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Algebra 2 Chapter 1 Notes Equations and Inequalities

Algebra 2 Chapter 1 Notes Equations and Inequalities. Algebra is the ability to move values from one side of the equation or inequality to the other side by doing the opposite operation. x + 2. 5. Common Assumptions with Numbers. + 1 n. 1 1. The sign of a number is positive, +

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Algebra 2 Chapter 1 Notes Equations and Inequalities

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  1. Algebra 2 Chapter 1 Notes Equations and Inequalities

  2. Algebra is the ability to move values from one side of the equation or inequality to the other side by doing the opposite operation. x + 2 5

  3. Common Assumptions with Numbers + 1 n. 1 1 • The sign of a number is positive, + • The coefficient is 1 • The decimal point is to the right of the number • As a whole number it is over 1 • The power of the number is 1

  4. 1.1 Real Numbers and Number Operations Whole numbers = 0, 1, 2, 3 … Integers = …, -3, -2, -1, 0, 1, 2, 3 … Rational numbers = numbers such as 3/4 , 1/3, -4/1 that can be written as a ratio of the two integers. When written as decimals, rational numbers terminate or repeat, 3/4 = 0.75, 1/3 = 0.333… Irrational numbers = real numbers that are NOT rational, such as, and π, When written as decimals, irrational numbers neither terminate or repeat. A Graph of a number is a point on a number line that corresponds to a real number The number that corresponds to a point on a number line is the Coordinate of the point. Origin  -3 -2 -1 0 1 2 3

  5. 1.1 Real Numbers and Number Operations Graph- 4/3, 2.7, • • • Graph- 2, 3 • • Graph- 1, - 3 • •

  6. 1.1 Real Numbers and Order of Operation Example: You can use a number line to graph and order real numbers. Increasing order (left to right): - 4, - 1, 0.3, 2.7 Properties of real numbers include the closure, commutative, associative, identity, inverse and distributive properties.

  7. 1.1 Using Properties of Real Numbers Properties of addition and multiplication [let a, b, c = real numbers] Property Addition Multiplication Closure a + b is a real number a • b is a real number Commutative a + b = b + a a• b = b • a Associative ( a + b ) + c = a + ( b + c ) ( a b ) c = a ( b c ) Identity a + 0 = a , 0 + a = a a• 1 = a , 1 • a = a Inverse a + ( -a ) = 0 a • 1/a = 1 , a  0 Distributive a ( b + c) = a b + a c Opposite = additive inverse, for example aand- a Reciprocal= multiplicative inverse (of any non-zero #) for example a and 1/a Definition of subtraction:a – b = a + ( - b ) Definition of division:a / b = a 1 / b , b  0

  8. 1.1 Real Numbers and Number Operations Identifying properties of real numbers & number operations ( 3 + 9 ) + 8 = 3 + ( 9 + 8 ) 14 • 1 = 14 [ Associative property of addition ] [Identity property of multiplication ] Operations with real numbers: Difference of 7 and – 10 ? 7 – ( - 10 ) = 7 + 10 = 17 • • Quotient of - 24 and 1/3 ?

  9. 1.1 Real Numbers and Number Operations • Give the answer with the appropriate unit of measure • A.) 345 miles – 187 miles=158 miles • B.) ( 1.5 hours ) ( 50 miles ) =75 miles • 1 hour • 24 dollars =8 dollars per hour • 3 hours • D) ( 88 feet ) ( 3600 seconds ) ( 1 mile ) =60 miles per hour • 1 second 1 hour 5280 feet “Per”means divided by

  10. 1.2 Algebraic Expressions and Models • Order of Operations • First, do operations that occur withingrouping symbols - 4 + 2 ( -2 + 5 ) 2 = - 4 + 2 (3 ) 2 • Next, evaluatepowers = - 4 + 2 ( 9 ) • Domultiplicationsanddivisionsfrom left to right = - 4 +18 • Doadditionsandsubtractions from left to right= 14 • Numerical expression: 25= 2 • 2 • 2 • 2 • 2 • [ 5 factors of 2 ] or [ 2 multiplied out 5 times ] • In this expression: • the number2is the base • the number5is theexponent • the expression is apower. • A variable is a letter used to represent one or more numbers. Any number used to replace variable is a value of the variable. An expression involving variables is called an algebraic expression. The value of the expression is the result when you evaluate the expression by replacing the variables with numbers. • An expression that represents a real-life situation is a mathematical model. See page 12.

  11. 1.2 Algebraic Expressions and Models Example: You can use order of operations to evaluate expressions. Numerical expressions: 8 (3 + 42) – 12 ÷2 = 8 (3 + 16) – 6 = 8 (19) – 6 = 152 – 6 = 146 Algebraic expression: 3 x2 – 1 when x = – 5 3 (– 5 )2 – 1 = 3 (25) – 1 = 74 Sometimes you can use the distributive property to simplify an expression. Combine like terms: 2 x2 – 4 x + 10 x – 1 = 2 x2 + (– 4 + 10 ) x – 1 = 2 x2 + 6 x - 1

  12. 1.2 Evaluating Powers • Example 1: ( - 3 ) 4 = ( - 3 ) ( - 3 ) ( - 3 ) ( - 3 ) = 81 • - 3 4 = - ( 3 3 3 3 ) = - 81 • Example 2: Evaluating an algebraic expression • - 3 x2 – 5 x + 7 when x = - 2 • - 3 ( - 2 ) 2 – 5 ( -2 )x + 7 [ substitute – 2 for x ] • - 3 ( 4 ) – 5 ( -2 )x + 7 [ evaluate the power, 2 2 ] • - 12 + 10 + 7 [ multiply ] • + 5 [ add ] • Example 3: Simplifying by combining like terms • 7 x + 4 x = ( 7 + 4 ) x [ distributive ] • = 11 x [ add coefficients ] • 3 n 2 + n – n 2 = ( 3 n 2 – n 2 ) + n [ group like terms ] • = 2 n 2 + n [ combine like terms ] • 2 ( x + 1 ) – 3 ( x – 4 ) = 2 x + 2 – 3 x + 12 [ distributive ] • = ( 2 x – 3 x ) + ( 2 + 12 ) [ group like terms ] • = - x + 14 [ combine like terms ]

  13. 1.3 Solving Linear Equations Transformations that produce equivalent equations Additional property of equality Add same number to both sides if a = b, then a + c = b + c Subtraction property of equality Subtract same number to both sides if a = b, then a - c = b - c Multiplication property of equalityMultiply both sides by the same number if a = b and c ǂ 0, then a • c = b • c Division property of equality Divide both sides by the same number if a = b and c ǂ 0, then a ÷ c = b ÷ c Linear Equations in one variable in form a x = b, where a & b are constants and a ǂ 0. A number is a solution of an equation if the expression is true when the number is substituted. Two equations are equivalent if they have the same solution.

  14. 1.3 Solve Linear Equations Solving for variable on one side [by isolating the variable on one side of equation] Example 1: 3 x + 9 = 15 7 3 x + 9 - 9 = 15 - 9 7 [ subtract 9 from both sides to eliminate the other term] 3 x = 6 7 7• 3 x = 7• 6 3 7 [ multiply both sides by 7/3, the reciprocal of 3/7, to get xby itself] x = 14 Example 2: 5 n + 11 = 7 n – 9 - 5 n - 5 n[ subtract 5 n from both sides to get the variable on one side ] 11 = 2 n – 9 + 9 + 9 [ add 9 to both sides to get rid of the other term with the variable] 20 = 2 n 2 2 [ divide both sides by 2to get the variablenby itself on one side] 10 = n

  15. 1.3 Solve Linear Equations Example: You can use properties of real numbers and transformations that produce equivalent equations to solve linear equations. Solve: – 2 ( x – 4 ) = 12 – 2 x + 8 = 12 – 2 x = 4 x = – 2

  16. 1.3 Solve Linear Equations Equations with fractions Example 3: 1 x + 1 = - 1 3 4 6

  17. 1.3 Solve Linear Equations Identifying Properties • − 8 + 8 = 0 • ( 3 • 5 ) • 10 = 3 • ( 5 • 10 ) • 7 • 9 = 9 • 7 • ( 9 + 2 ) + 4 = 9 + ( 2 + 4 ) • 12 (1) = 12 • 2 ( 5 + 11 ) = 2 • 5 + 2 • 11

  18. 1.3 Solve Word Problems Operations 43. What is the sum of 32 and – 7 ? 44. What is the sum of – 9 and – 6 ? 45. What is the difference of – 5 and 8 ? 46. What is the difference of – 1 and – 10 ? 47. What is the product of 9 and – 4 ? 48. What is the product of – 7 and – 3 ? 49. What is the quotient of – 5 and – ½ ? 50. What is the quotient of – 14 and 7/4 ?

  19. 1.3 Solve Unit Measures Unit Analysis • 8 1/6 feet + 4 5/6 feet = • 27 ½ liters – 18 5/8 liters = • 8.75 yards ( $ 70 ) = • 1 yard • ( 50 feet ) ( 1 mile ) ( 3600 seconds ) = • 1 second 5280 feet 1 hour

  20. 1.4 ReWriting Equations and Formulas Example: You can an equation that has more than one variable, such as a formula, for one of its variables. Solve the equation for y: 2 x – 3 y = 6 – 3 y = – 2 x + 6 y = 2 x – 2 3 Solve for the formula for the area of a trapezoid for h: A = 1 ( b1 + b2) h 2 2 A = ( b1 + b2) h 2 A = h ( b1 + b2)

  21. 1.4 ReWriting an Equation with more than 1 variable Solve : 7 x – 3 y = 8 for the variable y. 7 x – 3 y = 8 - 7 x - 7 x [ subtract 7 x from both sides to get rid of the other term] – 3 y = 8 – 7 x – 3 – 3 – 3 [divide both sides by – 3to get the variable x by itself on one side ] y = – 8 + 7 x 3 3 Calculating the value of a variable Solve: x + x y = 1 when x = – 1 and x = 3 x + x y = 1 [ first solve for yso that when you replace x with – 1 and 3, you also solve for y ] - x - x [ subtract x from both sides to get rid of the other term without yin it] x y = 1 – x x x [divide by xto get yby itself ] y = 1 – x when x = – 1, then y = – 2 and when x = 3, then y = – 2/3 x

  22. 1.4 Common Formulas Distance D = r td = distance, r = rate, t = time Simple interest I = p r tI = interest, p = principal, r = rate, t = time Temperature F = 9/5 C + 32F = degrees Fahrenheit, C = degrees Celsius Area of a Triangle A = ½ b hA = area, b = base, h = height Area of a Rectangle A = l wA = area, l = length, w = width Perimeter of Rectangle P = 2 l + 2 wP = perimeter, l = length, w = width Area of Trapezoid A = ½ ( b1 + b2 ) hA = area, b1 = 1 base, b2 = 2 base, h = height Area of Circle A = πr2A = area, r = radius Circumference of Circle C = 2 π rC = circumference, r = radius

  23. 1.4 ReWriting a Common Formula P = 2 l + 2 w [ solve for w ] – 2 l – 2 l[ subtract 2 l from both side to get rid of the other term] P – 2 1 = 2 w [ divide each side by 2to getwby itself on one side of the equation] 2 2 P – 2 l = w 2 Example 6: Applying a common formula Problem: You have 40 feet of fencing with which to enclose a rectangular garden. Express the garden’s area in terms of its length only. A= l w A = l ( P – 2 l ) [ substituteP – 2 l for w ] 2 2 A = l (40 – 2 l) [ substitute 40 for P as indicated in problem above ] 2 A = l ( 20 – l )

  24. 1.5 Problem Solving Using Algebraic Models Key steps in problem solving plan Write a verbal model Assign labels Write an algebraic model Solve algebraic model Answer question Verbal model is a word equation while an Algebraic model is a math statement Example 1, Writing and using a formula Verbal : Distance = Rate • Time Labels: 550 kilos = r kilos/hr • 2.25 hours Algebraic: 550 = r • 2.25 550 = r • 2.25 2.25 2.25 [ divide by 2.25 to solve for r ] 244 = r

  25. 1.5 Problem Solving Using Algebraic Models Example: create a problem solving plan in which you write a verbal model, assign labels, write and solve an algebraic model, and then answer the question. How far can you drive at 55 miles per hour for 4 hours? Verbal Model --- Distance = Rate  Time Labels --- Distance = d (miles), Rate= 55 (miles per hour), Time = 4 (hours) Algebraic Model --- d = 55 ●4 You can drive 220 miles

  26. 1.5 Problem Solving Using Algebraic Models Example 3, Highway Local Verbal : Total miles = fuel efficiency x amount of gas + fuel efficiency x amount of gas Labels: Total miles = 600 [miles] Fuel efficiency (highway) = 21 [miles per gallon] Amount of gas (highway) = x [gallons] Fuel efficiency (local) = 13 [miles per gallon] Amount of gas (local) = 40 – x [gallons] Algebraic 600 = 21 x + 13 (40 – x) 600 = 21 x + 520 – 13 x 600 = 8 x + 520 - 520 - 520 80 = 8 x 8 8 10 = x 10 gallons x cost of gas per gallon = total cost for gas for this trip

  27. 1.6 Solving Linear Inequalities Transformations that produce equivalent inequalities Add same number to both sides Subtract same number to both sides Multiply both sides by the same positive number Divide both sides by the same positive number Multiply both sides by the same negative number, then reverse the inequality Divide both sides by the same negative number, then reverse the inequality Linear inequalities are indicated by > , < , ≥, ≤ Solution of an inequality in one variable is a value which makes the inequality true. Graph of inequality consists of all points on a real # line that corresponds to solutions of the inequality. - 2 1 Compound inequality is two simple inequalities joined by “and” or “or” - 1 2

  28. 1.6 Solving an Inequality with a Variable on 1 side • Example 1: • 5 y – 8 < 12 • + 8 + 8 • 5 y < 20 • 5 • y < 4 0 4 Example 2: 2 x + 1 ≤6 x - 1 - 6 x - 6 x - 4 x + 1 ≤- 1 - 1 - 1 - 4 x≤- 2 -4 - 4 x½ 0 ½ 1 ≤ Divide both sides by the same negative number, then reverse the inequality

  29. 1.6 Solving Linear Inequalities • Example 3: • - 2 ≤3 t – 8 ≤10 • + 8 + 8 + 8 • 6≤3 t ≤18 • 3 3 3 • 2 ≤ t ≤6 2 6 Example 4: 2 x + 3 < 5 or 4 x – 7 > 9 - 3 - 3 + 7 + 7 2 x< 24 x> 16 2 2 4 4 x < 1 x > 4 0 1 4

  30. 1.6 Solving Linear Inequalties Example: You can use transformations to solve inequalities . Reverse the inequality when you multiply or divide both sides by a negative number. 4 x + 1 < 7 x – 5 – 3 x < – 6 x > 2 0 ≤6 – 2 n ≤ 10 – 6 ≤– 2 n ≤4 3 ≤n ≤– 2 2 – 2 3

  31. 1.6 Solving Absolute Value Equations Absolute Valueof a number x , | x | , is the distance the number is from 0 on a number line. Example: | - 3 | = 3 | 3 | = 3 3 units = 3 units –3 0 + 3 The absolute value equation: | a x + b | = c, where c > 0, is equivalent to the compound statement a x + b = c or a x + b = - c The value inside the absolute symbols has2solutions, + and – , | a x + b = c ora x + b = - c | = c But, remember that Absolute Value is always positive and has 2 solutions. “Distance has to be +”, so in this example the value of coutside the absolute value symbols is + even when one of the solutions inside the absolute value is negative. See also example above, where c = + 3.

  32. 1.6 Solving Absolute Value Inequalities Transformations of absolute value inequalities | a x + b | < c, where c > 0 means that a x + b is between – c and c, - c < a x + b < c [compound statement “and”, where both conditions must be met.] | a x + b | > c, where c > 0 means that a x + b is beyond – c and c, - c > a x + b or a x + b > c [two statements “or”, where only one condition must be met.] Remember: Absolute Value means distance ( + value) and always has 2 solutions. | a x + b | < c means “and”, - c < a x + b < c, where a x + b is between – c and + c | a x + b | > c means “or”, - c > a x + b or a x + b > c, where a x + b is beyond – c and + c < - c + c > - c + c

  33. 1.7 Solving Absolute Value Expressions and Inequalities Example: To solve an absolute value equation, rewrite it as two linear equations. To solve an absolute value inequality, rewrite it as a compound inequality. | x + 3 | = 5 x + 3 = 5 x + 3 = – 5 x = 2 x = – 8 | x – 7 |  2 – 2  x – 7  2 + 7 + 7 + 7 5  x  9 5 9

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