)2 = (4 + x)(2 + x) Solve for x: (x ¡ 3)2 = (4 + x)(2 + x) x2 ¡ 6x + 9 = 8 + 4x + 2x + x2 ) ) fexpanding each sideg x2 ¡ 6x + 9 ¡ x2 = 8 + 4x + 2x + x2 ¡ x2 ) ¡6x + 9 = 8 + 6x ) ¡6x + 9 + 6x = 8 + 6x + 6x ) 9 = 12x + 8 ) 9 ¡ 8 = 12x + 8 ¡ 8 ) 1 = 12x 12x 1 = ) 12 12 ) x= fsubtracting x2 from both sidesg fadding 6x to both sidesg fsubtracting 8 from both sidesg fdividing both sides by 12g 1 12 5 Solve for x: a x(x + 5) = (x ¡ 2)(x ¡ 3) c (x + 1)(x ¡ 2) = (4 ¡ x)2 e (x ¡ 2)(2x ¡ 1) = 2x(x + 3) x(2x + 1) ¡ 2(x ¡ 3) = 2x(x + 1) x2 ¡ 3 = (2 + x)(1 + x) (x + 4)2 = (x + 1)(x ¡ 3) b d f 6 Solve for x: a 2(3x + 1) ¡ 3 = 6x ¡ 1 b 3(4x + 1) = 6(2x + 1) c Comment on your solutions to a and b. D RATIONAL EQUATIONS Rational equations are equations involving fractions. We simplify them by writing all fractions with the same least common denominator (LCD), and then equating the numerators. Consider the following rational equations: x x = 2 3 LCD is 2 £ 3 i.e., 6 3x 5 = 2x 5 LCD is 2x £ 5 i.e., 10x cyan magenta 95 yellow Y:\HAESE\SA_10-6ed\SA10-6_01\020SA10-6_01.CDR Tuesday, 25 July 2006 4:43:02 PM PETERDELL 100 50 75 25 0 5 95 100 50 LCD is 3 £ (2x ¡ 1) i.e., 3(2x ¡ 1) 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 x¡7 x = 3 2x ¡ 1 black SA_10-6 ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1) Example 13 Self Tutor x 3+x = 2 5 Solve for x: 3+x x = 2 5 µ ¶ 2 3+x x 5 £ = £ 2 5 2 5 ) has LCD = 10 Notice the insertion of brackets here. fto create a common denominatorg ) 5x = 2(3 + x) ) 5x = 6 + 2x 5x ¡ 2x = 6 + 2x ¡ 2x ) 3x = 6 ) x=2 ) 21 fequating numeratorsg fexpanding bracketsg ftaking 2x from both sidesg fdividing both sides by 3g EXERCISE 1D 1 Solve for x: a x 4 = 2 7 b 5 x = 8 6 c x x¡2 = 2 3 d 2x ¡ 1 x+1 = 3 4 e 5¡x 2x = 3 2 f 2x ¡ 1 3x + 2 = 5 2 g 4¡x 2x ¡ 1 = 3 6 h 5¡x 4x + 7 = 7 2 i 4x ¡ 1 3x + 1 = 6 ¡2 Example 14 Self Tutor 3 4 = x 4 Solve for x: 4 3 = x 4 3 x 4 4 £ = £ x 4 4 x ) 16 = 3x ) ) x= has LCD = 4x fto create a common denominatorg fequating numeratorsg 16 3 fdividing both sides by 3g cyan magenta yellow 95 100 50 4 3 = 7x 2x 75 h 25 5 1 =¡ 4x 12 0 g 5 7 1 =¡ 3x 6 95 f 100 3 7 = 2x 3 50 e 75 3 7 = 2x 6 25 d 0 4 5 = 3 x 5 c 95 6 3 = x 5 100 b 50 5 2 = x 3 75 a 25 0 5 95 100 50 75 25 0 5 2 Solve for x: black Y:\HAESE\SA_10-6ed\SA10-6_01\021SA10-6_01.CDR Thursday, 3 August 2006 10:04:56 AM PETERDELL SA_10-6 22 ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1) Example 15 Self Tutor Solve for x: 3 2x + 1 = 3¡x 4 2x + 1 = 3¡x ¶ µ 4 2x + 1 = £ 4 3¡x ) ) ) 3 4 has LCD = 4(3 ¡ x) 3 £ 4 µ 3¡x 3¡x ¶ fto create a common denominatorg 4(2x + 1) = 3(3 ¡ x) ) ) Notice the use of brackets here. fequating numeratorsg 8x + 4 = 9 ¡ 3x fexpanding the bracketsg 8x + 4 + 3x = 9 ¡ 3x + 3x ) 11x + 4 = 9 11x + 4 ¡ 4 = 9 ¡ 4 ) 11x = 5 fadding 3x to both sidesg ) x= fsubtracting 4 from both sidesg 5 11 fdividing both sides by 11g 3 Solve for x: a 5 2x + 3 = x+1 3 b 2 x+1 = 1 ¡ 2x 5 c 3 2x ¡ 1 =¡ 4 ¡ 3x 4 d x+3 1 = 2x ¡ 1 3 e 4x + 3 =3 2x ¡ 1 f 3x ¡ 2 = ¡3 x+4 g 6x ¡ 1 =5 3 ¡ 2x h 5x + 1 =4 x+4 i 2+ Example 16 Self Tutor x 1 ¡ 2x ¡ = ¡4 3 6 x 1 ¡ 2x ¡ = ¡4 3 6 µ ¶ 6 x 2 1 ¡ 2x £ ¡ = ¡4 £ 3 2 6 6 fto create a common denominatorg 2x ¡ (1 ¡ 2x) = ¡24 2x ¡ 1 + 2x = ¡24 ) 4x ¡ 1 = ¡24 ) 4x ¡ 1 + 1 = ¡24 + 1 ) 4x = ¡23 fequating numeratorsg fexpandingg ) 95 yellow Y:\HAESE\SA_10-6ed\SA10-6_01\022SA10-6_01.CDR Friday, 18 August 2006 1:28:14 PM PETERDELL 100 50 75 0 95 fdividing both sides by 4g 100 50 75 25 0 95 50 75 25 0 5 95 100 50 75 25 0 5 100 magenta 5 x = ¡ 23 4 ) cyan fadding 1 to both sidesg 25 ) has LCD of 6 5 Solve for x: ) 2x + 5 = ¡3 x¡1 black SA_10-6 ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1) 4 Solve for x: x x a ¡ =4 2 6 b x 2x ¡3 = 4 3 c x x+2 + = ¡1 8 2 d x+2 x¡3 + =1 3 4 e 2x ¡ 1 5x ¡ 6 ¡ = ¡2 3 6 f x x+2 =4¡ 4 3 g x¡4 2x ¡ 7 ¡1= 3 6 h 2x ¡ 3 x+1 x ¡ = 3 6 2 i x 2x ¡ 5 3 ¡ = 5 3 4 j x+1 x¡2 x+4 + = 3 6 12 k x¡1 x ¡ 6 2x ¡ 1 ¡ = 5 10 2 l 3x + 7 2x + 1 1 ¡ 4x ¡ = 4 2 6 E 23 LINEAR INEQUATIONS The speed limit (s km/h) when passing roadworks is often 25 km/h. This can be written as a linear inequation, i.e., s 6 25 (reads ‘s is less than or equal to 25’). We can also represent the allowable speeds on a number line, 25 km/h i.e., 0 s 25 The number line shows that any speed of 25 km/h or less (24, 23, 18, 7 12 etc), is an acceptable speed. We say these are solutions of the inequation. RULES FOR HANDLING INEQUATIONS Notice that similarly 5>3 ¡3 < 2 3 < 5, and 2 > ¡3 and and This suggests that if we interchange the LHS and RHS of an inequation, then we must reverse the inequation sign. Note: > is the reverse of <; > is the reverse of 6. Recall also that: ² cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 50 25 0 5 95 100 50 75 25 0 5 ² 100 ² 75 If we add or subtract the same number to both sides the inequation sign is maintained, for example, if 5 > 3, then 5 + 2 > 3 + 2. If we multiply or divide both sides by a positive number the inequation sign is maintained, for example, if 5 > 3, then 5 £ 2 > 3 £ 2: If we multiply or divide both sides by a negative number the inequation sign is reversed, for example, if 5 > 3, then 5 £ ¡1 < 3 £ ¡1: black Y:\HAESE\SA_10-6ed\SA10-6_01\023SA10-6_01.CDR Tuesday, 5 September 2006 2:25:56 PM PETERDELL SA_10-6 24 ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1) Consequently, we can say that the method of solution of linear inequalities is identical to that of linear equations with the exceptions that: ² ² interchanging the sides reverses the inequation sign multiplying or dividing both sides by a negative number, reverses the inequation sign. GRAPHING SOLUTIONS Suppose our solution to an inequation is x > 4, so every number which is 4 or greater than 4 is a possible value for x. We could represent this on a number line by x 4 The filled-in circle indicates that 4 is included. The arrowhead indicates that all numbers on the number line in this direction are included. Likewise if our solution is x < 5 our representation would be x 5 The hollow circle indicates that 5 is not included. Example 17 Self Tutor Solve for x and graph the solutions: 3x ¡ 4 6 2 3x ¡ 4 + 4 6 2 + 4 ) 3x 6 6 6 3x 6 3 3 ) x62 magenta ftaking 3 from both sidesg Notice the reversal of the inequation sign in b line 4 as we are dividing by ¡2: fdividing both sides by ¡2, so reverse the signg x -2 yellow 95 100 50 75 3 ¡ 2x = 3 ¡ 2 £ 3 = ¡3 and ¡3 < 7 is true. 0 50 ) 75 0 5 95 50 25 0 5 95 100 50 75 25 0 75 Let x = 3 Check: 5 3x ¡ 4 = 3 £ 1 ¡ 4 = ¡1 and ¡1 < 2 is true. x > ¡2 25 ) 100 ) cyan ) 3 ¡ 2x < 7 3 ¡ 2x ¡ 3 < 7 ¡ 3 ) ¡2x < 4 4 ¡2x > ) ¡2 ¡2 b x 2 25 Check: Let x = 1 fdividing both sides by 3g 5 ) b 3 ¡ 2x < 7 fadding 4 to both sidesg 95 ) 100 a a 3x ¡ 4 6 2 black Y:\HAESE\SA_10-6ed\SA10-6_01\024SA10-6_01.CDR Monday, 21 August 2006 10:06:54 AM PETERDELL SA_10-6 ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1) 25 EXERCISE 1E 1 Solve for x and graph the solutions: a 3x + 2 < 0 b d 5 ¡ 2x 6 11 e 5x ¡ 7 > 2 2(3x ¡ 1) < 4 Example 18 2 ¡ 3x > 1 5(1 ¡ 3x) > 8 c f Self Tutor Solve for x and graph the solutions: ¡5 < 9 ¡ 2x ¡5 < 9 ¡ 2x ) ¡5 + 2x < 9 ¡ 2x + 2x ) 2x ¡ 5 < 9 ) 2x ¡ 5 + 5 < 9 + 5 ) 2x < 14 fadding 2x to both sidesg fadding 5 to both sidesg 14 2x < 2 2 ) fdividing both sides by 2g i.e., x < 7 x 7 Check: if x = 5, say ¡5 < 9 ¡ 2 £ 5, i.e., ¡5 < ¡1 is true. 2 Solve for x and graph the solutions: a 7 > 2x ¡ 1 b d ¡3 > 4 ¡ 3x e ¡13 < 3x + 2 3 < 5 ¡ 2x c f Example 19 20 > ¡5x 2 6 5(1 ¡ x) Self Tutor Solve for x and graph the solutions: 3 ¡ 5x > 2x + 7 3 ¡ 5x > 2x + 7 3 ¡ 5x ¡ 2x > 2x + 7 ¡ 2x ) 3 ¡ 7x > 7 ) 3 ¡ 7x ¡ 3 > 7 ¡ 3 ) ¡7x > 4 fsubtract 2x from both sidesg ) fsubtract 3 from both sidesg 4 ¡7x 6 ¡7 ¡7 ) fdivide both sides by ¡7; reverse the signg x 6 ¡ 47 ) x ¡ 47 cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 Check: if x = ¡1, say, 3 ¡ 5 £ ¡1 > 2 £ ¡1 + 7, i.e., 8 > 5 is true. black Y:\HAESE\SA_10-6ed\SA10-6_01\025SA10-6_01.CDR Wednesday, 26 July 2006 10:40:40 AM PETERDELL SA_10-6 26 ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1) 3 Solve for x and graph the solutions: a 3x + 2 > x ¡ 5 c 5 ¡ 2x > x + 4 e 3x ¡ 2 > 2(x ¡ 1) + 5x 2x ¡ 3 < 5x ¡ 7 7 ¡ 3x 6 5 ¡ x 1 ¡ (x ¡ 3) > 2(x + 5) ¡ 1 b d f 4 Solve for x: a 3x + 1 > 3(x + 2) b 5x + 2 < 5(x + 1) d Comment on your solutions to a, b and c. F c 2x ¡ 4 > 2(x ¡ 2) PROBLEM SOLVING Many problems can be translated into algebraic equations. When problems are solved using algebra, we follow these steps: Step Step Step Step Step Step 1: 2: 3: 4: 5: 6: Decide the unknown quantity and allocate a pronumeral. Decide which operations are involved. Translate the problem into an equation and check your translation is correct. Solve the equation by isolating the pronumeral. Check that your solution does satisfy the original problem. Write your answer in sentence form. Remember, there is usually no pronumeral in the original problem. Example 20 Self Tutor When a number is trebled and subtracted from 7 the result is ¡11. Find the number. Let x be the number. ) 3x is the number trebled. ) 7 ¡ 3x is this number subtracted from 7. ) So, 7 ¡ 3x = ¡11 7 ¡ 3x ¡ 7 = ¡11 ¡ 7 ) ¡3x = ¡18 ) x=6 fsubtracting 7 from both sidesg fdividing both sides by ¡3g So, the number is 6: Check: 7 ¡ 3 £ 6 = 7 ¡ 18 = ¡11 X EXERCISE 1F 1 When three times a certain number is subtracted from 15, the result is ¡6. Find the number. cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 2 Five times a certain number, minus 5, is equal to 7 more than three times the number. What is the number? black Y:\HAESE\SA_10-6ed\SA10-6_01\026SA10-6_01.CDR Wednesday, 26 July 2006 10:18:08 AM PETERDELL SA_10-6 ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1) 27 3 Three times the result of subtracting a certain number from 7 gives the same answer as adding eleven to the number. Find the number. 4 I think of a number. If I divide the sum of 6 and the number by 3, the result is 4 more than one quarter of the number. Find the number. 5 The sum of two numbers is 15. When one of these numbers is added to three times the other, the result is 27. What are the numbers? Example 21 Self Tutor What number must be added to both the numerator and the denominator of the fraction 13 to get the fraction 78 ? Let x be the number. 1+x = 3+x µ ¶ 8 1+x £ = 8 3+x ) ) ) 7 8 where the LCD is 8(3 + x) 7 £ 8 µ 3+x 3+x ¶ fto get a common denominatorg 8(1 + x) = 7(3 + x) fequating numeratorsg 8 + 8x = 21 + 7x fexpanding bracketsg ) 8 + 8x ¡ 7x = 21 + 7x ¡ 7x ) ) fsubtracting 7x from both sidesg 8 + x = 21 ) x = 13 So, 13 is added to both. 6 What number must be added to both the numerator and the denominator of the fraction 2 7 5 to get the fraction 8 ? 7 What number must be subtracted from both the numerator and the denominator of the fraction 34 to get the fraction 13 ? Example 22 Self Tutor Sarah’s age is one third her father’s age and in 13 years time her age will be a half of her father’s age. How old is Sarah now? Let Sarah’s present age be x years. ) father’s present age is 3x years. So, 3x + 13 = 2(x + 13) ) 3x + 13 = 2x + 26 ) 3x ¡ 2x = 26 ¡ 13 ) x = 13 Table of ages: cyan magenta yellow 95 100 50 75 25 0 ) Sarah’s present age is 13. 5 95 100 50 75 25 0 5 95 13 years time x + 13 3x + 13 Now x 3x 100 50 75 25 0 5 95 100 50 75 25 0 5 Sarah Father black Y:\HAESE\SA_10-6ed\SA10-6_01\027SA10-6_01.CDR Wednesday, 30 August 2006 11:12:11 AM DAVID3 SA_10-6 28 ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1) 8 Eli is now one-quarter of his father’s age and in 5 years time his age will be one-third the age of his father. How old is Eli now? 9 When Maria was born her mother was 24 years old. At present Maria’s age is 20% of her mother’s age. How old is Maria now? 10 Five years ago Jacob was one-sixth the age of his brother. In three years time his age doubled will match his brother’s age. How old is Jacob now? G MONEY AND INVESTMENT PROBLEMS Problems involving money are frequently made easier to understand by constructing a table and placing the given information into it. Example 23 Self Tutor Brittney has only 2-cent and 5-cent stamps with a total value of $1:78 and there are two more 5-cent stamps than there are 2-cent stamps. How many 2-cent stamps are there? Type 2-cent 5-cent If there are x 2-cent stamps then there are (x + 2) 5-cent stamps Number x x+2 Value 2x cents 5(x + 2) cents ) 2x + 5(x + 2) = 178 fequating values in centsg ) 2x + 5x + 10 = 178 ) 7x + 10 = 178 ) 7x = 168 ) x = 24 So, there are 24, 2-cent stamps. EXERCISE 1G 1 Michaela has 5-cent and 10-cent stamps with a total value of $5:75 . If she has 5 more 10-cent stamps than 5-cent stamps, how many of each stamp does she have? 2 The school tuck-shop has milk in 600 mL and 1 litre cartons. If there are 54 cartons and 40 L of milk in total, how many 600 mL cartons are there? 3 Aaron has a collection of American coins. He has three times as many 10 cent coins as 25 cent coins, and he has some 5 cent coins as well. If he has 88 coins with total value $11:40, how many of each type does he have? cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 4 Tickets at a football match cost $8, $15 or $20 each. The number of $15 tickets sold was double the number of $8 tickets sold. 6000 more $20 tickets were sold than $15 tickets. If the total gate receipts were $783 000, how many of each type of ticket was sold? black Y:\HAESE\SA_10-6ed\SA10-6_01\028SA10-6_01.CDR Wednesday, 26 July 2006 10:24:02 AM PETERDELL SA_10-6 ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1) 29 5 Kelly blends coffee. She mixes brand A costing $6 per kilogram with brand B costing $8 per kilogram. How many kilograms of each brand does she have to mix to make 50 kg of coffee costing her $7:20 per kg? Su Li has 13 kg of almonds costing $5 per kg. How many kg of cashews costing $12 per kg should be added to get a mixture of the two nut types which would cost $7:45 per kg? 6 Example 24 Self Tutor I invest in oil shares which earn me 12% yearly, and in coal mining shares which pay 10% yearly. If I invest $3000 more in oil shares than in coal mining shares and my total yearly earnings amount to $910, how much did I invest in each type of share? Let the amount I invest in coal mining shares be $x. Draw up a table: Type of Shares Coal Oil Amount invested ($) x (x + 3000) Interest 10% 12% Total Earnings ($) 10% of x 12% of (x + 3000) 910 From the table we can write the equation from the information about earnings: 10% of x + 12% of (x + 3000) = 910 ) 0:1x + 0:12(x + 3000) = 910 ) 0:1x + 0:12x + 360 = 910 ) 0:22x + 360 = 910 ) 0:22x = 550 550 ) x= 0:22 ) x = 2500 ) I invested $2500 in coal shares and $5500 in oil shares. 7 Qantas shares pay a yearly return of 9% and Telstra shares pay 11%. John invests $1500 more on Telstra shares than on Qantas shares and his total yearly earnings from the two investments is $1475. How much did he invest in Qantas shares? 8 I invested twice as much money in technology shares as I invested in mining shares. Technology shares earn me 10% yearly and mining shares earn me 9% yearly. My yearly income from these shares is $1450: Find how much I invested in each type of share. cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 9 Wei has three types of shares; A, B and C. A shares pay 8%, B shares pay 6% and C shares pay 11% dividends. Wei has twice as much invested in B shares as A shares and has $50 000 invested altogether. The yearly return from the share dividends is $4850. How much is invested in each type of share? black Y:\HAESE\SA_10-6ed\SA10-6_01\029SA10-6_01.CDR Wednesday, 26 July 2006 10:25:17 AM PETERDELL SA_10-6 30 ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1) 10 Mrs Jones invests $4000 at 10% annual return and $6000 at 12% annual return. How much should she invest at 15% return so that her total annual return is 13% of the total amount she has invested? H MOTION PROBLEMS (EXTENSION) Motion problems are problems concerned with speed, distance travelled and time taken. kilometres : hours Recall that speed is measured in kilometres per hour, i.e., speed = So, distance , time distance = speed £ time and Example 25 time = distance : speed Self Tutor A car travels for 2 hours at a certain speed and then 3 hours more at a speed 10 km/h faster than this. If the entire distance travelled is 455 km, find the car’s speed in the first two hours of travel. Let the speed in the first 2 hours be s km/h. Draw up a table. Speed (km/h) s (s + 10) First section Second section Time (h) 2 3 Total Distance (km) 2s 3(s + 10) 455 Using distance = speed £ time So, 2s + 3(s + 10) = 455 ) 2s + 3s + 30 = 455 ) 5s = 425 and so s = 85 ) the car’s speed in the first two hours was 85 km/h. EXERCISE 1H 1 Joe can run twice as fast as Pete. They start at the same point and run in opposite directions for 40 minutes and the distance between them is then 16 km. How fast does Joe run? 2 A car leaves a country town at 60 km per hour and 2 hours later a second car leaves the town and catches the first car in 5 hours. Find the speed of the second car. 3 A boy cycles from his house to a friend’s house at 20 km/h and home again at 25 km/h. 9 If his round trip takes 10 of an hour, how far is it to his friend’s house? 4 A motor cyclist makes a trip of 500 km. If he had increased his speed by 10 km/h, he could have covered 600 km in the same time. What was his original speed? cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 5 Normally I drive to work at 60 km/h. If I drive at 72 km/h I cut 8 minutes off my time for the trip. What distance do I travel? black Y:\HAESE\SA_10-6ed\SA10-6_01\030SA10-6_01.CDR Wednesday, 26 July 2006 10:25:59 AM PETERDELL SA_10-6 ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1) 31 6 My motor boat normally travels at 24 km/h in still water. One day I travelled 36 km against a constant current in a river and it took me the same time to travel 48 km with the current. How fast was the current? EXTENSION Click on the icon to access a set of problems called Mixture Problems. PRINTABLE PAGES OPENING PROBLEM 0 Revisit the Opening Problem on page 12. Answer the questions posed. REVIEW SET 1A 1 Write in algebraic form: a “3 more than the square of x” “the square of the sum of 3 and x” b 2 Write, in words, the meaning of: p p a+3 b a+3 a 3 If p = 1, q = ¡2 and r = ¡3, find the value of 4 Solve for x: a 5 ¡ 2x = 3x + 4 4q ¡ p : r 1 x ¡ 1 2 ¡ 3x ¡ = 2 7 3 b 5 Solve the following and graph the solutions: 5 ¡ 2x > 3(x + 6) 6 If a number is increased by 5 and then trebled, the result is six more than two thirds of the number. Find the number. cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 7 A drinks stall sells small, medium and large cups of fruit drink for $1:50, $2 and $2:50 respectively. In one morning three times as many medium cups were sold as small cups, and the number of large cups sold was 140 less than the number of medium cups. If the total of the drink sales was $1360, how many of each size cup was sold? black Y:\HAESE\SA_10-6ed\SA10-6_01\031SA10-6_01.CDR Tuesday, 5 September 2006 2:38:21 PM PETERDELL SA_10-6 32 ALGEBRA (NOTATION AND EQUATIONS) (Chapter 1) 8 Ray drives between towns A and B at an average speed of 75 km/h. Mahmoud drives the same distance at 80 km/h and takes 10 minutes less. What is the distance between A and B? REVIEW SET 1B 1 Write, in words, the meaning of: b a+b b a+ a 2 2 2 Write in algebraic form: a “the sum of a and the square root of b” b “the square root of the sum of a and b” 3 If a = ¡1, b = 4 and c = ¡6, find the value of 2b + 3c . 2a 4 Solve the following inequation and graph the solution set: 5x + 2(3 ¡ x) < 8 ¡ x 5 Solve for x: 2(x ¡ 3) + 4 = 5 a 2x + 3 3x + 1 ¡ =2 3 4 b 6 What number must be added to the numerator and denominator of with 13 ? 3 4 in order to finish 7 X shares pay 8% dividend and Y shares pay 9%. Reiko has invested $2000 more on X shares than she has on Y shares. Her total earnings for the year for the two investments is $2710. How much does she invest in X shares? cyan magenta yellow 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 95 100 50 75 25 0 5 8 Carlos cycles for 2 hours at a fast speed, then cycles for 3 more hours at a speed 10 km/h slower than this. If the entire distance travelled is 90 km, find Carlos’ speed in the first two hours of travel. black Y:\HAESE\SA_10-6ed\SA10-6_01\032SA10-6_01.CDR Monday, 21 August 2006 10:11:26 AM PETERDELL SA_10-6