Chapter 4 First Degree Equations and Inequalities in One Variable
Linear Equations and Inequalities in One Variable
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Transcript of Linear Equations and Inequalities in One Variable
- 1. LINEAR EQUATIONS and INEQUALITIES in ONE VARIABLE
2. Linear equations and Inequalities in One Variable Equation and Inequalities are relationsbetween two quantities. 3.
- Equationis a mathematical sentence indicating that two expressions are equal. The symbol = is used to indicate equality.
- Ex.
- 2x + 5 = 9is a conditional equation
- since its truth or falsity depends onthe value of x
- 2 + 9 = 11 is identityequation since both of itssides are identical to the samenumber 11.
4.
- Inequalityis 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 variableis 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 = 10this statement is either true offalse
- If x = 1, then4x + 2 = 10is false because 4(1) + 2 is 10
- If x = 2, then4x + 2 = 10is true because 4(2) + 2 = 10
7. B. x 4 < 3this 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 asolutionof 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, themust 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
- timesequalsplusequals
- 22x6
- Start
- 14End
- 286
- equalsdividedequalsminus
x 17. Example: 4y = 12
- timesequals
- 4
- Start12 End
- 4
- equalsdividedTherefore 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)
22. D. Addition Property
- Ifa = 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 .
27. A. Addition Property
- 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 + 5add 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 + 12add 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 4subtract 4 to both sides of
- x + 0 = 2the 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 12subtract 12 to both
- x + 0 = 25 12sides
- This problem also uses the subtraction property of equalities.
36.
- x/2 = 3
- x/2 . 2 = 3 . 2multiply 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 .7multiplyboth sides by 7
- x = -35
- This problem also uses multiplication property of equalities.
38.
- 7.5 x = 35
- 5x/5 = 35/5both sides of the equation is
- X = 7divided by the numericalcoefficient of x to makethe 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 = -72/12divide 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 3subtraction property
- 2x = 6
- 2x/2 = 6/2division property
- x = 3
41.
- 10.5y 4 = 12 y
- 5y 4 + 4 = 12 y + 4addition property
- 5y = 16 y
- 5y + y = 16 y + yaddition property
- 6y = 16
- 6y/5 = 16/5division 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
- Ox>6
- 01234567891011
- 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
- -2-101
- 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 + 2add 2 to both sides of the
- x + 0 > 8inequality
- x > 8
- Ox > 8
- 8
47.
- 2.x + 15 < -7
- x + 15 15 < -7 15subtract 15 from bothsides of the
- x +0 < - 22inequalities.
- x < -22
- x < -22o
- -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 = 23the 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 mothers present age?
- Solution:
- Step 1: Let x be Leilas age now
- 3x is Mrs. dela Rosas age now
- Step 2: x 6 is Leilas age 6 years ago
- 3x 6 is Mrs. dela Rosas 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 = 12Leilas age now
- 3x = 36Mrs. dela Rosas age now
- Step 5: Check: Thrice of Leilas present age, 12, is Mrs. dela Rosas presnt age, 36. Six years ago, Mrs. dela Rosa was 36 6 = 30years old which was five times Leilas age, 12 6 = 6.