A-Level Mthemtics Trnsition Tsk (compulsory for ll mths students nd ll further mths student) Due: st Lesson of the yer. Length: - hours work (depending on prior knowledge) This trnsition tsk provides revision for some of the sections from GCSE mthemtics tht re needed for A Level mthemtics. Ech section contins set of worked emples, followed by questions with nswers. All students re epected to nswer ll non-etension questions (with relevnt written clcultions) nd to tick/cross their questions to show tht they hve mrked them. This work will be hnded in during the first mthemtics lesson in September. There will lso be hour ssessment in the first lesson. This ssessment will cover the topics from the trnsition tsk. The mrked ssessment nd the trnsition tsk will be considered together. Students suitbility for the course will be judged bsed on both their ssessment nd the effort they hve mde on trnsition tsk. FAQ Help! Wht if I do bdly on my ssessment? The ssessment is designed to find gps in students knowledge. By finding these gps erly in the course, we cn offer support nd intervention. Help! I cn t nswer one (or more) sections: Copy out the emples from the section nd hve rel go t some of the questions. The trnsition tsk is used s mesure of how much effort hs been put in, so hve rel try nd effort will be considered. Help! This is tking long time. This tsk should tke bout -6 hours if student is lredy strong in ll sections. If student is not confident in these sections, it my tke considerbly longer. These topics re necessry prior knowledge for the course nd so will not be tught eplicitly in lessons. Mthemtics A-Level is very chllenging course, requiring lrge investment of time nd effort getting up to speed with these topics before September should gretly increse the likelihood tht you will succeed.
Epnding brckets nd simplifying epressions A LEVEL LINKS Scheme of work:. Algebric epressions bsic lgebric mnipultion, indices nd surds Key points When you epnd one set of brckets you must multiply everything inside the brcket by wht is outside. When you epnd two liner epressions, ech with two terms of the form + b, where 0 nd b 0, you crete four terms. Two of these cn usully be simplified by collecting like terms. Emples Emple Epnd ( ) ( ) = 8 Multiply everything inside the brcket by the outside the brcket Emple Epnd nd simplify ( + ) ( + ) ( + ) ( + ) = + 8 = Epnd ech set of brckets seprtely by multiplying ( + ) by nd ( + ) by Simplify by collecting like terms: 8 = nd = Emple Epnd nd simplify ( + )( + ) ( + )( + ) = ( + ) + ( + ) = + + + 6 = + + 6 Epnd the brckets by multiplying ( + ) by nd ( + ) by Simplify by collecting like terms: + = Emple Epnd nd simplify ( )( + ) ( )( + ) = ( + ) ( + ) = + 0 = 7 Epnd the brckets by multiplying ( + ) by nd ( + ) by Simplify by collecting like terms: 0 = 7
Prctice Epnd. ( ) b (pq + q ) c (y y ) Epnd nd simplify. 7( + ) + 6( 8) b 8(p ) (p + 9) c 9(s + ) (6s 0) d ( ) ( + ) Epnd. ( + 8) b k(k ) c h(6h + h ) d s(s 7s + ) Wtch out! When multiplying (or dividing) positive nd negtive numbers, if the signs re the sme the nswer is + ; if the signs re different the nswer is. Epnd nd simplify. (y 8) (y ) b ( + ) + ( 7) c p(p ) p(p ) d b(b ) b(6b 9) Epnd (y 8) 6 Epnd nd simplify. (m + 7) b p(p + 6p) 9p(p ) 7 The digrm shows rectngle. Write down n epression, in terms of, for the re of the rectngle. Show tht the re of the rectngle cn be written s 8 Epnd nd simplify. ( + )( + ) b ( + 7)( + ) c ( + 7)( ) d ( + )( ) e ( + )( ) f ( )( + ) g ( )( ) h ( )(7 + ) i ( + y)(y + 6) j ( + ) k ( 7) l ( y) Etend 9 Epnd nd simplify ( + )² + ( )² 0 Epnd nd simplify. b
Answers 6 b 0pq 8q c y + y + + 8 = b 0p 6 p 7 = 8p c 7s + 9 0s + 0 = s + 9 = 9 s d 8 6 = + b 0k 8k c 0h h h d s s 6s y b c p 7p d 6b
Surds nd rtionlising the denomintor A LEVEL LINKS Scheme of work:. Algebric epressions bsic lgebric mnipultion, indices nd surds Key points A surd is the squre root of number tht is not squre number, for emple,,, etc. Surds cn be used to give the ect vlue for n nswer. b b b b To rtionlise the denomintor mens to remove the surd from the denomintor of frction. To rtionlise b you multiply the numertor nd denomintor by the surd b To rtionlise b c you multiply the numertor nd denomintor by b c Emples Emple Simplify 0 0 Choose two numbers tht re fctors of 0. One of the fctors must be squre number Use the rule b b Use Emple Simplify 7 7 9 9 7 7 Simplify 7 nd. Choose two numbers tht re fctors of 7 nd two numbers tht re fctors of. One of ech pir of fctors must be squre number Use the rule b b Use 9 7 nd Collect like terms
Emple Simplify 7 7 7 7 = 9 7 7 = 7 = Epnd the brckets. A common mistke here is to write 7 9 Collect like terms: 7 7 7 7 0 Emple Rtionlise = = 9 Multiply the numertor nd denomintor by Use 9 = Emple Rtionlise nd simplify = Multiply the numertor nd denomintor by = Simplify in the numertor. Choose two numbers tht re fctors of. One of the fctors must be squre number = = 6 Use the rule b b Use Simplify the frction: simplifies to 6
Emple 6 Rtionlise nd simplify = = Multiply the numertor nd denomintor by Epnd the brckets = 6 Simplify the frction = 6 = 6 Divide the numertor by Remember to chnge the sign of ll terms when dividing by Prctice Simplify. b c 8 d 7 e 00 f 8 g 7 h 6 Hint One of the two numbers you choose t the strt must be squre number. Simplify. c e 7 6 b 0 8 d 7 8 8 8 f 7 Wtch out! Check you hve chosen the highest squre number t the strt. Epnd nd simplify. ( )( ) b ( )( ) c ( )( ) d ( )(6 8)
Rtionlise nd simplify, if possible. c 7 e b d f 8 g 8 h Rtionlise nd simplify. b c 6 Etend 6 Epnd nd simplify y y 7 Rtionlise nd simplify, if possible. 9 8 b y
Answers b c d 7 e 0 f 7 g 6 h 9 b c d e 6 7 f b 9 c 0 7 d 6 c 7 7 e f b d g h b ( ) c 6( ) 6 y 7 b y y y 6 m b p + p + 7p 7 7( ) = 8 + 9 + 0 b + 0 + c + d
e + f 6 g 0 + h + i 8 + 9y + 0y j + 0 + k 8 + 9 l 6 y + 9y 9 + 0 b
Rules of indices A LEVEL LINKS Scheme of work:. Algebric epressions bsic lgebric mnipultion, indices nd surds Key points m n = m + n m mn n ( m ) n = mn 0 = n n i.e. the nth root of m n n m n m m m The squre root of number produces two solutions, e.g. 6. Emples Emple Evlute 0 0 0 0 = Any vlue rised to the power of zero is equl to Emple Evlute 9 9 9 = Use the rule n n Emple Evlute 7 7 7 = = 9 m n n Use the rule Use 7 m
Emple Evlute 6 Use the rule Use 6 m m Emple Simplify 6 6 = 6 = nd use the rule give m n mn to Emple 6 Simplify 8 = 8 = Use the rule Use the rule m n m n m n mn Emple 7 Write s single power of Use the rule m m, note tht the frction remins unchnged Emple 8 Write s single power of Use the rule Use the rule n n m m
Prctice Evlute. 0 b 0 c 0 d 0 Evlute. 9 b 6 c d 6 Evlute. b 8 c 9 d 6 Evlute. b c d 6 Simplify. b 0 c y e y g y 0 d f h 7y y c c c Wtch out! Remember tht ny vlue rised to the power of zero is. This is the rule 0 =. 6 Evlute. d b 6 e 7 c 9 6 f 9 7 6 7 Write the following s single power of. d b 7 e c f
8 Write the following without negtive or frctionl powers. d b 0 c e f 9 Write the following in the form n. b d e c f Etend 0 Write s sums of powers of. b c
Answers b c d 7 b c d b c d 8 b 6 c d 6 b y c d e y f c g 6 h 6 d b e 9 c f 8 6 9 7 b 7 c d e f 8 d b c e f 9 d b c e f 0 0 b c 7
Fctorising epressions A LEVEL LINKS Scheme of work: b. Qudrtic functions fctorising, solving, grphs nd the discriminnts Key points Fctorising n epression is the opposite of epnding the brckets. A qudrtic epression is in the form + b + c, where 0. To fctorise qudrtic eqution find two numbers whose sum is b nd whose product is c. An epression in the form y is clled the difference of two squres. It fctorises to ( y)( + y). Emples Emple Fctorise y + 9 y y + 9 y = y(y + ) The highest common fctor is y. So tke y outside the brckets nd then divide ech term by y to find the terms in the brckets Emple Fctorise y y = ( + y)( y) This is the difference of two squres s the two terms cn be written s () nd (y) Emple Fctorise + 0 b =, c = 0 So + 0 = + 0 = ( + ) ( + ) = ( + )( ) Work out the two fctors of c = 0 which dd to give b = ( nd ) Rewrite the b term () using these two fctors Fctorise the first two terms nd the lst two terms ( + ) is fctor of both terms
Emple Fctorise 6 0 b =, c = 60 So 6 0 = 6 + 0 = ( ) + ( ) = ( )( + ) Work out the two fctors of c = 60 which dd to give b = ( nd ) Rewrite the b term ( ) using these two fctors Fctorise the first two terms nd the lst two terms ( ) is fctor of both terms Emple Simplify 99 99 Fctorise the numertor nd the denomintor For the numertor: b =, c = So = 7 + = ( 7) + ( 7) = ( 7)( + ) For the denomintor: b = 9, c = 8 So + 9 + 9 = + 6 + + 9 = ( + ) + ( + ) = ( + )( + ) So ( 7)( ) 99 ( )() = 7 Work out the two fctors of c = which dd to give b = ( 7 nd ) Rewrite the b term ( ) using these two fctors Fctorise the first two terms nd the lst two terms ( 7) is fctor of both terms 6 Work out the two fctors of c = 8 which dd to give b = 9 (6 nd ) 7 Rewrite the b term (9) using these two fctors 8 Fctorise the first two terms nd the lst two terms 9 ( + ) is fctor of both terms 0 ( + ) is fctor of both the numertor nd denomintor so cncels out s vlue divided by itself is
Prctice Fctorise. 6 y 0 y b b + b c y 0 y + y Fctorise + 7 + b + c + 0 d e 7 8 f + 0 g 0 h + 8 Hint Tke the highest common fctor outside the brcket. Fctorise 6 9y b 8y c 8 00b c Fctorise + b 6 + 7 + c + 7 + d 9 + e 0 + + 9 f 8 + 0 Simplify the lgebric frctions. c e 8 b d f 70 6 Simplify c Etend 9 6 78 0 6 b d 7 70 6 7 7 Simplify 0 8 Simplify ( ) ( )
Answers y ( y) b 7 b (b + ) c y ( + y) ( + )( + ) b ( + 7)( ) c ( )( 6) d ( 8)( + ) e ( 9)( + ) f ( + )( ) g ( 8)( + ) h ( + 7)( ) (6 7y)(6 + 7y) b ( 9y)( + 9y) c ( 0bc)( + 0bc) ( )( + ) b ( + )( + ) c ( + )( + ) d ( )( ) e ( + )( +) f ( )( ) ( ) b c d e f 6 7 b c d 7 ( + ) 8 ( )
Completing the squre A LEVEL LINKS Scheme of work: b. Qudrtic functions fctorising, solving, grphs nd the discriminnts Key points Completing the squre for qudrtic rerrnges + b + c into the form p( + q) + r If, then fctorise using s common fctor. Emples Emple Complete the squre for the qudrtic epression + 6 + 6 = ( + ) 9 = ( + ) Write + b + c in the form b b c Simplify Emple Write + in the form p( + q) + r + = = Before completing the squre write + b + c in the form b c Now complete the squre by writing in the form b b = = 8 7 8 Epnd the squre brckets don t forget to multiply by the fctor of Simplify
Prctice Write the following qudrtic epressions in the form ( + p) + q + + b 0 c 8 d + 6 e + 7 f + Write the following qudrtic epressions in the form p( + q) + r 8 6 b 8 6 c + 9 d + 6 8 Complete the squre. + + 6 b c + d + + Etend Write ( + 0 + ) in the form ( + b) + c.
Answers ( + ) b ( ) 8 c ( ) 6 d ( + ) 9 e ( ) + 6 f 7 ( ) b ( ) 0 c ( + ) d 9 8 b c 9 0 0 d 6 ( + ) +
Solving qudrtic equtions by fctoristion A LEVEL LINKS Scheme of work: b. Qudrtic functions fctorising, solving, grphs nd the discriminnts Key points A qudrtic eqution is n eqution in the form + b + c = 0 where 0. To fctorise qudrtic eqution find two numbers whose sum is b nd whose products is c. When the product of two numbers is 0, then t lest one of the numbers must be 0. If qudrtic cn be solved it will hve two solutions (these my be equl). Emples Emple Solve = = = 0 ( ) = 0 So = 0 or ( ) = 0 Therefore = 0 or = Rerrnge the eqution so tht ll of the terms re on one side of the eqution nd it is equl to zero. Do not divide both sides by s this would lose the solution = 0. Fctorise the qudrtic eqution. is common fctor. When two vlues multiply to mke zero, t lest one of the vlues must be zero. Solve these two equtions. Emple Solve + 7 + = 0 + 7 + = 0 b = 7, c = + + + = 0 ( + ) + ( + ) = 0 ( + )( + ) = 0 So ( + ) = 0 or ( + ) = 0 Therefore = or = Fctorise the qudrtic eqution. Work out the two fctors of c = which dd to give you b = 7. ( nd ) Rewrite the b term (7) using these two fctors. Fctorise the first two terms nd the lst two terms. ( + ) is fctor of both terms. When two vlues multiply to mke zero, t lest one of the vlues must be zero. 6 Solve these two equtions.
Emple Solve 9 6 = 0 9 6 = 0 ( + )( ) = 0 So ( + ) = 0 or ( ) = 0 or Fctorise the qudrtic eqution. This is the difference of two squres s the two terms re () nd (). When two vlues multiply to mke zero, t lest one of the vlues must be zero. Solve these two equtions. Emple Solve = 0 b =, c = So 8 + = 0 ( ) + ( ) = 0 ( )( + ) = 0 So ( ) = 0 or ( +) = 0 or Fctorise the qudrtic eqution. Work out the two fctors of c = which dd to give you b =. ( 8 nd ) Rewrite the b term ( ) using these two fctors. Fctorise the first two terms nd the lst two terms. ( ) is fctor of both terms. When two vlues multiply to mke zero, t lest one of the vlues must be zero. 6 Solve these two equtions. Prctice Solve 6 + = 0 b 8 = 0 c + 7 + 0 = 0 d + 6 = 0 e = 0 f + 0 = 0 g 0 + = 0 h 6 = 0 i + 8 = 0 j 6 + 9 = 0 k 7 = 0 l 0 = 0 Solve = 0 b = c + = d = e ( + ) = + f 0 = g ( + ) = + h ( ) = ( + ) Hint Get ll terms onto one side of the eqution.
Solving qudrtic equtions by completing the squre A LEVEL LINKS Scheme of work: b. Qudrtic functions fctorising, solving, grphs nd the discriminnts Key points Completing the squre lets you write qudrtic eqution in the form p( + q) + r = 0. Emples Emple Solve + 6 + = 0. Give your solutions in surd form. + 6 + = 0 ( + ) 9 + = 0 ( + ) = 0 ( + ) = + = = So = or = Write + b + c = 0 in the form b b c 0 Simplify. Rerrnge the eqution to work out. First, dd to both sides. Squre root both sides. Remember tht the squre root of vlue gives two nswers. Subtrct from both sides to solve the eqution. 6 Write down both solutions. Emple 6 Solve 7 + = 0. Give your solutions in surd form. 7 + = 0 7 = 0 Before completing the squre write + b + c in the form b c 7 7 7 9 = 0 8 7 7 = 0 8 = 0 Now complete the squre by writing 7 in the form b b Epnd the squre brckets. Simplify. (continued on net pge)
7 7 8 7 7 6 7 7 7 7 So 7 7 or 7 7 Rerrnge the eqution to work out. First, dd 7 8 6 Divide both sides by. to both sides. 7 Squre root both sides. Remember tht the squre root of vlue gives two nswers. 8 Add 7 to both sides. 9 Write down both the solutions. Prctice Solve by completing the squre. = 0 b 0 + = 0 c + 8 = 0 d 6 = 0 e + 8 = 0 f + = 0 Solve by completing the squre. ( )( + ) = b + 6 7 = 0 c + = 0 Hint Get ll terms onto one side of the eqution.
Solving qudrtic equtions by using the formul A LEVEL LINKS Scheme of work: b. Qudrtic functions fctorising, solving, grphs nd the discriminnts Key points Any qudrtic eqution of the form + b + c = 0 cn be solved using the formul b b c If b c is negtive then the qudrtic eqution does not hve ny rel solutions. It is useful to write down the formul before substituting the vlues for, b nd c. Emples Emple 7 Solve + 6 + = 0. Give your solutions in surd form. =, b = 6, c = b b c 6 6 ()() () 6 0 6 So or Identify, b nd c nd write down the formul. Remember tht b b c is ll over, not just prt of it. Substitute =, b = 6, c = into the formul. Simplify. The denomintor is, but this is only becuse =. The denomintor will not lwys be. Simplify 0. 0 Simplify by dividing numertor nd denomintor by. 6 Write down both the solutions.
Emple 8 Solve 7 = 0. Give your solutions in surd form. =, b = 7, c = b b c ( 7) ( 7) ()( ) () Identify, b nd c, mking sure you get the signs right nd write down the formul. Remember tht b b c is ll over, not just prt of it. Substitute =, b = 7, c = into the formul. 7 7 6 7 7 So or 6 7 7 6 Simplify. The denomintor is 6 when =. A common mistke is to lwys write denomintor of. Write down both the solutions. Prctice Solve, giving your solutions in surd form. + 6 + = 0 b 7 = 0 6 Solve the eqution 7 + = 0 Give your solutions in the form c 7 Solve 0 + + = Give your solution in surd form. b, where, b nd c re integers. Hint Get ll terms onto one side of the eqution. Etend 8 Choose n pproprite method to solve ech qudrtic eqution, giving your nswer in surd form when necessry. ( ) = b 0 = ( + ) c ( ) = 0
Answers = 0 or = b = 0 or = c = or = d = or = e = or = f = or = g = or = 6 h = 6 or = 6 i = 7 or = j = k = or = l = or = = or = b = or = c = 8 or = d = 6 or = 7 e = or = f = or = 7 g = or = h = or = = + 7 or = 7 b = + or = c = + or = d = + 7 or = 7 e = + 6. or = 6. f = 89 0 or = 89 0 = + or = b = c = or = or = = + or = b = + or = 6 = 7 or = 7 7 = 89 0 or = 89 0 8 = 7 7 8 or = 7 7 8 b = + 0 or = 0 c = or =
Sketching qudrtic grphs A LEVEL LINKS Scheme of work: b. Qudrtic functions fctorising, solving, grphs nd the discriminnts Key points The grph of the qudrtic function y = + b + c, where 0, is curve clled prbol. Prbols hve line of symmetry nd for > 0 for < 0 shpe s shown. To sketch the grph of function, find the points where the grph intersects the es. To find where the curve intersects the y-is substitute = 0 into the function. To find where the curve intersects the -is substitute y = 0 into the function. At the turning points of grph the grdient of the curve is 0 nd ny tngents to the curve t these points re horizontl. To find the coordintes of the mimum or minimum point (turning points) of qudrtic curve (prbol) you cn use the completed squre form of the function. Emples Emple Sketch the grph of y =. The grph of y = is prbol. When = 0, y = 0. = which is greter thn zero, so the grph hs the shpe: Emple Sketch the grph of y = 6. When = 0, y = 0 0 6 = 6 So the grph intersects the y-is t (0, 6) When y = 0, 6 = 0 ( + )( ) = 0 = or = So, the grph intersects the -is t (, 0) nd (, 0) Find where the grph intersects the y-is by substituting = 0. Find where the grph intersects the -is by substituting y = 0. Solve the eqution by fctorising. Solve ( + ) = 0 nd ( ) = 0. = which is greter thn zero, so the grph hs the shpe: (continued on net pge)
6 = 6 = When 0, nd y, so the turning point is t the point, 6 To find the turning point, complete the squre. 7 The turning point is the minimum vlue for this epression nd occurs when the term in the brcket is equl to zero. Prctice Sketch the grph of y =. Sketch ech grph, lbelling where the curve crosses the es. y = ( + )( ) b y = ( ) c y = ( + )( + ) Sketch ech grph, lbelling where the curve crosses the es. y = 6 b y = + c y = d y = + e y = 9 f y = + Sketch the grph of y = +, lbelling where the curve crosses the es. Etend Sketch ech grph. Lbel where the curve crosses the es nd write down the coordintes of the turning point. y = + 6 b y = + 7 c y = + 6 Sketch the grph of y = + +. Lbel where the curve crosses the es nd write down the eqution of the line of symmetry.
Answers b c b c d e f
b c 6 Line of symmetry t =.
Solving liner simultneous equtions using the elimintion method A LEVEL LINKS Scheme of work: c. Equtions qudrtic/liner simultneous Key points Two equtions re simultneous when they re both true t the sme time. Solving simultneous liner equtions in two unknowns involves finding the vlue of ech unknown which works for both equtions. Mke sure tht the coefficient of one of the unknowns is the sme in both equtions. Eliminte this equl unknown by either subtrcting or dding the two equtions. Emples Emple Solve the simultneous equtions + y = nd + y = + y = + y = = So = Using + y = + y = So y = Check: eqution : + ( ) = YES eqution : + ( ) = YES Subtrct the second eqution from the first eqution to eliminte the y term. To find the vlue of y, substitute = into one of the originl equtions. Substitute the vlues of nd y into both equtions to check your nswers. Emple Solve + y = nd y = simultneously. + y = + y = 6 = 8 So = Using + y = + y = So y = Check: eqution : + = YES eqution : = YES Add the two equtions together to eliminte the y term. To find the vlue of y, substitute = into one of the originl equtions. Substitute the vlues of nd y into both equtions to check your nswers.
Emple Solve + y = nd + y = simultneously. ( + y = ) 8 + y = 8 ( + y = ) + y = 6 7 = 8 So = Using + y = + y = So y = Check: eqution : + ( ) = YES eqution : + ( ) = YES Multiply the first eqution by nd the second eqution by to mke the coefficient of y the sme for both equtions. Then subtrct the first eqution from the second eqution to eliminte the y term. To find the vlue of y, substitute = into one of the originl equtions. Substitute the vlues of nd y into both equtions to check your nswers. Prctice Solve these simultneous equtions. + y = 8 + y = 7 + y = + y = + y = + y = 7 y = y = + y = 6 + y = y = 9 + y =
Solving liner simultneous equtions using the substitution method A LEVEL LINKS Scheme of work: c. Equtions qudrtic/liner simultneous Tetbook: Pure Yer,. Liner simultneous equtions Key points The subsitution method is the method most commonly used for A level. This is becuse it is the method used to solve liner nd qudrtic simultneous equtions. Emples Emple Solve the simultneous equtions y = + nd + y = + ( + ) = + 6 + = + = = So = Using y = + y = + So y = Check: eqution : = + YES eqution : + = YES Substitute + for y into the second eqution. Epnd the brckets nd simplify. Work out the vlue of. To find the vlue of y, substitute = into one of the originl equtions. Substitute the vlues of nd y into both equtions to check your nswers. Emple Solve y = 6 nd + y = simultneously. y = 6 + ( 6) = + 6 8 = 0 8 = 0 = So = Using y = 6 y = 6 So y = 7 Check: eqution : ( 7) = 6 YES eqution : + ( 7) = YES Rerrnge the first eqution. Substitute 6 for y into the second eqution. Epnd the brckets nd simplify. Work out the vlue of. To find the vlue of y, substitute = into one of the originl equtions. 6 Substitute the vlues of nd y into both equtions to check your nswers.
Prctice Solve these simultneous equtions. 7 y = 8 y = + y = y = 9 y = + 0 = y 9 + y = 8 y = + y = 8 y = 7 y = y = = y + y + = 0 y = y = 8 Etend Solve the simultneous equtions + y 0 = 0 nd ( y ) ( y).
Answers =, y = =, y = =, y = =, y = = 6, y = 6 =, y = 7 = 9, y = 8 =, y = 7 9 =, y = 0 =, y = =, y = =, y = =, y = =, y = =, y =
Solving liner nd qudrtic simultneous equtions A LEVEL LINKS Scheme of work: c. Equtions qudrtic/liner simultneous Key points Mke one of the unknowns the subject of the liner eqution (rerrnging where necessry). Use the liner eqution to substitute into the qudrtic eqution. There re usully two pirs of solutions. Emples Emple Solve the simultneous equtions y = + nd + y = + ( + ) = + + + + = + + = + = 0 ( )( + ) = 0 So = or = Using y = + When =, y = + = When =, y = + = Substitute + for y into the second eqution. Epnd the brckets nd simplify. Fctorise the qudrtic eqution. Work out the vlues of. To find the vlue of y, substitute both vlues of into one of the originl equtions. So the solutions re =, y = nd =, y = Check: eqution : = + YES nd = + YES eqution : + = YES nd ( ) + ( ) = YES 6 Substitute both pirs of vlues of nd y into both equtions to check your nswers.
Emple Solve + y = nd y + y = simultneously. y y y y y y y y yy y y 0 (y + 8)(y ) = 0 So y = 8 or y = Using + y = When y = 8, + ( 8) =, =. When y =, + =, = Rerrnge the first eqution. Substitute y for into the second eqution. Notice how it is esier to substitute for thn for y. Epnd the brckets nd simplify. Fctorise the qudrtic eqution. Work out the vlues of y. 6 To find the vlue of, substitute both vlues of y into one of the originl equtions. So the solutions re =., y = 8 nd =, y = Check: eqution :. + ( 8) = YES nd ( ) + = YES eqution : ( 8) +. ( 8) = YES nd () + ( ) = YES 7 Substitute both pirs of vlues of nd y into both equtions to check your nswers. Prctice Solve these simultneous equtions. y = + y = 6 + y = 0 + y = 0 y = y = 9 + y = + y = 7 y = 6 y = y = + y = 7 y = + 8 y = + y = + y = 9 y = 0 + y = y y = 8 y = Etend y = y = + y = + y =
Answers =, y = 9, y =, y = =, y = =, y = =, y = =, y = 6, y =, y = =, y = 6 = 7, y = =, y = 6 7 = 0, y = =, y = 0 8 = 8, y = =, y = 9 =, y = =, y = 9 0 =, y = 6 =, y = = =, y =, y = = = 7 7, y = 7, y = 7
Solving simultneous equtions grphiclly A LEVEL LINKS Scheme of work: c. Equtions qudrtic/liner simultneous Key points You cn solve ny pir of simultneous equtions by drwing the grph of both equtions nd finding the point/points of intersection. Emples Emple Solve the simultneous equtions y = + nd + y = grphiclly. y = y = hs grdient nd y-intercept. y = + hs grdient nd y-intercept. Rerrnge the eqution + y = to mke y the subject. Plot both grphs on the sme grid using the grdients nd y-intercepts. Lines intersect t = 0., y =. Check: First eqution y = + :. = 0. + YES Second eqution + y = : 0. +. = YES The solutions of the simultneous equtions re the point of intersection. Check your solutions by substituting the vlues into both equtions.
Emple Solve the simultneous equtions y = nd y = + grphiclly. 0 y Construct tble of vlues nd clculte the points for the qudrtic eqution. Plot the grph. Plot the liner grph on the sme grid using the grdient nd y-intercept. y = hs grdient nd y-intercept. The line nd curve intersect t =, y = nd =, y = Check: First eqution y = : = YES = YES Second eqution y = + : = + = + YES YES The solutions of the simultneous equtions re the points of intersection. Check your solutions by substituting the vlues into both equtions. Prctice Solve these pirs of simultneous equtions grphiclly. y = nd y = + b y = nd y = 7 c y = + nd y = Solve these pirs of simultneous equtions grphiclly. + y = 0 nd y = + 6 b + y = nd y = c + y + = 0 nd y = Hint Rerrnge the eqution to mke y the subject.
Solve these pirs of simultneous equtions grphiclly. y = nd y = + b y = nd y = c y = nd y = + + Solve the simultneous equtions + y = nd + y = grphiclly. Etend Solve the simultneous equtions + y = nd + y = i grphiclly ii lgebriclly to deciml plces. b Which method gives the more ccurte solutions? Eplin your nswer.
Answers =, y = b =, y = c = 0., y =. =, y = b = 0., y = 0. c =, y = =, y = 0 nd =, y = b =, y = 7 nd =, y = c =, y = nd =, y = =, y = nd =, y = i =., y = nd = 0., y = ii =., y =.8 nd = 0., y =.8 b Solving lgebriclly gives the more ccurte solutions s the solutions from the grph re only estimtes, bsed on the ccurcy of your grph.