Olympic College Topic 1 – Algebraic Expressions Page | 1 Algebraic Expressions 1. Introduction: Definition: In algebra, we use letters such as x, y and z to represent unknown numbers. For example 5x + 4y an algebraic expression. In this expression we call x the variable and 5 is called the coefficient of x. We can treat mathematical expressions in the same way as numbers provided we obey certain rules of algebra. We can also translate information that is given in English into an equivalent mathematical form using algebraic expressions? If a statement is given in English that uses the four basic arithmetic operations of addition, subtraction, multiplication and division we can translate this information into an equivalent algebraic expression. For example. p multiplied by 3 can be written as the expression 3p 11 added to x can be written as the expression 11 + x 8 subtracted from r can be written as the expression r – 8 m multiplied by 4 can be written as the expression 4m 4 multiplied by m can be written as the expression 4m x divided by 4 can be written as the expression 4 divided by x can be written as the expression When you combine more than one variable you can also express the information as algebraic expressions For example. x added to y can be written as the expression x + y x subtracted from y can be written as the expression y – x a multiplied by b can be written as the expression ab 4 multiplied by m squared can be written as the expression 4m 2 x divided by y can be written as

Olympic College Topic 1 – Algebraic Expressions Page | 2 Exercise 1A:Express each of the following statements as and algebraic expression. 1. p multiplied by added to x subtracted from t a multiplied by c r squared b multiplied by 2 p divided by q z cubed 9.a added to d multiplied by r 11. n squared added to b multiplied by x 15. w divided by e 17. r squared take away d to the power divided by h multiplied by m squared subtracted from q 18. y to the power multiplied by u plus 4 Exercise 1B: 1. Write an algebraic expression for the phrase 2 times the quantity q minus 3. A. 2q – 3B. 2(q – 3) C. 3 – 2q D. 2. Write an algebraic expression for the phrase 4 subtract x add y A. x – 4 + yB. 4 – x – y C. 4 – x+y D. 4 + x + y 3. Write an algebraic expression for the phrase x subtracted from y and the result is divided by 5 A.B.C.D. 4. Write an algebraic expression for the product of w and d plus 3 times g A. wd gB. w(d + 3)gC.w(d + 3g)D. wd + 3g 5. Write an algebraic expression for the phrase square the quantity x and then double it. A. (2x) 2 B. 2x 2 C. 2(2x) 2 D. 2(2x 2 )

Page | 3 Olympic College Topic 1 – Algebraic Expressions 2. Evaluation of an Algebraic Expression. If we are told the value of a variable it is then possible to calculate the value of any expression that contains this variable. The method used is to replace the variable with its given value and to evaluate the expression that is formed by using the usual rules of arithmetic, in particular PEMDAS will be used on some occasions. For example. If x = 4 The value of the expression 5x – 6 is. 5x – 6 = 5(4) – 6 Replace x with 4 = If x = – 4 20 – 6 14 The value of the expression 5x – 6 is 5x – 6 = 5(–4) – 6Replace x with – 4 ==== If a = 5 –20 – 6 –26 The value of the expression 2a 2 is Replace a with 5 PEMDAS states that we square 5 first then multiply by 2 2a 2 = = 2(5 2 ) 2(25) 50 If a = 2 and b = 3 and c = 4 the value of the expression a + 2b – 3c is Replace a with 2, b with 3 and c with 4. a + 2b – 3c = = 2 +2(3) – 3(4) – 4

Olympic College Topic 1 – Algebraic Expressions If a = 2 and b = –2 and c = 5 the value of the expression 5ac – b 2 = = 5(2)( –2) – (5) 2 –20 – 25 – 45 Replace a with 2, b with – 2 and c with 5. Use PEMDAS to perform the calculations in the appropriate order. Evaluate 4x 2 – 4x for x = 5 4x 2 – 4x Replace x with 5 Use PEMDAS to perform the calculations in the appropriate order. ======== 4(5) 2 – 4(5) 4(25) – – Evaluate x 2 – y 2 for x = – 2 and y = – 3 x 2 – y 2 = = (– 2 ) 2 – (– 3) 2 4 – 9 Replace x with – 2 and y with – 3 Using PEMDAS as a guide we square the numbers – 2 = – 5 Find the value of = and – 3 first then we subtract the result for a = 2, b = 3 and c = 4 Replace a with 2, b with 3 and c with 4. Using PEMDAS as a guide we calculate the Numerator =2.4 orand Denominators separately then divide the results Page | 4

Olympic College Topic 1 – Algebraic Expressions Page | 5 Exercise 2A If a = 2, b = 3, and c = 4 evaluate the following:– a 12b b 8c c – 3 9b – a a – 2b 10. 4a b – c–8 Exercise 2B If a = 2, b = 3, and c = 4 evaluate the following:– a 2 abc b2cab2ca ab 9c 4a ac 2c b – 5c10. 6bc ac – bc a + c 17. 2a bc + a 18. 3ab – ac 19. 9bc b + a 20. abc + 3 Exercise 2C If a = 2, b = 3, c = – 4 and d = 0, evaluate the following:– a – 3b 5a – 4c a 2 – b bcd d 2 – c bc – 10ab 4a 2 – 5b abd – d – ac 16. 3ab – 5b 11. abc + c 14. ac – bd 17. 9bc + 3c 12. bcd + a 15. 2a 2 c + b 18. ab + c Exercise 2DEvaluate each expression for the given value of the variables. (a) 2x 2 - 4x for x = 5 (c) x 2 – y 2 for x = – 5, y = 2 (b) 2x 2 - 4x for x = – 5 (d) x 2 – y 2 for x = – 2, y = – 1 (e) for x = 2, y = – 4 (f) 2x 2 – 4y 2 for x = 2, y =

Page | 6 Olympic College Topic 1 – Algebraic Expressions 3. Using and Evaluating a Formula In many situations in science and business we use formulas. these formula are essentially just an algebraic expression where the variables used have very specific meanings. In order to evaluate a formula we essentially just evaluate the algebraic expression that is the formula using the given values of the variables. Example 1: In electronics the formula V = IR will help you find the voltage V when you know the current I and the resistance R. So if you had a circuit with a current of 5 amps and a resistance of 10 ohms then the Voltage would be found by doing the following. VVVVVV ====== IR (5)(10) 50 volts Write down the formula being used Substitute I with 5 and R with 10 Perform the calculation Example 2: The Volume of a Cuboid V with a Length L, a width W and a height H is given by the formual V = LWH. What is the volume of a cuboid with length 10 cm, width 5 cm and height 8 cm? VVVVVV ====== LWH (10)(5)(8) 400 cm 3 Write down the formula being used Substitute L with 5, W with 5 and H with 8 Perform the calculation

Page | 7 Olympic College Topic 1 – Algebraic Expressions Example 3: The first formula of motion states that a= where a is the acceleration, v is the final velocity, u is the initial velocity and t is time. What is the acceleration of an object with an initial velocity of 10 ft/sec, a final velocity of 40 ft/sec that does this movement in 4 seconds? aaaa ==== Write down the formula being used Substitute v with 40, u with 10 and t with 4 a= 7.5 ft/sec 2 Perform the calculation Example 4: The second formula of motion states that s = ut + ½at 2 where s is the distance travelled, u is the initial velocity, a is the acceleration and t is time. What is the displacement (distance travelled) for an object with an initial velocity of 5m/sec travelling for 8 seconds and having an acceleration of 10 m/sec 2. ssssssss ======== ut + ½at 2 (5)(8) + ½(10)(8) ft Write down the formula being used Substitute u with 5, a with 10 and t with 8 Use PEMDAS to perform the calculations in the appropriate order

Olympic College Topic 1 – Algebraic Expressions Example 5: The simple interest I received when you invest an amount of money called the principal P, with an interest rate r (written as a decimal) for t years is I = Prt. What is the simple interest you would receive if you invested $400 for 3 years at a 2% interest rate? IIIIII ====== Prt (400)(0.02)(3) $24 Write down the formula being used Substitute P with 400, r with 0.02 and t with 3 Perform the calculation Notice that in this formula we needed to change the interest rate from 2% into the equivalent decimal 0.02 before we could make the appropriate calculation. Example 6: The simple interest I received when you invest an amount of money called the principal P, with an interest rate r (written as a decimal) for t years is I = Prt. What is the simple interest you would receive if you invested $400 for 18 months at a 3% interest rate? IIIIII ====== Prt (400)(0.03)(1.5) $18 Write down the formula being used Substitute P with 400, r with 0.03 and t with 1.5 Perform the calculation Notice that in this formula we needed to change the interest rate from 3% into the equivalent decimal 0.03 and that 18 months had to be changed into 1.5 years before we could make the appropriate calculation. The units used in much formula are important as they add meaning to the result and in many practical situations you will need to know the appropriate units to use. In the last two examples we needed to be careful to use the correct values in order to get a realistic result. Page | 8

Olympic College Topic 1 – Algebraic Expressions Exercise 3A In electronics the formula for calculating the voltage V across a circuit with a current I and a resistance R is V = IR. (a) What is the voltage across a circuit with a current of 12 amps and a resistance of 5 ohms? (b) What is the voltage across a circuit with a current of 4.5 amps and a resistance of 12.2 ohms? (c) What is the voltage across a circuit with a current of ½ amps and a resistance of 4¼ ohms? The Volume of a Cuboid V with a Length L, a width W and a height H is given by the Formula V = LWH. (a) What is the volume of a cuboid with length 4 cm, width 8 cm and height 2 cm? (b) What is the volume of a cuboid with length 15 cm, width 0.2 cm and height 1.5 cm? (c) What is the volume of a cuboid with length 1½ ft, width 2¼ ft and height 8 ft? The Surface area of a cuboid S with a Length L, a width W and a height H is given by the Formula S = 2(LW + LH + WH) (a) What is the Surface Area of a cuboid with length 6 ft, width 12 ft and height 4 ft? (b) What is the Surface Area of a cuboid with length 0.5 m, width 0.2 m and height 0.3 m? (c) What is the Surface Area of a cuboid with length 1½ ft, width 2 ft and height 3½ ft? Page | 9

Olympic College Topic 1 – Algebraic Expressions 4. The first formula of motion states that a =where a is the acceleration, v is the final velocity, u is the initial velocity and t is time. (a) What is the acceleration of an object with an initial velocity of 4 ft/sec, a final velocity of 8 ft/sec that does this movement in 2 seconds? (b) What is the acceleration of an object with an initial velocity of 10 m/s, a final velocity of 110 m/s that does this movement in 10 seconds? (c) What is the acceleration of an object with an initial velocity of 2.5 ft/sec, a final velocity of 12.5 ft/sec that does this movement in 2.5 seconds? The second formula of motion states that s = ut + ½at 2 where s is the distance travelled, u is the initial velocity, a is the acceleration and t is time. (a) What is the displacement (distance travelled) for an object with an initial velocity of 4m/sec travelling for 6 seconds and having an acceleration of 8 m/sec 2. (b) What is the displacement (distance travelled) for an object with an initial velocity of 12ft/sec travelling for 4 seconds and having an acceleration of 5 ft/sec 2. (c) What is the displacement (distance travelled) for an object with an initial velocity of 0.5 m/sec travelling for 2 seconds and having an acceleration of 1.2 m/sec 2. The simple interest I received when you invest an amount of money called the principal P, with an interest rate r, written as a decimal for t years is I = Prt. (a) What is the simple interest you would receive if you invested $600 for 2 years at a 6% interest rate? (b) What is the simple interest you would receive if you invested $200 for 6 mont6hs at a 3% interest rate? (c) What is the simple interest you would receive if you invested $600 for 30 months at a 1.5% interest rate? Page | 10

Page | 11 Olympic College Topic 1 – Algebraic Expressions 4. Simplifying an Expression By “simplifying” an algebraic expression, we mean writing it in the most compact or efficient manner, without changing the value of the expression. This mainly involves the process of collecting like terms and then adding and subtracting them What we mean when we say “like terms” is that we have terms which contain the same powers of same variables. They may have different coefficients, but that is the only difference. You probably know that if you have an expression like 4 + 3x you cannot add those terms to simplify it in any way. That’s because one term in a constant (the 4) and the other term has a variable (the x). We say that these are not like terms and cannot be combined. On the other hand it is possible to simplify the expression 4x + 5x to get 9x as we do have like terms and can therefore combine them in the usual way. For Example: 3x, x, and –2x are like terms. 2x 2, –5x 2, and are like terms. xy 2, 3y 2 x, and 3xy 2 are like terms. 2x + 3y are NOT like terms, because they have two different variables x and y xy 2 and x 2 y are NOT like terms, because the same variable is not raised to the same power. Exercise 4A. 1.Which of the following terms are like terms and which are not. (a) 6x, 3x, and –2 (c) xy 2, y 2 x, and 6xy 2 (e) x 2 y 2 and 2x 2 y 2 (g) y 2, y and y 3 (i) 2x, 4x and x (b) x 2, –x 2, and (d) 2x and 2y (f) 3x and– 7 (h) 3x, 2x and 5x (j) 3x and – 7x

Page | 12 Olympic College Topic 1 – Algebraic Expressions There is no standard procedure for simplifying all algebraic expressions because there are so many different kinds of expressions, but they can be grouped into three types:  Those that cannot be simplified at all.  Those that can be simplified by rearranging like terms and then simplifying.  Those that require preparation before being we can rearrange like terms. In some circumstances none of the terms are like any of the others, in theses circumstances we cannot simplify the algebraic expression. Example 1: The expression 4x – 5 y + 2z + 5 cannot be simplified as none of the terms are like any of the others. In other situations we can simplify the algebraic expression by Collecting like terms together by rearranging the terms one variable at a time. Example 2: Solution: Simplify the algebraic expression 3x + 4y + 2x – 5y Look at the expression 3x + 4y + 2x – 5y There are four terms 3x, 4y, 2x and – 2y. Two of the terms involve the variable x while the other two terms involve the variable y. We can re-order the terms in the expression so that the x terms are together and the y terms are together 3x + 4y + 2x – 5y. Now we can combine the x terms and combine the y terms to get: 3x + 4y + 2x – 5y = 3x + 2x + 4y – 5y = 5x – y Example 3: Simplify the algebraic expression 2x + 3y x + 6y + 7 Solution:This expression can be simplified by identifying like terms and then grouping and combining like terms, so we have three groups of like terms + 2x and +3x are like terms, and can be combined to give +5x, + 3y and +6y are like terms, and can be combined to give + 9y, and – 2 and +7 can be combined to give +5. So after simplifying, this expression becomes: 2x + 3y – 2 + 3x + 6y + 7 = 2x + 3x + 3y+ 6y – = 5x + 9y + 5

x  x  x Page | 13 Olympic College Topic 1 – Algebraic Expressions Exercise 4B. 1.Combine like terms where possible and simplify the result. (a) (b) (c) (d) 3x – 3x + 7x – x 2 + 3x – 5x + 1 9x – 5x + 5 – 6 2x + 5 – 5 – 2x (e) (f) 2x – 6y – c 2.Simplify the following algebraic expressions when possible. (a) (c) (e) (g) (j) (l) (n) 2a + 3a + 5a 8t d 7n + 5m – 8n + 3m 4k + 5a – 6a + 2k 32g – 30h + 7h + 2g 10a – 20b – 3c – 5a – 4c 0.7u – 9u – 0.5u + 1.6n + 7n (b) 4p + 5q + 6p + 3q (d) 9w + 2x + 7w + 3x (f) 3d + 8d + 6d (h) d + j + 25d + 2j (k) 3p + 6q + 5r + 10q + 2r (m) 5z + 2y + x + 8x + 3y + z (0) 1.5m + 2.4n + 3.3p – 4.4m+ 5.5n 3.Simplify the following algebraic expressions when possible. (a) (c) (e) (g) (i) 4p + 5q – 2p + 3q 8t + 6s – 2t + 4s 13b + 8a – 3a + 4b 4k + 5a + 6a + 2k 32g + 30h – 7h – 2g (b) 7r + 4p + 3p – r (d) 5I – 4I + 6j – 2j (f) 3d + 8d – 6d (h) 25d + j – 5d + 2j (j) 3p + 6q + 5r – 3q – 2r

Olympic College Topic 1 – Algebraic Expressions Page | 14 4.Simplify the following algebraic expressions when possible. (a) (c) (e) (g) (i) 10a + 20b + 3c – 7a + 4c – 6b 9y + 7c – 5y – 2c 23k + 8h – 13k – 4h 12h + 8y – 9h + 6y – 2h 20p + 6q – 8p – 4q +2p (b) 21w + 7f – 14w + f (d) 7u + 9u – 5u + 7n – 6n (f) 8a + 7b – 2a – 3b + 4c (h) 14t + 19u – 10t – 12u (j) 31r – 15r + 16e + 2e – 12e 5.Simplify the following algebraic expressions when possible. (a) 6p + 2r + 3p – 5r (c) 7m – 3e (e) 7ad + 8ty – 6ad – 9ty (g) 4x 3 + 2x 4 + 7x 3 – x 4 (i) 7y y 2 – 7y 2 – y 4 (k) 2x 2 + 3x 2 (m) 8x 4 – 3x 4 (o) 3j 5 +3j 5 (b) 7e + 3e – 6f + 2f (d) 5we + 6we – 5u + 2u (f) 4a 2 + 2a 3 + 5a 2 + 7a 3 (h) 10x 4 + 3x 2 – 7x 2 – 2x 4 (j) 8as – 2as + as (l) 5x x 3 (n) 7x 8 + 6x 8 – x 8 (p) 4r 8 +5r 8 – 6r 8

Page | 15 Olympic College Topic 1 – Algebraic Expressions 4. Using the Distributive Law to Simplify an Algebraic Expression The distributive law states that for all variables a,b and c a(b + c) = ab + ac The distributive law allows us a means of multiplying out parenthesis and to simplify the results. Example 1: Use the distributive law to remove the parenthesis form the expression 5(x + 3). Solution:Use the distributive Law5(x + 3) = = 5(x) + 5(3) 15x + 15 Example 2: Use the distributive law to remove the parenthesis form the expression 3(2x – 4y). Solution: 3(2x – 4y) = 3(2x) + 3(– 4y) Use the distributive Law = 6x – 12y Example 3: Use the distributive law to remove the parenthesis form the expression – 5(4x – 3). Solution:Use the distributive Law – 5(4x – 3) = = (– 5)(4x) + (– 5)(– 3) – 20x + 15 Example 4: Use the distributive law to remove the parenthesis form the expression – 2( 4x + 3). Solution:Use the distributive Law – 2( 4x + 3). = = (– 2)(4x) + (– 2)(+ 3) – 8x – 6 Example 5: Use the distributive law to remove the parenthesis form the expression – (2x – 4). Solution: – (2x – 4) ====== –1 (2x – 4) (– 1)(2x) + (– 1)(– 4) – 2x + 4

Olympic College Topic 1 – Algebraic Expressions Exercise 5A. 1.Which equation below shows a correct use of the distributive property? A. – 2(2x – 5) B. – 2(2x – 5) C. – 2(2x – 5) D. – 2(2x – 5) ======== – 4x – 10 – 4x + 5 – 4x + 10 – 4x – 5 2.Use the Distributive law to remove the parenthesis on the following expressions (a) 2(y + 5) (d) – 6(2g + 7) (g) 4(2y + 8) (j) –5(x + 6) (m) –3(3y – 1) (p) 4(2x + 4y) (b) 4(2x – 3) (e) 3(x + 5) (h) 6(3g – 7) (k) – 4(y –2) (n) – (x + 5) (q) 3(2c + 7d) (c) 5(2y + 8) (f) 4(x – 7) (i) 2(3g – 7k) (l) – 3(2x + 7y) (o) – (–x – 8) (r) – 4(–2x + 6) Page | 16

Olympic College Topic 1 – Algebraic Expressions When you simplify an algebraic expression you may also have to use the distributive law to rearrange the expression first and then you can simplify the expression. In some cases, you may have to perform other simplification before you can combine like terms. Example 6: Simplify 3x + 2(x – 4) Solution: 3x + 2(x – 4) = 3x + 2(x) + (2)( – 4) Use the distributive Law ==== 3x + 2x – 8 5x – 8 However, if there is only a plus sign in front of the parentheses, then in these situations you must recognize that this means that there is a + 1 in front of the parenthesis. However in these circumstances you can simply remove the parentheses and get the required result. Example 7: Simplify 4x + (5 – x) Solution: 4x + (5 – x) =4x + 1(5 – x) Use the distributive Law Rearrange like terms ======== 4x + 1(5) + 1(– x) 4x + 5 – x 4x – x + 5 3x + 5 This expression can also be simplified by just removing the parethesis as shown below. Solution:Remove the parenthesis Rearrange like terms 4x + (5 – x) = = 4x + 5 – x 4x – x + 5 3x + 5 A special case of the distributive law is when a minus sign appears in front of parentheses such as – (x – 6). At first glance, it looks as though there is no factor multiplying the parentheses, and you may be tempted to just remove the parentheses to obtain the wrong result – (x – 6) = – x – 6 What you need to remember is in these situations is that the minus sign in front of the parenthesis should always be thought of as the number – 1 and so – (x – 6) is equivalent to –1 (x – 6). Page | 17

Page | 18 Olympic College Topic 1 – Algebraic Expressions The following examples show us how to tackle similar problems. Example 8: Simplify 4x – (5 – x) Solution: 4x – (5 – x) =4x + (–1)(6 – x) Use the distributive Law Rearrange like terms ======== 4x + (–1)(6) + (–1)(–x) 3x – 6 + x 3x + x –6 4x – 6 Another common situation is to simplify an algebraic expression that contains the subtraction of two expressions such as (3x + 2) – (4x – 1). In these cases you can think of the subtraction as just multiplying the second parenthesis by – 1 so (3x + 2) – (4x – 1) = (3x + 2) – 1(4x – 1). Example 9: Simplify (7x + 6) – (2x – 5) Solution: (7x + 6) – (2x – 5) ========== 7x + 6 – 1(2x – 5) 7x (– 1)(2x) + (–1)(– 5) Use the distributive Law 7x + 6 – 2x + 5 7x – 2x Rearrange like terms 5x + 11 Example 10: Subtract (5x + 4) from (8x + 2) Solution: (8x + 2) – (5x + 4) ========== 8x + 2 – 1(5x + 4) 8x (– 1)(5x) + (–1)(+ 4) Use the distributive Law 8x + 2 – 5x – 4 8x – 5x + 2 – 4 Rearrange like terms 3x – 2

Olympic College Topic 1 – Algebraic Expressions Example 11: Simplify 4(2x – 3) – 4(x + 5) Solution: 4(2x – 3) – 4(x + 5) ======== (4)(2x) +(4)(– 3) – 4(x) + (– 4)(5) Use the distributive Law 8x – 12 – 4x – 20 8x – 4x – 12 – 20 Rearrange like terms 4x – 32 Example 12: Simplify the algebraic expression 3b – 2(4b – 6b + 2) + 7 Solution:The brackets in this expression can be removed first, and then the expression may be simplified. 3b – 2(4b – 6b + 2) + 7 Use the distributive Law ==== 3b – 8b + 12b – b + 3 Example 13: Simplify the algebraic expression 3 – (2x – 6 ) + x Solution:The brackets in this expression can be removed first, and then the expression may be simplified. 3 – (2x – 6 ) + x = = 3 – 2x x 9 – x Notice we can more efficiently process the subtraction by also changing the sign of the terms inside the parenthesis – (2x – 6 ) = – 2x + 6 Page | 19

Olympic College Topic 1 – Algebraic Expressions Exercise 5B. 1. Simplify the expression 4  x  3   5 x. A. 17x B. 21x C. 9x  3 D. 9x  12 2.Use the distributive property to remove parentheses and simplify the expression: (a) 2(3x – 7) + 6 (d) 10(3x – 7) + x (g) 2(3x – y) + 6(x + 3y) (j) 2 + 6(x – 7) (m) (3x – 4) (b) 4(3x – 7) + 4 (e) 2(3x – 7) + 2(x – 7) (h) 5(2x – 4) + 2(5x + 3) (k) 3x + 2(2x – 1) (n) 5(3x – 2y) + 4(2x – 3y) (c) 5(3x – 7) – 6 (f) 4(3x – 7) + 5(2x – 9) (i) 2(3x – 7) + 3(3x – 1) (l) 6x + 6(4 – x) (o) 4(3x – 7) + 5(2x – 1) 3.(a) 3.(c) 3.(e) Subtract (2x + 3) from (5x + 1) Subtract (6x – 7) from (x + 7) Simplify (5x – 1) – (2x + 7) (b) Subtract (5x + 1) from (2x + 3) (d) Subtract (4x – 4) from (4x + 2) (f) Simplify (2x – 5y) – (5x + 3y) 4.Use the distributive property to remove parentheses and simplify the expression: (a) 3(x + y) + 2y (d) 3 – 3(x + 6) (g) 2 – 5(2x – 2) (j) – (x + 2) – 3(x – 2) (b) (3y – 2) – 4(y – 3) (e) 2(3y – 2) + 5 (h) 2 – 3(3 + 4x) + x (k) –3x + 2(3 + 4x) + 1 (c) 5 – 1(2x + 7) (f) 5 – 4(2x + 6) (i) 6(2x + 3) – 2(2x – 7) (l) 6(2x + 3) – 2(2x – 7) Page | 20

Page | 21 Olympic College Topic 1 – Algebraic Expressions Solutions Exercise 1A: 1.10p2.5+x3.t-34.2b5.ac 6.7. r2r2 8. Z3Z3 9.a+d10. 12r 11. n 2 + b – q r 2 – x 18. y m d u + 4 Exercise 1B:1.A2.C3.A4.D5.B Exercise 2A Exercise 2B or Exercise 2C 1. – – – – 5 8. – – – – 815. – – – – 6 2 Exercise 2D(a) 30(b) 70(c) 21(d) 3(e) (f) – 56 Exercise 3Avolts m 2 m 1.(a) 60 volts 2.(a) 64 cm 3 3.(a) 288 ft 2 4.(a) 2 ft/s 2 5.(a) 168 m 6.(a) $72 1.(b) 54.9 volts 2.(b) 4.5 volts 3.(b) 0.62 m 2 4.(b) 10 ft/s 2 5.(b) 88 ft 6.(b) $3 1.(c) or 3 2.(c) 27 cm 3.(c) 30.5 or 4.(c) 4 ft/s 2 5.(c) 3.4 or 6.(c) $22.50

Olympic College Topic 1 – Algebraic Expressions Page | 22 Like Terms (b),(c),(e),(h) and (i) Not Like Terms (a),(d),(f),(g) and (j)Exercise 4A. Exercise 4B. 1.(a) 6x 1.(b) 3 – 2x1.(c) 4x – 1 1.(d) 0(e) x (f) 2x – 6y – c 2.(a) 10a 2.(e) – n + 8m 2.(j) 34g – 23h 2.(n) –8.8u – 2n 3.(a) 2p + 8q 3.(e) 17b + 5a 3.(i) 30g + 23h 4.(a) 3a + 14b + 7c 4.(e) 10k + 4h 4.(i) 18p + 2q 5.(a) 9p – 3r 5.(e) ad – ty 5.(i) 6y 4 + 3y 2 5.(m) 5x 4 2.(b) 10p + 8q 2.(f) 17d 2.(k) 3p + 16q + 7r 2.(o) –2. 9m + 7.9n + 3.3p 3.(b) 6r + 7p 3.(f) 5d 3.(j) 3p + 3q + 3r 4.(b) 7w + 8f 4.(f) 6a + 4b + 4c 4.(j) 16r + 6e 5.(b) 10e – 4f 5.(f) 9a 2 + 9a 3 5.(j) 7as 5.(n) 12x 8 2.(c) 8t d 2.(g) 6k – a 2.(l) 5a – 20b – 7c 3.(c) 10t + 10s 3.(g) 6k + 11a 4.(c) 4y + 5c 4.(g) h + 14y 5.(c) 7m – 3e 5.(g) 11x 3 + x 4 5.(k) 5x 2 5.(o) 6j 5 2.(d) 16w + 5x 2.(h) 26d + 3j 2.(m) 6z + 5y + 9x 3.(d) I + 4j 3.(h) 20d +3 j 4.(d) 11u + n 4.(h) 24t + 7u 5.(d) 11we + 7u 5.(h) 8x 4 – 4x 2 5.(l) 15x 3 5.(p) 3r 8 Exercise 5A. 1. C 2.(a) 2y (e) 3x (i) 6g – 14k 2.(b) 8x – 12 2.(f) 4x – 28 2.(j) –5x – 30 2.(c) 2.(g) 2.(k) 10y y + 32 – 4y (d) 2.(h) 2.(l) – 12g – 42 18g – 42 – 6x – 21y 2.(m) –9y + 32.(n) – x – 5 2.(o)x+82.(p)8x + 16y 2.(q)6c + 21d 2.(r) 8x – 24 Exercise 5B. 1. C 2.(a) 6x – 82.(b) 12x – 242.(c) 15x – 412.(d) 31x – 702.(e) 8x – 28 2.(f) 22x – 732.(g) 14x + 16y 2.(h) 20x – 142.(i) 15x – 172.(j) 6x – 40 2.(k) 5x – 2 2.(l) 242.(m) 9x 2. (n) 23x – 22y2.(o) 22x – 33 3.(a) 3x – 23.(b) – 3x + 23.(c) – 5x (d) 6 3.(e) 3x – 8 3.(f) – 3x – 8y 4.(a) 3x + 5y 4.(e) 6y (i) 8x (b) – y (f) – 8x – 19 4.(j) – 4x (c) – 2x – 2 4.(g) – 10x (k) 5x (d) – 3x – 15 4.(h) – 11x – 7 4.(l) 8x + 32