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ALGEBRA ALGEBRA(Topics) • Algebra Basics- Addition, Subtraction and Substitution. • Algebra Basics 2- Multiplication, Expansion of Brackets and Division • Indices • Equations and Inequalities • Simultaneous Equations • Simultaneous Equations in solving equations ALGEBRA(Topics) • Functions and Rearranging. • Quadratic equations ALGEBRA(Key Terms) • Substitution-replacing occurrences of some symbol or variable by a given value. • Evaluate-to determine or calculate the numerical value of • • • • • Variable-a quantity or function that may assume any given value or set of values Constant- A quantity assumed to have a fixed value in a specified mathematical context Simplify-to reduce (an equation, fraction, etc) to a simpler form by cancellation of common factors, regrouping of terms in the same variable. Coefficient-a number or quantity placed (generally) before and multiplying another quantity, as 3 in the expression 3x. => means ‘this gives’ nb: means ‘note well’ • Multiplication operator can be represented by a ‘*’ or a ‘ • So…… 2x3=6, 2*2=4, 5 4=20 • . . ’ or a ‘x ‘ sign. ALGEBRA(Basics) • The four basic operations for addition, subtraction, multiplication and division are carried out on variables much like on numbers. • One practical way to look at it is if I have a variable ‘a’ and I am adding ‘2a’ to it. Imagine a=‘apples'. Ask yourself if I have one apple and someone gives me two how many do I have. Questions in the exam normally come in the form:a+2a This gives 3a. ALGEBRA(Basics) • The same is done with subtraction • If I have 5 mangoes and my grandmother takes 3 mangoes how many do I have. • Let ‘m’ be equal to mangoes(step 1) • Represent it in a mathematical statement(step 2) 5m-3m= Then Evaluate the statement 5m-3m=2m.(step3) ALGEBRA(Basics) • Sometimes Variables have negative coefficients. • For example, You were playing cricket and you lost the ball by hitting a massive ‘6’ you now owe one of your friends a ball. Hence you have -balls(or negative 1 balls). • The next day your mother buys a three pack of balls but you still owe your friend, this situation can be represented in a mathematical statement as -b+3b This means you actually own only 2 of those balls, sine you owe you friend. Therefore –b+3b=2b. ALGEBRA(Basics) • This statement can be rewritten as 3b+-b which is equal to 3b-b which is equal to 2b. ALGEBRA(Basics) • When writing mathematical statements or expressions variables of different types cannot be combined or simplified further. • If you have three balls and two apples you can represent it as 3b+2a where b=‘balls’ and a=‘apples’ ALGEBRA(Basics) • You may be given an expression like eg.1 -b+3a+2b-4a And asked to simplify the expression. Bring similar(or like) variables together, that is ‘a’s with ‘a’s and ‘b’s with ‘b’s. Take whatever sign is in front of them and rewrite the statement. Do one variable first then the next.(step 1) -b+2b+3a-4a ALGEBRA(Basics) -b+2b+3a-4a Again negative one ‘b’ plus two b , is one b as if subtracting. Three ‘a’ minus four ‘a’ give negative one ‘a’. So we can simplify the expression as b+-a which is equal to b-a ALGEBRA(Basics) • eg.2 3p-4q+6p+2q Collect like terms with the same sign and rewrite =3p+6p-4q+2q simplify(remember-4p+2q =2q+-4q =2q-4q =-2q) =9p-2q ALGEBRA(Basics) • eg.3 5c-2d+7d-3c Collect terms and rewrite with the same signs in front of them =5c-3c-2d+7d Simplify(remember -2d+7d =7d+-2d =7d-2d =5d) =2c+5d ALGEBRA(Basics) • Sometimes we are given values for our variables we can Substitute the to get actual values for our expressions. • eg.1 Given x=3, y=2 Evalutate x+y • Replace x with the value given for x and y with the value given for y and rewrite(Step 1) • NB: it is best to use brackets. ALGEBRA(Basics) • • • • Evaluate (Step 2) x+y = (3)+(2) =5 We will show why brackets are important further down in this Topic. eg.2 Given a=2 and b=1 Evaluate 2a+2b Replace(Substitute) ‘a’ with 2 and ‘b’ with 1 in the expression.(Step 1) Expand Brackets and Evaluate(Step 2) =>2(2)+2(1) =4+2 =6 ALGEBRA(Basics) eg.3 Given c=-1 and d=2 Evaluate c+3d =(-1)+3(2) => -1+6 (NB:6+-1 = 6-1) =5 eg.4 Given e=2 and f=-3 evaluate e-2f Rewrite expression as it is just substituting the variables =>(2)-2(-3) =>2-(-6) =8 Expand Brackets, then evaluate We get 8 because if I take away some thing negative like debt it is as if I have added to what you have. Imagine you owe your father $4 that is -$4 you have. If your mother convinces him to clear the debt she has taken away the -$4. -$4-(-$4)=0 Like adding -$4+$4=0 ALGEBRA(Basics) • Algebra Basics Ends Here go to Algebra Exercise section and do Basic revision exercise. ALGEBRA(Basics 2) • There is a general sequence when working with Algebraic expressions involving which operations to conduct first. B.O.D.M.A.S • B.O.D.M.A.S – Brackets (operated on before),Division(then),Multiplication(then), Addition(then),Subtraction. ALGEBRA(Basics 2) eg.1 Simplify 3(x+2)= Expand brackets first by multiplying everything inside the bracket by the coefficient outside the bracket 3(x)+3(2)= 3x+6 ALGEBRA(Basics 2) eg.2 Simplify 3y-2(y+4)= According to BODMAS let us expand brackets first. That is take the coefficient in front the bracket and multiply it by everything inside the bracket. The coefficient is -2 so by expanding the bracket we get 3y+-2(y)+-2(4) {nb: rewrite sign the bracket} =>3y+-2y+-8 =>3y-2y-8 =>y-8 ALGEBRA(Basics 2) eg.3 Simplify 3x-2(x+5) Remember B.O.D.M.A.S! =>3x+-2(x)+-2(5) {nb: rewrite sign in front of the bracket} =>3x-2x-10 =>x-10 ALGEBRA(Basics 2) eg.4 Simplify 5(a-b)+2b =>5(a)-5(b)+2b {nb: rewrite sign the bracket} =>5a-5b+2b =>5a-3b ALGEBRA(Basics 2) eg.4 Simplify 3(a-b)-a =>3(a)-3(b)-a =>3a-3b-a By rearranging and collecting terms =>3a-a-3b =>2a-3b{Remember when collecting terms keep the sign in front of them} ALGEBRA(Basics 2) Variables are multiplied similarly to numbers a*b=ab. {Note * means multiply} Eg.5Expand the brackets a(b-c)= a(b)-a(c)= ab-ac Any variable by itself is equal to the variable squared eg b*b=b2 Eg.6 2b(a-b)= 2ab-2b2 ALGEBRA(Basics 2) When dividing we cancel similar terms and divide the numbers (coefficients) Eg.7 2𝑎 2𝑎 = 𝑎 𝑎 Eg.8 Eg.9 6𝑏 2𝑏 = 4𝑎𝑏 2𝑎 6𝑏 2𝑏 = =2 6 2 = =3 4𝑏 = 2 2b ALGEBRA(Basics 2)(Key Terms) Numerator- the number at the top of a fraction and so is being divided. Denominator- the number at the bottom of the fraction which is dividing the numerator. LCM-Lowest Common Multiple. ALGEBRA(Basics 2) Remember with numbers when adding two fractions with different denominators we found the LCM. For example 3 4 2 5 3 4 2 + The LCM is found by multiplying the denominators (4*5=20) this is the denominator of the resulting fraction. + 5 We want to convert both original fractions to equal values with the denominator of the LCM. 20 Step 1. divide the LCM by the denominator of the first fraction the multiply the quotient by the numerator of that fraction. Step 2.Repeat for the second fraction. ALGEBRA(Basics 2) This gives 15 and 8. 15+8 23 = 20 20 Same process would be done with a ‘-’ sign. This section was simply revision. Algebraic fractions undergo the same operations….. ALGEBRA(Basics 2) Ex.10 Simply 2𝑎 𝑏 + 3𝑐 𝑑 Step 1. Find LCM by multiplying denominators, b*d=bd Step 2. Divide denominators of both fractions and multiply the quotient by their numerators. Step 3. Place the result atop the LCM and simplify if possible. Remember the sign in the original question. ALGEBRA(Basics 2) Combining steps 1-3.Simply 2𝑎 𝑏 + 2𝑎𝑑 + 3𝑐𝑏 𝑏𝑑 This cannot be simplified any further. 3𝑐 𝑑 = ALGEBRA(Basics 2) Ex.11 Simplify 2𝑗 𝑥 2𝑘 + 𝑦 LCM=xy, After steps 2 and 3 we get…..try it yourself then go to the next page. ALGEBRA(Basics 2) = Ex.12 3𝑐 𝑑 − 2𝑗𝑦+2𝑘𝑥 𝑥𝑦 2𝑎 𝑏 LCM=?, Try it yourself before going to the next page. ALGEBRA(Basics 2) • Algebra Basics 2 Ends Here go to Algebra Exercise section and do Algebra Exercise 1 ALGEBRA(Basics 2) 3𝑐𝑏−2𝑎𝑑 = 𝑑𝑏 Remember the sign! ALGEBRA(Indices) Key Terms Index- An index is the power to which a number is ‘expressed’ that the number of times it is multiplied by itself. Eg. x2=x*x x3=x*x*x xn=x*x*x.…n times….x*x*x ALGEBRA(Indices) Laws of Indices. 1.am*an= am+n we add indices when multiplying similar terms. For example a2 * a3 = a5 Because….. since a2=a*a and a3=a*a*a then a2 * a3 =>(a*a)*(a*a*a) =>a*a*a*a*a =>a5 ALGEBRA(Indices) 𝑎𝑚 m-n 𝑛 =a 𝑎 2.am/an=am-n or we subtract indices when dividing similar terms. For example a6/ a4 =a6-4 =a2 Because…. a6 = a*a*a*a*a*a a4 =a*a*a*a a∗a∗a∗a∗a∗a a∗a 2 6 4 Therefore a / a = = =a a∗a∗a∗a 1 ALGEBRA(Indices) 1 -m 3.a = 𝑎 𝑚 For example 1 -2 4 = 4 2 = 1 16 ALGEBRA(Indices) 4.(am)n=amn For example (a2)3 = a6 Because… If a2=a*a Then =>(a 2)3 =>(a*a)*(a*a)*(a*a) =>a*a*a*a*a*a =>a6 ALGEBRA(Indices) Laws of Indices. Continued 5.a0=1 For example 30=1. Note any number or variable raised to the power zero is equal to one. Ex.1 x2 * x3 = x2+3 = x5 Ex.2 a6 /a3 = a6-3 =a3 Ex.3 2b-2 = 1 2* 2 𝑏 = 2 𝑏2 Ex.4 (2d2)3 = 23*1 * d2*3 = 23 * d6 =8d6 remember deal with brackets before anything else and a number with no power is to the power 1. ALGEBRA(Indices) Ex.5 This example combines two laws. 𝑏4∗𝑏6 Simplify 2 + 6 4 = 𝑏 𝑏2 = 𝑏 𝑏10 𝑏2 =b10-2 =b8 ALGEBRA(Indices) Ex.6 If x= 23 * 32 Evaluate x2. Step 1.Substitute 23 * 32 for x in the terms x2. (23 * 32 )2. In other words if x= 23 * 32 , Then x2=( 23 * 32 )2 Step 2. Evaluate by expanding brackets. =>23*2 * 32*2 Step 3. Simplify =>26 * 34 ALGEBRA(Indices) Ex.6 If x= 23 * 32 Evaluate x2. Step 1.Substitute 23 * 32 for x in the terms x2. (23 * 32 )2. In other words if x= 23 * 32 , Then x2=( 23 * 32 )2 Step 2. Evaluate by expanding brackets. =>23*2 * 32*2 Step 3. Simplify =>26 * 34 ALGEBRA(Indices) Ex.7 Simplify 3x2(x3+2xy2) Step 1. Always deal with brackets first remember BODMAS. Expand brackets. 3x2 *x3+ 3x2 * 2xy2 Step 2.Then simplify. 3x5 + 6x3y2 ALGEBRA(Indices) Simplify 𝑝2 𝑞 3 𝑝𝑞2 2 Step 1. Expand brackets 𝑝4𝑞6 𝑝𝑞2 Step 2. divide similar index terms(subtract)* =>p3q4 ALGEBRA(Indices) Algebra Indices Ends Here go to Algebra Exercise section and do Algebra Exercise 2 ALGEBRA(Equations and Inequalities) An equation is a mathematical statement which consists of an equal ‘=‘ sign and says what is on the left hand side of the equation is equivalent to what is on the right hand side of the equal sign. For example 4+5=9 or y=x2+2x+3 etc. ALGEBRA(Equations and Inequalities)(Key Terms) DifferenceSum- ALGEBRA(Equations and Inequalities) Ex.1 If x=-1, and y=x2-3x+9 find y. Simply substitute the value given for ‘x’ into the equation to find y, writing in the value given for ‘x’ instead of ‘x’ In this case x is given as -1. So.. y=(-1)2-3(-1)+9 NB: properties of negative numbers -any negative number by another negative number results in a positive number so (-1)2 =-1 *-1=+1. -subtracting a negative number is like adding, so 3(-1)=3*-1 =-3 And -(-3)=+3 y=1+3+9 y=13 ALGEBRA(Equations and Inequalities) Equations represents relationships which cannot be further simplified. Ex.2 The value of y is the difference between six times ‘a’ and 3 times ‘b’ squared. Represent this an equation. y=6a-3b2 ALGEBRA(Equations and Inequalities) Equations represents relationships which cannot be further simplified. Ex.3 The value of y is the sum of four times ‘a’ and two times ‘b’ squared. Represent this an equation. y=4a+2b2 ALGEBRA(Equations and Inequalities) Equations can used to find unknowns in situations where other information is given. Ex.4 Paul divides an undisclosed amount ($n) of money among his three daughters Sarah, Melanie and Paula in a ratio of 2:3:4.The greatest share is $20 what is the total value Paul gave out($n). Step.1 Find the value of one share the since greatest portion is 4 shares=$20, Let ‘s’ represent I share. $20 4 then 1 s = = $5. Because 4s=$20, by dividing both sides of the equation statement by 4 we get $5 as the value of one share. ALGEBRA(Equations and Inequalities) Ex.4 continued Step.2 So if one share is equal to $5 And there where 2+3+4=9(ratio given) shares Paul share out 9*$5= $40 which is the value of $n. ALGEBRA(Equations and Inequalities) Ex.5 Aisha has $12 less than Darlene. Kaila has twice as much as Aisha and Darlene together. If Darlene has ‘$x’, then Kaila has how much represented as an equation…… Darlene has $x Aisha has $x-$12 {$12 less than Darlene} Kaila has 2($x+$x-12){Twice what Darlene and Aisha have together.} Then Kaila has 2($2x-$12)= $4x-$24 =$4(x-6){when factorized} ALGEBRA(Equations and Inequalities)(Key Terms) An inequality a mathematical statement consisting of the following signs:< less than sign, what is on the left of the sign is greater than what is to the right of the sign but the value is never equal to the upper value. For example a<b says a is less than b but never equal to b. > more than sign, what is on the left of the sign is greater than what is to the right of the sign but the value ‘b’ is never equal to the upper value ‘a’. For example a>b says a is greater than b but never equal to b. ≤ less than or equal to sign, what is on the left of the sign is less than what is to the right of the sign or the value may be equal to the upper value. For example a ≤ b says a is less than b or may be equal to ‘b’. ≥ less than or equal to sign, what is on the left of the sign is greater than what is to the right of the sign or the value may be equal to the lower value. For example a ≥b says a is greater than b or may be equal to ‘b’. ALGEBRA(Equations and Inequalities) Ex.6 If 1≤x<5 it means ‘x’ may have a value between 1 and up to 5 but not including 5 since there isn’t a less than and equal sign before the 5. Ex.7 If 2<x≤4 this statement means ‘x’ may have a value greater than two and up to being equal to 4 but not inclusive of 2. ALGEBRA(Equations and Inequalities) Ex.8 The value of ‘x’ varies between more than -2 up to but not including 6 represent this in an inequality statement. -2≤x<6 Ex.9 The value of ‘x’ varies between more than -3 but less than 4 what is the range for these possible values. -3≤x<4 ALGEBRA(Equations and Inequalities) Ex.10 The value ‘x’ varies from greater than zero up to less than 5, represent this in an inequality. 0<x<5 Since it is never equivalent to either limit ALGEBRA(Equations and Inequalities) • Algebra Equations and Inequalities Ends Here go to Algebra Exercise section and do Algebra Exercise 4 ALGEBRA( Factorizing and Rearranging) Remember with all equations what is on the left hand side of the equal sign is equivalent to what is on the right hand side of the equal sign. RHS=LHS What ever you do to the right must be done to the right side to ensure the equation is true. ALGEBRA( Factorizing and Rearranging) Ex.1 x-2=6 find the value of ‘x’ ‘x’ is having 2 subtracted from it to give a difference of 6. To find the value of ‘x’ do the opposite to both sides of the equal sign. That is add to 2 to both sides. => x-2+2=6+2 => x=8 ALGEBRA( Factorizing and Rearranging) Ex.2 2a=6, Find the value of ‘a’. ‘a’ is being multiplied by 2 to produce 6, to get the value of ‘a’ we must do the opposite that is to divide. Divide both sides by the coefficient of ‘a’ which is 2. 2𝑎 6 = 2 2 Therefore…. a=3 ALGEBRA( Factorizing and Rearranging) Ex.3 6 =3 𝑎 , Find the value of ‘a’ The variable ‘a’ is dividing the number six, as with the other two examples do the opposite to both sides. That is multiply both sides by ‘a’ 6 𝑎 ∗ 𝑎 1 3 1 = * 𝑎 1 =>6=3a Divide both sides by 3 =>2=a ALGEBRA( Factorizing and Rearranging) Rearranging equations is useful to calculate unknown variables in formulae. 1 V= Ah 3 Ex.4 If (This the formula for calculating the volume where A=area and h=height.) Make ‘h’ the subject of the formula(equation). 1 3 ‘h’ has ‘A’ multiplied by it(hence its coefficient) and multiplied it(another coefficient). Do the opposite to both sides that is divide. Remember dividing a fraction is the same as multiplying the inverse. ALGEBRA( Factorizing and Rearranging) 1 V= Ah 3 RHS=LHS Divide both sides by 1 3 which is the same as multiplying both sides by 3,which is the inverse of want to get h by itself so divide both sides by A. 1 𝐴 3 1 ∗ V= 1 𝐴 3 1 1 3 ∗ ∗ Ah The A and the 1/3 are cancelled out. 3𝑉 =h 𝐴 1 . 3 We ALGEBRA( Factorizing and Rearranging) Ex.5 𝐸 = 𝐼2𝑅, make ‘I’ the subject of the formula. again RHS=LHS I2 is being multiplied by R, to remove R from this side do the opposite to both sides that is divide both sides by ‘R’. 𝐸 𝐼2𝑅 = 𝑅 𝑅 The R is cancelled out on the right side. ALGEBRA( Factorizing and Rearranging) 𝐸 This gives = 𝐼2 𝑅 But we want I alone, the opposite of squaring a number is finding the square root.( 𝐼) So 𝐼2 = I So we are doing the same to both sides that is square rooting. 𝐸 𝑅 = 𝐼2 Therefore 𝐸 =I 𝑅 ALGEBRA( Factorizing and Rearranging) Side note on roots. A square root is the opposite of raising a value to the power 2 eg. x2 = x. Just as a value can be raised to the power 2 or 3 or 4 as shown in ‘Indices’ it may be brought down to the root any number of times. These are called surds Eg. If x= 3 find x2 x2 =( 3)2 =3 Eg. If x= 3 find x3 Remember x3=x*x*x So when we substitute x3= 3* 3* 3= ( 3* 3)* 3 => ( 3)2 * 3 =>3 3 ALGEBRA( Factorizing and Rearranging) 3𝑥+2 Ex.6 If y= what is ‘x’. 5 As always with equations RHS=LHS. Multiply both sides by 5 (the opposite)to cancel out the fraction. 3𝑥+2 = *5 5 y*5 =>5y=3x+2 Subtract 2 (the opposite) from both sides to get 3x by itself. 5y-2=3x+2-2 =>5y-2=3x Divide both sides by 3 to get x by itself. 5y−2 3𝑥 = 3 3 5y−2 Therefore x= 3 ALGEBRA( Factorizing and Rearranging) Ex.7 If y=x2 + 9 what is ‘x’? The number 9 is added to x2 do the opposite to both sides to get x2 by itself that is to subtract. =>y-9=x2+9-9 =>y-9=𝑥2 Now x is squared that means raised to the power 2. We just want x so we do the opposite and sqaure root both sides. => y−9= x2 => y−9=x ALGEBRA( Factorizing and Rearranging) Ex.8Make the subject of the equation. 𝑎𝑦 =3 𝑥 =>ay=3x 3𝑥 =>y= 𝑎 𝑎𝑦 =3 𝑥 ALGEBRA( Factorizing and Rearranging) Make y the subject of the equation xy-y=4. Step.1 Factorize y out of the LHS This can be re-written as (x*y)-(1*y)=4 => y(x-1)=4 Step.2 Double check by re-expanding the bracket in your head. Step.3 Divide both sides by (x-1) to get y by itself 4 =>y= 𝑥−1 ALGEBRA( Factorizing and Rearranging) Factorizing is the process of rewriting mathematical statements in a shorter form by collecting factors of the same type. Ex.9 Factorize the following algebraic expression ax-ay2bx+2by. The first two terms are multiples of ’a’ hence ‘a’ is a common factor. They can be rewritten with ‘a’ outside the bracket giving the original result when expanded. =>a(x-y)-2bx+2by With the rest of the statement 2b is multiplies a ‘-x’ and y hence it is a common factor. =>a(x-y)+2b(-x+y) hence the statement is factorized and if re-expanded will give the original statement. ALGEBRA( Factorizing and Rearranging) Eg.10 3mx+6my+3mz All the terms in this algebraic expression are multiples of 3 and m hence 3m is the common factor of all these terms. 3mx+6my+3mz=(3*m*x)+(6*m*y)+(3*m*z) Step.1 Write the common factor outside a bracket. 3m() Step.2 Fill the bracket with terms that when expanded will give back the original expression. Remember 3mx+6my+3mz=(3*m*x)+(6*m*y)+(3*m*z) =(3*m*x)+(3*m*2*y)+(3*m*z) 3m(x+2y+z) When expanded gives the original expression. Note the coefficient of the variable ‘y’ was 6m not 3m so we put a 2 to give 6m is we re-expand.