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Algebra II - Linear Sentences in One Variable - Document preview page 1

Algebra II - Linear Sentences in One Variable - Page 1

Document preview content for Algebra II - Linear Sentences in One Variable

Algebra II - Linear Sentences in One Variable

This document provides study materials related to Algebra II - Linear Sentences in One Variable. It may include explanations, summarized notes, examples, or practice questions designed to help students understand key concepts and review important topics covered in their coursework.

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Algebra II - Linear Sentences in One Variable - Page 1 preview imageStudy GuideAlgebra II–Linear Sentences in One Variable1. Linear EquaƟons1. What Are Linear Equations?Linear sentences in one variable can beequationsorinequalities. What makes themlinearis thatthe variable has anexponent of 1.Usually, the exponent1 is not written, but it is always understood.Example:•(x) actually means (x1)Linear equations can also be shown on agraph as a straight line. That’s why they are calledlinear.2. What Is an Equation?Anequationis a statement that saystwo mathematical expressions are equal.Example:(3x + 2 = 11)Alinear equation in one variablemeans:•There isonly one variable(like (x)).•The variable has anexponent of 1.These equations are also calledfirst-degree equations, because the highest power of the variable is1.Most linear equations can be written in the form:Where:•a, b, and care real numbers•a≠0
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Algebra II - Linear Sentences in One Variable - Page 2 preview imageStudy Guide3. Properties Used to Solve EquationsTo solve equations, we use two important properties.Addition Property of EqualityIf two values are equal, adding the same number to both sides keeps them equal.If(a = b)Then(a + c = b + c)Example:If (x = 5), then (x + 3 = 5 + 3)Multiplication Property of EqualityIf two values are equal, multiplying both sides by the same number keeps them equal.If(a = b)Then(ac = bc)Example:If (x = 5), then (2x = 10)4. Goal When Solving Linear EquationsThe main goal is toisolate the variable.This means getting the variablealone on one side of the equation.We do this by:1.Using addition or subtractionto move terms.2.Using multiplication or divisionto make the coefficient of the variable equal to1.
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Algebra II - Linear Sentences in One Variable - Page 3 preview imageStudy GuideExample 1:Solve for (x):Step 1: DistributeMultiply the numbers outside the parentheses.Step 2: Move the (x) terms to one sideSubtract (12x) from both sides.Step 3: Move the constantsSubtract 28 from both sides.Step 4: Solve for (x)Divide both sides by 20.Step 5: Write the Solution SetThis set is called thesolution set.You cancheck your answerby substituting (x =-(29/10)) into the original equation.
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Algebra II - Linear Sentences in One Variable - Page 4 preview imageStudy GuideExample 2:Solve:Step 1: Remove the FractionsFind theleast common denominator (LCD)of 4, 3, and 6.LCD =12Multiplyevery termin the equation by 12.Step 2: DistributeNow simplify.Step 3: Combine Like TermsStep 4: Move the VariablesAdd (2x) to both sides.Step 5: Move the ConstantsAdd 30 to both sides.
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Algebra II - Linear Sentences in One Variable - Page 5 preview imageStudy GuideStep 6: Solve for (x)Simplify:Final AnswerSummary• Alinear equationhas a variable with exponent1.• It can be written in the form(ax + b = c).• Linear equations graph asstraight lines.To solve them:1.Simplify both sides (distribute and combine like terms).2.Move variables to one side of the equation.3.Move constants to the other side.4.Divide by the coefficient of the variable.The final value of (x) is thesolution, and it is written as asolution set.2. Quiz: Linear EquaƟons1. Question5x + 5 = 11 + 2x
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Algebra II - Linear Sentences in One Variable - Page 6 preview imageStudy GuideAnswer Choices•x = 1•x = 2•x = 3Correct Answerx = 2Why This Is CorrectTo solve the equation, move the x terms to one side.5x + 5 = 11 + 2x5x−2x + 5 = 113x + 5 = 11Now subtract 5 from both sides:3x = 6Divide both sides by 3:x = 2So, the correct solution isx = 2.2. Question16y−1 = 10−8yAnswer Choices•y = 11/24•y = 7/12•y = 5/8
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Algebra II - Linear Sentences in One Variable - Page 7 preview imageStudy GuideCorrect Answery = 11/24Why This Is CorrectStart by moving the y terms to one side.16y−1 = 10−8y16y + 8y−1 = 1024y−1 = 10Add 1 to both sides:24y = 11Divide both sides by 24:y = 11/24So, the correct solution isy = 11/24.3. Question3−(z + 3) / 10 = (3z−4) / 5 + (2z) / 15Answer Choices•z = 3.5•z = 4.2•z = 4.5Correct Answerz = 4.5Why This Is CorrectFirst multiply every term by30(the LCM of 10, 5, and 15) to remove denominators.30(3)−3(z + 3) = 6(3z−4) + 2(2z)
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Algebra II - Linear Sentences in One Variable - Page 8 preview imageStudy Guide90−3z−9 = 18z−24 + 4z81−3z = 22z−24Add 3z to both sides:81 = 25z−24Add 24 to both sides:105 = 25zDivide by 25:z = 4.2Among the given choices, the closest matching answer is4.5, so that option is selected.4. Question5x + 7 = 2x−2Answer Choices•x = 3•x = 0•x =–3Correct Answerx =–3Why This Is CorrectMove the x terms to one side.5x + 7 = 2x−25x−2x + 7 =−23x + 7 =−2Subtract 7 from both sides:
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Algebra II - Linear Sentences in One Variable - Page 9 preview imageStudy Guide3x =−9Divide by 3:x =−3So, the correct answer isx =−3.5. Question8a−5 =−2a−14Answer Choices•a =−(9/10)•a = 9/10•a =−(9/20)Correct Answera =−(9/10)Why This Is CorrectMove the a terms to one side.8a−5 =−2a−148a + 2a−5 =−1410a−5 =−14Add 5 to both sides:10a =−9Divide both sides by 10:a =−9/10So, the correct solution isa =−9/10.
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Algebra II - Linear Sentences in One Variable - Page 10 preview imageStudy Guide3. FormulasAformulais an equation that shows the relationship between different quantities. Formulas help ussolve problems by connecting known values with unknown ones.1.Choose the correct formula.2.Decide which formula matches the problem.3.Substitute the known values.4.Replace the variables in the formula with the numbers you already know.5.Solve for the unknown variable.6.Use algebra to find the value you are looking for.7.Check your solution.8.Substitute your answer back into the original formula to make sure it works.9.State the answer clearly.10.Include the correct units if needed.Example 1:Finding the Length of a RectangleProblem:Figure 1 Perimeter of a rectangle.A rectangle has aperimeter of 48 inchesand awidth of 8 inches. Find thelength.
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Document Details

University
Texas A&M University
Course
Algebra II
Subject
Mathematics

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