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Educational Level Secondary school, High school and College
Program Purpose Provide step by step solutions of your problems using online calculators (online solvers)
Problem Source Your textbook, etc

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 - c2
     = 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)2
4. (3Y - 2X)3
5. (2X + 3Y)2
6. (X + Y)2 - (X - Y)2
7. (X + Y + Z)2 + (X + Y - Z)2
8. (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 + AB
Here X=X,A=2, B=3
= (X)2 + (2 + 3)X + (2)×(3)
= X2+5X+6

2. Expand an Algebraic Expressions (2X+3Y+4Z)2

Using The Identity,
(A + B + C)2 = A2 + B2 + C2 + 2AB + 2BC + 2CA
Here 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, 9.82
2. 123, 12.23
3. 93, 8.93
4. 102 × 106
5. 342 - 322, 3.42 - 3.22
6. 103 - 73 - 33
7. 43 + 33, 3.43 - 2.93

1. Find Value of 102 × 106

Using, The Identity (X+A)(X+B) = X2 + (A+B)X + AB
Here X = 100, A = 2 and B = 6
102 × 106 = (100 + 2)(100 + 6)
= 1002 + (2 + 6)100 + 2 × 6
= 10000 + 800 + 12
= 10812

2. Find Value of 342 - 322

Using, The Indentity A2 - B2 = (A-B)(A+B)
Here A = 34 and B = 32
Here 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 = 2`


Now, 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 = 0`

`X - 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 = 1`


Now, We know that
`4 XY = ( X + Y )^2 - ( X - Y )^2`

`4 XY = 5^2 - 1^2`

`4 XY = 25 - 1`

`4 XY = 24`

`XY = 24/4`

`XY = 6`


Now, We know that
`X^2 + Y^2 = ( X + Y )^2 - 2 XY`

`X^2 + Y^2 = 5^2 - 2 * 6`

`X^2 + Y^2 = 25 - 12`

`X^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 = 29`


Now, 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

1. Addition

(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)2


GCD = 2 × 5 (X - 1)
GCD = 10(X - 1)

LCM = 2 × 3 × 5 × 5 (X - 2)(X - 1)2
LCM = 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 have

`3X-Y-5 = 0`

`=> Y = 3X-5 ->(3)`

Putting `Y = 3X-5` in Eqn. `(1)`, we get

`2X+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 get

`Y = 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 = 11`

Eqn.`(2) xx 7 => 21X-7Y = 35`

Adding ` => 23X = 46`

`=> X = 46 / 23`

`=> X = 2`

Putting `X = 2` in Eqn. `(2)`, we have

`3(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=-11`

`a_2=3, b_2=-1, c_2=-5`

`X = (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)`

`= 2`

`Y = (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 get

`5X+6Y-16 = 0`

`5X+6Y-16 = 0 ->(iii)` (On simplifying)

Subtracting Equation `(ii)` from `(i)`, we get

`-X+8Y-6 = 0`

`X-8Y+6 = 0 ->(iv)` (On simplifying)

`5X+6Y = 16 ->(1)`

`X-8Y = -6 ->(2)`

Eqn.`(1) xx 1 => 5X+6Y = 16`

Eqn.`(2) xx 5 => 5X-40Y = -30`

Substracting `=> 46Y = 46`

`=> Y = 46 / 46`

`=> Y = 1`

Putting `Y = 1` in Eqn. `(2)`, we have

`X-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) = 0`

Here `12x+5y=7`, `x+y=7`

Now 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=0`

`x+y-7=0`

Use 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=7`

`x+y=7`

The point of intersection of the linear equations
Consider `12x+5y=7`

`x`-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=7`

`x`-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 method

Here `2x+y+z=5`, `3x+5y+2z=15`, `2x+y+4z=8`

Now 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 method

Total Equations are `3`

`2 x + y + z = 5`

`3 x + 5 y + 2 z = 15`

`2 x + y + 4 z = 8`


Converting 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 = 1`

`y = 2`

`z = 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*Y`

Now, `X=4,Y=2`

`=> 4 = K * 2`

`=> K = 2`

Hence, `X=2*Y`

`X=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/Z`

Now, `X=8,Y=4,Z=3`

`=> 8 = K * 64/3`

`=> K = 3/8`

Hence, `X=3/8*Y^3/Z`

`X=?,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 C`

`A 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.
 

 
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