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#Swaraj - Author on ShareChat
#Swaraj
641 views • 3 months ago
Maths Tricks ##Maths
METHOD ) BILOPAN Math Notes ( Elimination Method Used to solve simultaneous linear equations two variables and J) Steps to Solve using General Form Bilopan Method bi) = Ct Q1* + equations  Write the two 02y = C2 02*+ Make coefficients of equal X O (by multiplying ) C102 - C2 01 . X = Add or Subtract the equations 01 02 _ 020( One variable eliminated gets 01C2 - 02 C1 remaining variable Solve for the 0102 0201 Substitute value in any equation 10 Tind other  variable @mwharshit Example 1 Exomple 2 : Solve Solve  2x +y = 7 2x + Y = 7 +) = 5 X = Y =1 both equations Add Solution 2* + Y+ *-Y = 7+1 (2x+ ٧)-(x+ ٧) = 7 - 5 3X २x-x+)-)= 2 X= 2 X = 8 in (2x+y = 7) equation Put X= 2 in 2nd Put X  (8) +y = 7 +)= २ +)= 5 1 23 +)= _)= 3 21 1 Answer  = 2 , ) = 3` 5 Trick Formula Tor Exams Remember C102 - C2 01 If ax + by = C Make coeTTicients equal q102  02 01 02X + 02/ = C2 3 Add/Subtrqct  cqrefully 01C2 = 02C1 d Substitute to Find other value 0102 = 0201 METHOD ) BILOPAN Math Notes ( Elimination Method Used to solve simultaneous linear equations two variables and J) Steps to Solve using General Form Bilopan Method bi) = Ct Q1* + equations  Write the two 02y = C2 02*+ Make coefficients of equal X O (by multiplying ) C102 - C2 01 . X = Add or Subtract the equations 01 02 _ 020( One variable eliminated gets 01C2 - 02 C1 remaining variable Solve for the 0102 0201 Substitute value in any equation 10 Tind other  variable @mwharshit Example 1 Exomple 2 : Solve Solve  2x +y = 7 2x + Y = 7 +) = 5 X = Y =1 both equations Add Solution 2* + Y+ *-Y = 7+1 (2x+ ٧)-(x+ ٧) = 7 - 5 3X २x-x+)-)= 2 X= 2 X = 8 in (2x+y = 7) equation Put X= 2 in 2nd Put X  (8) +y = 7 +)= २ +)= 5 1 23 +)= _)= 3 21 1 Answer  = 2 , ) = 3` 5 Trick Formula Tor Exams Remember C102 - C2 01 If ax + by = C Make coeTTicients equal q102  02 01 02X + 02/ = C2 3 Add/Subtrqct  cqrefully 01C2 = 02C1 d Substitute to Find other value 0102 = 0201
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  • #Swaraj - Author on ShareChat
    #Swaraj
    Maths Tricks ##Maths
    0F LAWS EXPONENTS Example Lawl २० = 1 =1 Q० al = a 171 17 = 14 = 42 02 Va = 273 an W0 = V27 = am 92 9-2 ಕ  ೆಕ  a~m 67 5 m 5) 3 మ 52 X 54 = 52+4 am+n Xan 3 am 45 24 45-3 Qm-n E ' 43 an (25)3 25*3 (am)n amxn { 3 25 X 35 = (2*3)5 X b1 = ( X b)7 @ m Vam 812 = V813 an S 2 3 3 bm 45 Q-n 3-2 ಕಕ S b-m 32 an 4-5  0F LAWS EXPONENTS Example Lawl २० = 1 =1 Q० al = a 171 17 = 14 = 42 02 Va = 273 an W0 = V27 = am 92 9-2 ಕ  ೆಕ  a~m 67 5 m 5) 3 మ 52 X 54 = 52+4 am+n Xan 3 am 45 24 45-3 Qm-n E ' 43 an (25)3 25*3 (am)n amxn { 3 25 X 35 = (2*3)5 X b1 = ( X b)7 @ m Vam 812 = V813 an S 2 3 3 bm 45 Q-n 3-2 ಕಕ S b-m 32 an 4-5
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  • #Swaraj - Author on ShareChat
    #Swaraj
    Maths Tricks ##Maths
    Maths Tricks, Maths
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  • #Swaraj - Author on ShareChat
    #Swaraj
    Maths Tricks ##Maths
    Algebraic Ldentities TC Basic_Squore Tdentities a२ + २ab + b२ (a+6)2 २ab + b२ (a - b)2 a2 62 (a + b)(a - b) 02 Binomials Product ofl Two (a+b)x+abl 2 + (x+a)(x + b) X (a+b)x+ab (xza) (x-b) +२ (X+0) (X-0) = 0 +२ of_Three_Terms Square (a+b+c)2 02 + 62 + C2 + 20b + 2bc + 2co Cube of Binomial 0 3a7b + 3ab2 (a + b)3 +63 + 63 3a2b + 3ab2 (a-b)3 ofl Cubes Difference Sum and (a+b)(az ab + b२  b3' + (a -b)(a? + ab + b2) b3 3 Three Terms of Cube (a+b+c)3 b3 + C3 + 3(q + b) (b + c)(c + ೦) + Q3 Tdentity Condirion - Based Tmportant 0+b + C =0 Tr +63 3 3abc = Then +C3 Advonced_Toentity 63 + C3 _ 3qbc 3 + (a + b + c) (a२ + b२ + c२ - ab ca) bc [ Algebraic Ldentities TC Basic_Squore Tdentities a२ + २ab + b२ (a+6)2 २ab + b२ (a - b)2 a2 62 (a + b)(a - b) 02 Binomials Product ofl Two (a+b)x+abl 2 + (x+a)(x + b) X (a+b)x+ab (xza) (x-b) +२ (X+0) (X-0) = 0 +२ of_Three_Terms Square (a+b+c)2 02 + 62 + C2 + 20b + 2bc + 2co Cube of Binomial 0 3a7b + 3ab2 (a + b)3 +63 + 63 3a2b + 3ab2 (a-b)3 ofl Cubes Difference Sum and (a+b)(az ab + b२  b3' + (a -b)(a? + ab + b2) b3 3 Three Terms of Cube (a+b+c)3 b3 + C3 + 3(q + b) (b + c)(c + ೦) + Q3 Tdentity Condirion - Based Tmportant 0+b + C =0 Tr +63 3 3abc = Then +C3 Advonced_Toentity 63 + C3 _ 3qbc 3 + (a + b + c) (a२ + b२ + c२ - ab ca) bc [
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  • #Swaraj - Author on ShareChat
    #Swaraj
    Maths Tricks ##Maths
    SQUARE AND SQUARE ROOT TABLE V1=1 162= 256 |256 =16 12 =1 ४ = 2 22= 4 172 = 289 V289 =17 V9 = 3 1324 182= 324 32= 9 18 Y16 = 4 Y36] = 19 192 = 361 42=16 125= 5 1400 = 20 52= 25 202=400 |36 = 6 62= 36 Y44] = 21 212 = 441 149 = 7 72 = 49 222= 484 |484 22 164= 8 82 = 64 232= 529 1529 = 23 81 92= 81 242 = 576 9 |576 24 = Y100 = 10 252= 625 1625 102=100 25 |676 26 ११२ = १२१ V121 = Il 262= 676 = १४४ = १२ 122=144 272 = 729 |729 = 27 1784 = 28 132 = 169 169 =13 282= 784 V841 V196 =14 142 196 292= 841 29 302= 900 900 = 30 152= 225 1225= 15 SQUARE AND SQUARE ROOT TABLE V1=1 162= 256 |256 =16 12 =1 ४ = 2 22= 4 172 = 289 V289 =17 V9 = 3 1324 182= 324 32= 9 18 Y16 = 4 Y36] = 19 192 = 361 42=16 125= 5 1400 = 20 52= 25 202=400 |36 = 6 62= 36 Y44] = 21 212 = 441 149 = 7 72 = 49 222= 484 |484 22 164= 8 82 = 64 232= 529 1529 = 23 81 92= 81 242 = 576 9 |576 24 = Y100 = 10 252= 625 1625 102=100 25 |676 26 ११२ = १२१ V121 = Il 262= 676 = १४४ = १२ 122=144 272 = 729 |729 = 27 1784 = 28 132 = 169 169 =13 282= 784 V841 V196 =14 142 196 292= 841 29 302= 900 900 = 30 152= 225 1225= 15
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  • #Swaraj - Author on ShareChat
    #Swaraj
    Maths Tricks ##Maths
    Example Rulel 25 X an X 23  = 28 =am+n am 57 : 53 =5 am an =am-n (103)7 = 1021 (am)n = a m Xn 171 al 17 a E F 340 = 1 = 1 al m a" 25 5 a 3 bm 36 b 6 1 m 2 9 a ಕ 5 am 81 Va 49 2 X 49 =7 a 5 3 Example Rulel 25 X an X 23  = 28 =am+n am 57 : 53 =5 am an =am-n (103)7 = 1021 (am)n = a m Xn 171 al 17 a E F 340 = 1 = 1 al m a" 25 5 a 3 bm 36 b 6 1 m 2 9 a ಕ 5 am 81 Va 49 2 X 49 =7 a 5 3
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  • #Swaraj - Author on ShareChat
    #Swaraj
    Maths Tricks ##Maths
    7.33 ?83 64% Illl Matrix Subtroctiorc  Definition : Subtracting two matrices by subtracting corresponding elements  Condition : Orders must be SQy1e  Definition : [4j] and B= [bijl then, Tf A= A-B= [aij - b!j] २. Example : A=[ః <] and B= 3 4 A-3 = [5-1 6-2 > 8-3 7-4 Another_Example: | A-[7 9 11] and B-[2 4 6] 11-6->5 5 5] > 5 5 5 A-B = ([7-2 9-4, Properties  of Motrix Subtroction : A-BfB-A Non-Commutative (A-8)-C# A-(8-C) Associative 3. Subtractive Identity: A+o = A (0= Zro Mntrix ) A-A= Subtractive Inverse  Subtract corresponding elements! Eg (1,1)-(1,1), (1,2)-(1,2) 7.33 ?83 64% Illl Matrix Subtroctiorc  Definition : Subtracting two matrices by subtracting corresponding elements  Condition : Orders must be SQy1e  Definition : [4j] and B= [bijl then, Tf A= A-B= [aij - b!j] २. Example : A=[ః <] and B= 3 4 A-3 = [5-1 6-2 > 8-3 7-4 Another_Example: | A-[7 9 11] and B-[2 4 6] 11-6->5 5 5] > 5 5 5 A-B = ([7-2 9-4, Properties  of Motrix Subtroction : A-BfB-A Non-Commutative (A-8)-C# A-(8-C) Associative 3. Subtractive Identity: A+o = A (0= Zro Mntrix ) A-A= Subtractive Inverse  Subtract corresponding elements! Eg (1,1)-(1,1), (1,2)-(1,2)
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  • #Swaraj - Author on ShareChat
    #Swaraj
    Maths Tricks ##Maths
    7:33 ?83 64% 111 CONERTING FRACTIONS 3 TOPERGEN 3  fRACIION 2 3 950 2- $  04 0+=100==0 5 TOPERCEN ORAGION 0.75 DECIMAL 0751100  75  075' 0,75 *100 = 75% 100 100 TOFRACON TODECIMAL 80% 0 PERCEN '80 5 809 - 100 ' 08 80% 100 CONVERTING FRACTIONS 2 O23RC N 0 n fRACIION 0 0 2 2 2*5*04 2 - 5 = 04 5 5 04'I040* 1OpERCENI .9 075  +0 DEGMAE {[ 0.75 *100  75 075' 0,75*100 = 75% 00 iod a MODaSLA  80% 0  0~4100` a PERCEN 80 80% 80% - I00 = 0.8 Joo 7:33 ?83 64% 111 CONERTING FRACTIONS 3 TOPERGEN 3  fRACIION 2 3 950 2- $  04 0+=100==0 5 TOPERCEN ORAGION 0.75 DECIMAL 0751100  75  075' 0,75 *100 = 75% 100 100 TOFRACON TODECIMAL 80% 0 PERCEN '80 5 809 - 100 ' 08 80% 100 CONVERTING FRACTIONS 2 O23RC N 0 n fRACIION 0 0 2 2 2*5*04 2 - 5 = 04 5 5 04'I040* 1OpERCENI .9 075  +0 DEGMAE {[ 0.75 *100  75 075' 0,75*100 = 75% 00 iod a MODaSLA  80% 0  0~4100` a PERCEN 80 80% 80% - I00 = 0.8 Joo
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  • #Swaraj - Author on ShareChat
    #Swaraj
    Maths Tricks ##Maths
    7.33 64% 80.0 Illl KBls -imit Properties Limit Propertles (constant; r() = A) lim 4= A (linear function) lim 1 lim g(x) Vim|f (x) L g(x)l lim f() _ (additionlsubtraclion) limlk  I()l = k lim /() k) (constant multiple limlr() ' g(7)| 9() Imuiltiplication ) 1im / () lim lim ](r) 9() + 0 m (division) lim given that lm p(1)  Ilim F(x)]" limlr()I" 7. (exponential, na [limT(x)]" "((x))" (radicals/roots) lim lim x7 using the limit properties Exammple 1: Find Find lm v३t using the limit properties Lxample 7.33 64% 80.0 Illl KBls -imit Properties Limit Propertles (constant; r() = A) lim 4= A (linear function) lim 1 lim g(x) Vim|f (x) L g(x)l lim f() _ (additionlsubtraclion) limlk  I()l = k lim /() k) (constant multiple limlr() ' g(7)| 9() Imuiltiplication ) 1im / () lim lim ](r) 9() + 0 m (division) lim given that lm p(1)  Ilim F(x)]" limlr()I" 7. (exponential, na [limT(x)]" "((x))" (radicals/roots) lim lim x7 using the limit properties Exammple 1: Find Find lm v३t using the limit properties Lxample
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  • #Swaraj - Author on ShareChat
    #Swaraj
    Maths Tricks ##Maths
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