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# Linear Equations and Inequalities in One Variable

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### Transcript

• 1. LINEAR EQUATIONS and INEQUALITIES in ONE VARIABLE
• 2. Linear equations and Inequalities in One Variable Equation and Inequalities are relations between two quantities.
• 3.
• Equation is a mathematical sentence indicating that two expressions are equal. The symbol “=“ is used to indicate equality.
• Ex.
• 2x + 5 = 9 is a conditional equation
• since its truth or falsity depends on the value of x
• 2 + 9 = 11 is identity equation since both of its sides are identical to the same number 11.
• 4.
• Inequality is a mathematical sentence indicating that two expressions are not equal. The symbols <, >, are used to denote inequality.
• Ex.
• 3 + 2 ≠ 4 is an inequality
• If two expressions are unequal, then their relationship can be any of the following, >, ≥, < or ≤.
• 5.
• Linear equation in one variable is an equation which can be written in the form of ax + b = 0, where a and b are real-number constants and a ≠ 0.
• Ex.
• x + 7 = 12
• 6. Solution Set of a Linear Equation
• Example
• 4x + 2 = 10 this statement is either true of false
• If x = 1, then 4x + 2 = 10 is false because 4(1) + 2 is ≠ 10
• If x = 2, then 4x + 2 = 10 is true because 4(2) + 2 = 10
• 7. B. x – 4 < 3 this statement is either true or false If x =6, then x – 4 is true because 6 – 4 < 3 If x = 10 , then x – 4 is false because 6 – 4 is not < 3
• When a number replaces a variable in an equation (or inequality) to result in a true statement, that number is a solution of the equation (or inequality). The set of all solutions for a given equation (or inequality) as called the solution set of the equation (or inequality).
• 8. Solution Set of Simple Equations and Inequalities in One Variable by Inspection
• To solve an equation of inequality means to find its solution set. There are three(3) ways to solve an equation or inequality by inspection
• 9. A. Guess-and-Check
• In this method, one guesses and substitutes values into an equation of inequality to see if a true statement will result.
• 10. Consider the inequality x – 12 < 4 If x = 18, then 18 – 12 is not < 4 If x = 17, then 17 – 12 is not < 4 If x = 16, then 16 – 12 is not < 4 If x = 15, then 15 – 12 < 4 If x = 14, then 14 – 12 < 4
• Inequality x - 12 < 4 is true for all values of x which are less than 16. Therefore, solution set of the given inequality is x < 16.
• 11. Another example
• X + 3 = 7
• If x = 6, then 6 + 3 ≠ 7
• If x = 5, then 5 + 3 ≠ 7
• If x = 4, then 4 + 3 = 7
• Therefore x = 4
• 12. B. Cover-up
• In this method , one covers up the term with the variable.
• 13. Example
• Consider equation x + 9 = 15
• x + 9 = 15
• + 9 = 15
• To result in a true statement, the must be 6. Therefore x = 6
• 14.
• Another example
• X – 1 = 3
• – 1 = 3
• x = 4
• 15. C. Working Backwards
• In this method, the reverse procedure is used
• 16. Consider the equation 2x + 6 = 4
• times equals plus equals
• 2 2x 6
• Start
• 14 End
• 2 8 6
• equals divided equals minus
x
• 17. Example: 4y = 12
• times equals
• 4
• Start 12 End
• 4
• equals divided Therefore y = 3
y
• 18. Properties of Equality and Inequality
• 19. Properties of Equality
• Let a, b, and c be real numbers.
• Reflexive Property
• a = a
• Ex. 3 = 3, 7 = 7 or 10.5 = 10.5
• 20. B. Symmetric Property
• If a = b, then b = a
• Ex. If 3 + 5 = 8, then 8 = 3 + 5
• If 15 = 6 + 9, then 6 + 9 = 15
• If 20 = (4)(5), then (4)(5) = 20
• 21. C. Transitive Property
• If a = b and b = c, then a = c
• Ex. If 8 + 5 = 13 and 13 = 6 + 7
• then 8 + 5 = 6 + 7
• If (8)(5) = 40 and 40 = (4)(10)
• then (8)(5) = (4)(10)
• If a = b, then a + c = b + c
• Ex. If 3 + 5 = 8, then (3 + 5) = 3 = 8 +3
• 23. E. Subtraction Property
• If a = b, then a – c = b – c
• Ex. 3 + 5 = 8, then (3 + 5) – 3 = 8 - 3
• 24. F. Multiplication Property
• If a = b, then ac = bc
• Ex. (4)(6) = 24, then (4)(6)(3) = (24)(3)
• 25. G. Division Property
• If a = b, and c ≠ 0, then a/c = b/c
• Ex. If (4)(6) = 24, then (4)(6)/3 = 24/3
• 26. Properties of Inequality
• Let a, b and c be real numbers.
• Note: The properties of inequalities will still hold true using the relation symbol ≤ and ≥.
• If a < b, then a + c < b + c
• Ex. If 2 < 3, then 2 + 1 < 3 + 1
• 28. B. Subtraction Property
• If a < b, then a – c < b – c
• Ex. If 2 < 3, then 2 – 1 < 3 – 1
• 29. C. Multiplication Property
• If a < b and c > 0, then ac < bc
• IF a < b and c < 0, then ac > bc
• Ex. If 2 < 3, then (2)(2) < (3)(2)
• If 2 < 3, then (2)(-2) > (3)(-2)
• 30. D. Division Property
• If a < b and c > 0, then a/c < b/c
• If a < b and c < 0, then a/c > b/c
• Ex. If 2 < 3, then 2/3 < 3/3
• If 2 < 3, then 2/-3 > 3/-3
• 31. Solving Linear Equations in One Variable
• 32.
• Example:
• Solve the following equations:
• x – 5 = 8
• x – 5 + 5 = 8 + 5 add 5 to both sides
• x + 0 = 13 of the equation
• x = 13
• Recall that if the same number is added to both sides of the equation, the resulting sums are equal.
• 33.
• x – 12 = -18
• x – 12 + 12 = -18 + 12 add 12 to both sides
• x + 0 = -6
• x = -6
• This problem also uses the addition property of equalities.
• 34.
• x + 4 = 6
• x + 4 – 4 = 6 – 4 subtract 4 to both sides of
• x + 0 = 2 the equation
• x = 2
• Recall that if the same number is subtracted to both sides of the equation, the differences are equal.
• 35.
• x + 12 = 25
• x + 12 – 12 = 25 – 12 subtract 12 to both
• x + 0 = 25 – 12 sides
• This problem also uses the subtraction property of equalities.
• 36.
• x/2 = 3
• x/2 . 2 = 3 . 2 multiply both sides by 2
• x = 6
• Recall that if the same number is multiplied to both sides of the equation, the products are equal.
• 37.
• 6. x/7 = -5
• x/7 . 7 = -5 .7 multiply both sides by 7
• x = -35
• This problem also uses multiplication property of equalities.
• 38.
• 7. 5 x = 35
• 5x/5 = 35/5 both sides of the equation is
• X = 7 divided by the numerical coefficient of x to make the coefficient of x equals to 1
• Recall the if both sides of the equation is divided by a non-zero number, the quotients are equal.
• 39.
• 8. 12y = -72
• 12y/12 = -72/12 divide both sides by 12
• y = -6
• This problem also uses the division property of equalities.
• 40.
• Other equations in one variable are solved using more than on property of equalities.
• 9. 2x + 3 = 9
• 2x+ 3 – 3 = 9 – 3 subtraction property
• 2x = 6
• 2x/2 = 6/2 division property
• x = 3
• 41.
• 10. 5y – 4 = 12 – y
• 5y – 4 + 4 = 12 – y + 4 addition property
• 5y = 16 – y
• 5y + y = 16 – y + y addition property
• 6y = 16
• 6y/5 = 16/5 division property
• y = 2 4/6
• 42. Solving Linear Inequalities in One Variable
• 43.
• The solution set o inequalities maybe represented on a number line.
• Recall that a solution of a linear inequality in one variable is a real number which makes the inequality true.
• Example:
• 1. Graph x > 6 on a number line
• O x>6
• 0 1 2 3 4 5 6 7 8 9 10 11
• The ray indicates the solution set of x > 6
• 44.
• The ray indicates the that he solution set, x > 6 consist of all numbers greater than 6. The open circle of 6 indicates that 6 is not included.
• 45.
• 2. Graph the solution set x ≤ -1 on a number line.
• x ≤ -1
• -2 -1 0 1
• The ray indicates that the solution set of x ≤ -1 consist of all the numbers less than or equal to -1. The solid circle of -1 indicates that -1 is included in the solution set.
• 46.
• Applying the Properties of Inequalities in Solving Linear Inequalities:
• 1. Solve x – 2 > 6 and graph the solution set.
• x – 2 > 6
• x – 2 + 2 > 6 + 2 add 2 to both sides of the
• x + 0 > 8 inequality
• x > 8
• O x > 8
• 8
• 47.
• 2. x + 15 < -7
• x + 15 – 15 < -7 – 15 subtract 15 from both sides of the
• x +0 < - 22 inequalities.
• x < -22
• x < -22 o
• -22
• 48. Solving Word Problems Involving Linear Equations
• 49.
• Steps in solving word problems:
• Read and understand the problem. Identify what is given and what is unknown. Choose a variable to represent the unknown number.
• Express the other unknown, if there are any., in terms of the variable chosen in step 1.
• Write a equation to represent the relationship among the given and unknown/s.
• Solve the equation for the unknown and use the solution to find for the quantities being asked.
• Check by going back to the original statement.
• 50.
• Example:
• One number is 3 less than another number. If their sum is 49, find the two numbers.
• Step 1: Let x be the first number.
• Step 2: Let x – 3 be the second number.
• Step 3: x + ( x – 3) = 49
• Step 4: x + x – 3 = 49
• 2x – 3 = 49
• 2x = 49 + 3
• 2x = 52
• x = 26 the first number
• x – 3 = 23 the second number
• Step 5: Check: The sum of 26 and 23 is 49,
• and 23 is 3 less than 26.
• 51.
• 2. Six years ago, Mrs. dela Rosa was 5 times as old as her daughter Leila.
• How old is Leila now if her age is one-third of her mother’s present age?
• Solution:
• Step 1: Let x be Leila’s age now
• 3x is Mrs. dela Rosa’s age now
• Step 2: x – 6 is Leila’s age 6 years ago
• 3x – 6 is Mrs. dela Rosa’s age 6 years ago
• Step 3: 5(x – 6) = 3x – 6
• Step 4: 5(x – 6) = 3x – 6
• 5x – 30 = 3x – 6
• 5x – 30 + 30 = 3x – 6 + 30
• 5x = 3x + 24
• 5x – 3x = 3x +24 – 3x
• 2x = 24
• 2x/2 = 24/ 2
• X = 12 Leila’s age now
• 3x = 36 Mrs. dela Rosa’s age now
• Step 5: Check: Thrice of Leila’s present age, 12, is Mrs. dela Rosa’s presnt age, 36. Six years ago, Mrs. dela Rosa was 36 – 6 = 30years old which was five times Leila’s age, 12 – 6 = 6.