Bridging the gap: GCSE AS Level


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1 Bridging the gp: GCSE AS Level CONTENTS Chpter Removing rckets pge Chpter Liner equtions Chpter Simultneous equtions 8 Chpter Fctors 0 Chpter Chnge the suject of the formul Chpter 6 Solving qudrtic equtions 6 Chpter 7 Indices 8 Chpter 8 Surds Chpter 9 Stright lines Solutions to the exercises 7 Pge
2 Chpter : REMOVING BRACKETS To remove single rcket, we multiply every term in the rcket y the numer or the expression on the outside: Exmples (x + y) = x + 6y ) (x  ) = ()(x) + ()() = x + 6 To expnd two rckets, we must multiply everything in the first rcket y everything in the second rcket. We cn do this in vriety of wys, including * the smiley fce method * FOIL (Fronts Outers Inners Lsts) * using grid. Exmples: (x + (x + ) = x(x + ) + (x + ) or (x +(x + ) = x + + x + x = x + x + or x x x x x (x +(x + ) = x + x + x + = x + x + ) (x  )(x + ) = x(x + )  (x +) = x + x x  6 = x x 6 or (x  )(x + ) = x 6 + x x = x x 6 or x  x x x x 6 (x +)(x  ) = x + x  x  6 = x  x  6 Pge
3 EXERCISE A Multiply out the following rckets nd simplify.. 7(x + ). (x  7). ( . y + y( + y). x (x + ) 6. (x  (x  ) 7. (x + )(x + ) 8. (t  )(t  ) 9. (x + y)(x y) 0. (x  )(x + ). (y  (y +. ( + x)( x) Two Specil Cses Perfect Squre: Difference of two squres: (x + ) = (x + )(x + ) = x + x + (x  )(x + ) = x (x  ) = (x )(x ) = x x + 9 (x  )(x + ) = x = x 9 EXERCISE B Multiply out. (x . (x + ). (7x  ). (x + )(x  ). (x + (x  6. (y  )(y + ) Pge
4 Chpter : LINEAR EQUATIONS When solving n eqution, you must rememer tht whtever you do to one side must lso e done to the other. You re therefore llowed to dd the sme mount to oth side sutrct the sme mount from ech side multiply the whole of ech side y the sme mount divide the whole of ech side y the sme mount. If the eqution hs unknowns on oth sides, you should collect ll the letters onto the sme side of the eqution. If the eqution contins rckets, you should strt y expnding the rckets. A liner eqution is n eqution tht contins numers nd terms in x. A liner eqution does not contin ny x or x terms. Exmple : Solve the eqution 6 x = Solution: There re vrious wys to solve this eqution. One pproch is s follows: Step : Add x to oth sides (so tht the x term is positive): 6 = x + Step : Sutrct from oth sides: Step : Divide oth sides y : 9 = x = x So the solution is x =. Exmple : Solve the eqution 6x + 7 = x. Solution: Step : Begin y dding x to oth sides 8x + 7 = (to ensure tht the x terms re together on the sme side) Step : Sutrct 7 from ech side: 8x =  Step : Divide ech side y 8: x = ¼ Exercise A: Solve the following equtions, showing ech step in your working: x + = 9 ) x = ) x = ) 7x = 9 ) + x = 8 x 6) 7x + = x Pge
5 Exmple : Solve the eqution (x ) = 0 (x + ) Step : Multiply out the rckets: 6x = 0 x 6 (tking cre of the negtive signs) Step : Simplify the right hnd side: 6x = x Step : Add x to ech side: 9x = Step : Add : 9x = 8 Step : Divide y 9: x = Exercise B: Solve the following equtions. (x ) = ) ( x) = (x 9) ) 8 (x + ) = ) (x + ) = EQUATIONS CONTAINING FRACTIONS When n eqution contins frction, the first step is usully to multiply through y the denomintor of the frction. This ensures tht there re no frctions in the eqution. y Exmple : Solve the eqution Solution: Step : Multiply through y (the denomintor in the frction): y 0 Step : Sutrct 0: y = Exmple : Solve the eqution (x Solution: Step : Multiply y (to remove the frction) x Step : Sutrct from ech side x = Step : Divide y x = 7 When n eqution contins two frctions, you need to multiply y the lowest common denomintor. Pge
6 This will then remove oth frctions. x x Exmple 6: Solve the eqution Solution: Step : Find the lowest common denomintor: The smllest numer tht oth nd divide into is 0. Step : Multiply oth sides y the lowest common denomintor Step : Simplify the left hnd side: 0( x 0( x) 0 0 ( x 0 ( x ) 0 (x + + (x + ) = 0 Step : Multiply out the rckets: x + + x + 8 = 0 Step : Simplify the eqution: 9x + = 0 Step 6: Sutrct 9x = 7 Step 7: Divide y 9: x = Exmple 7: Solve the eqution x x x 6 Solution: The lowest numer tht nd 6 go into is. So we multiply every term y : ( x) ( x) x 6 Simplify x ( x ) ( x) Expnd rckets x x 6 6 0x Simplify x 6 8 0x Sutrct 0x x 6 8 Add 6 x = Divide y x =.8 Exercise C: Solve these equtions ( ) x ) x x ) y y ) Pge 6 x x 7
7 Exercise C (continued) ) 7x x y y y 6) 6 7) x x x 8) 0 x x FORMING EQUATIONS Exmple 8: Find three consecutive numers so tht their sum is 96. Solution: Let the first numer e n, then the second is n + nd the third is n +. Therefore n + (n + + (n + ) = 96 n + = 96 n = 9 n = So the numers re, nd. Exercise D: Find consecutive even numers so tht their sum is 08. ) The perimeter of rectngle is 79 cm. One side is three times the length of the other. Form n eqution nd hence find the length of ech side. ) Two girls hve 7 photogrphs of celerities etween them. One gives to the other nd finds tht she now hs hlf the numer her friend hs. Form n eqution, letting n e the numer of photogrphs one girl hd t the eginning. Hence find how mny ech hs now. Pge 7
8 Chpter : SIMULTANEOUS EQUATIONS An exmple of pir of simultneous equtions is x + y = 8 x + y = In these equtions, x nd y stnd for two numers. We cn solve these equtions in order to find the vlues of x nd y y eliminting one of the letters from the equtions. In these equtions it is simplest to eliminte y. We do this y mking the coefficients of y the sme in oth equtions. This cn e chieved y multiplying eqution y, so tht oth equtions contin y: x + y = 8 0x + y = = To eliminte the y terms, we sutrct eqution from eqution. We get: 7x = i.e. x = To find y, we sustitute x = into one of the originl equtions. For exmple if we put it into : 0 + y = y = Therefore the solution is x =, y =. Rememer: You cn check your solutions y sustituting oth x nd y into the originl equtions. Exmple: Solve x + y = 6 x y = Solution: We egin y getting the sme numer of x or y ppering in oth eqution. We cn get 0y in oth equtions if we multiply the top eqution y nd the ottom eqution y : 8x + 0y = 6 x 0y = As the SIGNS in front of 0y re DIFFERENT, we cn eliminte the y terms from the equtions y ADDING: x = 69 + i.e. x = Sustituting this into eqution gives: 6 + y = 6 y = 0 So y = The solution is x =, y =. Pge 8
9 Exercise: Solve the pirs of simultneous equtions in the following questions: x + y = 7 ) x + y = 0 x + y = 9 x + y = 7 ) x y = ) 9x y = x + y = 6 x y = 7 ) + = 6) p + q = = p + q = Pge 9
10 Chpter : FACTORISING Common fctors We cn fctorise some expressions y tking out common fctor. Exmple : Fctorise x 0 Solution: 6 is common fctor to oth nd 0. We cn therefore fctorise y tking 6 outside rcket: x 0 = 6(x ) Exmple : Fctorise 6x xy Solution: is common fctor to oth 6 nd. Both terms lso contin n x. So we fctorise y tking x outside rcket. 6x xy = x(x y) Exmple : Fctorise 9x y 8x y Solution: 9 is common fctor to oth 9 nd 8. The highest power of x tht is present in oth expressions is x. There is lso y present in oth prts. So we fctorise y tking 9x y outside rcket: 9x y 8x y = 9x y(xy ) Exmple : Fctorise x(x (x Solution: There is common rcket s fctor. So we fctorise y tking (x out s fctor. The expression fctorises to (x (x ) Exercise A Fctorise ech of the following x + xy ) x xy ) pq p q ) pq  9q ) x 6x 6) 8 7) y(y + (y Pge 0
11 Fctorising qudrtics Simple qudrtics: Fctorising qudrtics of the form x x c The method is: Step : Form two rckets (x )(x ) Step : Find two numers tht multiply to give c nd dd to mke. These two numers get written t the other end of the rckets. Exmple : Fctorise x 9x 0. Solution: We need to find two numers tht multiply to mke 0 nd dd to mke 9. These numers re 0 nd. Therefore x 9x 0 = (x 0)(x +. Generl qudrtics: Fctorising qudrtics of the form x x c The method is: Step : Find two numers tht multiply together to mke c nd dd to mke. Step : Split up the x term using the numers found in step. Step : Fctorise the front nd ck pir of expressions s fully s possile. Step : There should e common rcket. Tke this out s common fctor. Exmple : Fctorise 6x + x. Solution: We need to find two numers tht multiply to mke 6  = 7 nd dd to mke. These two numers re 8 nd 9. Therefore, 6x + x = 6x  8x + 9x = x(x ) + (x ) (the two rckets must e identicl) = (x )(x + ) Difference of two squres: Fctorising qudrtics of the form x Rememer tht x = (x + )(x ). Therefore: x x x x 9 ( )( ) 6x ( x) (x )(x ) Also notice tht: nd x 8 ( x ) ( x )( x ) x 8xy x( x 6 y ) x( x y)( x y) Fctorising y piring We cn fctorise expressions like x xy x y using the method of fctorising y piring: x xy x y = x(x + y) (x + y) (fctorise front nd ck pirs, ensuring oth rckets re identicl) = (x + y)(x Pge
12 Exercise B Fctorise ) ) x x x 6 6x 6 x x ) ) x x (fctorise y tking out common fctor) x x 6) y 7y 7) 7y 0y 8) 0x x 0 9) x 0) x x xy y x x 8 6m 8n y 9 y 8( x ( x 0 Pge
13 Chpter : CHANGING THE SUBJECT OF A FORMULA We cn use lger to chnge the suject of formul. Rerrnging formul is similr to solving n eqution we must do the sme to oth sides in order to keep the eqution lnced. Exmple : Mke x the suject of the formul y = x +. Solution: y = x + Sutrct from oth sides: y = x Divide oth sides y ; y x y So x is the sme eqution ut with x the suject. Exmple : Mke x the suject of y = x Solution: Notice tht in this formul the x term is negtive. y = x Add x to oth sides y + x = (the x term is now positive) Sutrct y from oth sides x = y Divide oth sides y y x Exmple : ( F ) The formul C is used to convert etween Fhrenheit nd 9 Celsius. We cn rerrnge to mke F the suject. ( F ) C 9 Multiply y 9 9C ( F ) (this removes the frction) Expnd the rckets 9C F 60 Add 60 to oth sides 9C60 F Divide oth sides y 9C 60 F 9C 60 Therefore the required rerrngement is F. Exercise A Mke x the suject of ech of these formule: y = 7x ) y x x (x ) ) y ) y 9 Rerrnging equtions involving squres nd squre roots Pge
14 Exmple : Mke x the suject of x y w Solution: Sutrct y from oth sides: Squre root oth sides: x y w x w y (this isoltes the term involving x) x w y Rememer tht you cn hve positive or negtive squre root. We cnnot simplify the nswer ny more. Exmple : Mke the suject of the formul t h Solution: Multiply y Squre oth sides Multiply y h: Divide y : t h t h 6t h 6t h 6th Exercise B: Mke t the suject of ech of the following wt P ) r wt P r ) V t h ) P t g ) w( v t) P 6) g r t Pge
15 More difficult exmples Sometimes the vrile tht we wish to mke the suject occurs in more thn one plce in the formul. In these questions, we collect the terms involving this vrile on one side of the eqution, nd we put the other terms on the opposite side. Exmple 6: Mke t the suject of the formul xt yt Solution: xt yt Strt y collecting ll the t terms on the right hnd side: Add xt to oth sides: yt xt Now put the terms without t on the left hnd side: Sutrct from oth sides: yt xt Fctorise the RHS: t( y x) Divide y (y + x): So the required eqution is t y x t y x W Exmple 7: Mke W the suject of the formul T W Solution: This formul is complicted y the frctionl term. We egin y removing the frction: Multiply y : T W W Add W to oth sides: T W W (this collects the W s together) Fctorise the RHS: T W( ) Divide oth sides y + : W T Exercise C Mke x the suject of these formule: x x c ) ( x ) k( x ) ) x y x x x ) Pge
16 Chpter 6: SOLVING QUADRATIC EQUATIONS A qudrtic eqution hs the form x x c 0. There re two methods tht re commonly used for solving qudrtic equtions: * fctorising * the qudrtic formul Note tht not ll qudrtic equtions cn e solved y fctorising. The qudrtic formul cn lwys e used however. Method : Fctorising Mke sure tht the eqution is rerrnged so tht the right hnd side is 0. It usully mkes it esier if the coefficient of x is positive. Exmple : Solve x x + = 0 Fctorise (x (x ) = 0 Either (x = 0 or (x ) = 0 So the solutions re x = or x = Note: The individul vlues x = nd x = re clled the roots of the eqution. Exmple : Solve x x = 0 Fctorise: x(x ) = 0 Either x = 0 or (x ) = 0 So x = 0 or x = Method : Using the formul Recll tht the roots of the qudrtic eqution x x x c 0 re given y the formul: c Exmple : Solve the eqution x 7 x Solution: First we rerrnge so tht the right hnd side is 0. We get We cn then tell tht =, = nd c = . Sustituting these into the qudrtic formul gives: x x 0 ( 0 x (this is the surd form for the solutions) If we hve clcultor, we cn evlute these roots to get: x =.8 or x = . Pge 6
17 EXERCISE Use fctoristion to solve the following equtions: ) x + x + = 0 ) x x = 0 c) x = x ) Find the roots of the following equtions: ) x + x = 0 ) x x = 0 c) x = 0 ) Solve the following equtions either y fctorising or y using the formul: ) 6x  x = 0 ) 8x x + 0 = 0 ) Use the formul to solve the following equtions to significnt figures. Some of the equtions cn t e solved. ) x +7x +9 = 0 ) 6 + x = 8x c) x x 7 = 0 d) x x + 8 = 0 e) x + x + = 0 f) x = x 6 Pge 7
18 Chpter 7: INDICES Bsic rules of indices y mens y y y y. is clled the index (plurl: indices), power or exponent of y. There re sic rules of indices: ) ) ( ) m n m n e.g. m n m n e.g. m n mn e.g. 0 Further exmples y y y 7 6 (multiply the numers nd multiply the s) c c 6 6c 8 (multiply the numers nd multiply the c s) 7 7 d d d 8d d (divide the numers nd divide the d terms i.e. y sutrcting the powers) Exercise A Simplify the following: ) ) ) ) 6) = (Rememer tht c c = c c = 6 n ( 6 n ) = 8 8n n = d d = 9 ) 7) = 8) d = More complex powers Zero index: Recll from GCSE tht Pge 8 0.
19 This result is true for ny nonzero numer. 0 Therefore Negtive powers A power of  corresponds to the reciprocl of numer, i.e. Therefore nd (you find the reciprocl of frction y swpping the top ottom over) This result cn e extended to more generl negtive powers: This mens: n. n Frctionl powers: Frctionl powers correspond to roots: / / / In generl: /n n Therefore: / 8 8 / / m / n / n A more generl frctionl power cn e delt with in the following wy: So / 8 / / / / m Pge 9
20 Exercise B: Find the vlue of: ) / / 7 ) / 9 ) ) 6) 7) 8) 9) / 7 8 / 0) 0.0 / / / Simplify ech of the following: x / / x / xy Pge 0
21 Pge Chpter 8: SURDS Surds re squre roots of numers which don t simplify into whole (or rtionl) numer: e.g.... ut it is more ccurte to leve it s surd: Generl rules But you cnnot do: These re NOT equl ) )( ( Simplifying Surds Find the lrgest squre numers nd simplify s fr s possile Worked Exmples Creful  this is times the squre root of NOT the cue root of Rtionlising the Denomintor This is fncy wy of sying getting rid of the surd on the ottom of frction. We multiply the frction y the denomintor (or the denomintor with the sign swpped) Worked Exmples. e Rtionlis we multiply y which is the sme s multiplying y, which mens we don t fundmentlly chnge the frction.. 0 e Rtionlis. 0 0 ) ( ) ( ) ( ) ( Rtionlis e ) ( ( ( ( e Rtionlis
22 Exercise A: Simplify the surds ) ) 8 ) 7 ) 7 Exercise B: Expnd nd simplify ) 6 8 ) ( ) ) ( )( ) ) ( )( ) 6) ( )( ) 7) ( )( ) 8) ( 8 )(8 ) 9) ( )( ) Exercise C: Rewrite the following expressions with rtionl denomintors 7) ) 8) 8 6 ) 9 7 9) 8 7 ) 0) ) 6)
23 Chpter 9: Stright line grphs Liner functions cn e written in the form y = mx + c, where m nd c re constnts. A liner function is represented grphiclly y stright line, m is the grdients nd c is the y intercept of the grph. Exmple : Drw the grph of y = x + Solution: Step : Mke tle of vlues Step : Use your tle to drw the stright line grph Exmple : Plot the stright line using the grdient nd y intercept Solution: Step : Mrk on the y xis the yintercept = Step : The grdient= so strt from the y intercept for every unit cross to the right go down y hlf unit nd mrk second point there. Step : Join the y intercept with the new point with line nd extend form oth sides.. Here re some exmples of liner functions not ll of them in the form y = mx + c. You need to e confident into rerrnging the functions mking y the suject in order to identify the grdient nd y intercept. y = x + x  y + = 0 y  x = so y x so y x grdient= grdient= grdient= yintercept= yintercept= yintercept=
24 To find the yxis crossing, sustitute x = 0 into the liner eqution nd solve for y. To find the xxis crossing, sustitute y = 0 into the liner eqution nd solve for x. Exmple : Rewrite the eqution y  x = into the form y = mx + c, find the grdient nd the y intercept Solution: Step : Add x to oth sides (so tht the x term is positive): y = + x Step : Divide y oth sides: y x Step : Identify the grdient nd yintercept grdient= yintercept= Exmple : Find the grdient of the line which psses through the points A (, ) nd B (, ) Solution: Step : Use the x nd y vlues of A x, ) nd B x, ) Step : find the grdient m y x y x ( y ( y m Finlly you need to e le to find the eqution of line from grph. Exmple : Find the eqution of the stright line which psses through the point (, ) nd hs grdient
25 Solution: Step : Find where the line crosses the y xis. This is the y intercept, c. Line crosses y xis t, so yintercept c= Step : Drw tringle elow the line from the intercept to point you know y y And work out the grdient etween the two points m x x Grdient tringle from (6,) to (0,) so m 0 6 Step : Write in the form y = mx + c y x Exercise A: Plot the grph of ech function tking the given vlues ) y= x  ( x =  to ) ) y= x + ( x =  to ) c) y = x ( x =  to ) d) y= x + ( x =  to ) 6 Exercise B: Rewrite the equtions elow into the form y = mx + c, find the grdient nd the yintercept )x y = 0 ) x + y 8 =0 c) = x y Then plot the grph of ech eqution Exercise C: Work out the grdient etween the sets of coordintes ) A ( 0, ) nd B(, 6) ) A (, 0) nd B(, ) c) A (, ) nd B(, ) d) A ( , ) nd B(, ) e) A (, 0.) nd B(, ) f) A ( 7, ) nd B( , 6)
26 Exercise D: Find the eqution of these lines in the form 6
27 SOLUTIONS TO THE EXERCISES CHAPTER : Ex A 8x + ) x + ) 7 + ) 6y + y ) x 6) 7x 7) x + x + 6 8) t t 0 9) 6x + xy y 0) x + x y + 7x x Ex B x x + ) 9x + 0x + ) 9x 8x + ) x ) 9x  6) y 9 CHAPTER Ex A 7 ) ) ½ ) ) / 6) 7/ Ex B. ) ) ) ½ Ex C 7 ) ) /7 ) / ) 6) 7) 9/ 8) Ex D, 6, 8 ) 9.87, 9.6 ), 8 CHAPTER x =, y = ) x = , y = ) x = 0, y =  ) x =, y = ) = 7, = 6) p = /, q = / CHAPTER Ex A x( + y) ) x(x y) ) pq(q p) ) q(p q) ) x (x  ) 6) ( ) 7) (y (y + ) Ex B (x )(x + ) ) (x + 8)(x ) ) (x + (x + ) ) x(x ) ) (x  )(x + ) 6) (y + )(y + 7) 7) (7y )(y 8) (x )(x + ) 9) (x + )(x ) 0) (x )(x y) (x )(x (m 9n)(m + 9n) y(y )(y + ) (x + )(x ) CHAPTER Ex A y 9y 0 x ) xy ) x( y ) ) x 7 Ex B rp rp V Pg Pg t ) t ) t ) t ) t v 6) w w h w Ex C c k y x ) x ) x ) x k y CHAPTER 6 ) ,  ) , c) , ) ) 0,  ) 0, c),  ) ) /, / ) 0.,. ) ) .0, .70 ).07, c) .0,. d) no solutions e) no solutions f) no solutions t r CHAPTER 7 Ex A 6 ) 6c 7 ) c ) n 8 ) n 6) d 7) 6 8) d Ex B ) ) / ) / ) 6) /7 7) 9 8) 9/ 9) ¼ 0) 0. /9 6 6 x xy CHAPTER 8 7
28 8 ExA ) 6 ) ) ) ExB 8 9) ) 6 7) 6) ) ) ) $ 8 ) 0 Ex C
29 9 6 6) ( ) ) ) ) stys the sme 8 ) ( ( ( ( 0) ) 7 7( 7 7 9) ) 6 ( ) 6 ( 6 8) 7) CHAPTER 9 ExB : ) ) ) x y c x y x y ExC : ) 8. ) 7 ) ) ) ) grdient f grdient e grdient d grdient c grdient grdient
30 Ex D : ). y = x+. y = 0.x+ c. y = x+ ). y =  x. y = 6x c. y = x c). y = 0.x . y =  x + c. y = x + d). y = x +. y = x + c. y = 0.x  e). y =  x. y = 0.x + c. y = x  f). y = x . y = x + c. y = 6x 0
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