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Solving Linear Equations. Algebra I Ms. Wiggins. Content Standards. HSA.CED.A.1 Create equations and inequalities in one variable and use them to solve problems.
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Solving Linear Equations Algebra I Ms. Wiggins
Content Standards • HSA.CED.A.1 Create equations and inequalities in one variable and use them to solve problems. • HSA.REI.A.1 Explain each step in solving a simple equation as following from the quality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. • HAS.REI.B.3 • Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.
Do Now Solve the equation x + 2 = 8 using all the vocabulary words (equivalent equations, inverse operations, isolate, solution of an equation and variable).
Objective • Students will be able to explain that each step in solving a linear equation follows from the equality in the previous step. Create and solve linear equations with one variable using the properties of equality.
Essential Understanding • Linear equations can be used to solve mathematical and real-world problems. You can solve linear equations by using the properties of equality.
Vocabulary • Equivalent equations • Inverse operations • Isolate • Solution of an equation • Variable
Properties of Equality • Addition property • Subtraction property • Multiplication property • Division property
Use inverse operations • Use the opposite operation of what’s happening to isolate variables. Example: x – 2 = 4 • To solve this problems use the addition property since the subtraction property is being used. Add 2 to both sides. X = 6 Example: x + 2 = 4 • To solve this problems use the subtraction property since the addition property is being used. Subtract 2 from both sides. X = 2
PEMDAS • Don’t forget you still must apply PEMDAS when solving equations • Please (parenthesis) • Excuse (exponents) • My (multiplication) • Dear (division) • Aunt (addition) • Sally (subtraction)
Students try: 4 + = 9 Step 1: multiply each side by 3 (3)(3) 2(x + 4) – 24 = 96 Step 2: Distribute 2 into parenthesis and simplify. 2x -16 = 96 Step 3: inverse operation, add 16 to both sides of equation. 2x = 112 Step 4: inverse operation, divide 2 by both sides of equation. X= 56
Example 2: The sum of three consecutive integers is 132. What are the three integers? • First integers = x; Second integer = x+1; Third integer = x+2 x + (x + 1) + (x + 2) = 132 Step 1: Combine like terms 3x + 3 = 132 Step 2: inverse operation, subtract three from both sides. 3x = 129 Step 3: inverse operation, divide three by both sides. X = 43 The other two consecutive numbers after 43 is 44 & 45.
Example 3: • A lab technician needs 25 liters of a solution that is 15% acid for a certain experiment but she has only a solution that is 10% acid and a solution that is 30% acid. How many liters of the 10% and the 30% solutions should she mix to get what she needs? • Solution: Write an equation relating the number of liters of acid in each solution. Represent the total number of liters of one solution with a variable, like x. Then the total number of liters of the other solution must be 25 – x.
25L of 15% solution = x L of 10% solution + (25-x) L of 30% solution (.15)(25) = 0.10x + 0.30(25 - x) 3.75 = 0.1x + 7.5 – 0.3x 3.75 = (0.1x – 0.3x) + 7.5 combine like terms 3.75 - 7.5 = -0.2x + 7.5 - 7.5 subtract 7 both sides -3.75 = -0.2xsimplified -3.75/-0.2 = x divide both sides by -0.2 18.75 = x solution
Essential Questions • When writing an equation to represent a real-world problem situation, how do you determine what the variable will represents? • How are the properties of equality useful when solving the equation? • What do you need to know in order to write a percent as a decimal?
