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 Algebra Calculators 1. Addition, subtraction, multiplication, division of two polynomials eg.(X^3-4X^2+4X-8) xx (X-2) 2. Factoring Polynomials eg. (1) 25x^2-36, (2) 4x^2+12xy+9y^2, (3) x^3-3x^2-6x+8, (4) a^2(b-c)+b^2(c-a)+c^2(a-b) 3. Expand and simplify polynomial eg. (1) (X+2)(X+3), (2) (2X+3Y)^2, (3) (2X+3Y+4Z)^2, (4) 102^2, (5) 102 xx 106 (Expand using the identity) 4. Complete square, Is perfect square, Find missing term 1. Completing the square for quadratic equation eg. 9x^2+6x+1= 9( x+1/3 )^2 2. Determining if the polynomial is a perfect square eg. (1) x^2-4xy+4y^2, (2) 3x^2+5x+2 3. Find the missing term in a perfect square trinomial eg. (1) 9x^2 - __ + 16, (2) __ + 12x^2 + 9, (3) 49x^2 + 56 xy + __ 5. HCF(GCD)-LCM of Polynomials eg. Find GCD, LCM of (2X^2-4X), (3X^4-12X^2), (2X^5-2X^4-4X^3) 6. Rational Expression of Polynomials 1. Reduced terms of rational expressions eg. (4X^2-25)/(8X^3-125) 2. Adding, subtracting, multiplying, dividing of rational expressions polynomials eg. (X-3)/(X+1)-(X-6)/(X) 7. Simplifying Algebraic Expressions eg. (1) (4x+1)/(3x-2)=3/2, (2) (2x-1)/(3x+1)+(4x-1)/(6x)=0 8. Quadratic Equation 1.1 Solving quadratic equations by factoring, eg. (1) 25x^2-30x+9=0, (2) x^2+10x-56=0 1.2 Solving quadratic equations by Completing the Square, eg. (1) 25x^2-30x+9=0, (2) x^2+10x-56=0 1.3 Discriminant & Nature of Roots eg. (1) 25x^2-30x+9=0, (2) x^2+10x-56=0 2. Find the quadratic equation whose roots are alpha and beta eg. (1) alpha=3, beta=-4, (2) alpha=1+3sqrt(2), beta=1-3sqrt(2) 3. Roots for non-zero denominator eg. (1) (5x-18)/(x+2)=(2x-6)/(x-1), (2) (x)/(x+1)+(x+1)/(x)=5/2, (3) 4((4x+1)/(4x-1))^(2)+(4x+1)/(4x-1)=3, (4) (4x+1)/(4x-1)+(4x-1)/(4x+1)=3 4. Roots of non-quadratic equation eg. (1) 6(x^2+1/x^2)-25(x-1/x)+12=0, (2) (x^2+1/x^2)-8(x+1/x)+14=0 9. Solve linear equation in two variables by (eg. Solve 7y+2x-11=0 and 3x-y-5=0 using Substitution method) 1. Substitution method 2. Elimination method 3. Cross multiplication method 4. Addition-Subtraction method 5. Inverse matrix method 6. Cramer's Rule method 7. Graphical method 10. Solve linear equation of any number of variables (simultaneous equations) using 1. Inverse Matrix method 2. Cramer's Rule method 3. Gauss Elimination (Jordan) method 4. Gauss Elimination (Back Substitution) method 5. Gauss Seidel method 6. Gauss Jacobi method 11. Variation Equations 1. Find value of variation using given value (1) x prop y and x=6 when y=3. Find y=? when x=18, (2) x prop y/z. x=8 when y=4 and z=3. Find x=? when y=6 and z=4. 2. Prove results for given variation (1) If x prop y then prove that x^3+y^3 prop x^2y-xy^2, (2) If 3x-5y prop 5x+6y then prove that x prop y. 12. If X+1/X=2 then find X-1/X, X^2-1/X^2, X^3+1/X^3 1. If x-1/x=6 then find (1) x^2+1/x^2 (2) x+1/x (3) x^2-1/x^2 2. If x+y=5 and xy=6 then find x^2+y^2 3. If x^2+y^2+z^2=29 and xy+yz+zx=-14 then find x+y+z 13. Interval notation and set builder notation eg. (1) 3 <= x <=7, x is odd. (2) |x^3-2| <= 25, x in Z. (3) x in[2,8) 14. Set Theory eg. A={x<=5; x in N}, B={2<=x<=8; x in N}, C={x^3-3x^2-4x=0}, Find 1. Union eg. A uu (B uu C)=(A uu B) uu C 2. Intersection eg. A nn (B uu C)=(A nn B) uu (A nn C) 3. Complement eg. (A uu B)'=A' nn B' 4. Power set(Proper Subset) eg. P(A) 5. Difference eg. (1)A-B, (2) A-(B uu C)=(A-B) nn (A-C) 6. Symmetric difference eg. (1)A Delta B, (2) B Delta C, (3) A Delta C 7. Cross Product eg. A xx B 8. Prove that any two expression is equal or not eg. A-(B uu C)=(A-B) nn (A-C) 9. Cardinality of a set eg. n(A) 10. is Belongs To a set eg. 2inB ? 11. is Subset Of a set eg. AsubB ? 12. is two set Equal or not eg. A=B ? 15. Functions 1. Find Range of f:A->B eg. 1. f(x)=5x+2 where A={1<=x<5}, 2. f(x)=sqrt(x) where A={1,4,16,36} 2. Composite functions and Evaluating functions eg. f(x)=2x+1, g(x)=x+5. Find fog(x), gof(x), fog(2), f(-3), g(4) 3. Find value eg. 1. f(x)=x(x+1)(2x+1). Find f(x)-f(x-1), 2. f(x)=x^2-2^x. Find f(2)-f(0) 16. Descartes' rule of signs eg. x^5-x^4+3x^3+9x^2-x+5 17. Mathematical Logic, truth tables, logical equivalence eg. Prepare the truth table p^^(qvvr)=(p^^q)vv(p^^r) 18. Boolean Algebra eg. D_(6), D_(9) is a boolean algebra ?
 1. Factoring Polynomials Type-1 (Taking common) Examples... (1) ax + a + 2x + 2     =(ax + a) + (2x + 2)     =a(x + 1) + 2(x + 1)     =(x + 1)(a + 2) Type-2 (Difference of squares) Examples... (1) 25x2 - 36     =(5x)2 - (6)2     =(5x - 6)(5x + 6) Type-3 (Sum and Difference of cubes) Examples... (1) x3 + 27     =(x)3 + (3)3     =(x + 3)(x2 - (x)(3) + (3)2)     =(x + 3)(x2 - 3x + 9) Type-4 (Whole square of a bionomial) Examples... (1) 4x2 + 12xy + 9y2     =(2x)2 + 2(2x)(3y) + (3y)2     =(2x + 3y)2 Type-5 (Splitting the middle term of a Quadratic Equation) Examples... (1) x2 + 10x + 24     =x2 + 4x + 6x + 24     =x(x + 4) + 6(x + 4)     =(x + 4)(x + 6) Type-6 (Whole square of a trinomial) Examples... (1) 4x2 + y2 + 1 + 4xy + 4x + 2y     =(2x)2 + (y)2 + (1)2 + 2(2x)(y) + 2(2x)(1) + 2(y)(1)     =(2x + y + 1)2 Type-7 (Factorization with the help of factor theorem) Examples... (1) x3 - 3x2 - 6x + 8     Here p(x)=x3 - 3x2 - 6x + 8     sum of coefficient of all the terms of p(x) = 1 - 3 - 6 + 8 = 0     \ (x-1) is a factor of p(x).     Now, p(x) = x3 - 3x2 - 6x + 8     =x3 - x2 - 2x2 + 2x - 8x + 8     =x2(x - 1) - 2x(x - 1) - 8(x - 1)     =(x - 1)(x2 - 2x - 8)     =(x - 1)(x - 4)(x + 2) Type-8 Cyclic Expressions Examples... (1) a2(b - c) + b2(c - a) + c2(a - b)      = a2b - a2c + b2c - b2a + c2a - c2b       = a2b - a2c + cb2 - ab2 + ac2 - bc2       = a2b - a2c - ab2 + ac2 + b2c - bc2       = a2(b - c) - a(b2 + c2) + bc(b - c)       = a2(b - c) - a(b - c)(b + c) + bc(b - c)       = (b - c)(a2 - a(b + c) + bc)       = (b - c)(a2 - ab - ac + bc)       = (b - c)(a(a - b) - c(a - b))       = (b - c)(a - b)(a - c)       = -(a - b)(b - c)(c - a)
 2. Expand and simplify polynomial Option-1 => Expression Like 1. (X + 2)(X + 3)2. (X + Y)(X2 - XY + Y2)3. (2X + 3Y + 4Z)24. (3Y - 2X)35. (2X + 3Y)26. (X + Y)2 - (X - Y)27. (X + Y + Z)2 + (X + Y - Z)28. (3X - 5Y)3 + (5Y - 9Z)3 + (9Z - 3X)3 1. Expand an Algebraic Expressions (X+2)(X+3)Using The Identity,(X + A)(X + B) = X2 + (A + B)X + ABHere X=X,A=2, B=3 = (X)2 + (2 + 3)X + (2)×(3) = X2+5X+6 2. Expand an Algebraic Expressions (2X+3Y+4Z)2Using The Identity,(A + B + C)2 = A2 + B2 + C2 + 2AB + 2BC + 2CAHere A = 2X, B = 3Y, C = 4Z = (2X)2 + (3Y)2 + (4Z)2 + 2(2X)(3Y) + 2(3Y)(4Z) + 2(4Z)(2X) = 4X2+9Y2+16Z2+12XY+24YZ+16XZ Option-2 => Numeric Expression Like 1. 1022, 0.232, 9.822. 123, 12.233. 93, 8.934. 102 × 1065. 342 - 322, 3.42 - 3.226. 103 - 73 - 337. 43 + 33, 3.43 - 2.93 1. Find Value of 102 × 106Using, The Identity (X+A)(X+B) = X2 + (A+B)X + ABHere X = 100, A = 2 and B = 6102 × 106 = (100 + 2)(100 + 6) = 1002 + (2 + 6)100 + 2 × 6 = 10000 + 800 + 12 = 10812 2. Find Value of 342 - 322Using, The Indentity A2 - B2 = (A-B)(A+B)Here A = 34 and B = 32Here 342 - 322 = (34-32)(34+32) = (2)(66) = 132

 3. If X+1/X=2 then find X-1/X, X^2-1/X^2, X^3+1/X^3 1. If x + 1/x = 2 then find x^2 + 1/x^2 2. If x - 1/x = 6 then find (1) x^2 + 1/x^2 (2) x^4 + 1/x^4 (3) x^3 + 1/x^3 (4) x + 1/x (5) x^2 - 1/x^2 (6) x^4 - 1/x^4 (7) x^3 - 1/x^3 3. If x^2 + 1/x^2 = 23 then find (1) x + 1/x (2) x^2 + 1/x^2 4. If x + y = 5 and xy = 6 then find (1) x^2 + y^2 (2) x^3 + y^3 (3) x^4 + y^4 (4) x - y (5) x^2 - y^2 (6) x^3 - y^3 (7) x^4 - y^4 5. If x + y = 3 and x - y = 15 then find x^2 + y^2 and xy 6. If x + y + z = 9 and x^2 + y^2 + z^2 = 29 then find xy + yz + zx 7. If x^2 + y^2 + z^2 = 29 and xy + yz + zx = -14 then find x + y + z 1. If X+1/X = 2, then find X-1/X Here X + 1/X = 2Now, We know that( X - 1/X )^2 = ( X + 1/X )^2 - 4( X - 1/X )^2 = 2^2 - 4( X - 1/X )^2 = 4 - 4( X - 1/X )^2 = 0X - 1/X = 0 2. If X+Y = 5 and X-Y = 1, then find X^2+Y^2 Here X + Y = 5 and X - Y = 1Now, We know that4 XY = ( X + Y )^2 - ( X - Y )^24 XY = 5^2 - 1^24 XY = 25 - 14 XY = 24XY = 24/4XY = 6Now, We know thatX^2 + Y^2 = ( X + Y )^2 - 2 XYX^2 + Y^2 = 5^2 - 2 * 6X^2 + Y^2 = 25 - 12X^2 + Y^2 = 13 3. If X+Y+Z = 1 and X^2+Y^2+Z^2 = 29, then find XY+YZ+ZX Here X+Y+Z = 1 and X^2 + Y^2 + Z^2 = 29Now, We know that(X+Y+Z)^2 = (X^2 + Y^2 + Z^2) + 2(XY+YZ+ZX)2 (XY+YZ+ZX) = (X+Y+Z)^2 - (X^2 + Y^2 + Z^2)2 (XY+YZ+ZX) = 1^2 - (29)2 (XY+YZ+ZX) = 1 - (29)2 (XY+YZ+ZX) = -28(XY+YZ+ZX) = -28/2(XY+YZ+ZX) = -14
4. Addition, subtraction, multiplication, division of two polynomials
Find addition, Subtraction, multiplication, division 4X3-3X2+2X-4 and X+1

(X4 - 4X3 - 4X + 4) + (X2 + 2X - 2)
= X4 - 4X3 - 4X + 4 + X2 + 2X - 2
= X4 - 4X3 + X2 - 4X + 2X + 4 - 2
= X4 - 4X3 + X2 - 2X + 2

OR
 X4 - 4X3 - 4X + 4
+
 X2 + 2X - 2

 X4 - 4X3 + X2 - 2X + 2

2. Subtraction

(4X3 - 3X2 + 2X - 4) - (X + 1)
= 4X3 - 3X2 + 2X - 4 - X - 1
= 4X3 - 3X2 + 2X - X - 4 - 1
= 4X3 - 3X2 + X - 5

OR
 4X3 - 3X2 + 2X - 4
-
 -X + -1

 4X3 - 3X2 + X - 5

3. Multiplication

(4X3 - 3X2 + 2X - 4) (X + 1)
= + 4X3(X + 1) - 3X2(X + 1) + 2X(X + 1) - 4(X + 1)
= 4X4 + 4X3 - 3X3 - 3X2 + 2X2 + 2X - 4X - 4
= 4X4 + 4X3 - 3X3 - 3X2 + 2X2 + 2X - 4X - 4
= 4X4 + 4X3 - 3X3 - 3X2 + 2X2 + 2X - 4X - 4
= 4X4 + X3 - X2 - 2X - 4

4. Division

 4X2 - 7X + 9

X + 1|
 4X3 - 3X2 + 2X - 4

|
 -4X3 + -4X2
(X + 1) (4X2)
|

|
 - 7X2 + 2X - 4

|
 - +7X2 - +7X
(X + 1) (- 7X)
|

|
 9X - 4

|
 -9X + -9
(X + 1) (9)
|

|
 - 13

Here, Divisor = X + 1
Dividend = 4X3 - 3X2 + 2X - 4
Quotient = 4X2 - 7X + 9
Remainder = - 13

 5. HCF(GCD)-LCM of Polynomials Gives you Greatest Common Divisor (Or Highest Common Factor) and Least Common Multiple. 1. Find GCD, LCM of 30(X^2-3X+2) and 50(X^2-2X+1)Factor of 30(X^2-3X+2)= 30(X2 - 3X + 2)= 30(X2 - X - 2X + 2)= 30(X(X - 1)+(- 2)(X - 1))= 30(X - 2)(X - 1)= 2 × 3 × 5 (X - 2)(X - 1)= 2 × 3 × 5 (X - 2)(X - 1)Factor of 50(X^2-2X+1)= 50(X2 - 2X + 1)= 50(X - 1)(X - 1)= 50(X - 1)2= 2 × 5 × 5 (X - 1)2= 2 × 5 × 5 (X - 1)2GCD = 2 × 5 (X - 1)GCD = 10(X - 1)LCM = 2 × 3 × 5 × 5 (X - 2)(X - 1)2LCM = 150(X - 2)(X - 1)2 2. Find GCD, LCM of (2X^2-4X), (3X^4-12X^2), (2X^5-2X^4-4X^3) Factor of (2X^2-4X)= (2X2 - 4X)= 2X(X - 2)= 2 X(X - 2)Factor of (3X^4-12X^2)= (3X4 - 12X2)= 3X2(X2 - 4)= 3X2(X - 2)(X + 2)= 3 X2(X - 2)(X + 2)Factor of (2X^5-2X^4-4X^3)= (2X5 - 2X4 - 4X3)= 2X3(X2 - X - 2)= 2X3(X2 + X - 2X - 2)= 2X3(X(X + 1)+(- 2)(X + 1))= 2X3(X - 2)(X + 1)= 2 X3(X - 2)(X + 1)GCD = X(X - 2)LCM = 2 × 3 X3(X - 2)(X + 2)(X + 1)LCM = 6X3(X - 2)(X + 2)(X + 1)
 6. Rational Expression of Polynomials 1. Find Reduced Terms of Rational Expressions (1) (X^2+2X+1)/(2(X+1)) (2) (2(X^2-Y^2))/(3(X^3-Y^3)) (3) (X^4-1)/(X^2+1) (4) (X^2-9)/(X^3-27) 2. Adding, subtracting, multiplying, dividing of rational expressions polynomials. (1) (X-3)/(X+1) - (X-6)/(X) (2) (4x+1)/(4x-1) - (4x-1)/(4x+1) - 3 (3) (1)/(6X^2-13X+6) + (1)/(8X^2-14X+3) - (1)/(12X^2-11X+2) (4) (X+2)/(X^2-4X+3) xx (X+3)/(X^2-3X+2) - (X+1)/(X^2-5X+6) (5) (X^3-1)/(X-1) + (X^3+1)/(X+1) - (2(X^4-1))/(X^2-1) 1. Find Reduced Term of (4X^2-25)/(8X^3-125)  = ((4X^2 - 25))/((8X^3 - 125)) = ((2X-5)(2X+5))/((2X-5)(4X^2+10X+25))"Now cancel the common factor " (2X - 5) = ((2X+5))/((4X^2+10X+25)) 2. Find (X+2)/(X^2-4X+3) - (X+3)/(X^2-3X+2)  = ((X + 2))/((X^2 - 4X + 3)) - ((X + 3))/((X^2 - 3X + 2)) = ((X+2))/((X-3)(X-1)) - ((X+3))/((X-2)(X-1)) = ((X + 2) × (X - 2) - (X + 3) × (X - 3))/((X-1)(X-3)(X-2)) = ((X^2 - 4) - (X^2 - 9))/((X-1)(X-3)(X-2)) = (5)/((X-1)(X-3)(X-2))

 7. Complete Square, Is perfect Square, Find Missing Term 1. Convert the given equation into perfect square form (1) 3x2+5x+2 (2) 9x2+6x+1 (3) x2+6x+9 2. Determining if the polynomial is a perfect square (1) 4x2 + 4x + 4 (2) x2 - 4xy + 4y2 (3) 30xy - 25x2y2 + 9 3. Find the missing term in a perfect square trinomial (1) x2 + ____ + 4 (2) 9x2 - ____ + 16 (3) 81x2 + ____ + 49 (4) x2 + ____ + 9 (4) ____ + 12x2 + 9 (5) 49x2 + 56 xy + ____ (6) 25x2 - ____ + 121y2 (7) 9x2 + ____ + y2 (8) x4 + 6x2 + ____ (9) x4y2 - 10x2yz + ____ 1.1 Convert the given equation 3X^2+5X+2 into perfect square form = (3X^2+5X+2)= 3 (X^2+5/3X+2/3)= 3 (X^2+5/3X + 25/36 - 25/36 + 2/3)= 3 [(X^2+5/3X+25/36) -1/36]= 3 [( X + 5/6 )^2 -1/36] 2. Check the given equation 3X^2+5X+2 is perfect square or not  3X^2+5X+2 "Here "F.T. = 3X^2, M.T. = 5X and L.T. = 2  (M.T.)^2 = (5X)^2 = 25X^2 4(F.T.)(L.T.) = 4 * 3X^2 * 2 = 24X^2 :. (M.T.)^2 != 4(F.T.)(L.T.) :. "given polynomial is not a perfect square." 3. Find Missing First Term of a given equation X^2+4  X^2+4 "Here "L.T. = 4, M.T. = X^2 and F.T. = ?  (M.T.)^2 = 4(F.T.)(L.T.) F.T. = (M.T.)^2 / (4(L.T.)) F.T. = (X^2 * X^2) / (4 * 4) F.T. = (X^4) / (16)
 8. Simplifying Algebraic Expressions Solve linear equation 1. 5 - 3/5 + 7/(3*3) 2. 4 + a^2 * 3 * (x+y) where a=4,x=4 3. x/5 + 3/5 - 8/3 4. (2x-1)/(3x+1) + (4x-1)/(6x) 5. (2x+8)/(3x+1) + 3 - (x+4)/(7x+1) - 17/4 6. (3x+1)/2 - (3x-1)/3 - (5(x+2))/6 Simplify linear equation 1. 4 + 3 * (x+y) = 0 where x=4 2. (4x+1)/(3x-2) = 3/2 3. (2x-1)/(3x+1) = (4x-1)/(6x) 4. (3x+1)/2 - (3x-1)/3 = (5(x+2))/6 1. Solve linear equation (3x+1)/5-(3x-1)/3 (3x+1)/5-(3x-1)/3 = ((3x + 1))/(5) - ((3x - 1))/(3) = ((3x+1))/(5) - ((3x-1))/(3) = ((3x + 1) × 3 - (3x - 1) × 5)/(15) = ((9x + 3) - (15x - 5))/(15) = ((-6x+8))/(15) 2. Solve linear equation 4+3*(x+y)=0 and given values are x=4 4+3*(x+y)=0 where x=4 => (4+3(4+y)) = 0 => (3y+16) = 0 => 3y = -16 => y = -16/3

 1. Substitution Method Solve linear equations 7Y+2X-11=0 and 3X-Y-5=0 using Substitution Method 2X+7Y-11 = 0 ->(1)3X-Y-5 = 0 ->(2)Taking Eqn. (2), we have3X-Y-5 = 0=> Y = 3X-5 ->(3)Putting Y = 3X-5 in Eqn. (1), we get2X+7(3X-5)-11 = 0=> 2X+21X-35-11 = 0=> 23X-46 = 0=> 23 (X-2) = 0=> X-2 = 0=> X = 2 ->(4)Now, Putting X = 2 in Eqn. (3), we getY = 3(2)-5=> Y = 6-5=> Y = 1:. Y = 1" and "X = 2
 2. Elimination Method Solve linear equations 7Y+2X-11=0 and 3X-Y-5=0 using Elimination Method 2X+7Y = 11 ->(1)3X-Y = 5 ->(2)Eqn.(1) xx 1 => 2X+7Y = 11Eqn.(2) xx 7 => 21X-7Y = 35Adding  => 23X = 46=> X = 46 / 23=> X = 2Putting X = 2 in Eqn. (2), we have3(2)-Y = 5=> -Y = 5-6=> -Y = -1=> Y = 1:. X = 2" and "Y = 1

 3. Cross Multiplication Method Solve linear equations 7Y+2X-11=0 and 3X-Y-5=0 using Cross Multiplication Method 2X+7Y-11 = 0 ->(1)3X-Y-5 = 0 ->(2)Here,a_1=2, b_1=7, c_1=-11a_2=3, b_2=-1, c_2=-5X = (b_1*c_2-b_2*c_1)/(a_1*b_2-a_2*b_1)= ((7)(-5)-(-1)(-11))/((2)(-1)-(3)(7))= ((-35)-(11))/((-2)-(21))= (-46)/(-23)= 2Y = (c_1*a_2-c_2*a_1)/(a_1*b_2-a_2*b_1)= ((-11)(3)-(-5)(2))/((2)(-1)-(3)(7))= ((-33)-(-10))/((-2)-(21))= (-23)/(-23)= 1:. X = 2" and "Y = 1
 4. Addition-Subtraction Method Solve linear equations 7Y+2X-11=0 and 3X-Y-5=0 using Addition-Substriaction Method 2X+7Y-11 = 0 ->(i)3X-Y-5 = 0 ->(ii)Adding Equation (i) and (ii), we get5X+6Y-16 = 05X+6Y-16 = 0 ->(iii) (On simplifying)Subtracting Equation (ii) from (i), we get-X+8Y-6 = 0X-8Y+6 = 0 ->(iv) (On simplifying)5X+6Y = 16 ->(1)X-8Y = -6 ->(2)Eqn.(1) xx 1 => 5X+6Y = 16Eqn.(2) xx 5 => 5X-40Y = -30Substracting => 46Y = 46=> Y = 46 / 46=> Y = 1Putting Y = 1 in Eqn. (2), we haveX-8(1) = -6=> X = -6+8=> X = 2=> X = 2:. X = 2" and "Y = 1

 5. Inverse Matrix Method Solve linear equations 12x+5y=7 and x+y=7 using Using Matrix Method => (12x+5y-7) = 0=> (x+y-7) = 0Here 12x+5y=7, x+y=7Now converting given equations into matrix form [[12,5],[1,1]] [[ x ],[ y ]] = [[7],[7]]Now, A = [[12,5],[1,1]], X = [[ x ],[ y ]] and B = [[7],[7]]:. AX = B:. X = A^-1 B| A |=|[12,5],[1,1]| = 12 × 1 - 5 × 1 = 12 - 5 = 7"Here, " | A | = 7 != 0 :. A^(-1)" is possible."Adj(A)=Adj[[12,5],[1,1]]=[[+(1),-(1)],[-(5),+(12)]]^T=[[1,-1],[-5,12]]^T=[[1,-5],[-1,12]]"Now, "A^(-1)=1/| A | × Adj(A)"Here, "X = A^(-1) × B:. X =1/| A | × Adj(A) × B=1/7 × [[1,-5],[-1,12]] × [[7],[7]] =1/7 ×[[1*7 + -5*7],[-1*7 + 12*7]] =1/7 ×[[-28],[77]] =[[-4],[11]]:.[[ x ],[ y ]] = [[-4],[11]]:. x = -4, y = 11
 6. Cramer's Rule Method Solve linear equations 12x+5y=7 and x+y=7 using Using Cramer's Rule Method The equations can be expressed as 12x+5y-7=0x+y-7=0Use Cramer’s Rule to find the values of x, y, z.(x)/D_x=(-y)/D_y=(1)/D D_x =|[5,-7],[1,-7]| = 5 × -7 - -7 × 1 = -35 + 7 = -28 D_y =|[12,-7],[1,-7]| = 12 × -7 - -7 × 1 = -84 + 7 = -77 D =|[12,5],[1,1]| = 12 × 1 - 5 × 1 = 12-5 = 7(x)/D_x=(-y)/D_y=(1)/D:. (x)/-28=(-y)/-77=(1)/7:. (x)/-28=(1)/7,(-y)/-77=(1)/7:. x=(-28)/(7),y=(77)/(7):. x=-4,y=11

 7. Graphical Method Solve linear equations 12x+5y=7 and x+y=7 using Graphical Method 12x+5y=7x+y=7The point of intersection of the linear equationsConsider 12x+5y=7x-intercept: put y = 0, we get x = 0.5833 :. A(0.5833,0)y-intercept: put x = 0, we get y = 1.4 :. B(0,1.4)Consider x+y=7x-intercept: put y = 0, we get x = 7 :. C(7,0)y-intercept: put x = 0, we get y = 7 :. D(0,7)Intersection Point = P(-4,11)

Solve linear equation of any number of variables Inverse matrix method, Gauss elimination method
Solve equations like
1. 2x + y + z = 10, 3x + 2y + 3z = 18, x + 4y + 9z = 16
2. 2x + 5y = 16, 3x + y = 11
3. 2x + 5y = 21, x + 2y = 8
4. 2x + y = 8, x + 2y = 1
5. 2x + 3y - z = 5, 3x + 2y + z = 10, x - 5y + 3z = 0
6. x + y + z = 3, 2x - y - z = 3, x - y + z = 9
7. x + y + z = 7, x + 2y + 2z = 13, x + 3y + z = 13
8. 2x - y + 3z = 1, -3x + 4y - 5z = 0, x + 3y - 6z = 0
 1. Inverse matrix method 1. Solve Equations 2x+y+z=5,3x+5y+2z=15,2x+y+4z=8 using Inverse Matrix methodHere 2x+y+z=5, 3x+5y+2z=15, 2x+y+4z=8Now converting given equations into matrix form [[2,1,1],[3,5,2],[2,1,4]] [[ x ],[ y ],[ z ]]=[[5],[15],[8]]Now, A = [[2,1,1],[3,5,2],[2,1,4]], X = [[ x ],[ y ],[ z ]] and B = [[5],[15],[8]]:. AX = B:. X = A^-1 B| A |=|[2,1,1],[3,5,2],[2,1,4]| = 2 (5 × 4 - 2 × 1) - 1 (3 × 4 - 2 × 2) + 1 (3 × 1 - 5 × 2) = 2 (20 - 2) - 1 (12 - 4) + 1 (3 - 10) = 2 (18) - 1 (8) + 1 (-7) = 36 - 8 - 7 = 21"Here, " | A | = 21 != 0 :. A^(-1)" is possible."Adj(A)=Adj[[2,1,1],[3,5,2],[2,1,4]]=[[+(5 × 4 - 2 × 1),-(3 × 4 - 2 × 2),+(3 × 1 - 5 × 2)],[-(1 × 4 - 1 × 1),+(2 × 4 - 1 × 2),-(2 × 1 - 1 × 2)],[+(1 × 2 - 1 × 5),-(2 × 2 - 1 × 3),+(2 × 5 - 1 × 3)]]^T=[[18,-8,-7],[-3,6,0],[-3,-1,7]]^T=[[18,-3,-3],[-8,6,-1],[-7,0,7]]"Now, "A^(-1)=1/| A | × Adj(A)"Here, "X = A^(-1) × B:. X =1/| A | × Adj(A) × B=1/21 × [[18,-3,-3],[-8,6,-1],[-7,0,7]] × [[5],[15],[8]] =1/21 ×[[18*5 + -3*15 + -3*8],[-8*5 + 6*15 + -1*8],[-7*5 + 0*15 + 7*8]] =1/21 ×[[21],[42],[21]] =[[1],[2],[1]]:.[[ x ],[ y ],[ z ]]=[[1],[2],[1]]:. x=1, y=2, z=1
 2. Gauss elimination method 1. Solve Equations 2x+y+z=5,3x+5y+2z=15,2x+y+4z=8 using Gauss Elimination methodTotal Equations are 32 x + y + z = 53 x + 5 y + 2 z = 152 x + y + 4 z = 8Converting given equations into matrix form[[2,1,1,|,5],[3,5,2,|,15],[2,1,4,|,8]]Dividing R_1 by 2[[1,1/2,1/2,|,5/2],[3,5,2,|,15],[2,1,4,|,8]]R_2 larr R_2 - 3 * R_1[[1,1/2,1/2,|,5/2],[0,7/2,1/2,|,15/2],[2,1,4,|,8]]R_3 larr R_3 - 2 * R_1[[1,1/2,1/2,|,5/2],[0,7/2,1/2,|,15/2],[0,0,3,|,3]]Dividing R_2 by 7/2[[1,1/2,1/2,|,5/2],[0,1,1/7,|,15/7],[0,0,3,|,3]]R_1 larr R_1 - 1/2 * R_2[[1,0,3/7,|,10/7],[0,1,1/7,|,15/7],[0,0,3,|,3]]Dividing R_3 by 3[[1,0,3/7,|,10/7],[0,1,1/7,|,15/7],[0,0,1,|,1]]R_1 larr R_1 - 3/7 * R_3[[1,0,0,|,1],[0,1,1/7,|,15/7],[0,0,1,|,1]]R_2 larr R_2 - 1/7 * R_3[[1,0,0,|,1],[0,1,0,|,2],[0,0,1,|,1]]Solution By Gauss Elimination Method.x = 1y = 2z = 1

 1. Solving quadratic equations by factoring 1. Find the roots of Quadratic Equation x^2+8x+12=0 by Factoring  x^2+8x+12=0 => x^2+8x+12 = 0 => (x^2+8x+12) = 0 => (x^2+2x+6x+12) = 0 => x(x+2)+6(x+2) = 0 => (x+6)(x+2) = 0 => (x+6) = 0" or "(x+2) = 0 => x = -6" or "x = -2 2. Find the roots of Quadratic Equation X^2-25=0 by Factoring  X^2-25=0 => X^2-25 = 0 => (X^2-25) = 0 => (X-5)(X+5) = 0 => (X-5) = 0" or "(X+5) = 0 => X = 5" or "X = -5
2. Roots For Non-Zero Denominator
 Method Example 1. (5X-18)/(X+2) = (2X-6)/(X-1) (X-3)/(X+1) = (X-6)/(X) 2. (x)/(x+1) + (x+1)/(x) = 5/2 3. 4((4x+1)/(4x-1))^(2) + (4x+1)/(4x-1) = 3 4. (4x+1)/(4x-1) + (4x-1)/(4x+1) = 3

1. Find roots of the equation (5X-18)/(X+2) = (2X-6)/(X-1)

 (5X-18)/(X+2) = (2X-6)/(X-1)

 => (5X-18)*(X-1) = (2X-6)*(X+2)

 => 5X^2-23X+18 = 2X^2-2X-12

 => 3X^2-21X+30 = 0

 => (3X^2-21X+30) = 0

 => 3X^2-21X+30 = 0

 => 3(X^2-7X+10) = 0

 => 3(X^2-2X-5X+10) = 0

 => 3(X(X-2)+(-5)(X-2)) = 0

 => 3(X-5)(X-2) = 0

 => (X-5) = 0" or "(X-2) = 0

 => X = 5" or "X = 2

2. Find roots of the equation (X)/(X+1) + (X+1)/(X) = (5)/(2)

 (X)/(X+1) + (X+1)/(X) = (5)/(2)

 => (X)*(X)*(2) + (X+1)*(X+1)*(2) = (5)*(X+1)*(X)

 => 2X^2 + (2X^2+4X+2) = (5X^2+5X)

 => 2X^2 + (2X^2+4X+2) + (-5X^2-5X) = 0

 => -X^2-X+2 = 0

 => (-X^2-X+2) = 0

 => -X^2-X+2 = 0

 => (-1)(X^2+X-2) = 0

 => (-1)(X^2-X+2X-2) = 0

 => (-1)(X(X-1)+2(X-1)) = 0

 => (-1)(X+2)(X-1) = 0

 => (X+2) = 0" or "(X-1) = 0

 => X = -2" or "X = 1

3. Find roots of the equation 12((2X+1)/(X-1))^2 + -5((2X+1)/(X-1)) + -2 = 0

 12((2X+1)/(X-1))^2 - 5((2X+1)/(X-1)) - 2 = 0

 "Let " (2X+1)/(X-1) = m

 => (12m^2-5m-2) = 0

 => 12m^2-5m-2 = 0

 => (12m^2-5m-2) = 0

 => (12m^2+3m-8m-2) = 0

 => 3m(4m+1)+(-2)(4m+1) = 0

 => (3m-2)(4m+1) = 0

 => (3m-2) = 0" or "(4m+1) = 0

 => 3m = 2" or "4m = -1

 => m = 2/3" or "m = -1/4

 "Now, " (2X+1)/(X-1) = 2/3

 => 3(2X+1) = 2(X-1)

 => 3(2X+1) - 2(X-1) = 0

 => (3(2X+1)-2(X-1)) = 0

 => (4X+5) = 0

 => 4X = -5

 => X = -5/4

 "Now, " (2X+1)/(X-1) = -1/4

 => 4(2X+1) = -1(X-1)

 => 4(2X+1) + 1(X-1) = 0

 => (4(2X+1)+(X-1)) = 0

 => (9X+3) = 0

 => 9X = -3

 => X = -3/9

 => X = -1/3

4. Find roots of the equation 12((X)/(X-1)) + 12((X-1)/(X)) = 25

 12((X)/(X-1)) + 12((X-1)/(X)) = 25

 "Let " (X)/(X-1) = m

 => 12m + 12/m = 25

 => 12m^2 - 25m + 12 = 0

 => (12m^2-25m+12) = 0

 => 12m^2-25m+12 = 0

 => (12m^2-25m+12) = 0

 => (12m^2-9m-16m+12) = 0

 => 3m(4m-3)+(-4)(4m-3) = 0

 => (3m-4)(4m-3) = 0

 => (3m-4) = 0" or "(4m-3) = 0

 => 3m = 4" or "4m = 3

 => m = 4/3" or "m = 3/4

 " Now," (X)/(X-1) = 4/3

 => 3(X) = 4(X-1)

 => 3(X) - 4(X-1) = 0

 => (3X-4(X-1)) = 0

 => (-X+4) = 0

 => X = 4

 " Now," (X)/(X-1) = 3/4

 => 4(X) = 3(X-1)

 => 4(X) - 3(X-1) = 0

 => (4X-3(X-1)) = 0

 => (X+3) = 0

 => X = -3

 3. Discriminant & Nature of Roots Rules: 1. If D > 0 then the roots are real and distinct.     (i) If D is a perfect square then the roots are rational and distinct.     (ii) If D is not a perfect square then the roots are irrational and distinct. 2. If D = 0 then the roots are real and equal. 3. If D < 0 then the quadratic equation has no real roots. Example : 1 Find the discriminant of Quadratic Equation x^2-5x+6=0 and discuss the nature of its roots x^2-5x+6=0 => x^2-5x+6 = 0 Comparing the given equation with the standard quadratic equation ax^2 + bx + c = 0, we get, a = 1, b = -5, c = 6. :. Delta = b^2 - 4ac  = (-5)^2 - 4 (1) (6)  = 25 - 24  = 1  = (1)^2 Here, Delta > 0 and is a perfect square. Also a and b are rational. Hence, the roots of the equation are distinct and rational.
 4. Solving quadratic equations by Completing the Square 1. Find the roots of Quadratic Equation X^2+10X-56=0 by the method of perfect square X^2+10X-56=0 => X^2+10X-56 = 0 Comparing the given equation with the standard quadratic equation ax^2 + bx + c = 0, we get, a = 1, b = 10, c = -56. :. Delta = b^2 - 4ac  = (10)^2 - 4 (1) (-56)  = 100 + 224  = 324 :. sqrt(Delta) = sqrt(324) = 18 Now, alpha = (-b + sqrt(Delta)) / (2a)  = (-(10) + 18) / (2 * 1)  = 8 / 2  = 4 and, beta = (-b - sqrt(Delta)) / (2a)  = (-(10) - 18) / (2 * 1)  = -28 / 2  = -14

 5. Find the quadratic equation whose roots are alpha and beta 1. Find the quadratic equation whose roots are Alpha = 2, Beta = 5 Let alpha = 2 and beta = 5 Then, the sum of the roots = alpha + beta = (2)+(5) = 7 and the Product of the roots = alpha * beta = (2)*(5) = 10 The Equation with roots alpha and beta is given by X^2 - (alpha+beta)X + alpha*beta = 0 :. The required equation X^2 - (7)X + (10) = 0 2. Find the quadratic equation whose roots are Alpha = -1/2, Beta = +2/3 Let alpha = -1/2 and beta = +2/3 Then, the sum of the roots = alpha + beta = (-1/2)+(+2/3) = 1/6 and the Product of the roots = alpha * beta = (-1/2)*(+2/3) = -1/3 The Equation with roots alpha and beta is given by X^2 - (alpha+beta)X + alpha*beta = 0 :. The required equation X^2 - (1/6)X + (-1/3) = 0
 6. Roots of Non-Quadratic Equation 1. Solve the equation 1 (X^2 + 1/X^2) - 8 ( X + 1/X ) + 14 = 0 1 * (X^2+1/X^2) + (-8) * (X+1/X) + (14) = 0 Let X + 1/X = m => (X + 1/X)^2 = m^2 => X^2 + 1/X^2 + 2 = m^2 => X^2 + 1/X^2 = m^2 - 2 Substituting this values in the given equation, we get (m^2 - 2) - 8m + 14 = 0 m^2 - 8m + 12 = 0 => (m^2-8m+12) = 0 => m^2-8m+12 = 0 => (m^2-8m+12) = 0 => (m^2-2m-6m+12) = 0 => m(m-2)+(-6)(m-2) = 0 => (m-6)(m-2) = 0 => (m-6) = 0" or "(m-2) = 0 => m = 6" or "m = 2 Now, X + 1/X = 6 => X^2 + 1 = 6X => X^2 - 6X + 1 = 0 => (X^2-6X+1) = 0 => X^2-6X+1 = 0 Comparing the given equation with the standard quadratic equation ax^2 + bx + c = 0, we get, a = 1, b = -6, c = 1. :. Delta = b^2 - 4ac  = (-6)^2 - 4 (1) (1)  = 36 - 4  = 32 :. sqrt(Delta) = sqrt(32) = 4 * sqrt(2) Now, alpha = (-b + sqrt(Delta)) / (2a) = (-(-6) + 4 * sqrt(2)) / (2 * 1) = (6 + 4 * sqrt(2)) / 2 = 3 + 2 * sqrt(2) and, beta = (-b - sqrt(Delta)) / (2a) = (-(-6) - 4 * sqrt(2)) / (2 * 1) = (6 - 4 * sqrt(2)) / 2 = 3 - 2 * sqrt(2) Now, X + 1/X = 2 => X^2 + 1 = 2X => X^2 - 2X + 1 = 0 => (X^2-2X+1) = 0 => X^2-2X+1 = 0 => (X^2-2X+1) = 0 => (X-1)(X-1) = 0 => (X-1) = 0" or "(X-1) = 0 => X = 1" or "X = 1

 1. Find Value Of Variation 1. X prop Y and X=4,Y=2. Find X=18,Y=? X prop Y=> X=K*YNow, X=4,Y=2=> 4 = K * 2=> K = 2Hence, X=2*YX=18,Y=?=> 18 = 2 * Y=> 9 = Y=> Y = 9 2. X prop Y^3/Z and X=8,Y=4,Z=3. Find X=?,Y=6,Z=4 X prop Y^3/Z=> X=K*Y^3/ZNow, X=8,Y=4,Z=3=> 8 = K * 64/3=> K = 3/8Hence, X=3/8*Y^3/ZX=?,Y=6,Z=4=> 1 * X = 3/8 * 54=> X = 81/4
 2. Prove Results For Given Variation 1. If X prop Y, then prove that X^3+Y^3 prop X^2Y-XY^2 X prop Y=> X=M*Y (where constant M != 0)Now (X^3+Y^3) / (X^2Y-XY^2)= (M^3Y^3+Y^3) / (M^2Y^3-MY^3)= (Y^3(M^3+1)) / (MY^3(M-1))= ((M^3+1)) / (M(M-1))= non-zero constant:. X^3+Y^3 prop X^2Y-XY^2 2. If 5X-7Y prop 6X+3Y, then prove that X prop Y 5X-7Y prop 6X+3Y=> 5X-7Y = M(6X+3Y) (where constant M != 0)=> 5X-7Y = 6MX+3MY=> 5X-6MX = 7Y+3MY=> X(5-6M) = Y(7+3M)=> X/Y = (7+3M) / (5-6M)=> X/Y = non-zero constant=> X prop Y

 1. Union (1) A = {1,2,3,4,5}, B = {3,4,5,6}Find A U B ...Here A = {1,2,3,4,5}, B = {3,4,5,6}A uu B = {1,2,3,4,5} uu {3,4,5,6}= {1,2,3,4,5,6}
 2. Intersection (1) A = {1,2,3,4,5}, B = {3,4,5,6}Find A n B ...Here A = {1,2,3,4,5}, B = {3,4,5,6}A nn B = {1,2,3,4,5} nn {3,4,5,6}= {3,4,5}

 3. Complement (1) U = {1,2,3,4,5,6,7,8,9,10}, A = {1,2,3,4,5}Find A' ...Here U = {1,2,3,4,5,6,7,8,9,10}, A = {1,2,3,4,5}A' = {1,2,3,4,5}'= {6,7,8,9,10}
 4. Power set(Proper Subset) (1) C = {7,8,9}Find Proper subset of C ...Here C = {7,8,9}P(C) = {{7},{8},{9},{7,8},{7,9},{8,9},{7,8,9}}

 5. Minus (1) A = {1,2,3,4,5}, B = {3,4,5,6}Find A - B ...Here A = {1,2,3,4,5}, B = {3,4,5,6}A - B = {1,2,3,4,5} - {3,4,5,6}= {1,2}
 6. Cross Product (1) A = {1,2,3}, B = {4,5}Find A × B ...Here A = {1,2,3}, B = {4,5}A × B = {1,2,3} × {4,5}= { (1,4), (1,5), (2,4), (2,5), (3,4), (3,5) }

 7. Prove that any two expression is equal or not A = {1,2,3,4,5}, B = {3,4,5,6}, C = {7,8,9}, Prove that A U (B U C) = (A U B) U C ...Here A={1,2,3,4,5},B={3,4,5,6},C={7,8,9}To find LHS = (A uu B) uu CA uu B = {1,2,3,4,5} uu {3,4,5,6}= {1,2,3,4,5,6}(A uu B) uu C = {1,2,3,4,5,6} uu {7,8,9}= {1,2,3,4,5,6,7,8,9}:. (A uu B) uu C = {1,2,3,4,5,6,7,8,9} ->(1)To find RHS = A uu (B uu C)B uu C = {3,4,5,6} uu {7,8,9}= {3,4,5,6,7,8,9}A uu (B uu C) = {1,2,3,4,5} uu {3,4,5,6,7,8,9}= {1,2,3,4,5,6,7,8,9}:. A uu (B uu C) = {1,2,3,4,5,6,7,8,9} ->(2)From (1) and (2):. (A uu B) uu C = A uu (B uu C) (proved)

14. Descartes' rule of signs
1. Find Descartes' rule of signs for x^5-x^4+3x^3+9x^2-x+5

Here f(x)=x^5-x^4+3x^3+9x^2-x+5

f(x)=x^5color{red}{-}x^4color{blue}{+}3x^3color{blue}{+}9x^2color{red}{-}xcolor{blue}{+}5

look first at f(x): (positive case)

 f(x)= color{blue}{+} x^5 color{red}{-} x^4 color{blue}{+} 3x^3 color{blue}{+} 9x^2 color{red}{-} x color{blue}{+} 5 Sign change count color{blue}{+} to color{red}{-}1 color{red}{-} to color{blue}{+}2 color{blue}{+} to color{blue}{+} color{blue}{+} to color{red}{-}3 color{red}{-} to color{blue}{+}4

There are 4 sign changes, so there are 4 or counting down in pairs, 2 or 0 positive roots.

Now look at f(–x): (negative case)

f(-x)=(-x)^5color{red}{-}(-x)^4color{blue}{+}3(-x)^3color{blue}{+}9(-x)^2color{red}{-}(-x)color{blue}{+}5

f(-x)=color{red}{-}x^5color{red}{-}x^4color{red}{-}3x^3color{blue}{+}9x^2color{blue}{+}xcolor{blue}{+}5

 f(-x)= color{red}{-} x^5 color{red}{-} x^4 color{red}{-} 3x^3 color{blue}{+} 9x^2 color{blue}{+} x color{blue}{+} 5 Sign change count color{red}{-} to color{red}{-} color{red}{-} to color{red}{-} color{red}{-} to color{blue}{+}1 color{blue}{+} to color{blue}{+} color{blue}{+} to color{blue}{+}

There is 1 sign change, so there is exactly 1 negative roots.