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Mythila Publishers, Puduvayal 41 X std Mathematics Made Easy
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1-=y6-5=y
5= y+6
5= 4+ y+2
(1)in 4=zand2xapplying By
2=x6=3x
8-14=3x
12=8+3x
14=2(4)+3x
(4),in 4=zapplying By
)()(
).....()()(
)(
).....()()(
Þ
Þ
=Þ-=--
=+
=-
=--´
=+-
=+-´
=++
=+-
=++
412345
1423
23
523322
1632
1822422
4142321
92
5
zz
zx
zx
zx
zyx
zyx
zx
zyx
zyx
a) 1432
0111
0421
=+=+-=+-xzzyyx
;;
4
1z4=
z
14=cIf
3
1=y3=
13=bIf
2
12=
12,=aIf
3=b6-=2b-
4-=2b-2
(1),in2=aapplying By
4=c8-=2c-
6-=2c-2
(4),in2=aapplying By
2=a8=4a
14=2c+3a
6-=2c-a(3)+(4)
-(4)---62
2220
402,2(2)+(1)
-(3)---14=2c+3a
-(2)---1-=c-b
-(1)---4-=2b-a
1;
1,
1
=Þ
Þ
=Þ
Þ
Þ
Þ
===
y
xx
c = -a -
c = -b - a +
c = -b + a -
cz
by
ax
b)15
22
4
5
11
3
11
4
1
3
1
4
1
2
1=+-==-+
zyxyxzyx;;
., z , y x
ba
aa
c a
c a
c a
c b a
cb
a
ba
c b a
c
ba
cz
by
ax
213is,That
1(2)in3
1Subs
3
13215,
,(4)in2
1cSubs
2
1c77 – 154c– (5),,(4)Solve
......(5)163032)(
32606
......(4)3415
(3)and(1)inSub.(2)
.......(3)3260315.
15
224
5
......(2);3
.......(1)3436
;4
1
3
1
42
1,
1,
1
===
=Þ=
=Þ=-
=
=Þ=
=+¸
=+
=-
=+-
=+-
=
=-+
=-+
===
c) 1323
27
16
12
=++=+=-zyzyx
;
.14z,18y,10x Therefore,
1442270
(2),in10x ng Substituti
181230
(1),in10x ng.Substituti
1033033)(
484214,(1)
282421)(
156642,(3)
126621,3(2)
)3.....(.783z2y
1323
)2.......(4227
372
27
12
)1.......(123
262
16
12
1323
;27
16
12
===
=Þ=-
=
=Þ=-
=
=Þ=+
=-´
=++
=+´
=-´
=+
=+
=-
=-Þ+=-
=-
=-Þ+=-
=++=+=-
z z
y y
x x
y x
y x
z y
z x
zy
z x
zxzx
y x
yxyx
zyzyx
Mythila Publishers, Puduvayal 42 X std Mathematics Made Easy
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d) )( zyzy
x +-=+=+=+ 11052102
320
25.=z30,=y35,=x Solutions,
30=y105=y+75
105=y+3(25)
(3),in25=zApplying
35=x 30-=100-2x
30-=4(25)-2x
(4),in25=zApplying
25=z325-=13z-
295=9z+2x
30-=4z-2x (5),-(4)
-(5)---295=9z+2x
315=9z+3y+0x 3,(3)
20-=0z+3y-2x (1),
(4)---30-=4z-2x
10-=4z-3y+0x
20-=0z+3y-2x (2).+(1)
-(3)---105=y+3z
z)+(y-110=5+2z
(2)-----10-=4z-3y
52102
3
(1)-----20-=3y-2x
102
320
Þ
Þ
Þ
´
+=+
+=+
zy
yx
2. Discuss the nature of solutions of the following system of equations a) 1145,2,52 -=+---=+-=-+ zyxzyxzyx
00(4)(5)
)4.......(32
)5......(323),(
936(2)(1)
)3......(1145
)2........(2
)4.......(32(2)(1)
)2.......(2
)1......(.52,
)3......(1145
)2........(2
)1.........(,52
=-
=+
=+¸
=+-
-=+--
-=+-
=++
=+-
=-+
-=+--
-=+-
=-+
y x
y x
y x
zyx
zyx
y x
. z y x
z y x
zyx
zyx
zyx
system has an infinite number of solutions. b) 2122133 =+--=+--=-+ zyxzyxzyx ;;
10),(
4222)4(
)5.......(322),3()1(
)3.......(2
)1.......(133
)4.......(2),2()1(
)2.......(122
)1.......(133
)3.......(2
)2.......(122
)1.......(133
-=-
=-´
=-+
=+--
=-+
=-+
=+--
=-+
=+--
=+--
=-+
zx
zx
zyx
zyx
zx
zyx
zyx
zyx
zyx
zyx
system is inconsistent and has no solution. c) 321252362 =--=+--=-+ zxzyxzyx ;;
solution.manyyIinfinitel
0=0
6-=4z+4z-6-
6-=4z+2z)+2(3-
(4)inx ofvalueApply
2z+3=x (3),From
(4)-----6-=4z+2x -
12=5z+2y3x-
6=z2y+x (2),+(1)
(3)-----3=2zx
(2)-----12=5z+2y3x -
(1)-----6=z2y+x
--
-
-
--
-
d)2
12343132
-=++-=-+-+-=+ zyxzyxxzy ;);(
solution.nohassystem
7=0
9-=10y+4x
2-=10y+4x (5).-2(4)
(5)---9-=10y+4x
1-=2z+4y+6x
8=2z6y+2x (3),+2(2)
(4)---1-=5y+2x
4=z3y+x
3=z+2y+3x (2),+(1)
-(3)-----1-=2z+4y+6x
2
1=z+2y+3x
-(2)-----4=z3y+x
-(1)-----3=z+2y+3x
1)+x 3(=z+2y
\
---
---
-
---
-
e) 27234
=+++
=+
=+
zyxyxxzzy
;
Mythila Publishers, Puduvayal 43 X std Mathematics Made Easy
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solution.unique3=x
81=27x
81=4z+7x
0=4z-20x (5),+4(4)
(5)---81=4z+7x
81=3z+3y+3x
0=z+3y-4x 3(3),-(1)
-(4)---0=z-5x
0=2z-3y+x
0=z+3y-4x (2),+(1)
(3)---27=z+y+x
-(2)---0=2z-3y+x
3y+3x =2x +2z
23
(1)---0=z+3y-4x
4x+4z=3z+3y
x)+4(z=z)+3(y
34
\
+=
+
+=
+
yxxz
xzzy
3. In an interschool atheletic meet, with 24 individual events, securing a total of 56 points, a first place secures 5 points, second place secures 3 points, and a third place secures 1 point. Having as many third place finishers as first and second place finishers, find how many athletes finished in each place
No. of I, II and III place finishers = x , y, z
12.finishersplacethirdofNo.
8finishersplacesecondofNo.
4finishersplacefirstofNo.
8y12y8
(3)in12z,4x Sub.
4x 82x )(
363y3x ,(3)3
443y5x
12yx
(3),in12zSub.
12z24zz,(1)in(3)Sub.
(3).........zyx
(2).......56z3y5x
(1).......24zyx
56pointsofnumber Total
24eventsofnumber Total
=
=
=
=Þ=+
==
=Þ=-
=+´
=+
=+
=
=Þ=+
=+
=++
=++
=
=
4. Sum of thrice the first number, second number and twice the third number is 5. If thrice the second number is subtracted from sum of first
number and thrice the third we get 2. If the third number is subtracted from the sum of twice the first, thrice the second, we get 1. Find numbers.
3.z2,y.1,x Therefore,
.1x 5623x
(1),in3zand2ySub.
3z77z14
(5),.in2ySub.
2y63y)(
17z10y
(5)7......7z7y)(
33z9y6x ,3(3)
104z2y6x ,2(1)
(4)......17z.10y),(
69z9y3x ,3(2)
52zy3x ,1(1)
(3).......1z3y2x
(2).......23y3zx
......(1).52zy3x
===
=Þ=++
==
=Þ=+-
=
=Þ=+
-=-
=+--
=-+´
=++´
-=--
=+-´
=++´
=-+
=-+
=++
5. Vani, her father and her grand father have an average age of 53. One-half of her grand father’s age plus one-third of her father’s age plus one fourth of Vani’s age is 65. Four years ago if Vani’s grandfather was four times as old as Vani then how old are they all now ?
511598424
(1)in84,24sub.
841296
(3)in24Sub
241687x )(
(4).........14423
24282,(3)
(4).........14423)(
)2........(780643
6364444,(1)
)3....(.124
)4(4)4(
)2........(780643
654
1
3
1
2
1
)1........(159
533
zfather grand
fatherher
, Vani,ofagePresent
y y
z x
z z
x
x
. z x
zx
. z x
zyx
z yx
zx
xz
zyx
x yz
z yx
z yx
y
x
=Þ=++
==
=Þ=-
=
=Þ=-
=-
=-´
=--
=++
=++´
=-
-=-
=++
=++
=++
=++
=
=
=
Mythila Publishers, Puduvayal 44 X std Mathematics Made Easy
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∴ Vani' present age = 24 years Father's present age = 51 years Grand father's age = 84 years
6. Sum of the digits of a three-digit number is 11. If the digits are reversed, the new number is 46 more than fve times the former number. If the hundreds digit plus twice the tens digit is equal to the units digit, Find original 3 digit number ?
731.numberdigitReversed
137
7(3)10100(1)
z10y100x nodigit3
7z11z31
11zyx
(1)in3y1,Subx
1x 1192x
(4)in3ySub
38428y,)(
375y22x ,1(5)
12133y22x ,11(4)
(5)...........375y22x ,27by
999135y94x )(
4695z40y499x ,1(2)
104595z95y95x ,95(1)
(4).........113y2x (3),(1)
(3)...........0z-2yx
z2yx Given(3)
(2)...........46-95z-40y499x
46z)y10x (1005x y10z100Given(2)
(1)...........11zyx Given(1),
x10y100znumber Reversed
z10y100x number Rqd
digitsunit'
digits10'
digits100'
=
=
++=
++=
=Þ=++
=++
==
=Þ=+
=
=Þ=-
=+´
=+´
=+¸
=+-
-=-+´
=++´
=++
=+
=+
=+
+++=++
=++
++=
++=
=
=
=
y
z
y
x
7. There are 12 pieces of five, ten and twenty rupee
currencies whose total value is ₹105. When first 2 sorts are interchanged in their numbers its
value will be increased by ₹20. Find the number of currencies in each sort.
(3).......12520z5y10x
(2).......10520z10y5x
(1).........12zyx
currenciesrupee5ofNo.
currenciesrupee10ofNo.
currenciesrupee5ofNo.
=++
=++
=++
=
=
=
z
y
x
2notesrupee20ofNo.
3notesrupee10ofNo.
7notesrupee5ofNo.
7x 1223x
(1)in3y2,zSub
3y96y
(4)in2zSub
2z105z.,)(
279z3y3,(4)
174z3y,1(5)
(5).....174z3y
8520z15y.,)(
12520z5y10x ,1(3)
21040z20y10x 2,(2)
(4)..........93zy
4515z5y.,)(
10520z10y5x ,1(2)
605z5y5x 5,(1)
=
=
=
=Þ=++
==
=Þ=+
=
=Þ=-
=+´
=+´
=+
=+-
=++´
=++´
=+
-=---
=++´
=++´
3.2 GCD and LCM of Polynomials
1. Find the GCD of the polynomials
a) 3552and2 2323 -+-+-+ xxxxxx
3552g(x)
2f(x)23
23
-+-=
+-+=
xxx
xxx
2
223 +-+ xxx 3552 23 -+- xxx
4222 23 +-+ xxx
777 2 x x -+-
−7 ( 12 x + x - ) x
12 x + x - 223 +-+ xxx
xxx +- 23
222 2 +- xx
222 2 +- xx
0
GCD= 12 x + x -
b) 1261263333 2323 ++++++ xxxxxx and
)1(3
3333)(
)22(6
126126)(
23
23
23
23
+++=
+++=
+++=
+++=
xxx
xxxxg
xxx
xxxxf
1
123 +++ xxx 22 23 +++ xxx
123 +++ xxx
12 +x
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1+x
12 +x 123 +++ xxx
003 +++ xx
12 +x
12 +x
0
G.C.D = 3( 12 +x )
c) 1821123and4860306 2323 -+--+- xxxxxx
)8105(6
4860306)(
)674(3
1821123)(
23
23
23
23
-+-=
-+-=
-+-=
-+-=
xxx
xxxxg
xxx
xxxxf
1
8105 23 -+- xxx 674 23 -+- xxx
8105 23 -+- xxx
232 +- xx
2-x
232 +- xx 8105 23 -+- xxx
xxx 23 23 +-
882 2 -+- xx
462 2 -+- xx42 -x
= )2(2 -x
1-x
2-x 232 +- xx
xx 22 -
2+- x2+- x
0
GCD coefficients = 3, )2(3GCD -= x
2. Find the GCD of the polynomials
a) 3523 2334 +-+--+ xxxxxx and
)32(325)(
23)(223
34
-+=+-+=
--+=
xxxxxxg
xxxxf
2+x
3523 +-+ xxx 230 324 --++ xxxx
xxxx 35 234 +-+
3452 23 --+ xxx
61022 23 +-+ xxx
963 2 -+ xx )32(3 2 -+= xx
1-x
322 -+ xx 3523 +-+ xxx
xxx 32 23 -+
322 +-- xx
322 +-- xx
0
GCD = 322 -+ xx
b) 11111 234 -+-- xxxx and
1111)(
1)(23
4
-+-=
-=
xxxxg
xxf
11+x
1111 23 -+- xxx 1000 234 -+++ xxxx
xxxx 1111 234 -+-
11111 23 -+- xxx
1211112111 23 -+- xxx
120120 2 +x
= )1(120 2 +x
11-x
12 +x 1111 23 -+- xxx
xxx +- 23 0
1111 2 -- x
1111 2 -- x
0
GCD= 12 +x
c) xxxxxxxx 88144241263 234234 -++--+ and
)842(3
241263)(
)4472(2
88144)(
23
234
23
234
--+=
--+=
-++=
-++=
xxxx
xxxxxg
xxxx
xxxxxf
2
842 23 --+ xxx 4472 23 -++ xxx
16842 23 --+ xxx
12123 2 ++ xx
)44(3 2 ++ xx
2-x
442 ++ xx 842 23 --+ xxx
xxx 44 23 ++
882 2 --- xx
882 2 --- xx
0
GCD = 442 ++ xx3. Find the LCM of the following
a) 4224 48,8 yxyx
23
444
424
4242
243
2424
yx48=
y32=L.C.M
yx32=
yx32222=48
2=
222=8
´´´
´´´
´´´´´´
´´
´´´´
x
yx
yx
yxyx
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b) 232 84 yxyx ,
23
3
233
223
22
22
yx8=
y2x32=L.C.M
yx2=
yx3222=y8x
yx2=
yx22=y4x
´´
´´
´´´´
´´
´´´
c) cbaba 2223 129 ,-
cb12a=
c..b.a3.2=L.C.M
c.b?.3.a2=
.cb.a3..2=cb12a
b.a.2=
b.a.3.3-=b9a
22
222
222
22222
232
2323
d) 222 81216 nnmm ,,-
22
224
232
22222
4
n48m=
n.3m2=L.C.M
n.2=8n
nm3.2-=n12m-
m.2=16m
4. Find the LCM of the following
a) 205105 2 -- xx ,
)4(5
)2)(2(5LCM
)2)(2(5205
)2(5105
2
2
-=
+-=
+-=-
-=-
x
xx
xxx
xx
b) 121 24 +-- xxx ,
)1)(1(
)1)(1)(1(LCM
)1)(1(12
)1)(1)(1(1
4
2
2
24
--=
+-+=
--=+-
+-+=-
xx
xxx
xxxx
xxxx
c) 423 22 -+- ppp ,
1)2)(p2)(p+(p=L.C.M
2)-(p2)+(p=4-p
2)1)(p(p232
2
--
--- = p + p
d) 364352 22 --- xxx ,
3)3)(x +1)(x +4(2x =L.C.M
3)+3)(x 4(x =
9)4(x=36-4x
3)1)(x +(2x =35x 2x22
2
-
-
-
---
5. Find the LCM of the following
a) 9327 223 --- xxx ,)(,
)27)(3(
)93)(3)(3(LCM
)3()3(9
)3()3()3(
)93)(3(27
3
2
2
2
23
-+=
+++-=
+-=-
--=-
++-=-
xx
xxxx
xxx
xxx
xxxx
b) 33323 2786432 yxyxxyx --- ,)(,)(
)9y+6xy-(4x3y)-(2x8x=L.C.M
)9y+6xy-(4x3y)-(2x=
)(3y)+(2x)(3y)-((2x)3y)-(2x=
(3y)-(2x)=27y-8x
3y)-8(2x=
3y)]-[2(2x=6y)-(4x
3y)-(2xx=
3y)]-[x(2x=3xy)-(2x
2232
22
22
3333
3
33
22
222
3.3 Relationship between LCM and GCD
1. Find LCM ,GCD. Verify GCD LCMx gxf .)().( =
a) 22 3521 xyyx ,
verified.
735735
)7)(105(=)(35xy21x2y
GCD×LCM=g(x)×f(x)
7xy=
y·x ·7
105
537=L.C.M
7535=g(x)
7321=f(x)
3333
222
22
22
22
22
yx = yx
xy y x
GCD =
y x =
· y · x · ·
· x y · = xy
y · x · y = x
b) )(),( 3322 812 yxyx --
verified.
))((96
y)-4(x)()(24=GCD×LCM
))((96
)(8)(12=g(x)×f(x)
)()(24
))(()(24=L.C.M
y)-4(x=GCD
)y)(-2.2.2(x
)(8=g(x)
))(.(3.2.2
),(12=f(x)
3322
2222
3322
3322
2222
22
22
33
22
yxyx
yxyxyx
yxyx
yxyx
yxyxyx
yxyxyxyx
yxyx
yx
yxyx
yx
--=
+--
--=
--
+--=
++--
+-=
-
-+=
-
c) )(),)(( 111 33 ++- xxx
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verified.Hence
1)+(x )y(x
1)+(x )y(x=GCD×LCM
1)+(x )y(x
1)+(x1)+1)(x (x=g(x)×f(x)
yx=
)(y)(x=
)y+)(xy(x=
1)+x-1)(x+x+1)(x+1)(x-(x=L.C.M
1+x =GCD
1)+x -1)(x+(x =
1)+(x=g(x)
1)+1)(x +x +1)(x-(x =
1)+1)(x -(x=f(x)
66
66
66
33
66
2323
3333
22
2
3
2
3
-=
-
-=
-
-
-
-
d) )(),( xyxxyyx ++ 222
verified.iprelationshtheHence
y)+y(xx= y)+y(xx
y)+x(x· y)+xy(x= y)+x(x· y)+(xxy
GCD×LCM=g(x)×f(x)
y)+xy(x=L.C.M
y)+x(x=GCD
y)+x(x=xy)+(x=g(x)
y)+(xxy=)xy+y(x=(f(x)
2222
2
22
2. Find the LCM
a) axaxxax 3327 234 --- isGCDwhose)(,
222
222
2
22
22
33
3a)-(x)9a+3ax-x(x=
3a)-(x
3a)-(x)9a+3ax-3a)(x-(xx=
GCD
g(x)×f(x)=LCM
GCD×LCM=g(x)(x).f
3a)-(xGCD
3a)-(x=g(x)
)9a+3ax-3a)(x-(xx=
)(3a)+x(3a)-3a)(x-(xx=
)(3a)-x(x=f(x)
=
b) 265124 22 -+--+ aaaaa isGCDwhose,
GCD×LCM=g(x)×(x)f
2-aGCD
2)-3)(a-(a=65=g(x)
2)-6)(a+(a=124=f(x)2
2
=
+-
-+
aa
aa
3)-(a2)-6)(a+(a=LCM
2-a
2)-3)(a-(a2)-6)(a+(a=
GCD
g(x)×f(x)=LCM
3. Find the GCD
)2)(1(24x:LCM)23(8),(12 323434 --+-- xxxxxxx
1)-(x 4x=GCD
2)-1)(x-(x24x
2)+3x-(x8x1)-(x12x=
LCM
g(x)×f(x)=GCD
GCD×LCM=g(x)×(x)f
2)-1)(x-(x24x=GCD
2)+3x-(x8x=
)2x+3x-8(x=g(x)
1)-(x12x=
)x-12(x=f(x)
2
3
223
3
22
234
3
34
2. ))((LCM),(),( 2233422433 yxyxyxyyxxyx ++++++
) y+xy-(x=
) y+xy+)(x y+(x
) y+xy+)(x y2+xy-)(x y+(x=
LCM
g(x)×f(x)=GCD
GCD×LCM=g(x)×(x)f
) y+xy+)(x y+(x=LCM
) y+xy+)(x y+xy-(x=
(xy)-) y+(x=
) y+x2y+(x=g(x)
) y+(x=f(x)
22
2233
22233
2233
2222
2222
424
33
4. Given the LCM and GCD of two polynomials p(x) and q(x) find the unknown polynomial
i) LCM : 701110 23 ++- aaa GCD : 7-a
p(x) : 35122 +- aa q(x) = ?
7)-(a2)+(a=
35)+12a-(a
7)-a70)(+11a+10a-(a=
(x)p
GCD×LCM=q(x)
GCD×LCM=q(x)×(x)p
2
23
ii) LCM: )()( 422422 yyxxyx +++
GCD : )( 22 yx -
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q(x) = ))(( 2244 yxyxyx +--
` p(x) :?
) y+xy+(x=
xy)-y+(x
) y+xy+)(x y+xy-(x=
xy)-y+(x
(xy)-) y+(x=
xy)-y+)(xy-(x
)y+yx+(x)y-(x=
xy)-y+)(xy-(x
)y-)(xy+yx+(x)y+(x=
q(x)
GCD×LCM=p(x)
22
22
2222
22
2222
2244
422444
2244
22422422
3.4 Rational Expressions
1. Reduce the rational expressions to its lowest form
i) 9
32 -
-
x
x
3
1
33
3
9
32
+=
-+
-=
-
-
x
xx
x
x
x
))((
ii)168
162
2
++
-
xx
x
4
4
44
44
168
162
2
+=
++
-+=
++
-
x
x-
xx
xx
xx
x
))((
))((
iii)xx
x
+
-2
2 1
x
1)-(x
1)+x(x
1)-(x1)+(x
=
=+
-
xx
x2
2 1
iv)44
18112
2
+-
+-
xx
xx
2)-(x
9)-(x
2)-2)(x-(x
2)-9)(x-(x
=
=+-
+-
44
18112
2
xx
xx
v)xxx
xx
98
81923
2
-+
+
1)-9)(x +x(x
9)+9x(x
)98x(
9)+9x(x
98
819223
2
=
-+=
-+
+
xxxxx
xx
1-x
9=
vi)ppp
pp
64242
40323
2
+-
--
4)-(p2p
5+p
4)-(p8)-(p2p
5)+8)(p-(p
32)+12p-2p(p
5)+8)(p-(p2
=
=
=+-
--
ppp
pp
64242
40323
2
2. Find the excluded values, if any
i) x
x
8
10+
0 valueexcludedThus,
when,undefinedis
=
=
=+
0
088
10
x
xx
x
ii)5138
272 ++
+
pp
p
1,8
5
0)1)(58(ifundefinedis)1)(58(
27
--=
=++++
+
p
pppp
p
iii)12 +x
x
valuesexcludedrealnobecan there
01
101
0
2
2
2
¹+
+³+
³
x
x
x
iv)252 -y
y
5and-5 valuesexcluded
or
))((
0rDenominato
))((
=
-==
=+-
=
+-=
-
55
055
55252
yy
yy
yy
y
y
y
v) 652 +- tt
t
3.and2 valuesexcluded
3tor2t
03)-2)(t-(t
=
==
=
=+- 0652 tt
vi)2
862
2
-+
++
xx
xx
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1 valueexcluded
0rdenominato
1)-2)(x+(x
4)+2)(x+(x
=
=-
=-
+=
=-+
++
01
1
4
2
862
2
x
x
x
xx
xx
vii)xxx
x
6
2723
3
-+
-
3and20, valueexcluded
,,
2)-3)(x+x(x
0rdenominato
2)-3)(x+x(x
9)+3x+3)(x-(x
6)-x+x(x
)3+x(3)+3)(x-(x
2
2
22
=
=-==
=
=
=
=-+
-
230
0
6
2723
3
xxx
xxx
x
3.5. Operations of Rational Expressions
1. Simplify
i) 52
3 27
9 x
y
y
x´
2
3
yx=
ii) 34
224 1
1 ba
x
x
bx -´
-
ba
xx
ba
xx
x
bx
4
4
34
24
1
11
1
)(
))((
+=
+-´
-=
iii)4
2
2
2
20
6
2
4
y
xz
z
yx´
3
3
5
3
y
x=
iv) t
t
t
t
10
126
84
5 3 -´
-
4
3
10
26
24
5
2
3
t
t
t
t
t
=
-´
-=
)(
)(
v) 3
22
3
12
7
2110
)( -
-+´
-
+-
p
pp
p
pp
3
4
333
34
7
37
-
+=
---
-+´
-
--=
p
p
ppp
pp
p
pp
))()((
))(())((
vi)2032
169
43
42
22
-+
-´
+
+
xx
yx
yx
x
5)-(2x
)(
5)- 4)(2x+(x
))((
yx
yxyx
yx
x
43
4343
43
4
+=
-+´
+
+=
vii)22
22
22
33 2
693 yx
yxyx
yxyx
yx
-
++´
++
-
)(
))((
))((
))((
))((
yx
yxyx
yxyx
yxyx
yxyx
yxyxyx
23
2322
22
+
++=
-+
++´
++
++-=
2. Simplify
i) 4
4
3
714
y
x
y
x¸
33
44
6
7
314
yx
x
y
y
x
=
´=
ii)4
4
4
162
+
-¸
+
-
x
x
x
x
44
4
4
44
+=-
+´
+
+-=
xx
x
x
xx ))((
iii) 145
49
44
2832
2
2
2
--
-¸
++
-+
bb
b
bb
bb
2+b
4-b
7)+7)(b-(b
2)+(b7)-(b
2)+2)(b+(b
))((
=
´+-
=74 bb
iv)2
2
12
6
4
2
y
xx
y
x --¸
+
3
3
23
12
4
2 2
-=
+-´
+=
x
y
xx
y
y
x
))((
v)4113
3118
123
32162
2
2
2
--
++¸
--
--
xx
xx
xx
xx
1
492
11
412
138
413
113
1238
3118
4113
123
3216
2
2
2
2
2
2
-
+-=
+-
--=
++
-+´
-+
-+=
++
--´
--
--=
x
xx
xx
xx
xx
xx
xx
xx
xx
xx
xx
xx
))((
))((
))((
))((
))((
))((
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vi)50355
56
672
3522
2
2
2
---
++¸
++
++
aa
aa
aa
aa
55)+1)(a+(a
5)+2)(a+5(a-
2)+3)(a+(2a
3)+1)(2a+(a-=´=
vii)tt
tt
t
tt
42
823
3
822122
22
+
-+¸
+-
)(
))((
)())((
123
4
432
22
3
12432
-=
-+
+´
--=
t
tt
tt
t
tt
3. If 422
82
33
432
2
2
2
--
-+=
-
-+=
aa
aay
a
aax , find 22 -yx
9
4
3
2
3
2
2)-4)(a+(a
1)+2)(a-2(a
1)-1)(a+3(a
1)-4)(a+(a
422
82
33
43
22
22
2
2
2
2
=
úû
ùêë
é=ú
û
ùêë
é
=
´=
--
-+¸
-
-+=
-yx
y
x
y
x
aa
aa
a
aa
y
x
4. If a polynomial 1452 --= xxxp )( is divided by
another polynomial q(x) we get 2
7
+
-
x
x, find q(x)
44
222
7
2
7145
2
7
2
2
++=
++=
=+
-´
+
-=
--
+
-=
xx
xxx
xxq
x
x
xq
xx
x
x
xq
xp
))((q(x)
2)+7)(x-(x)(
)(
)(
)(
3.6. Addition and Subtraction
1. Simplify
a)2
1
2
1
-
-+
-
+
x
xx
x
xx )()(
2
2
2
2
)1()1(
22
-=
-
-++=
-
-++=
x
x
x
xxxx
x
xxxx
b)283
412
283
36202
2
2
2
--
+++
--
++
xx
xx
xx
xx
)4)(7(
328
283
41236202
22
+-
+=
--
---++=
xx
x
xx
xxxx
7
8
)4)(7(
)4(8
-=
+-
+=
x
xx
x
c)2
1
3
2
-
-+
+
+
x
x
x
x
)3)(2(
722
)3)(2(
324
)3)(2(
)1)(3()2)(2(
2
22
+-
-+=
+-
-++-=
+-
-+++-=
xx
xx
xx
xxx
xx
xxxx
d)xy
y
yx
x
-+
-
33
22
22
33
33
yxyx
yx
yxyxyx
yx
yx
yx
y
yx
x
++=
-
++-=
-
-=
--
-=
))((
e)4
252
4
212 2
-
+--
-
-+
x
xx
x
xx )())((
4
42
4
)252()232(
4
)252(
4
)232(
22
22
-
-=
-
+----=
-
+--
-
--=
x
x
x
xxxx
x
xx
x
xx
f)1
1
1
42 -
+-
- x
x
x
x
1
1
)1)(1(
)1)(1(
)1)(1(
)12(
)1)(1(
)12(4
)1)(1(
)1)(1(4
1
1
)1)(1(
4
2
2
+
-=
+-
---=
+-
+--=
+-
++-=
+-
++-=
-
+-
+-=
x
x
xx
xx
xx
xx
xx
xxx
xx
xxx
x
x
xx
x
g)2
1
2
32222
23
+-
+
++
xx
xx
)(
22
3
22
223
)2(
12
)2(
)2()32(
+
+=
+
+-++=
x
x
x
xxx
h)158
1
23
1
65
1222 +-
-+-
++- xxxxxx
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))()((
)(
))()()((
))((
))()()((
))()()((
))()()((
))(())(())((
))(())(())((
513
9
5123
29
5123
1811
5123
2315856
5123
125351
35
1
12
1
23
1
2
222
---
-=
----
--=
----
+-=
----
-+-+-++-=
----
-----+--=
---
--+
--=
xxx
x
xxxx
xx
xxxx
xx
xxxx
xxxxxx
xxxx
xxxxxx
xxxxxx
2. Which rational expression should be subtracted
from 8
868
2
+
++
x
xx to get
42
32 +- xx
42
1
42
3
42
4
42
3
)42)(2(
)2)(4(
42
3
8
86)(
42
3)(
8
86
2
22
22
28
2
23
2
+-
+=
+--
+-
+=
+--
+-+
++=
+--
+
++=
+-=-
+
++
xx
x
xxxx
x
xxxxx
xx
xxx
xxxp
xxxp
x
xx
3. 22
21
12
12
12
12
BA
B
BAx
xB
x
xA
--
-+
-=
-
+= find,If
14
28
14
144144
12
12
12
12
B)+(A
121
B)-(AB)+(A
B-A
B)-(AB)+(A
2B-B)+(A
B-(A)(
2121
2
2
2
22
22
22
+
+=
-
+-+++=
+
-+
-
+=+
=-
--
=
=
+-
-=
--
-
x
x
x
xxxx
x
x
x
xBA
BA
B
BA
BA
B
BABA
B
BA
28
1421
28
14
B)+(A
1
2
2
22
2
2
+
+=
--
-
+
+=
x
x
BA
B
BA
x
x
4. If ,1
1,
1 A
+=
+=
xB
x
x then prove that
2
222
)1(
)1(2)()(
+
+=
¸
--+
xx
x
BA
BABA
2
2
2
222
2
2
22
22
222222
)1(
)1(2
1
)1(
)1(2)()(
1
1
1
)1(
)1(2
1
1
12
)B+2(A=
)B+2AB-(A+)B+2AB+(A)()(
+
+=
´+
+=
¸
--+
=+
´+
=
+
+=
úúû
ù
êêë
é÷ø
öçè
æ
++÷
ø
öçè
æ
+=
=--+
xx
x
xx
x
BA
BABA
xx
x
x
B
A
x
x
xx
x
BABA
5. Pari needs 4 hours to complete a work. His
friend Yuvan needs 6 hours to complete the same work. How long will it take to complete if they work together?
minutes.24hours2=5
12=both bytaken Hrs
12
56
1
4
1=togetherbydonework
6
1=hr1in doneWork
hours6=Yuvan bytaken Hrs4
1=hr1in doneWork
hours 4=paribytaken Hrs
=
+
6. Iniya bought 50 kg of of apples and bananas. She
paid twice as much per kg for the apple as she did for the banana. If Iniya bought ₹ 1800 worth
of apples and ₹ 600 worth bananas, then how many kgs of each fruit did she buy?
20,30503
2xin(1),.sub.
3
2600
900in(3),Sub.
900
getwe(2),in2sub.By
)3......(600
)2....(1800
)1(..........50
,banana&appleofprice
,banana&appleofweight
==Þ=+
=Þ=
=
=
=
=
=+
=
=
yxx
xyy
xq
xq
qp
qy
px
y x
qp
yx
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3.7 Square Root of Polynomials
1. Find the square root of the following expressions
a) 201648256 )()()()( dxcxbxax ----
10824
201648
16
256
)()()()(
)()()()(
dxcxbxax
dxcxbxax
----=
----=
b)4124
888
81
121
)()()(
)()()(
cbbacb
cbyxba
---
-++
262
444
4124
888
)()()(9
)()()(11
)()()(81
)()()(121
cbbacb
cbyxba
cbbacb
cbyxba
---
-++=
---
-++
c) 14412
16128
81
144
hgf
cba
726
864
726
864
14412
16128
3
4
9
12
81
144
hgf
cba
hgf
cba
hgf
cba
=
=
d)448
16124
100
400
zyx
zyx
2
64
224
862
448
16124
2
10
20
100
400
x
zy
zyx
zyx
zyx
zyx
=
=
2. Find the square root of the following expressions
a) 25204 2 ++ xx
|5+2x|=
5)+(2x=
)5+2(2x)(5)+((2x)
2
22=
++= 25204 2 xx
b)
16
1
2
1
21427
2
2
+-
++
xx
xx
÷ø
öçè
æ-
+=
÷ø
öçè
æ-
+=
÷ø
öçè
æ+-
+++=
+-
++
4
1
147
4
1
147
4
1
4
12
214147
16
1
2
1
21427
2
2
22
2
2
2
x
x
x
x
xx
xxx
xx
xx
)(
)(
.
c)36
211
xx++
3
2
3
323
2
36
11
11
112
11
211
x
x
xx
xx
+=
÷ø
öçè
æ+=
++=
++=
..
d) ))()(( 13282137294 222 ----+- xxxxxx
|1)+2)(7x-1)(x-(4x|
1)+(7x1)-2)(4x-1)(x+2)(7x-1)(x-(4x
))()((
=
=
----+-= 13282137294 222 xxxxxx
e) ))()(( 13212316 222 ++-+-+ xxxxxx
|)1)(12)(13(|
)1)(12)(1)(13)(12)(13(
)132)(123)(16( 222
++-=
+++-+-=
++-+-+=
xxx
xxxxxx
xxxxxx
f) ))()(( 23
11
3
424
2
31
6
172 222 ++++++ xxxxxx
6
|2)+3)(x+2)(4x+(3x|
3
3)+2)(4x+(x
2
2)+2)(3x+(x
6
3)+2)(4x+(3x
)()()(
))()((
=
=
++++++=
++++++=
3
6114
2
483
6
61612
23
11
3
424
2
31
6
172
222
222
xxxxxx
xxxxxx
g) ]2)152(5][2)103(15[ 22 ++++++ xxxx
]22)322(3[ 2 +++ xx
)2)(23(
)23(2)23(
22322322)322(3
)2)(15(
)2()2(5
25252)152(5
)15)(23(
)15(2)15(3
2103152)103(15
22
22
22
++=
+++=
+++=+++
++=
+++=
+++=+++
++=
+++=
+++=+++
xx
xxx
xxxxx
xx
xxx
xxxxx
xx
xxx
xxxxx
)2)(15)(23(
)2)(23)(2)(15)(15)(23(
+++
++++++
xxx
xxxxxx
3. Find the square root of the following expressions
a) 9182424916 22 +-+-+ yxxyyx
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334
)334(
3.3.24.3.23.4.23)3()4(
2
222
+-=
+-=
-+-+-+=
yx
yx
yxyxyx
b) 222 16254030249 yzyzxzxyx ++-+-
|5z)+ 4y-(3x|=
5z)+ 4y-(3x=
2.4y.5z2.3x.5z+2.3x.4y(-4y)+(5z)+(3x)=
2
222 ++
++-+-= 222 16254030249 yzyzxzxyx
3.8. Square Root by Division Method
1. Find the square root of
a) 12171664 234 +-+- xxxx
18 2 +- xx28x
4
234
64
12171664
x
xxxx +-+-
xx -21623
23
16
1716
xx
xx
+-
+-
1216 2 +- xx
1216
12162
2
+-
+-
xx
xx
0
12171664 234 +-+- xxxx = 18 2 +- xx
b) 9364212 234 +-+- xxxx
362 +- xx2x
4
234 9364212
x
xxxx +-+-
xx 62 2 -23
23
3612
4212
xx
xx
+-
+-
3122 2 +- xx
9366
93662
2
+-
+-
xx
xx
0
369364212 2234 +-=+-+- xxxxxx
c) 94242837 432 +++- xxxx
372 2 -- xx22x
4
234
4
94237284
x
xxxx +++-
xx 74 2 -23
23
4928
3728
xx
xx
+-
+-
3144 2 -- xx
94212
942122
2
++-
++-
xx
xx
37294237284 2234 --=+++- xxxxxx
d) 1816 24 ++ xx
104 2 ++ xx24x
4
234
16
108016
x
xxxx ++++
xx 08 2 +23
23
00
80
xx
xx
+
+
108 2 ++ xx
108
1082
2
++
++
xx
xx
0
141816 224 +=++ xxx
e) 144216183198121 234 ++-- xxxx
12911 2 ++ xx211x
4
234
121
144216183198121
x
xxxx ++--
xx 922 2 +23
23
81198
183198
xx
xx
+-
--
121811 2 ++ xx
144216264
1442162642
2
++-
++-
xx
xx
0
12911144216183198121 2234 ++=++-- xxxxxx
2. Find the square root of
a)2
2
2
2 93013
204
x
y
x
y
y
x
y
x+-++
Multiply Nr. & Dr by x2y2
x
y
y
x
xy
yxyx
xy
yyxyxx
35
2
352913204 2242234
--=
--=
+++
22 352 yxyx --22x
4
42234
4
913204
x
yyxyxx +++
xyx 54 2 -
223
223
2520
1320
yxyx
yxyx
+
+
22 3104 yxyx --422
422
912
912
yyx
yyx
+
+
b)2
2
2
2 1027
10
x
y
x
y
y
x
y
x+-+-
Multiply Nr. & Dr by x2y2
xy
+ yxyyxy + xx 432234 102710 --
x
y
y
x
xy
yxyx+-=
+-5
5 22
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22 5 yxyx +-2x
4
432234 102710
x
+ yxyyxy + xx --
xyx 52 2 -223
223
2510
2710
yxy + x
yxy + x
-
-
22 102 yxyx +-4322
4322
102
102
+ yxyyx
+ yxyyx
-
-
3. Find a and b if the following are perfect squares
a) baxxxx ++++ 234 28129
423 2 ++ xx23x
4
234
9
28129
x
baxxxx ++++
xx 26 2 +23
23
412
2812
xx
xx
+
+
446 2 ++ xx
161624
242
2
++
++
xx
baxx
0
16
016
16
016
=
=-
=
=-
b
b
a
xax
b) abxxxx +++- 234 37124
732 2 +- xx22x
4
234
4
37124
x
abxxxx +++-
xx 34 2 -23
23
912
3712
xx
xx
+-
+-
764 2 +- xx
494228
282
2
+-
++
xx
abxx
0
49
049
42
042
=
=-
-=
=+
a
a
b
xbx
c) 100220361 234 ++++ xxbxax2121110 xx ++
10
100
361220100 432 axbxxx ++++
x1120 +
2
2
121220
361220
xx
xx
+
+
2122220 xx ++432
432
144264240
240
xxx
axbxx
++
++
0
144
0144
264
0264 4433
=
=-
=
=-
a
xax
b
xbx
d) 168 234 +++- bxaxxx
442 +- xx2x
4
234 168
x
bxaxxx +++-
xx 42 2 -23
23
168
8
xx
axx
+-
+-
482 2 +- xx
16328
16)16(2
2
+-
++-
xx
bxxm
0
32
032
032
2
0816
08)16( 22
-=
=+
=+
=
=--
=--
b
b
xbx
m
m
xxm
e) bx
a
xxx+++-
234
1361
Multiply Nr. & Dr by x2
2
4321361
x
bxaxxx +++-
2231 xx +-
1
1
1361 432 bxaxxx +++-
x32 -
2
2
96
136
xx
xx
+-
+-
2262 xx +-432
432
4124
4
xxx
bxaxx
+-
++
0
4
04
12
012 4433
=
=-
-=
=+
b
xbx
a
xax
3.9 Quadratic Equations
1. Find zeroes of the expression 1282 ++ xx
zerosare6and2
01248366)(p
0121642)(p
2,6
02
,06
0)2)(6(
01282
--
=+-=-
=+-=-
-=-=
=+
=+
=++
=++ xx
x
x
xx
xx
2. Determine the quadratic equations, whose sum and product of roots are General form of the quadratic equation
0rootsofproductroots)theof(sum2 =+- xx
a) 149, 01492 =+- xx
b) 209,- 02092 =++ xx
c) 43
5,
01253
043
5
2
2
=+-
=+-
xx
xx
d) 12
3-
-,
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022
0)1(2
3
2
2
=-+
=-+÷ø
öçè
æ --
xx
xx
e)2
5
2
7,
-
0572
02
5
2
7
2
2
=++
=+÷ø
öçè
æ --
xx
xx
f)2
1
5
3 --,
05610
02
1
5
3
2
2
=-+
=÷ø
öçè
æ -+÷
ø
öçè
æ --
xx
xx
g) 22 )5(,)2( +-- aa
0)5()2(
0)5()2(222
222
=++-+
=++---
axax
axax
3. Find the sum and product of the roots
a
c
a
b
cbxax
=
-=+
=++
ab
ba
ba
roots,ofproduct
roots,ofsum
,0ofrootsbeandLet 2
a) 06582 =-+ xx
651
65,8
1
8-=
-=-=
-=+ abba
b) 0752 2 =++ xx
2
7,
2
5=
-=+ abba
c) 02832 =-+ xx
281
28,3
1
3-=
-=-=
-=+ abba
d) 032 =+ xx
01
0,3
1
3==-=
-=+ abba
e)2
1013
aa=+
3
10,
3
1
0103
1013
1013
2
2
2
-=
-=+
=-+
=+
=+
abba
aa
aa
a
aa
f) 043 2 =-- yy
3
4,
3
1
3
)1( -==
--=+ abba
g) 02 322 =-- kxkkx
23
22
,1)(
kk
k
k
k-=
-==
--=+ abba
3.10 Solving Quadratic Equations
1. Solve by factorization method
a) 0274 2 =-- xx
2,4
1
02,014
0)2)(14(
-=
=-=+
=-+
x
xx
xx
b) 030192 2 =++ mm
2
5,2
2
5,2
0152,02
0)152)(2(
0)2(15)2(2
--=
-=-=
=+=+
=++
=+++
Roots
mm
mm
mm
mmm
c) 03622 2 =+- xx
2
3
032or032
0)32)((32(
0)32(3)32(2
03662 2
=
=-=-
=--
=---
=+--
x
xx
xx
xxx
xxx
d) 02572 2 =++ xx
2,2
5:Roots
052,02
0)52)(2(
0)2(5)2(2
025522 2
-
=+=+
=++
=+++
=+++
xx
xx
xxx
xxx
e) 04213 24 =+- xx
6,,7areroots
6or7Thus
6or7a
06)()7(
04213
.Let
22
2
2
±±=
==
=
=--
=+-
=
x
xx
aa
a a
ax
f) )5()6(3 2 +=- ppp
2,2
9
0)2)(92(
01852
51832
22
-=
=+-
=--
+=-
p
pp
pp
ppp
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g) 23)7( =-aa
9,2are,Roots
0)2)(9(
0187
18)7(
23)7(
2
22
-=
=+-
=--
=-
=-
a
aa
aa
aa
aa
h) 08
12 2 =+- xx
4
1,
4
1are,Roots
0)14)(14(
01816 2
=
=--
=+-
x
xx
xx
i)2
12
1
1=
-+
- x
x
x
x
2,1are,roots
0)1)(2(
01
][5]12[2
2
5)1(
2
222
2
22
-=
=+-
=--
-=+-+
=-
-+
x
xx
xx
xxxxx
xx
xx
2. Solve by completing the square method
a) 012 2 =-- xx
16
1
4
1
2
12
2
1,1
4
2,
4
4
4
3
4
1
16
9
4
1
16
9
4
1
16
1
2
1
16
1
2
2
1
2
02
1
2
2
2
2
2
2
=
=
=
-=Þ
-=
±=
±=-
=÷ø
öçè
æ-
+=+-
=-
=--
b
b
b
xx
x
x
x
xx
xx
xx
b) 04129 2 =+- xx
9
4
3
2
9
122
3
20
3
2
03
2
9
4
9
4
9
4
9
12
9
4
9
12
2
2
2
2
=
=
=
=Þ=-
=÷ø
öçè
æ-
+-=+-
-=-
b
b
b
xx
x
xx
xx
c) 0232 =-- xx
2
9
2
3
32
2
133
2
13
2
3
2
13
2
3
2
13
2
3
2
92
2
93
23
2
2
2
2
=
=
=
±=
±=
±=-
=÷ø
öçè
æ-
+=+-
=-
b
b
b
x
x
x
x
xx
xx
d) 231
75+=
-
+x
x
x
1
13
21
21
4
1
12375
=
=
=
-==
±=
±=-
=
+=+
=
-+=+
22
2
2
2
2
b
1b
22b
,
1)-(x
132-
32-
0=3-2-
0=9-6-3
))((
xx
x
x
xx
xx
xx
xx
xxx
3. Solve by Formula method
a) 0222 =-+ xx
1313
2
322
2
322
2
322
2
842
12
21422
2
4
221
0
022
2
2
2
2
---=
--+-=
±-=
+±-=
--±-=
-±-=
-===
=++
=-+
or
or
)(
))((
,,
with compare
x
x
a
acbbx
cba
cbxax
xx
b) 0252 2 =+- xx
0withcompare
02522
2
=++
=+-
cbxax
xx
a
acbbx
cba
2
4
2,5,2
2 -±-=
=-==
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2
1or2
4
35or
4
35
4
35
4
16255
)2(2
)2)(2(4)5()5( 2
=
-+=
±=
-±=
--±--=
x
x
x
c) 0332 2 =-- xx
4
333or
4
333
4
333
4
2493
)2(2
)3)(2(4)3()3(
2
4
3,3,2
0withcompare
0332
2
2
2
2
-+=
±=
+±=
---±--=
-±-=
-=-==
=++
=--
x
x
a
acbbx
cba
cbxax
xx
d) 023203 2 =-- yy
13
236
2620
6
26206
67620
6
27640020
32
23342020
2
4
23203
0
023203
2
2
2
2
-=
-+=
±=
+±=
---±--=
-±-=
-=-==
=++
=--
or
or
)(
))(()()(
,,
with compare
x
x
x
a
acbbx
cba
cbxax
yy
e) 02362 2 =+- ff
0withcompare
023622
2
=++
=+-
cbxax
ff
2
1or2
2
33or
2
33
22
326
22
24366
)2(2
)23)(2(4)6()6(
2
4
23,6,2
2
2
=
-+=
±=
-±=
--±--=
-±-=
=-==
x
x
x
a
acbbx
cba
f) 05523 2 =-+ pp
5or3
5
6
5452or
6
5452
6
5452
6
602052
)3(2
)5)(3(4)52()52(
2
4
5,52,3
0withcompare
2
2
2
-=
--+-=
±-=
+±-=
--±-=
-±-=
-===
=++
p
p
p
a
acbbp
cba
cbxax
g) 01236 222 =-+- )( baayy
6or
6
72
1212or
72
1212
72
14412
72
14414414412
)36(2
))(36(4)12()12(
)(,12,36
0withcompare
2
222
222
22
2
babax
babax
ba
baaa
baaa
bacaba
cbxax
-+=
-+=
±=
+-±=
---±--=
-=-==
=++
h) 0222 =+++- )()( qpxqppqx
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p
qp
q
qpx
pq
qpqpqp
pq
qpqpqp
pq
qpqpqp
pq
qpqpqp
pq
pqqpqpqp
pq
qppqqpqpx
qpcqpbpqa
cbxax
++=
-±++=
-±++=
-+±+=
-+±+=
-++±+=
+-+-±+=
+=+-==
=++
,
2
)]())[((
2
)]())[((
2
))(()(
2
][][)(
2
}]4)[(][)(
)(2
))((4])([)(
)(,)(,
0withcompare
2
222
222
2222
22
2
3.11Problems Involving Quadratic Equations
1. Product of Kumaran’s age two years ago and his age four years from now is one more than twice his present age. What is his present age?
years3agepresents'Kumaran
3negative,betcan'Age
3
03)3)((
2182
214)(2)(Given,
years.4)(agehisnow,fromyears4
years.2)(agehisago,yearsTwo
years.KumaranofagePresent
2
=
-¹
±=
=+-
+=-+
+=+-
+=
-=
=
x
x
xx
x xx
xxx
x
x
x
2. A girl is twice as old as her sister. Five years hence, the product of their ages (in years) will be 375. Find their present ages.
years.20=2(10)=girl ofAge
years10sister ofageHence
10,=0=10)-35)(x +(2x
0=350-15x +2x
0=375-25+10x +5x +2x
375=)5+5)(2x +(x
375=agestheir ofProduct
5+2x =girl ofAge
5+x =sisterofagehence,years5
2xgirlofAge,sisterofAge
235
2
2
=
=Þ
==
-xx
x
3. If the difference between a number and its
reciprocal is 5
24 , find the number.
1/5.and5numbersRequired
55
1
05015
0515
05245
5
241
5
241
1reciprocalits
xnumber required
2
2
=
--
-+
--
-
-
=
=
= or xx =
= or x = x
) = ) (xx (
= x x
= x
x
= x
x
x
4. A ladder 17 feet long is leaning against a wall. If the ladder, vertical wall and the floor from the bottom of the wall to the ladder form a right triangle, find the height of the wall where the top of the ladder meets if the distance between bottom of the wall to bottom of the ladder is 7 feet less than the height of the wall?
ft15wall ABofHeight
8ABve,betcan'Height
8(or)15
08)(15(
01207
4914289
7)((17)
BC ABAC
:theoremPythagorasBy
feet7)(BC
ft,17 ACABC,rt.In
feet7)(BC
feetwall ABofHeight
2
22
222
222
=
-¹-
-=
=+-
=--
+-+=
-+=
+=
-=
=D
-=
=
x
x)x
xx
xxx
xx
x
x
x
5. Hypotenuse of right angled triangle is 25 cm Perimeter 56 cm. Find length of smallest side.
cm.7measuresideSmallest
25and247,triangletheofSides
24=731=x 31
7=x and24=x
0=7)24)((
016831
62961625
)-(31+x=theorem,25PythogorasBy
312556=BC
56=ACBCAB
ABcm,25= ACHypotenuse
2
22
222
=
=
--
--
-
-
-=--
++
=
xx
= x + x
x+ x+ = x
x
xx
x
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6. A pole has to be erected at a point on the boundary of a circular ground of diameter 20 m in such a way that the difference of its distances from two diametrically opposite fixed gates P and Q on the boundary is 4 m. Is it possible to do so? If answer is yes at what distance from two gates should pole be erected?
m.16andm12idistanceRequired
12.=xand16-=x
0=12-xand0=16+x
0=12)-16)(x+(x
0=192- 4x+x
0= 400-16+8x+2x
400= 4+2x(4)+x+x
20= 4)+(x+x
2
2
222
222
=
7. A passenger train takes 1 hr more than an express train to travel a distance of 240 km from Chennai to Virudhachalam. The speed of passenger train is less than that of an express train by 20 km per hour. Find the average speed of both the trains.
0km/hr8=trainexpressofspeedAverage
km/hr60trainpassenger ofspeedAverage
80negative,betcan'speed
60.or 80
0=80)+(x )06(x
0480020
120
4800
1=20)+x(x
])20240[(
120
240240Given,
20
240=trainexpressbytakenTime
240=trainpassenger bytakenTime
Speed
Distance=Time
20+=trainexpressofspeedAverage
trainpassenger ofspeedAverage
km24=coveredDistance
2
2
=
-¹
-=
-
=-
-
-
=
x
x
x + x
= x + x
xx +
= x + x
x +
x
x
x
8. A bus covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hour more it would have taken 30 minutes less for the journey. Find the original speed of the bus.
km/hr45bustheofspeedOriginal
45
0=60)+(x 45)(x
0270015
152700
180
1
15
15
2
1=
15)+x(x
])1590[(
2
1
15
9090
15
90=speedincreasedintakenTime
90=speedoriginal intakenTime
Speed
Distance=Time
15+x =speedIncreased
xbustheofspeedoriginal
km90=coveredDistance
2
2
2
=
-
-
-
-
=
x =
= x + x
x + = x
= x + x
xx +
= x + x
x +
x
9. A garden measuring 12m by 16m is to have a pedestrian pathway that is ‘w’ meters wide installed all the way around so that it increases the total area to 285 m2. What is the width of the pathway?
m.'1.5width,betcan'
2
3,
2
310)32)(312(
0=93-56w+4w
0=285-192+56w+4w
285=4w+32w+24w+192
285=2w)+(16·2w)+(12
m285=Area
2w.+16width
2w+12pathwayincluding
gardenoflength
wwidthpathway
2
2
2
2
=-
-=Þ=-+
=
=þýü
=
vew
www
10. There is a square field whose side is 10 m. A square flower bed is prepared in its centre leaving a gravel path all round the flower bed. Total cost of laying the flower bed
and gravelling the path at ₹3
, ₹4 per square metre respectively is ₹364. Find the width of the gravel path.
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6mbedflower ofLength
636
36444003
364)x(10043x.
./mRs.4pathgravel laying ofCost
./mRs.3bedflower laying ofcost
m)x(100pathgravel theofArea
mxbedflower of Area
mbedflower theofSide
m100fieldsquareofArea
10mfieldsquareoflength
2
22
22
2
2
22
22
2
=
±=Þ=
=+
=-+
=
=
-=
=
=
=
=
x x
x – x
ei
x
11. A flock of swans contained x2 members. As the clouds gathered, 10x went to a lake and one-eighth of the members flew away to a garden. The remaining three pairs played about in the water. How many swans were there in total?
14412=swansofnumber Total
)(7
412
0=)47)(12(
048807
x=6+8
+10.
623=swansofnumber Remaining
x8
1=gardentoflewswansofNo.
10=laketowentswansofNo.
=swansofnumber Total
2
2
22
2
2
=
´-
+-
--
=´
, x = x =
xx
= xx
xxei
x
x
12. From a group of 2x2 black bees , square root of half of the group went to a tree. Again eight-ninth of the bees went to the same tree. The remaining two got caught up in a fragrant lotus. How many bees were there in total?
72=2(6)=beesofnumber Total
)(2
3-=x 6,=x
0=6)-3)(x +(2x
01892
229
16.
2=beesofnumber Remaining
9
16x=2x
9
8=treetoreturnedbeesofNo.
x=x
1=treetowentbeesofNo.
2x=beesofnumber Total
2
2
22
22
2
2
´
--
´
= xx
x = + x
x + ei
13. Music is been played in two opposite galleries with certain group of people. In the first gallery a group of 4 singers were singing and in the second gallery 9 singers were singing. The two galleries are separated by the distance of 70 m. Where should a person stand for hearing the same intensity of the singers voice?
m42=9ofgalleryfromDistance
m284ofgalleyfromDistance
28(or)140d
028)(d140)(d
03920112
0196005605
9456019600
9d)1404900(4
9=d)²4(70
)70(9
4
d-70=9ofgalleryfromDistance
d4ofgalleyfromDistance
ratiodistanceofsq.intensitysoundofRatio
2
2
22
22
=
-=
=-+
=-+
=-+
=+-
=+-
-
-
=
=
dd
dd
ddd
dd
d²
²d
d² =
x
x
14. Two women together took 100 eggs to a market, one had more than the other. Both sold them for the same sum of money. The first then said to the second: “If had your eggs, I would have
earned ₹15”, to which the second replied: “If I
had your eggs, I would have earned ₹3
26 . How
many eggs did each had in the beginning?
40,60=woman1,2ineggsofNo.
0)40)(200(
08000160
4800400009
)100(49(1),From
49
3
2015(2),in,Sub
3
20
3
26
1515Given,
)2.........(
:moneyequal for eggssold
=priceselling their
)1.........(100
100
woman1,2ineggsofNo.
2
22
22
22
=-+
=-+
+-=
-=
=
=
=Þ
=Þ=
=
-=
=+
=
xx
xx
xxx
xx
yx
yx
xy
qp
xqqx =
yppy
qypx
p,q.
xy
yx
x,y
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15. A ball rolls down a slope and travels a distance d = t2 −0.75t feet in t seconds. Find the time when the distance traveled by ball is 11.25 feet.
)3(sec753
3,753
030753
03753
02511750
2511750
2511where
750
2
2
2
-¹=
-==
=+=-
=+-
=--
=-
=
-=
t . t
t .
, t .t
) ) (t.(t
. t.t
.t .t
ft. d
t . td
3.12. Nature of Roots
1. Determine the nature of roots
rootReal No
rootsEqual andReal
rootsUnequal andReal
RootsofNature
0
0
0
42
<D
=D
>D
-=D ac b
a) 0202 =-- xx
rootsunequal andReal
081
)20)(1(4)1(
20112
.
, c , b a
>=D
---=D
-=-==
b) 016249 2 =+- xx
rootsEqual andReal
0
)16)(9(4)24(
162492
.
, c , b a
=D
--=D
=-==
c) 0922 2 =+- xx
rootsReal No
068
)9)(2(4)2(
9222
.
, c , b a
<-=D
--=D
=-==
d) 021115 2 =++ xx
UnequalandReal
01
)2)(15(4)11(
211152
, c , b a
>=D
-=D
===
e) 012 =-- xx
UnequalandReal
05
)1)(1(4)1(
1112
, c , b a
>=D
---=D
-=-==
f) 02332 2 =+- tt
RootsReal No
015
)23)(2(4)3(
2332
2
, c , ba
<-=D
--=D
=-==
g) 02269 2 =+- yy
rootsEqual andReal
0
)2)(9(4)26(
2269
2
.
, c , b a
=D
--=D
=-==
h) 016249 22222 =+- dcabcdxxba
rootsEqual andReal
0
)16)(9(4)24(
1624922222
2222
.
dcbaabcd
dcabcd, c, b baa
=D
--=D
=-==
2. Find the value(s) of ‘k’ for which the roots of the following equations are real and equal.
a) 081)48(2 =++- kkx
4,16
1
0)4)(116(
046516
0324166464
0)81)((4)]48[(
04
0.
andareRoots
81)48(
2
2
2
2
=
=--
=+-
=-++
=-+
=-
=D
=+-==
k
kk
k k
k kk
kk
ac b
equalreal
, c k k, b a
b) 012)65( 2 =++- kxxk
2,30)2)(3(
065
024204
0)1)(65(4]2[
04
12)65(
2
2
2
2
=Þ=--
=+-
=+-
=--
=-
==-=
kkk
k k
k k
kk
ac b
k, c , b ka
c) 016)26(2 =+++ xkkx
16)26( , c kk, b a =+==
9
1,10)1)(19(
01109
044036
0)16)((4)]26[(
04
2
2
2
2
=Þ=--
=+-
=+-
=-+
=-
kkk
k k
k k
kk
ac b
3. Find the values of ‘k’ such that quadratic equation has no real roots?
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0119 2 =++++ xkxk )()(
.75Thus,
0)7()5(
0352
0(1))94()1(
04
0roots,real No
119
2
2
2
k
kk
kk
k k
ac b
, c k , b k a
<<-
<-+
<--
<+-+
<-
<D
=+=+=
4. i) 02 22222 =+++++ srqsprxqpx )()( has no real
roots - Prove ii) If qrps = , show that the roots real and equal.
equal.andreal bewill rootso0,
])[(4
])[(4
thenqr psIf
equal.notareoots,0As
])[(4
]2[4
]
2[4
)()(4)][2(
4
)(2
0)(2)(
2
2
2
2222
2222
22222222
22222
2
2222
22222
S
qrps
qrps
r
qrps
r qpqrs sp
sqrq
sprpsqpqrsrp
srqpqspr
acb
s r, c qspr, b q pa
srqsprxqpx
=D
--=
--=D
=
<D
--=D
-+-=
--
--++=
++-+=
-=D
+=+=+=
=+++++
5. If the roots of 02 =-+-+- )()()( acxcbxba are
real and equal, prove that b, a, c are in A.P
acb
acb
cba
cba
ac bcabcba
ababcacbccb
ababcacbccb
acbacb
accbba
=+
=+
=--
=--
=-+-++
=-++--+
=+----+
=----
=-
-=-=-=
2
2
02
0)(2
04244
044442
0)(4)2(
0)()(4)(
04ACB
C,B, AHere
2
222
222
222
2
2
b, a, c are in A.P.6. If a, b are real, show that roots of the equation
0962 =--+-- )()()( baxbaxba are real and
unequal
4ACB
)(9c),(6B,A2 -=D
--=+-=-= bababa
unequal.&real areRoots
realareb&a0,)(72
)][2(36
])()[(36
)(36)(36
0b))(a9()(4b)(a36
22
22
22
22
2
>+=
+=
-++=
-++=
=----+=
ba
ba
baba
baba
ba
7. 0)(2)( 2222 =-+--- acbxbcaxabc roots are real
& equal prove: either abccbaa 30 333 =++= or
acbbcaabc
acbxbcaxabc
-=--=-=
=-+---222
2222
C)(2B,A
0)(2)(
abccba
abccba
a
abccbaa
ac abbc aa
bcaacabcbbcacba
acbabcbca
acbabcbca
3
03or
0
0)3(
03
0)()2(
0)()()(
0)()(4)(4
04ACB
333
333
333
3324
233222224
2222
2222
2
=++
=-++
=
=-++
=++-
=+----+
=----
=----
=-
3.13 Relation b/w Roots and Coefficients
1. Write each of the following expression in terms of abba and+
ab
baabba
ab
baba
ab
bbaa
a
b
b
a
baab
baabba
ba
ba
abba
ab
abba
ab
ba
a
b
b
a
)(32)(
)(3
3333iii)
1)(39
1339)13)(13(ii)
11
2)(
33i)
2
22
22
2222
2
22
++-+=
+++=
+++=
++
+
++-=
+--=--
+=+
-+=
+=+
2. The roots of the equation 0572 2 =+- xx are α and β. Without solving for the roots, find
i) ba
11+ ii)
a
b
b
a+ iii)
2
2
2
2
+
++
+
+
a
b
b
a
2
5,
2
7
2
)7(
then,,be0572ofRoots 2
==--
=+
=+-
abba
baxx
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6
13
42
72
2
5
82
74
2
52
2
7
4)(2
8)(42)(
422
4444
2
2
2
2iii)
10
29
2
5
2
52
2
7
2)(
ii)
5
7
2
52
7
11i)
2
2
22
2
2
22
=
+÷ø
öçè
æ+
+÷ø
öçè
æ+÷
ø
öçè
æ-÷
ø
öçè
æ
=
+++
+++-+=
+++
+++++=
+
++
+
+
=
÷ø
öçè
æ-÷
ø
öçè
æ
=
-+=
+=+
=
=
+=+
baab
baabba
baab
bbaa
a
b
b
a
ab
abba
ab
ba
a
b
b
a
ab
ba
ba
3. If α,β are roots of 01072 =++ xx find
values of
i) ba - ii) 22 ba + iii) 33 ba -
iv) 44 ba + v)a
b
b
a+ vi)
a
b
b
a 22
+
10,7
then,,be0107ofRoots 2
=-=+
=++
abba
baxx
641)10(229
)(2)(iv)
117
)3)((10(3)3(
)(3)(iii)
29
)10(2)7(
2)(ii)
3)10(4)7(
4)(i)
22
222244
3
333
2
222
2
2
=-=
-+=+
=
+=
-+-=-
=
--=
-+=+
=--=
-+=-
abbaba
baabbaba
abbaba
abbaba
10
133
10
)7)(10(3343
)(3)(vi)
10
29
10
2049
2)(v)
322
2
-=
---=
+-+=+
=-
=
-+=+
ab
baabba
a
b
b
a
ab
abba
a
b
b
a
4. If α,β are roots of 0273 2 =-+ xx find values of
i)a
b
b
a+ ii)
a
b
b
a 22
+
3
2,
3
7
then,,be0273ofRoots 2
-=
-=+
=-+
abba
baxx
9
67
3
2
3
2
3
73
3
7
)(3)(
ii)
6
61
3
2
3
22
3
7
2)(
i)
3
3
3322
2
2
22
=-
÷ø
öçè
æ -÷ø
öçè
æ --÷
ø
öçè
æ -
=
+-+=
+=+
-=
-
÷ø
öçè
æ --÷
ø
öçè
æ -
=
-+=
+=+
ab
baabba
ab
ba
a
b
b
a
ab
abba
ab
ba
a
b
b
a
5. If one root of the equation 0642 2 =+- ayy is
twice the other then find the values of a.
322
64,
2
then,,be0642ofRoots 2
===+
=+-
abba
ba
a
ayy
242
84roots,ofSum
8,416
32)2(roots,ofProduct
(given)2
2
±=Þ=±±
±=±=Þ=
=
=
aa
abb
bb
ba
6. If one root of the equation 0813 2 =++ kxx(having real roots) is the square of other, find k.
273
81,
3
then,,be0813ofRoots 2
==-
=+
=++
abba
ba
k
kxx
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363
93roots,ofSum
9,327
27)(roots,ofProduct
(given)
3
2
2
-=Þ-
=+
==Þ=
=
=
kk
abb
bb
ba
7. If the difference between the roots of the
equation 0132 =+- kxx is 17 find k.
k
kxx
==+
=+-
abba
ba
,13
then,,be013ofRoots 2
30)2)(15(
2,15302)2()1(
.....(2),13
.(1)given)....(17
-=Þ=-
=
-==Þ=+
=+
=-
kk
kab
baa
ba
ba
8. If α, β are the roots of 027 2 =++ axx and if
7
13-=-ab . Find the values of a.
7
2,
7=
-=+ abba
a
15or1577
12
7roots,ofsum
2,7
1,
7
1,2
0)2)(17(
02137
7
2
7
137
7
2)
7
13(
7
13
7
13
2
2
-=Þ-
=±±
-=+
-=Þ-
=
=-+
=--
=-
=-
-=Þ-
=-
aa
aba
ba
aa
aa
aa
aa
abab
9. If α and β are roots of 012 2 =-- xx form the equation whose roots are
2
1,
2
1 -==+ abba
i) ba
1,
1 ii) abba 22 , iii) abba ++ 2,2
4
2
1
2
1
1
1rootsofProduct
1
2
12
1
rootsofSum
1,
1EqnrqdofRootsi)
=
÷ø
öçè
æ÷ø
öçè
æ -=
=
-=-
=
+=
=
abab
ba
ba
0rootsofProductroots)ofSum(
:EquationQuadraticofForm2 =+- xx
04EquationRequired 2 =++ xx
8
1
2
1
)(
Product
4
1
2
1
2
1
)(
Sum
,EqnrqdofRootsii)
3
3
2222
22
-=
÷ø
öçè
æ -=
=
´=
-=
-´=
+=
+=
=
ab
abba
baab
abba
abba
0128
08
1
4
1Equation,Required
2
2
=-+
=÷ø
öçè
æ -+÷
ø
öçè
æ --
xx
xx
02
1
2
1
)(2
4)(25
)(25
)2)(2(Product
2
3
2
13
)(3
22Sum
2,2EqnrqdofRootsii)
2
2
22
=-
+=
-+=
-++=
++=
++=
=
÷ø
öçè
æ=
+=
+++=
++=
abba
abbaab
baab
abbaba
abba
abba
032
002
3Equation,Required
2
2
=-
=+÷ø
öçè
æ-
xx
xx
10. The roots of the equation 0462 =-+ xx are α, β. Find the quadratic equation whose roots are
i) 2ba ,2 ii) abba 22 , iii) ba
2,
2
4,6 -=-=+ abba
01644Equation,Required
164
)(
Product
44)4(26
2)(
Sum
,EqnrqdofRootsi)
2
2
2
22
2
2
22
22
=+-
==
=
´=
=---=
-+=
+=
=
xx
ab
ba
abba
ba
ba
06424Eqn,Rqd
64)4(
)(
Product
24)6)(4(
)(
Sum
,EqnrqdofRootsii)
2
3
3
2222
22
=--
-=-=
=
´=
=--=
+=
+=
=
xx
ab
abba
baab
abba
abba
1)4(
4
4Product
34
)6(2
)(2rootsofiii)Sum
-=-
=
=
=-
-=
+=
abab
ba
013Eqn,Rqd 2 =-- xx
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3.14. MATRICES
1. Following information regarding No. of men
and women workers in 3 factories I, II and III.
Factory Men Women I 23 18 II 47 36 III 15 16
i) Represent above information in matrix form
÷÷÷
ø
ö
ççç
è
æ
=
1615
3647
1823
A
ii) What does the entry in the second row and
first column represent?
There are 47 men workers in factory II.
2. If a matrix has 16 elements, what are the
possible orders it can have?
1 × 16, 16 × 1
2 × 8, 8 × 2
4 × 4,
3. i) If a matrix has 18 elements, what are the possible orders it can have?
1 × 18, 18 × 1,
2 × 9, 9 × 2,
3 × 6 6 × 3,
ii) What if it has 6 elements? 1 × 6, 6 × 1, 3 × 2, 2 × 3
4. In the matrix, write
úúúúú
û
ù
êêêêê
ë
é
-
-
11186
0341
52
371
3498
(i) The number of elements
A has 4 rows and 4 columns
Number of elements = 4 × 4= 16
(ii) The order of the matrix
4 × 4
(iii) Write the elements
1110
52
3,7
444334
242322
=-==
===
aaa
aaa
5. Construct a 33´ matrix whose elements are
÷÷÷
ø
ö
ççç
è
æ
=´
333231
232221
131211
AFormMatrix 33Rqd
aaa
aaa
aaa
i. 22 jiaij =
÷÷÷
ø
ö
ççç
è
æ
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
81369
36164
941
81
)3()3(
36
)2()3(
9
)1()3(
36
)3()2(
16
)2()2(
4
)1()2(
9
)3()1(
4
)2()1(
1
)1()1(
2233
2232
2231
2223
2222
2221
2213
2212
2211
A
aaa
aaa
aaa
ii. || jiaij 2-=
÷÷÷
ø
ö
ççç
è
æ
=
=
-=
=
-=
=
-==
-=
=
-=
=
-==
-=
=
-=
=
-=
311
420
531
3
)3(23
1
)2(23
1
)1(234
)3(22
2
)2(22
0
)1(225
)3(21
3
)2(21
1
)1(21
333231
232221
131211
A
aaa
aaa
aaa
iii.3
)( 3jiaij
+=
÷÷÷÷÷÷
ø
ö
çççççç
è
æ
=
=
+=
=
+=
=
+=
=
+=
=
+=
==
+=
=
+=
==
+=
=
+=
723
125
3
643
125
3
649
3
649
3
8
72
3
)33(
3
125
3
)23(
3
64
3
)13(3
125
3
)32(
3
64
3
)22(
93
27
3
)12(3
64
3
)31(
93
27
3
)21(
3
8
3
)11(
3
33
3
32
3
31
3
23
3
22
3
21
3
13
3
12
3
11
A
aaa
aaa
aaa
6. Find the transpose of A.
i.÷÷÷
ø
ö
ççç
è
æ
-=
283
971
345
A÷÷÷
ø
ö
ççç
è
æ
-=Þ
293
874
315TA
ii.÷÷÷
ø
ö
ççç
è
æ
-
-
=
574
982
135
A÷÷÷
ø
ö
ççç
è
æ
-
-
=Þ
591
783
425TA
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7. Find the transpose of -A.
i. ÷÷ø
öççè
æ
--
-=
135
942A
÷÷÷
ø
ö
ççç
è
æ
-
--
=-Þ÷÷ø
öççè
æ
-
--=-
19
34
52
135
942 TAA
ii.
÷÷÷÷
ø
ö
çççç
è
æ
-
-
-
=
53
25
37
A
÷÷
ø
ö
çç
è
æ
-
--=-Þ
÷÷÷÷
ø
ö
çççç
è
æ
-
-
-
=-523
357
53
25
37T
AA
8. Verify AATT =
i.
÷÷÷÷
ø
ö
çççç
è
æ
-=
1382
57.017
225
A
AA
AA
TT
TTT
=
÷÷÷÷
ø
ö
çççç
è
æ
-=Þ
÷÷÷÷
ø
ö
çççç
è
æ-
=
1382
57.017
225
,
12
52
37.02
8175
ii.÷÷÷
ø
ö
ççç
è
æ
=
34
98
51
A
AA
AA
TT
TTT
=
÷÷÷
ø
ö
ççç
è
æ
=Þ÷÷ø
öççè
æ=
34
98
51
395
481
9. Find the values of x, y and z from equations
i. ÷÷ø
öççè
æ=
÷÷
ø
ö
çç
è
æ
532
3312 zy
x
equal.areelementsing Correspond
equal.arematricesgivenIf
3,3,12
Elements,Corres.Equating
=== xzy
ii. ÷÷ø
öççè
æ=÷÷
ø
öççè
æ
+
+
85
26
5
2
xyz
yx
Elements,Corres.Equating
)2.....(8
8
)1....(.6
0.....55
yxxy
yx
zz
=Þ=
=+
=Þ=+
4or 224
0)2)(4(
086
68
(1),in2)(Sub.
2
=Þ=
=--
=-+-
=+
xory
yy
yy
yy
iii.÷÷÷
ø
ö
ççç
è
æ
=÷÷÷
ø
ö
ççç
è
æ
+
+
++
7
5
9
zy
zx
zyx
474(1),in''ofvalueSub.
39z(1)5in2)(Sub.
297(1),in3)(Sub.
)3..(..........7
)2..(..........5
)1..(..........9
=Þ=+
=Þ=+
=Þ=+
=+
=+
=++
yyz
z
xx
zy
zx
zyx
10. Find the value of a, b, c, d from the equation
÷÷ø
öççè
æ=÷÷
ø
öççè
æ
+-
+-
20
51
32
2
dcba
caba
19getwe),4(in7Sub
7getwe),3(in1Sub
2getwe),2(in1Sub
,1gives)2()1(
)4.........(23
)3.........(52
)2.........(02
)1.........(1
-==
=-=
-=-=
-=-
=+
=+
=-
=-
dc
ca
ba
a
dc
ca
ba
ba
3.17. Operations on Matrices
1. Find A+B.
a)÷÷÷
ø
ö
ççç
è
æ
=÷÷÷
ø
ö
ççç
è
æ
=
042
131
071
987
654
321
BA ,
÷÷÷
ø
ö
ççç
è
æ
=÷÷÷
ø
ö
ççç
è
æ
+++
+++
+++
=
÷÷÷
ø
ö
ççç
è
æ
+÷÷÷
ø
ö
ççç
è
æ
=+
9129
785
392
094827
163514
037211
042
131
071
,
987
654
321
BA
b)÷÷÷
ø
ö
ççç
è
æ
=÷÷÷
ø
ö
ççç
è
æ
=
18524781
80171218
40153820
40328053
30216250
23141522
BA ,
÷÷÷
ø
ö
ççç
è
æ
=
÷÷÷
ø
ö
ççç
è
æ
+÷÷÷
ø
ö
ççç
è
æ
=+
5884127134
110387468
63295342
18524781
80171218
40153820
40328053
30216250
23141522
BA
Mythila Publishers, Puduvayal 67 X std Mathematics Made Easy
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c)÷÷÷
ø
ö
ççç
è
æ
=÷÷÷
ø
ö
ççç
è
æ
-
-
-
=
69
43
81
923
645
231
BA ,
It is not possible to add A and B Because they have different orders
2. If ÷÷÷
ø
ö
ççç
è
æ
-
=
38
43
91
A Verify that
0)()( =+-=-+ AAAA
0)()(
00
00
00
38
43
91
38
43
91
)(
00
00
00
38
43
91
38
43
91
)(
=+-=-+
÷÷÷
ø
ö
ççç
è
æ
=÷÷÷
ø
ö
ççç
è
æ
-
+÷÷÷
ø
ö
ççç
è
æ
-
--
--
=+-
÷÷÷
ø
ö
ççç
è
æ
=÷÷÷
ø
ö
ççç
è
æ
-
--
--
+÷÷÷
ø
ö
ççç
è
æ
-
=-+
AAAA
AA
AA
3. verify that ABBA +=+ if
÷÷÷
ø
ö
ççç
è
æ
=÷÷÷
ø
ö
ççç
è
æ
-
=
01
33
75
,
38
43
91
BA
ABBA
AB
BA
+=+
÷÷÷
ø
ö
ççç
è
æ
-
=÷÷÷
ø
ö
ççç
è
æ
-
+÷÷÷
ø
ö
ççç
è
æ
=+
÷÷÷
ø
ö
ççç
è
æ
-
=÷÷÷
ø
ö
ççç
è
æ
+÷÷÷
ø
ö
ççç
è
æ
-
=+
39
76
166
38
43
91
01
33
75
39
76
166
01
33
75
38
43
91
4. Verify that CBACBA ++=++ )()( if
÷÷÷
ø
ö
ççç
è
æ
-
-=÷÷÷
ø
ö
ççç
è
æ
--
=÷÷÷
ø
ö
ççç
è
æ
-
-=
142
321
438
117
291
432
401
132
134
CBA ,,
÷÷÷
ø
ö
ççç
è
æ
--
=++
÷÷÷
ø
ö
ççç
è
æ
--
=+
654
4104
9914
)(
255
572
8610
CBA
CB
÷÷÷
ø
ö
ççç
è
æ
--
=+
516
1123
566
BA
CBACBA
CBA
++=++
÷÷÷
ø
ö
ççç
è
æ
--
=++
)()(
654
4104
9914
)(
5. Find 2A+B
if÷÷÷
ø
ö
ççç
è
æ
-
-
=÷÷÷
ø
ö
ççç
è
æ
--
=
057
421
3114
134
931
687
BA ,
÷÷÷
ø
ö
ççç
è
æ
--
=
÷÷÷
ø
ö
ççç
è
æ
-
-
+÷÷÷
ø
ö
ççç
è
æ
--
=
÷÷÷
ø
ö
ççç
è
æ
-
-
+÷÷÷
ø
ö
ççç
è
æ
--
=+
2111
2281
92718
057
421
3114
268
1862
121614
057
421
3114
134
931
687
22 BA
6. Find 4A - 3B
÷÷÷÷
ø
ö
çççç
è
æ
-
--
=
÷÷÷÷
ø
ö
çççç
è
æ -
=
965
32
7
4
1347
491
24
3
2
1245
BA ,
÷÷÷÷
ø
ö
çççç
è
æ
--
--
=
÷÷÷÷
ø
ö
çççç
è
æ
--
----
+÷÷÷
ø
ö
ççç
è
æ -
=
÷÷÷÷
ø
ö
çççç
è
æ
-
--
-
÷÷÷÷
ø
ö
çççç
è
æ -
=-
115411
9242
15
4
51441
271815
92
21
4
391221
16364
2432
8620
965
32
7
4
1347
3
491
24
3
2
1245
434 BA
7. Find the value of (i) B -5A (ii) 3A - 9B if
÷÷ø
öççè
æ=÷÷
ø
öççè
æ=
941
837
738
940BA ,
÷÷ø
öççè
æ
--
---=
÷÷ø
öççè
æ-÷÷
ø
öççè
æ=-
÷÷ø
öççè
æ
---
--=
÷÷ø
öççè
æ
---
--+÷÷
ø
öççè
æ=
÷÷ø
öççè
æ-÷÷
ø
öççè
æ=-
602715
456563
941
8379
738
94039Bii)3A
261139
37177
351540
45200
941
837
738
9405
941
8375i) AB
Mythila Publishers, Puduvayal 68 X std Mathematics Made Easy
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8. Compute BAiiCBAi2
3
2
123 --+ ))
÷÷÷
ø
ö
ççç
è
æ
--=÷÷÷
ø
ö
ççç
è
æ
-
--
=÷÷÷
ø
ö
ççç
è
æ
=
341
271
035
,
510
3112
468
,
678
053
381
CBA
CBA -+ 23i)
÷÷÷
ø
ö
ççç
è
æ
÷÷÷
ø
ö
ççç
è
æ
÷÷÷
ø
ö
ççç
è
æ -
÷÷÷
ø
ö
ççç
è
æ
÷÷÷
ø
ö
ççç
è
æ
÷÷÷
ø
ö
ççç
è
æ
÷÷÷
ø
ö
ççç
è
æ
-=
---
--
+-
-
+=
----
--
+=
251923
84414
11314
341
271
035
1020
6224
8
182124
0159
928
341
271
035
510
3112
468
2
678
053
381
3
12163
÷÷÷÷÷÷
ø
ö
çççççç
è
æ
-
--
-
=
úúú
û
ù
êêê
ë
é
÷÷÷
ø
ö
ççç
è
æ
--
--
-
+÷÷÷
ø
ö
ççç
è
æ
=
úúú
û
ù
êêê
ë
é
÷÷÷
ø
ö
ççç
è
æ
-
--
-÷÷÷
ø
ö
ççç
è
æ
=
-=-
2
924
2
914
2
32
1513
2
23
1530
9336
121824
678
053
381
2
1
510
3112
468
3
678
053
381
2
1
]3[2
1
2
3
2
1ii) BABA
9. Find X and Y if ÷÷ø
öççè
æ=-÷÷
ø
öççè
æ=+
40
03
53
07YXYX ,
÷÷
ø
ö
çç
è
æ=Þ÷÷
ø
öççè
æ=+
÷÷
ø
ö
çç
è
æ=Þ÷÷
ø
öççè
æ=+
÷÷ø
öççè
æ=-
÷÷ø
öççè
æ=+
2
1
2
302
13
042,)2()1(
2
9
2
305
93
0102,)2()1(
)2......(40
03
)1.......(,53
07
YY
XX
YX
YX
10. Find the value of a, b, c, d, x, y from the following matrix equation
÷÷ø
öççè
æ
-+÷÷
ø
öççè
æ=÷÷
ø
öççè
æ
--+÷÷
ø
öççè
æ
05
10
4
22
42
3
3
8
cb
aa
ab
d
:ElementsCorres.Equate
45
122
423
83÷÷ø
öççè
æ
-
+=÷÷
ø
öççè
æ
--
++
cb
a
ab
ad
4
3
447
44
2
3
523
7
128
1
23
=
=-
=--
=
-=-
=
+=+
-=
=+
c
c
ca
b
bb
a
aa
d
d
11. Find the values of x, y, z if
÷÷ø
öççè
æ=÷÷
ø
öççè
æ
++++
--
61
01
7
33
zyxyx
zxx
10
64
6
17
12
012
03
4
13
-=
-=+
-=+
=++
=
=-
=-
=
=-
y
y
y x
y x
z
z
z x
x
x
12. Find the values of x, y, z if 1684343 =++- yzzyx
10
166
14y
410
4
84
10
414
4
z
z
y
z y
z y
x
x
y x
=
=+
=
=-
=-
=+-
-=
=+
=+
13. Find values of x, y if ÷÷ø
öççè
æ=÷÷
ø
öççè
æ-+÷÷
ø
öççè
æ
- 6
4
3
2
3
4yx
6
24
(2)in4Sub
4)(
23)2(
22,2)1(
)2.........(633
)1(..........424
6
4
33
24
6
4
3
2
3
4
y
y
x
x
yx
y x
yx
y x
y x
y x
y
y
x
x
=
=+-
=
=+
=+-¸
=-¸
=+-
=-
÷÷ø
öççè
æ=÷÷
ø
öççè
æ
+-
-
÷÷ø
öççè
æ=÷÷
ø
öççè
æ-+÷÷
ø
öççè
æ
-
14. Solve for x, y ÷÷ø
öççè
æ-=÷÷
ø
öççè
æ
-
-+÷
÷ø
öççè
æ
8
522
2
2
y
x
y
x
2,4
0)2)(4(
082
1,5
0)1)(5(
054
8
5
2
4
22
2
2
-=
=+-
=--
-=
=+-
=--
÷ø
öçè
æ=÷÷
ø
öççè
æ
-
-
y
yy
yy
x
xx
xx
yy
xx
15. Find the non-zero values of x
Mythila Publishers, Puduvayal 69 X std Mathematics Made Easy
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÷÷ø
öççè
æ +=÷÷
ø
öççè
æ+÷÷
ø
öççè
æ
x
x
x
x
x
xx
610
2482
44
582
3
22 2
4of,valuezeroNon
42083
4,00)4(
128
44812
1220
48162
883
12162
610
2482
88
1016
3
22
2
2
2
2
2
2
2
=
=Þ=+
=Þ=-
=+
=Þ=
÷÷ø
öççè
æ +=÷
÷ø
öççè
æ
++
+
÷÷ø
öççè
æ +=÷÷
ø
öççè
æ+÷
÷ø
öççè
æ
x
xx
xxx
xxx
xx
x
x
xxx
xx
x
x
x
x
xx
xx
3.18.Multiplication Of Matrices
1. Find order of the product matrix AB if
(i) (ii) (iii) (iv) (v) Orders of A 3 × 3 4 × 3 4 × 2 4 × 5 1 × 1
Orders of B 3 × 3 3 × 2 2 × 2 5 × 1 1 × 3
i) A 3 × 3 B3 × 3
∴ Order of AB = 3 × 3
ii) A 4 × 3 B 3 × 2
∴ Order of AB = 4 × 2
iii) A4 × 2 B 2 × 2
∴ Order of AB = 4 × 2
iv) A 4 × 5 B 5 × 1
∴ Order of AB = 4 × 1
v) A 1 × 1 B 1 × 3
∴ Order of AB = 1 × 3
2. If A is of order qp ´ and B is of order
rq ´ what is the order of AB and BA?
A p × q B q × r
∴ Order of AB = p × r
B q × r A p × q
No. of columns in B ¹ No. of rows in A
∴ Order of BA is not defined
3. A has ‘a’ rows and 3a + columns. B has‘b’
rows and –b17 columns, and if both
products AB, BA exist, find a, b?
10b7,
10b7b17
(1)in7aSub
7a142a(2),(1)
(2)........17ba
(1)........3ba
(2).......ab17
AinrowsofNo.BincolumnsofNo.
(1)b........3a
inrowsofNo. AincolumnsofNo.
exists,BA,AB
b)(17bB
3)(aaA
==
=Þ=-
=
=Þ=+
=+
-=-
=-
=
=+
=
-´+´
a
B
4. Find AB ,If ÷÷÷
ø
ö
ççç
è
æ
=÷÷ø
öççè
æ=
135
142
138
,513
021BA
÷ø
öçè
æ=
÷ø
öçè
æ
++++++
++++++=
÷ø
öçè
æ=
úúúúúúú
û
ù
êêêêêêê
ë
é
÷÷÷
ø
ö
ççç
è
æ
÷÷÷
ø
ö
ççç
è
æ
÷÷÷
ø
ö
ççç
è
æ
÷÷÷
ø
ö
ççç
è
æ
÷÷÷
ø
ö
ççç
è
æ
÷÷÷
ø
ö
ççç
è
æ
=
÷÷÷
ø
ö
ççç
è
æ
÷ø
öçè
æ=
++++++
++++++
91851
31112
51354925224
021083048
1
1
1
513
1
4
3
513
5
2
8
513
1
1
1
021
1
4
3
021
5
2
8
021
135
142
138
513
021
)1(5)1(1)1(31)4(1)3(3)5(5)2(1)8(3
0)1(2)1(10)4(2)3(10)2(2)8(1
AB
5. Show that BA, satisfy commutative
property w.r. to matrix multiplication (or) Find AB, BA . Check if BAAB = ,if
i) ÷÷ø
öççè
æ=÷÷
ø
öççè
æ=
31
02,
31
12BA
BAAB
BA
BA
AB
AB
¹\
÷ø
öçè
æ=
÷ø
öçè
æ++
++=
÷ø
öçè
æ÷ø
öçè
æ=
÷ø
öçè
æ=
÷ø
öçè
æ++
++=
÷ø
öçè
æ÷ø
öçè
æ=
105
24
9132
0204
31
12
31
02
95
35
9032
3014
31
02
31
12
Mythila Publishers, Puduvayal 70 X std Mathematics Made Easy
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ii) ÷÷ø
ö
ççè
æ
-=
÷÷ø
ö
ççè
æ -=
22
222,
22
222BA
BAAB
BA
AB
=\
÷ø
öçè
æ=÷÷
ø
öççè
æ
++-
+-+=
÷÷ø
öççè
æ
-÷÷ø
öççè
æ -=
÷ø
öçè
æ=÷÷
ø
öççè
æ
+-
-+=
÷÷ø
öççè
æ
-÷÷ø
öççè
æ -=
80
08
442222
242444
22
222
22
222
80
08
442222
242444
22
222
22
222
iii) ÷÷ø
öççè
æ -=÷÷
ø
öççè
æ=
52
31,
34
52BA
BAAB
BAAB
¹\
÷ø
öçè
æ --=
÷ø
öçè
æ++
--=
÷ø
öçè
æ÷ø
öçè
æ -=
÷ø
öçè
æ=
÷ø
öçè
æ+-+
+-+=
÷ø
öçè
æ -÷ø
öçè
æ=
2524
410
1510204
95122
34
52
52
31
310
1912
151264
256102
52
31
34
52
iv) ÷÷ø
öççè
æ
-
-=÷÷
ø
öççè
æ=
13
21,
13
21BA
true.ispropertyeCommutativ
BA AB
50
05
1633
2261
13
21
13
21
50
05
1633
2261
13
21
13
21
=\
÷ø
öçè
æ-
-=
÷ø
öçè
æ+-+-
+--=
÷ø
öçè
æ÷ø
öçè
æ-
-=
÷ø
öçè
æ-
-=
÷ø
öçè
æ+--
+--=
÷ø
öçè
æ-
-÷ø
öçè
æ= BAAB
6. Solve ÷÷ø
öççè
æ=÷÷
ø
öççè
æ÷÷ø
öççè
æ
5
4
21
12
y
x
1x(1)in2Sub
263)(
522)2(
.421)1(
)2.....(52
)1......(42
5
4
2
2
=Þ=
=Þ-=--
=+´
=+´
=+
=+
÷ø
öçè
æ=÷
ø
öçè
æ
+
+
y
yy
yx
yx
yx
yx
yx
yx
7. Show that A(BC)(AB)C = , If
i) ÷÷ø
öççè
æ
-=
÷÷÷
ø
ö
ççç
è
æ -
=-=12
21,
31
12
11
,111 CBA
)()(
29
1
3
3
111
7
4
1
111
17
34
31
111)(
17
34
31
1
2)31(
2
1)31(
1
2)12(
2
1)12(
1
2)11(
2
1)11(
12
21
31
12
11
)29(
1
241
2
141
12
2141)(
41
611221
3
1
1
111
1
2
1
111
31
12
11
111
BCACAB
BCA
BC
CAB
AB
AB
=
-=
úúú
û
ù
êêê
ë
é
÷÷÷
ø
ö
ççç
è
æ-
÷÷÷
ø
ö
ççç
è
æ -
-=
÷÷÷
ø
ö
ççç
è
æ
-
-
-=
÷÷÷
ø
ö
ççç
è
æ
-
-
=
úúúúúúúú
û
ù
êêêêêêêê
ë
é
÷ø
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æ
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æ
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æ
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æ
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ø
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ø
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æ
-=
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ø
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ii) ÷÷ø
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æ=÷÷
ø
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æ=÷÷
ø
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æ=
21
02,
51
04,
31
21CBA
(AB)CA(BC)
3029
2022
107
08
31
21A(BC)
107
08
21
02
51
04BC
3029
2022
21
02
157
106(AB)C
157
106
51
04
31
21AB
=
÷ø
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æ=
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æ÷ø
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æ=
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æ=
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æ=
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æ=
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æ÷ø
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æ=
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æ=
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æ÷ø
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æ=
Mythila Publishers, Puduvayal 71 X std Mathematics Made Easy
Download Question Bank From https://winglishcoachings.weebly.com/
8. Verify that AC. AB C) A (B +=+
i) ÷÷ø
öççè
æ-=÷÷
ø
öççè
æ
-=÷÷
ø
öççè
æ
-=
23
67,
24
21,
31
21CBA
AC. AB C) A (B
AC
AB
C) A (B
C B
+=+\
÷ø
öçè
æ-=÷
ø
öçè
æ-+÷
ø
öçè
æ
-
-=+
÷ø
öçè
æ-=÷
ø
öçè
æ-÷ø
öçè
æ
-=
÷ø
öçè
æ-
-=÷
ø
öçè
æ-
÷ø
öçè
æ-
=
÷ø
öçè
æ-=÷
ø
öçè
æ-
-÷ø
öçè
æ-
=+
÷ø
öçè
æ-
-=÷
ø
öçè
æ-+÷
ø
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æ-
=+
.43
127
016
84
413
43 AC AB
016
84
23
67
31
21
413
43
24
21
31
21
43
127
41
86
31
21
41
86
23
67
24
21
ii) ÷÷ø
öççè
æ
-=÷÷
ø
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æ -=÷÷
ø
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æ
-=
314
231,
253
211,
15
31CBA
÷ø
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æ-+÷÷
ø
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æ
-=+
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æ-=
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æ-
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æ-
=
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æ
-=
úúúú
û
ù
êêêê
ë
é
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æ-÷
ø
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æ --÷
ø
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æ÷ø
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æ -÷ø
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æ
=
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æ -÷ø
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æ-
=
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æ-=
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æ-
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æ-
=+
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æ-
=
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æ-
+÷ø
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æ -=+
15411
19201
7149
11611
8102
81410
7149
11611
314
231
15
31
8102
81410
2
2)15(
5
1)15(
3
1)15(
2
231(
5
131(
3
1)31(
253
211
15
31
15411
19201
561
422
15
31)(
561
422
314
231
253
211)(
ACAB
AC
AB
CBA
CB
9. Show that . A B(AB) TTT = If
i) ÷÷÷
ø
ö
ççç
è
æ
-
-
=÷÷ø
öççè
æ
-=
40
41
12
,112
121BA
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æ
-=
÷÷÷
ø
ö
ççç
è
æ
-÷ø
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æ
-
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æ
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æ
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ø
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æ
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æ
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æ
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ø
ö
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è
æ
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-
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æ
-=
49
50
11
12
21
441
012
49
50)(
45
90
11
12
21
441
012
242014
281022
40
41
12
112
121
TATB
TAB
TA
TBAB
ii) ÷÷÷
ø
ö
ççç
è
æ
-
=÷÷ø
öççè
æ=
15
21
71
,821
925BA
÷ø
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æ=÷
ø
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æ-+-+
++++=
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ø
ö
ççç
è
æ
÷ø
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æ
-=
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æ=
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æ=÷
ø
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æ
-+++
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ø
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æ
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æ=
330
4352
8479435
40214525
89
22
15
127
511
330
4352)(
343
3052
8474021
94354525
15
21
71
821
925
TT
T
AB
AB
AB
10.Prove that IA.AT = ,If ÷÷ø
öççè
æ
-=
cossin
sincosA
IAA
θθθθθθ
θθθθθθ
θθ
θθ
θθ
θθAA
θθ
θθA
θθ
θθA
T
T
T
=\÷ø
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æ
+-
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æ -÷ø
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æ-
=
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æ -=Þ÷
ø
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æ-
=
10
01
sincossincoscossin
sincoscossinsincos
cossin
sincos
cossin
sincos
cossin
sincos
cossin
sincos
22
22
11.Verify that IA =2 when ÷÷ø
öççè
æ
-
-=
56
45A
IA
AA
=\÷ø
öçè
æ=
úúúú
û
ù
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ø
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ø
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=
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æ-
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æ-
-=´
2
10
01
5
4)56(
6
5)56(
5
4)45(
6
5)45(
56
45
56
45
Mythila Publishers, Puduvayal 72 X std Mathematics Made Easy
Download Question Bank From https://winglishcoachings.weebly.com/
12.Show that I.BA =+ 22 If
÷÷ø
öççè
æ=÷÷
ø
öççè
æ=
q
q
q
q
sin0
0sin,
cos0
0cosBA
IBA
BA
B
A
=+\÷ø
öçè
æ=
÷÷ø
öççè
æ
+
+=
÷÷ø
öççè
æ+÷÷
ø
öççè
æ=+
÷÷ø
öççè
æ=
÷ø
öçè
æ÷ø
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æ=
÷÷ø
öççè
æ=
÷÷ø
öççè
æ
++
++=
÷ø
öçè
æ÷ø
öçè
æ=
22
22
22
2
2
2
222
2
2
2
2
2
2
2
2
10
01
sincos0
0sincos
sin0
0sin
cos0
0cos
sin0
0sin
sin0
0sin
sin0
0sin
cos0
0cos
cos000
000cos
cos0
0cos
cos0
0cos
qqqq
13.Show that 22 I)( A)(A adbc da -=+- ,If
÷÷ø
öççè
æ=÷÷
ø
öççè
æ=
10
01, I
dc
baA
RHS adbc
adbc
adbc
adbc
daddbccdcacdac
bdabbdabadabca
dadcdca
bdabada
dbccdac
bdabbca
dc
ba da
dc
ba
dc
ba
da
=-=
÷ø
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æ-=
÷ø
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æ
-
-=
÷÷ø
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æ
--+--+
--+--+=
÷÷ø
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æ
++
++-÷÷
ø
öççè
æ
++
++=
÷ø
öçè
æ+-÷
ø
öçè
æ÷ø
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æ=
+-=
2
22
22
2
2
2
2
2
I)(
10
01)(
0
0
)(
A)(ALHS
14. ÷ø
öçè
æ=÷
ø
öçè
æ=÷
ø
öçè
æ=
21
02,
51
04,
31
21CBA Show that
TT BA -=- TB)(Ai.
÷ø
öçè
æ-
-=-
÷ø
öçè
æ=
÷ø
öçè
æ=
÷ø
öçè
æ-
-=-
÷ø
öçè
æ-
-=
÷ø
öçè
æ--
--=-
22
03
50
14
32
11
22
03B)(A
20
23
5311
0241
T TT
T
T
BA
B
ABA
BC ACB) C (A -=-ii.
÷ø
öçè
æ--
-=
÷ø
öçè
æ-÷
ø
öçè
æ=
÷ø
öçè
æ÷ø
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æ-÷
ø
öçè
æ÷ø
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æ=
-=
÷ø
öçè
æ--
-=
÷ø
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æ÷ø
öçè
æ
-
-=
÷ø
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æ÷ø
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æ
--
--=
-=
42
44
107
08
65
44
21
02
51
04
21
02
31
21
RHS
42
44
21
02
20
23
21
02
5311
0241
LHS
BC AC
B) C(A
15.Show that 075 22 IAA =+- if ÷
ø
öçè
æ-
=21
13A
RHS
IAA
=
÷ø
öçè
æ=
÷ø
öçè
æ
+-++-
+-+-=
÷ø
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æ+÷
ø
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æ
--÷
ø
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æ
-=
÷ø
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æ+÷
ø
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æ-
-÷ø
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æ+---
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æ+÷
ø
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æ
--÷
ø
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æ
-÷ø
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æ
-=
+-=
00
00
7103055
0557158
70
07
105
515
35
58
70
07
105
515
4123
2319
10
017
21
135
21
13
21
13
75LHS 22
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