Equations and Inequalities by Edu Pristine

video locked

About the Lecture

The lecture Equations and Inequalities by Edu Pristine is from the course ARCHIV Foundations of Mathematics . It contains the following chapters:

  • Linear Equation in one variable
  • Equations and Inequations
  • Solving System of linear equations
  • Quadratic equations

Author of lecture Equations and Inequalities

 Edu Pristine

Edu Pristine


Customer reviews

(1)
5,0 of 5 stars
5 Stars
5
4 Stars
0
3 Stars
0
2 Stars
0
1  Star
0


Excerpts from the accompanying material

... and logarithms, Equations and inequalities, Functions and graphs, Compounding ...

... Equations with one dependent variable, one independent variable and a constant  Denoted by equation Y = mX+c, where m is the slope of the line. Capital asset pricing model is a good example, where, coefficient of the market return represents the so-called beta factor. The intersection with the axis is the market-independent return. Here R is the stock return (dependent variable), R mis the market return (independent variable). ...

... 8 from both sides we get, -3X-15 Step 3: Divide both sides by 3, we get X>-5 Hence on solving 8-3X<23, we get X>-5 "Example: Solve 16-4x8 Step 3: Divide both sides by 4, we get X>2 Step 4: Solving 8-3X<23, we get X>-5 (from example above) Step 5: X>2 and X>-5, hence X>2 ...

... System of Equations: Used to solve equations with more than 1 unknown variables. Number of equations must be equal to the number of unknowns. Solving using elimination method. Step 1: Write both equations in term of 1 variable. Step 2: Equate the both equations. Step 3: Solve for unknown variable. Step 4: Put the variable solved in step 3 in any equation to find other unknown variable ...

... Step 4: Put the value of one of the variable in original equation to get the value of another unknown variable "Example: Solve 2x+3y =13 and 6x+y =15 Step 1: From 2x+3y =13, we get x=(13-3y)/2 Step 2 : 6((13-3y)/2) + y = 15 Step 3: Solving for y we get y = 3 Step 4: Putting y in equation 1, we get x = 2 ...

... Quadratic equation is of form where, a, b, c are constants. Solution is given by "Nature of roots The solution depends on, known as discriminant of the equation. If the discriminant is negative then there are no real solutions, only imaginary solutions. If the discriminant is zero there is only one solution, x =  b/2a, equal roots If the discriminant is positive there are two different real solutions, real distinct ...

... Example: The equation mentioned below has real, imaginary, or equal roots. Solution: For equation D = 6*6  4*1*4 = 20 Since, D = 20 > 0, roots are real and distinct ...