Linear equations Lecture for Maths1.pptx

Consisting of Linear Equations

1. Applied Mathematics for Business and Social Sciences (8405)

2. Chapter No. 2 Linear Equations

3. Linear Equations in One Variable Objectives Decide whether a number is a solution of a linear equation. Solve linear equations using the addition and multiplication properties of equality. Solve linear equations using the distributive property. Solve linear equations with fractions or decimals. Linear equation in two variables. Solving simultaneous equations. Slope and intercept. Representing lines in slope intercept form.

4. Using Linear Equations of One Variable Algebraic Expressions vs. Equations algebraic expressions : are the statements without any sign of =, > or < – 9 y + 5 , 10 k , and Equations are statements that two algebraic expressions are equal : 3 x – 13 = 29 , 2 + y = – 11 , and 3 m = 4 m – 2 An equation always contains an equals sign, but an expression does not.

5. Using Linear Equations of One Variable Linear Equations in One Variable Linear Equation in One Variable A linear equation is also called a first-degree equation since the greatest power on the variable is one . 5 x + 10 = 13 A linear equation in one variable can be written in the form ax + b = c where A, B, and C are real numbers, with A = 0. /

6. Using Linear Equations of One Variable Linear Equations in One Variable Determine whether the following equations are linear or nonlinear. 8 x + 3 = – 9 9 x 3 – 8 = 15 x 7 = –12 Yes, x is raised to the first power. No, x is not raised to the first power. No, x is not raised to the first power. No, x is not raised to the first power.

7. 2.1 Using Linear Equations of One Variable Deciding Whether a Number is a Solution If a variable can be replaced by a real number that makes the equation a true statement, then that number is a solution of the equation, x – 10 = 3 . x – 10 = 3 13 13 – 10 = 3 x – 10 = 3 8 8 – 10 = 3 (true) (false) 13 is a solution 8 is not a solution

8. 2.1 Using Linear Equations of One Variable Finding the Solution Set of an Equation An equation is solved by finding its solution set – the set of all solutions . The solution set of x – 10 = 3 is {13}. Equivalent equations are equations that have the same solution set . These are equivalent equations since they all have solution set { –3}. 3 x + 5 = – 4 3 x = – 9 x = – 3

9. 2.1 Using Linear Equations of One Variable Solving Linear Equations An equation is like a balance scale, comparing the weights of two quantities. Expression-1 Expression-2 We apply properties to produce a series of simpler equivalent equations to determine the solution set. Variable Solution = =

10. C 2.1 Using Linear Equations of One Variable Addition Property of Equality The same number may be added to both sides of an equation without changing the solution set. A = = C + A = B + A B B

11. C 2.1 Using Linear Equations of One Variable Multiplication Property of Equality Each side of an equation may be multiplied by the same nonzero number without changing the solution set. A = = C A = B A B B

12. 2.1 Using Linear Equations of One Variable Addition and Multiplication Properties of Equality For all real numbers A , B , and C , the equation A = B and A + C = B + C are equivalent. Addition Property of Equality For all real numbers A , B , and for C = , the equation A = B and A C = B C are equivalent. Multiplication Property of Equality /

13. 2.1 Using Linear Equations of One Variable Addition and Multiplication Properties of Equality Because subtraction and division are defined in terms of addition and multiplication, we can extend the addition and multiplication properties of equality as follows: The same number may be subtracted from each side of an equation, and each side of an equation may be divided by the same nonzero number, without changing the solution set.

14. 2.1 Using Linear Equations of One Variable Solving Linear Equations in One Variable Step 1 Clear fractions. Eliminate any fractions by multiplying each side by the least common denominator. Step 2 Simplify each side separately. Use the distributive property to clear parentheses and combine like terms as needed. Step 3 Isolate the variable terms on one side. Use the addition property to get all terms with variables on one side of the equation and all numbers on the other. Step 4 Isolate the variable. Use the multiplication property to get an equation with just the variable (with coefficient of 1) on one side. Step 5 Check. Substitute the proposed solution into the original equation.

15. 2.1 Using Linear Equations of One Variable Solving Linear Equations Solve 3 x + 2 = 10. 3 x + 2 = 10 3 x + 2 – 2 = 10 – 2 3 x = 8 Subtract 2. Combine like terms. Divide by 3. Proposed solution.

16. 2.1 Using Linear Equations of One Variable Solving Linear Equations 3 x + 2 = 10 3 • + 2 = 10 3 8 Check by substituting the proposed solution back into the original equation. 8 + 2 = 10 Since the value of each side is 10, the proposed solution is correct. The solution set is .

17. 2.1 Using Linear Equations of One Variable Solving Linear Equations Solve 2 x – 5 = 5 x – 2 . 2 x – 5 = 5 x – 2 2 x – 5 – 5 x = 5 x – 2 – 5 x – 3 x – 5 = –2 Subtract 5 x . Combine like terms. Add 5. Divide by –3 . – 3 x – 5 + 5 = –2 + 5 – 3 x = 3 Combine like terms. x = – 1 Proposed solution.

18. 2.1 Using Linear Equations of One Variable Solving Linear Equations 2 x – 5 = 5 x – 2 Check by substituting the proposed solution back into the original equation. – 2 – 5 = – 5 – 2 Since the value of each side is –7 , the proposed solution is correct. The solution set is {–1} . 2 ( – 1 ) – 5 = 5 ( – 1 ) – 2 –7 = –7

19. 2.1 Using Linear Equations of One Variable Solving Linear Equations Solve 5(2 x + 3) = 3 – 2(3 x – 5). 5(2 x + 3) = 3 – 2(3 x – 5) 10 x + 15 = 3 – 6 x + 10 10 x + 15 – 15 = 3 – 6 x + 10 – 15 10 x = – 6 x – 2 10 x + 6 x = –6 x – 2 + 6 x 16 x = –2 Distributive Prop. Add –15. Collect like terms. Add 6 x . Collect like terms.

20. Sec 2.1 - 2.1 Using Linear Equations of One Variable Solving Linear Equations Divide by 16. Proposed solution. 16 x = –2

21. 2.1 Using Linear Equations of One Variable Solving Linear Equations Check proposed solution:

22. 2.1 Using Linear Equations of One Variable Solving Linear Equations with Fractions Solve .

23. 2.1 Using Linear Equations of One Variable Solving Linear Equations with Fractions continued

24. 2.1 Using Linear Equations of One Variable Solving Linear Equations with Decimals Solve .

25. 2.1 Using Linear Equations of One Variable Solving Linear Equations with Decimals continued

26. 2.1 Using Linear Equations of One Variable Conditional, Contradiction, and Identity Equations Linear equations can have exactly one solution, no solution, or an infinite number of solutions. Type of Linear Equation Number of Solutions Indication When Solving Conditional One Final results is x = a number. Identity Infinite; solution set {all real numbers} Final line is true, such as 5 = 5 . Contradiction None; solution set is Final line is false, such as –3 = 11 .

27. 2.1 Using Linear Equations of One Variable Conditional, Contradiction, and Identity Equations A contradiction has no solutions. Since 0 = –5 is never true, and this equation is equivalent to x + 7 = x + 2, the solution set is empty.

28. 2.1 Using Linear Equations of One Variable Conditional, Contradiction, and Identity Equations An identity has an infinite number of solutions. Since 0 = 0 is always true, and this equation is equivalent to 2 x + 2 = 2( x + 1), the solution set is all real numbers.

29. LINEAR EQUATION: A Linear Equation is an algebraic equation in which terms are a constants or the product of a constants and variables. Linear Equations can have one or more variables.Ex:2x-3=5(linear equation in one variable) Ex:2x+3y=7(linear equation in two variables)

30. Representing a linear equation Y=mx + c

31. Slope of a Line (m) Slope basically describes the steepness of a line

32. If a line goes up from left to right, then the slope has to be positive Conversely, if a line goes down from left to right, then the slope has to be negative

33. Definitions of Slope Slope is simply the change in the vertical distance over the change in the horizontal distance

34. The formula above is the one which we will use to find the slope of specific lines In order to use that formula we need to know, or be able to find 2 points on the line

35. If a line is in the form Ax + By = C, we can use the following formula to find the slope:

36. Examples

37. Horizontal lines have a slope of zero while vertical lines have no slope Horizontal y= Vertical x= m = 0 m = no slope

38. The World Of Linear Equations Writing Linear Equations In Slope-Intercept Form y = mx + b

39. If you are given: The slope and y-intercept Finding the equation of the line in y= mx + b form. Given: slope and y-intercept. Just substitute the “m” with the slope value and the “b” with the y-intercept value. Slope = ½ and y-intercept = -3 y= mx + b ½ -3 y= ½x – 3

40. If you are given: A Graph Find the: y – intercept = b = the point where the line crosses the y axis. Slope = = m = y – intercept = b = -3 Slope = = m = ½ y= mx + b ½ -3 y= ½x – 3

41. If you are given: The slope and a point Given: slope (m) and a point (x,y). To write equations given the slope and a point using Point-Slope Form. Slope =½ and point (4,-1) ½ 4 -1 y= ½x – 3 Point-Slope Form -1 -1

42. If you are given: Two points Finding the equation of the line in y= mx + b form. Given: Two points. First find the slope (m) and then substitute one of the points x and y values into Point-Slope Form. Point-Slope Form Point (-2, -4) & Point (2, -2) Find the: Slope = = m = Slope =½ and point (2, -2) ½ -2 2 -2 -2 y= ½x – 3

43. Write the equation of a line that has a y-intercept of -3 and a slope of -4. y = -3x – 4 y = -4x – 3 y = -3x + 4 y = -4x + 3

44. Write an equation of the line that goes through the points (0, 1) and (1, 4). y = 3x + 4 y = 3x + 1 y = -3x + 4 y = -3x + 1

45. To find the slope and y-intercept of an equation, write the equation in slope-intercept form: y = mx + b. Find the slope and y-intercept. y = 3x – 7 y = m x + b m = 3, b = -7

46. Find the slope and y-intercept. 2) y = x y = m x + b y = x + 0 3) y = 5 y = m x + b y = 0x + 5 m = b = 0 m = 0 b = 5

47. -3 -3 -3 Find the slope and y-intercept. 4) 5x - 3y = 6 Write it in slope-intercept form. (y = mx + b) 5x – 3y = 6 -3y = -5x + 6 y = x - 2 m = b = -2

48. Write it in slope-intercept form. (y = mx + b) 2y + 2 = 4x 2y = 4x - 2 y = 2x - 1 Find the slope and y-intercept. 5) 2y + 2 = 4x 2 2 2 m = 2 b = -1

49. Find the slope and y-intercept of y = -2x + 4 m = 2; b = 4 m = 4; b = 2 m = -2; b = 4 m = 4; b = -2