ONE and two-STEP EQUATIONS Ms. Harrison Math 8

Before we begin… In previous lessons we solved simple one-step linear equations… In this lesson we will look at solving multi-step equations…as the name suggests there are steps you must do before you can solve the equation… To be successful here you have to be able to analyze the equation and decide what steps must be taken to solve the equation…

Let’s review ONE STEP EQUATIONS An equation is like a balance scale because it shows that two quantities are equal. What you do to one side of the equation must also be done to the other side to keep it balanced.

ONE STEP EQUATIONS To solve one step equations, you need to ask three questions about the equation: What is the variable? What operation is performed on the variable? What is the inverse operation?

ONE STEP EQUATIONS Example 1 Solve x + 4 = 12 x + 4 = 12 - 4 - 4 x = 8 What is the variable? The variable is x. What operation is being performed on the variable? Addition. What is the inverse operation (the one that will undo what is being done to the variable)? Subtraction. x + 4 = 12 - 4 - 4 x = 8

ONE STEP EQUATIONS Example 2 Solve y - 7 = -13 y - 7 = -13 + 7 + 7 y = What is the variable? The variable is y. What operation is being performed on the variable? Subtraction. What is the inverse operation (the one that will undo what is being done to the variable)? Addition. y - 7 = -13 + 7 + 7 y = -6

ONE STEP EQUATIONS Example 3 Solve –6a = 12 –6a = 12 -6 -6 a = -2 What is the variable? The variable is a. What operation is being performed on the variable? Multiplication. What is the inverse operation (the one that will undo what is being done to the variable)? Division –6a = 12 -6 -6 a = -2

ONE STEP EQUATIONS Example 4 Solve = -10 = -10 2 • = -10 • 2 b = -20 What is the variable? The variable is b. What operation is being performed on the variable? Division. What is the inverse operation (the one that will undo what is being done to the variable)? Multiplication = -10 2 • = -10 • 2 b = -20

Two-Step Equations Think a little harder, do a little more; but either way, follow the rules! Ms. H 

What is a Two-Step Equation? An equation written in the form Ax + B = C

Examples of Two-Step Equations y/3 + 5 = 12 5b+ 4 = 6 w/2 – 6 = 4

Steps for Solving Two-Step Equations Solve for any Addition or Subtraction on the variable side of equation by “undoing” the operation from both sides of the equation. Solve any Multiplication or Division from variable side of equation by “undoing” the operation from both sides of the equation.

Opposite Operations Addition  Subtraction Multiplication  Division

Helpful Hints? Identify what operations are on the variable side. (Add, Sub, Mult, Div) “Undo” the operation by using opposite operations. Whatever you do to one side, you must do to the other side to keep equation balanced.

Solve 4x – 5 = 11 4x – 5 = 15 +5 +5 (Add 5 to both sides) 4x = 20 (Simplify) 4 4 (Divide both sides by 4) x = 5 (Simplify)

Try These Examples 2m + 5 = 19 3y - 7 = -25 -5n + 2 = 42 12b - 4 = -28

Ready to Move on?

Ex. 2: Solve x/3 + 4 = 9 x/3 + 4 = 9 - 4 - 4 (Subt. 4 from both sides) x/3 = 5 (Simplify) (x/3)  3 = 5  3 (Mult. by 3 on both sides) x = 15 (Simplify)

Try these examples! x/5 +3 = -8 c/7 + 4 = 9 r/3 + 4 = -2 d/-9 + 4 = 5

Time to Review! Make sure your equation is in the form Ax + B = C Keep the equation balanced. Use opposite operations to “undo” Follow the rules: Undo Addition or Subtaction Undo Multiplication or Division

Multi-Step Equations This lesson will look at: Solving linear equations with 2 operations Combining like terms first Using the distributive property Distributing a negative Multiplying by a reciprocal first Your ability to be organized and lay out your problem will enable you to be successful with these concepts

Linear Equations with 2 Functions Sometimes linear equations will have more than one operation In this instance operations are defined as add, subtract, multiply or divide The rule is First you add or subtract Second you multiply or divide.

Linear Equations with 2 Operations You will have to analyze the equation first to see what it is saying… Then, based upon the rules, undo each of the operations… The best way to explain this is by looking at an example….

Example # 1 2x + 6 = 16 In this case the problem says 2 times a number plus 6 is equal to 16 Times means to multiply and plus means to add The first step is to add or subtract. To undo the addition of 6 I have to subtract 6 from both sides which looks like this: 16 minus 6 equals 10 The 6’s on the left side cancel each other out 2x + 6 = 16 - 6 -6 2x = 10 What is left is a one step equation 2x = 10

Example #1 (Continued) 2x = 10 The second step is to multiply or divide To undo the multiplication here you would divide both sides by 2 which looks like this: 10 divided by 2 is equal to 5 2x = 10 The 2’s on the left cancel out leaving x 2 2 x = 5 The solution to the equation 2x + 6 = 16 is x = 5

Combining Like Terms First When solving multi-step equations, sometimes you have to combine like terms first. The rule for combining like terms is that the terms must have the same variable and the same exponent. Example: You can combine x + 5x to get 6x You cannot combine x + 2x2 because the terms do not have the same exponent

Example # 2 I begin working on the left side of the equation 7x – 3x – 8 = 24 I begin working on the left side of the equation On the left side I notice that I have two like terms (7x, -3x) since the terms are alike I can combine them to get 4x. 7x – 3x – 8 = 24 4x – 8 = 24 After I combine the terms I have a 2-step equation. To solve this equation add/subtract 1st and then multiply/divide

Example # 2 (continued) Step 1: Add/Subtract Since this equation has – 8, I will add 8 to both sides 4x – 8 = 24 +8 +8 24 + 8 = 32 The 8’s on the left cancel out 4x = 32 I am left with a 1-step equation

The solution that makes the statement true is x = 8 Example # 2 (continued) 4x = 32 Step 2: Multiply/Divide In this instance 4x means 4 times x. To undo the multiplication divide both sides by 4 32  4 = 8 The 4’s on the left cancel out leaving x 4x = 32 4 4 x = 8 The solution that makes the statement true is x = 8

Solving equations using the Distributive Property When solving equations, sometimes you will need to use the distributive property first. At this level you are required to be able to recognize and know how to use the distributive property Essentially, you multiply what’s on the outside of the parenthesis with EACH term on the inside of the parenthesis Let’s see what that looks like…

Example #3 In this instance I begin on the left side of the equation 5x + 3(x +4) = 28 In this instance I begin on the left side of the equation I recognize the distributive property as 3(x +4). I must simplify that before I can do anything else 5x + 3(x +4) = 28 5x +3x +12 = 28 After I do the distributive property I see that I have like terms (5x and 3x) I have to combine them to get 8x before I can solve this equation

I am now left with a 2-step equation Example # 3(continued) 8x + 12 = 28 I am now left with a 2-step equation Step 1: Add/Subtract The left side has +12. To undo the +12, I subtract 12 from both sides 8x + 12 = 28 28 – 12 = 16 The 12’s on the left cancel out leaving 8x -12 -12 8x = 16

Example # 3 (continued) Step 2: Multiply/Divide On the left side 8x means 8 times x. To undo the multiplication I divide both sides by 8 8x = 16 16  8 = 2 The 8’s on the left cancel out leaving x 8 8 x = 2 The solution that makes the statement true is x = 2

Distributing a Negative Distributing a negative number is similar to using the distributive property. However, students get this wrong because they forget to use the rules of integers Quickly the rules are…when multiplying, if the signs are the same the answer is positive. If the signs are different the answer is negative

Example #4 4x – 3(x – 2) = 21 I begin by working on the left side of the equation. In this problem I have to use the distributive property. However, the 3 in front of the parenthesis is a negative 3. When multiplying here, multiply the -3 by both terms within the parenthesis. Use the rules of integers 4x – 3(x – 2) = 21 4x – 3x + 6 = 21 After doing the distributive property, I see that I can combine the 4x and the -3x to get 1x or x

The 6’s cancel out leaving x Example # 4 (continued) 4x – 3x + 6 = 21 x + 6 = 21 After combining like terms you are left with a simple one step equation. To undo the +6 subtract 6 from both sides of the equation x + 6 = 21 21 – 6 = 15 The 6’s cancel out leaving x -6 -6 x = 15 The solution is x = 15

Multiplying by a Reciprocal First Sometimes when doing the distributive property involving fractions you can multiply by the reciprocal first. Recall that the reciprocal is the inverse of the fraction and when multiplied their product is equal to 1. The thing about using the reciprocal is that you have to multiply both sides of the equation by the reciprocal. Let’s see what that looks like…

Example # 5 In this example you could distribute the 3/10 to the x and the 2. The quicker way to handle this is to use the reciprocal of 3/10 which is 10/3 and multiply both sides of the equation by 10/3 On the left side of the equation, after multiplying by the reciprocal 10/3 you are left with 120/3 which can be simplified to 40 On the right side of the equation the reciprocals cancel each other out leaving x + 2 The new equation is: 40 = x + 2

Example #5 (continued) 40 = x + 2 After using the reciprocals you are left with a simple one-step equation To solve this equation begin by working on the right side and subtract 2 from both sides of the equation The 2’s on the right side cancel out leaving x 40 = x + 2 40 – 2 = 38 - 2 -2 38 = x The solution to the equation is x = 38

Comments On the next couple of slides are some practice problems…The answers are on the last slide… Do the practice and then check your answers…If you do not get the same answer you must question what you did…go back and problem solve to find the error… If you cannot find the error bring your work to me and I will help…

Your Turn 6x – 4(9 –x) = 106 2x + 7 = 15 6 = 14 – 2x 3(x – 2) = 18

Your Turn 6. 7. 8. 9. 10. 5m – (4m – 1) = -12 55x – 3(9x + 12) = -64

Your Turn Solutions 14.2 4 8 1 ½ 3 -52 -13 -1 5

Warm Up Solve each equation. 1. 2x – 5 = –17 2. –6 14 Solve each inequality and graph the solutions. 3. 5 < t + 9 t > –4 4. a ≤ –8

Solving Multi-Step Inequalities Solve the inequality and graph the solutions. 45 + 2b > 61 1. Add/ Subtract 45 + 2b > 61 –45 –45 2. Multiply/Divide 2b > 16 b > 8 The solution set is {b:b > 8}. 2 4 6 8 10 12 14 16 18 20

Solving Multi-Step Inequalities Solve the inequality and graph the solutions. 8 – 3y ≥ 29 1. Add/Subtract 8 – 3y ≥ 29 –8 –8 2. Multiply/Divide –3y ≥ 21 The solution set is {y:y  –7}. y ≤ –7 –10 –8 –6 –4 –2 2 4 6 8 10 –7

You try Solve and Graph the inequality

To solve more complicated inequalities, you may first need to simplify the expressions on one or both sides.

Simplifying Before Solving Inequalities Solve the inequality and graph the solutions. 2 – (–10) > –4t 1. Combine like terms. 12 > –4t 2. Multiply/Divide –3 < t (or t > –3) The solution set is {t:t > –3}. –3 –10 –8 –6 –4 –2 2 4 6 8 10

Simplifying Before Solving Inequalities Solve the inequality and graph the solutions. –4(2 – x) ≤ 8 1. Distributive Property −4(2 – x) ≤ 8 −4(2) − 4(−x) ≤ 8 2. Add/Subtract –8 + 4x ≤ 8 +8 +8 3. Multiply/Divide 4x ≤ 16 x ≤ 4 The solution set is {x:x ≤ 4}. –10 –8 –6 –4 –2 2 4 6 8 10

Now you try… 1. 3x – 7 > 2 4. x – 4 < 3 5 2. 4x + 1  -3

Lesson Quiz: Part I Solve each inequality and graph the solutions. 1. 13 – 2x ≥ 21 x ≤ –4 2. –11 + 2 < 3p p > –3 3. 23 < –2(3 – t) t > 7 4.

3.3 Absolute Value Equations and Inequalities Summary: Solving Absolute Value Equations and Inequalities Let k be a positive real number, and p and q be real numbers. 1. To solve |ax + b| = k, solve the following compound equation. ax + b = k or ax + b = –k. The solution set is usually of the form {p, q}, which includes two numbers. p q Copyright © 2010 Pearson Education, Inc. All rights reserved.

3.3 Absolute Value Equations and Inequalities Summary: Solving Absolute Value Equations and Inequalities Let k be a positive real number, and p and q be real numbers. 2. To solve |ax + b| > k, solve the following compound inequality. ax + b > k or ax + b < –k. The solution set is of the form (-∞, p) U (q, ∞), which consists of two separate intervals. p q Copyright © 2010 Pearson Education, Inc. All rights reserved.

3.3 Absolute Value Equations and Inequalities Summary: Solving Absolute Value Equations and Inequalities Let k be a positive real number, and p and q be real numbers. 3. To solve |ax + b| < k, solve the three-part inequality –k < ax + b < k The solution set is of the form (p, q), a single interval. p q Copyright © 2010 Pearson Education, Inc. All rights reserved.

3.3 Absolute Value Equations and Inequalities EXAMPLE 1 Solving an Absolute Value Equation Solve |2x + 3| = 5.

3.3 Absolute Value Equations and Inequalities EXAMPLE 2 Solving an Absolute Value Inequality with > Solve |2x + 3| > 5.

3.3 Absolute Value Equations and Inequalities EXAMPLE 3 Solving an Absolute Value Inequality with < Solve |2x + 3| < 5.

3.3 Absolute Value Equations and Inequalities EXAMPLE 4 Solving an Absolute Value Equation That Requires Rewriting Solve the equation |x – 7| + 6 = 9.

3.3 Absolute Value Equations and Inequalities Special Cases for Absolute Value Special Cases for Absolute Value The absolute value of an expression can never be negative: |a| ≥ 0 for all real numbers a. The absolute value of an expression equals 0 only when the expression is equal to 0. Copyright © 2010 Pearson Education, Inc. All rights reserved.

3.3 Absolute Value Equations and Inequalities EXAMPLE 6 Solving Special Cases of Absolute Value Equations Solve each equation. (a) |2n + 3| = –7 See Case 1 in the preceding slide. Since the absolute value of an expression can never be negative, there are no solutions for this equation. The solution set is Ø. (b) |6w – 1| = 0 See Case 2 in the preceding slide. The absolute value of the expres- sion 6w – 1 will equal 0 only if 6w – 1 = 0. 1 6 The solution of this equation is . Thus, the solution set of the original equation is { }, with just one element. Check by substitution. 1 6 Copyright © 2010 Pearson Education, Inc. All rights reserved.

3.3 Absolute Value Equations and Inequalities EXAMPLE 7 Solving Special Cases of Absolute Value Inequalities Solve each inequality. (a) |x| ≥ –2 The absolute value of a number is always greater than or equal to 0. Thus, |x| ≥ –2 is true for all real numbers. The solution set is (–∞, ∞). (b) |x + 5| – 1 < –8 Add 1 to each side to get the absolute value expression alone on one side. |x + 5| < –7 There is no number whose absolute value is less than –7, so this inequality has no solution. The solution set is Ø. Copyright © 2010 Pearson Education, Inc. All rights reserved.

3.3 Absolute Value Equations and Inequalities EXAMPLE 7 Solving Special Cases of Absolute Value Inequalities Solve each inequality. (c) |x – 9| + 2 ≤ 2 Subtracting 2 from each side gives |x – 9| ≤ 0 The value of |x – 9| will never be less than 0. However, |x – 9| will equal 0 when x = 9. Therefore, the solution set is {9}. Copyright © 2010 Pearson Education, Inc. All rights reserved.

This can be written as 1  x  7. Solving an Absolute-Value Inequality Solve | x  4 | < 3 x  4 IS POSITIVE x  4 IS NEGATIVE | x  4 |  3 | x  4 |  3 x  4  3 x  4  3  x  7 x  1 Reverse inequality symbol. The solution is all real numbers greater than 1 and less than 7. This can be written as 1  x  7.

Solve | 2x  1 | 3  6 and graph the solution. Solving an Absolute-Value Inequality | 2x  1 |  3  6 | 2x  1 |  9 2x  1  +9 x  4 2x  8 | 2x  1 | 3  6 2x  1  9 2x  10 x  5 2x + 1 IS POSITIVE 2x + 1 IS NEGATIVE Solve | 2x  1 | 3  6 and graph the solution. 2x + 1 IS POSITIVE 2x + 1 IS NEGATIVE | 2x  1 |  3  6 | 2x  1 | 3  6 | 2x  1 |  9 | 2x  1 |  9 2x  1  +9 2x  1  9  2x  8 2x  10 The solution is all real numbers greater than or equal to 4 or less than or equal to  5. This can be written as the compound inequality x   5 or x  4. x  4 Reverse inequality symbol. x  5  5 4.  6  5  4  3  2  1 0 1 2 3 4 5 6

Strange Results True for All Real Numbers, since absolute value is always positive, and therefore greater than any negative. No Solution Ø. Positive numbers are never less than negative numbers.

Examples or or Check and verify on a number line. Numbers above 6 or below -1 keep the absolute value greater than 7. Numbers between them make the absolute value less than 7.

Key Skills Solve absolute-value inequalities. Solve |x – 4|  5. Case 1: Case 2: x – 4 is positive x – 4 is negative x – 4  5 x – 4  –5 x  9 x  –1 solution: –1  x  9

Key Skills Solve absolute-value inequalities. Solve |4x – 2|  -18. Exception alert!!!! When the absolute value equals a negative value, there is no solution.

Key Skills TRY THIS Solve absolute-value inequalities. Solve |2x – 6|  18. Case 2: Case 1: 2x – 6 is negative 2x – 6 is positive 2x – 6  18 2x - 6  –18 2x  24 2x  –12 x  12 x  –6 Solution: –6  x  12

Key Skills TRY THIS Solve absolute-value inequalities. Solve |3x – 2|  -4. Exception alert!!!! When the absolute value equals a negative value, there is no solution.