Mr. Istik Grade 9.  The Student Will Be Able To Solve One- Variable Equations using addition, subtraction, multiplication, division, and a combination.

Slides:



Advertisements
Similar presentations
ALGEBRA 1 BASICS CHEAT SHEET THINGS YOU SHOULD KNOW . . .
Advertisements

Warm Up Lesson Presentation Lesson Quiz.
Solving Linear Equations
The Fundamental Property of Rational Expressions
Factors, Fractions, and Exponents
Solving Linear Equations
Solving Equations Please view this tutorial and answer the follow-up questions on loose leaf to turn in to your teacher.
Solving One Step Equations using inverse operations. 2-2
Math is a language, learn the words!
Introduction Solving inequalities is similar to solving equations. To find the solution to an inequality, use methods similar to those used in solving.
Solving Algebraic Equations
Gateway Quiz Reminders: The next Gateway will be given in class next week or the week after )check your course calendar.) Before then, each student who.
Integers and Introduction to Solving Equations
To Start: 10 Points.
Lesson 2-4. Many equations contain variables on each side. To solve these equations, FIRST use addition and subtraction to write an equivalent equation.
Working with Fractions
ALGEBRAIC EQUATIONS. EQUATIONS AND SOLUTIONS  A correct equation is like a balance scale.  In order to determine if a given value for a variable is.
Introduction While it may not be efficient to write out the justification for each step when solving equations, it is important to remember that the properties.
Properties of Equality, Identity, and Operations.
Algebra Notes.
Basic Concepts of Algebra
Mrs. Martinez CHS MATH DEPT.
Copyright © 2010 Pearson Education, Inc
Operations With Decimals
Module 6 Test Review.
Algebraic Expressions
MATH 31 LESSONS PreCalculus 2. Powers. A. Power Laws Terminology: b x.
Solving Equations II Lesson After completing this lesson, you will be able to say: I can solve one-variable equations containing multiplication.
Objectives: To evaluate and simplify algebraic expressions.
Variables Tutorial 3c variable A variable is any symbol that can be replaced with a number to solve a math problem. An open sentence has at least one.
Warm Up  – Evaluate.  (0.29)
Warm up Lesson 2.2 Use the distributive property to simplify each expression:
Distributive Property 2.2 LESSON DO NOW: IF YOU WERE ASKED TO DISTRIBUTE MATERIALS IN CLASS, EXPLAIN WHAT YOU THINK YOUR JOB MIGHT REQUIRE YOU TO DO?
Unit 1, Day 4.  Compare and contrast your slip of paper with other class mates.  Your slips of paper can be separated into three distinct groups. 
RATIONAL EXPRESSIONS. Definition of a Rational Expression A rational number is defined as the ratio of two integers, where q ≠ 0 Examples of rational.
Solving Linear Equations To Solve an Equation means... To isolate the variable having a coefficient of 1 on one side of the equation. Examples x = 5.
Chapter 1 Review College Algebra Remember the phrase “Please Excuse My Dear Aunt Sally” or PEMDAS. ORDER OF OPERATIONS 1. Parentheses - ( ) or [ ] 2.
Unit 6 Math Vocab By: Marshall Lockyer. Constant Term A constant term is a term in an equation that does not change Example: a = 6 + b : In this case,
Solving Equations. The equations are equivalent If they have the same solution(s)
Chapter 1 Review Advanced Algebra 1.
Unit 0 Lessons 1-3 Evaluating expressions using order of operations By R. Portteus and By Miss Klien modified by LHope.
Real numbers In algebra, we work with the set of real numbers, which we can model using a number line. Real numbers describe real-world quantities such.
 Term- number or product of a number and a variable  Constant- term with no variable  Coefficient- number that multiplies to a variable  Like Terms-
Chapter 2 Real Numbers and algebraic expressions ©2002 by R. Villar All Rights Reserved Re-engineered by Mistah Flynn 2015.
2.1 Solving One Step Equations: Equivalent Equations: equations that have the same solutions. Isolate: Get a variable with a coefficient of 1 alone on.
Solving equations with Rational Coefficients
Gateway Quiz Reminders: The next Gateway will be given in class next week (check your course calendar.) Before then, each student who did not score 8/8.
Revision on Matrices Finding the order of, Addition, Subtraction and the Inverse of Matices.
Algebra 1: Topic 1 Notes.
You have seen positive exponents
Single-Step Equations Addition/Subtraction Multiplication/Division Substitution Simplifying Expressions.
Week 1 Real Numbers and Their Properties (Section 1.6, 1.7, 1.8)
Section 6.2 Solving Linear Equations Math in Our World.
1.7: Adding Like Terms TERM: A number, a variable, or the product of the two.Ex: a, 3x, 2x, 5, CONSTANT: a term with no variable (number) Ex: 4, -1, 6,
Copyright 2013, 2009, 2005, 2002 Pearson, Education, Inc.
Solving One-Step Equations
CHAPTER 1.3 Solving Equations.
Tuesday September 22, 2015 Algebra I.
Lesson 1.1 Pattern: orderly and predictable way (rule) that items appear. Could be numbers, letters, images, figures. Describe the rule and name next.
Introduction While it may not be efficient to write out the justification for each step when solving equations, it is important to remember that the properties.
Solving Equations with the Variable on Both Sides
Splash Screen.
Equations Containing Decimals
Solving Equations: The Addition and Multiplication Properties
Introduction Solving inequalities is similar to solving equations. To find the solution to an inequality, use methods similar to those used in solving.
Algebra
Solving Equations Containing Fractions
Equations …. are mathematical sentences stating that two expressions are equivalent.
Presentation transcript:

Mr. Istik Grade 9

 The Student Will Be Able To Solve One- Variable Equations using addition, subtraction, multiplication, division, and a combination of them, each to an accuracy of 90%  The Student Will Be Able To Relate Algebra Problems to real-life by solving word problems to an accuracy of 90%

 Navigate the following PowerPoint by clicking on the links that look like this: This Link will take you from one page to the next.  Occasionally, question slides will pop up. On these slides, answer the question. If your answer is correct, you’ll be sent to a “correct” slide, and will be able to continue on. If your answer is incorrect, you will be sent to an “incorrect” slide, and will either be directed to go back to the beginning of the content, or just sent back to the question slide to give another answer.

 The following pages and links combine to form the next unit. This unit is called Solving One-Variable Equations. The lesson goes in conjunction with Chapter 3 in your Algebra 1 book.  This lesson is extremely important in understanding math from now until you graduate from high school and eventually college, so please take this seriously.

 Before getting into actually solving the equations, we will first discuss the importance of symbols and variables.  SYMBOL - something that is used to represent something else  Go to the next slide and take note of the symbols!

 Note what the following pictures represent: America ReligionPirates

Money STOP The Number 4

a VARIABLE is just a letter that represents a number! For instance, x = 4 makes x a symbol that represents 4 in this example! All letters can be used as a symbol in algebra.

 Identity Property Says that 1 a = a  Commutative Property Says that a b = b a  Distributive Property Says that a(b + c) = ab + ac  Associative Property a (b c) = (a b) c

 Coefficient The number multiplied by the variable in a term Examples The Coefficient Here is 4. The Coefficient Here is 3.

 Constant The number added to or subtracted to the variable in a term  This number has no variable attached to it (like a coefficient does). Examples The Constant Here is -1. The Constant Here is 7.

 Equivalent Equations with the same solutions as the original one. Example: x + 7 and y – 3 both equal 5 in this case, making the expressions (x + 7) and (y – 3) equivalent.

 Reciprocal You will get 1 if you multiply a number by its reciprocal (or all of the terms will cancel out. Ex. a/b has the reciprocal b/a Ex2:  Are reciprocals of one another because:  3/2 times 2/3 = 1  And because one is the “inverse” of the other

 By the Associative Property, x (y z) equals which of the following x (y z) (x y z) (x y) z ((x ((y) z)))

Please Continue

Back to Content Back to Question

Now, let’s move on to actually SOLVING these equations!

It is easiest to proceed by example. Try the following example: Solve x – 5 = -13 To solve this equation, you want to get x by itself. How might you do this?

 Solve x – 5 = -13  x – 5 = -13  Rewrite the original equation  x – =  Add 5 to each side.  x = -8.  Simplify, and our solution is x = -8.

 Now Check Your Answer  x – 5 = -13  Rewrite the original equation  = -13  Substitute -8 in for x.  -13 = -13  The Solution is Correct!

 Did you understand what was meant by the word “simplify?”  Definition: Simplify  To Take an equation down to the simplest terms.  Ex: we had x – = in the earlier example.  Obviously = 0, so those terms essentially CANCEL.  Additionally, = -8  Therefore you could write x – = more simply as x = -8.

 Now Try Another Example!  Solve r + 3 = 2  r + 3 = 2  Rewrite the original equation  r + 3 – 3 = 2 – 3  Subtract 3 from each side.  r = -1.  Simplify, and our solution is r = -1.

 Now Check Your Answer  r + 3 = 2  Rewrite the original equation  = 2  Substitute -1 in for r.  2 = 2  The Solution is Correct!

 x + 5 = 29 x = 24 x = 34 x = 5.4 x = 22

Please Continue

Back to Content Back to Question

Again, it is best to proceed by example. Try the following example: Solve -4x = 16 You want to get x by itself. What would you do?

 Solve -4x = 16  -4x = 16  Rewrite the original equation  -4x =  Divide each side by -4  x = -4.  Simplify, and our solution is x = -4.

 Now Check Your Answer  -4x = 16  Rewrite the original equation  -4 (-4) = 16  Substitute -4 in for x.  -16 = -16  The Solution is Correct!

 Try Another! Solve ½x = 5  ½x = 5  Rewrite the original equation  ½x (2) = 5 (2)  Multiply each side by 2.  x = 10.  Simplify, and our solution is x = 10.

 Now Check Your Answer  ½x = 5  Rewrite the original equation  ½(10) = 5  Substitute 10 in for x.  5 = 5  The Solution is Correct!

 3x= 12 x = 36 x = 9 x = 15 x = 4

Back to Content Back to Question

Please Continue

Multi-Step Equations are equations that include solving by addition and/or subtraction as well as by multiplication and/or division. Once again, we will proceed by example. Try the following: Solve 7x + 6 = -8

 We will start manipulating the equation by doing the addition and subtraction part.  7x + 6 = -8  Rewrite the original equation  7x + 6 – 6 = -8 – 6  Subtract 6 from each side.  7x = -14  Simplify.

 Now we’ll do the multiplication part. Remember, from the last slide we were left with 7x = -14  7x =  Divide each side by 7  x = -2.  Simplify once more, and our solution is x = -2.

 Now Check Your Answer  7x +6 = -8  Rewrite the original equation  7(-2) +6 = -8  Substitute -2 in for x.  = -8  Distribute (multiply the 7 and -2).  -8 = -8  The Solution is Correct!

 Now Try Another Example  Solve 7x – 3x – 8 = 24  7x – 3x – 8 = 24  Rewrite the original equation  Notice that there are 2 “x” terms.  4x – 8 = 24  Combine Like Terms  4x – =  Add 8 to each side.

 4x = 32  Simplify.  4x = 32 4  Divide each side by 4.  x = 8.  Simplify once more to get x = 8.

 Now Check Your Answer.  7x – 3x – 8 = 24  Rewrite the original equation  7(8) – 3(8) – 8 = 24  Substitute 8 in for EACH x.  56 – 24 – 8 = 24  Distribute.  24 = 24  The Solution is Correct!

 5 – 4x = 8 – x x = 5 x = -1 x = 3 x = 8

Back to Content Back to Question

Please Continue

 A word problem is just a paragraph version of an equation.  The tricky part is to determine what the word problem means in equation terms.  Once you figure out how to do that part, it becomes either an addition/subtraction, multiplication/division problem or a multi-step problem.

 In order to solve word problems, the best way, again, is to proceed by example.  Try this one: Jason sells chocolate bars for his baseball team. Each chocolate bar costs 2 dollars. If at the end of the day he had 38 dollars, how many chocolate bars did Jason sell?

WORD PROBLEMHOW TO SOLVE  Jason sells chocolate bars for his baseball team. Each chocolate bar costs 2 dollars. If at the end of the day he had 38 dollars, how many chocolate bars did Jason sell?  First, identify all of the important facts/numbers in the problem.  We see that Jason sold each chocolate bar for 2 dollars and ended up with 38 dollars

WORD PROBLEMHOW TO SOLVE  Jason sells chocolate bars for his baseball team. Each chocolate bar costs 2 dollars. If at the end of the day he had 38 dollars, how many chocolate bars did Jason sell?  Now that we have these facts, we can derive the equation  Since each bar costs 2 dollars, then we know 2 will be in the problem somehow

WORD PROBLEMHOW TO SOLVE  Jason sells chocolate bars for his baseball team. Each chocolate bar costs 2 dollars. If at the end of the day he had 38 dollars, how many chocolate bars did Jason sell?  We want to MULTIPLY the 2 by our variable (call it x).  Therefore, a 2x will appear somewhere in the equation that we want to solve

WORD PROBLEMHOW TO SOLVE  Jason sells chocolate bars for his baseball team. Each chocolate bar costs 2 dollars. If at the end of the day he had 38 dollars, how many chocolate bars did Jason sell?  Finally, 38 is the result of our equation.  Combing the two together, we can derive the following equation: 2x = 38

 Now that we have our equation, we can solve it just like we did earlier.  2x = 38  Rewrite the original equation  2x = 38 2  Divide each side by 2  x = 19.  Simplify, and our solution is x = 19.

 Now Check Your Answer  2x = 38  Rewrite the original equation  2 (19) = 38  Substitute 19 in for x.  38 = 38  The Solution is Correct!

 The last component of a word problem is a “summary sentence”  Definition: Summary Sentence  A sentence that goes with word problems that sums up your answer in sentence or word form.  So for our example, x = 19 would be answered as follows: Jason sold a total of 19 chocolate bars.

 Now You Try:  Kelly has 15 DVDs in her case. After buying more DVDs at The Exchange, she now has 23. How many more CDs did Kelly buy? Kelly Bought 4 CDsKelly Bought 23 CDs Kelly Bought 15 CDsKelly Bought 8 CDs

Please Continue

Back to Content Back to Question

 Now we’ll try a more difficult word problem, a multi-step problem.  Solve this: Dan and Amber are adding up their ages. Dan is 2 times Amber’s age then minus 5. Amber is 9 years old. How old is Dan?

WORD PROBLEMHOW TO SOLVE  Dan and Amber are adding up their ages. Dan is 2 times Amber’s age then minus 5. Amber is 9 years old. How old is Dan?  First, identify all of the important facts/numbers in the problem.  We see that Amber is 9 years old.  We see that Dan is 2 times Amber’s age then minus 5.

WORD PROBLEMHOW TO SOLVE  Dan and Amber are adding up their ages. Dan is 2 times Amber’s age then minus 5. Amber is 9 years old. How old is Dan?  Since Dan is 2 times Amber’s age then minus 5, and Amber is 9, our equation is then 2x – 5 = 9

 Now solve the equation we just got.  2x – 5 = 9 Rewrite the original equation  2x – = Add 5 to each side  2x = Divide each side by 2  x = 7. Simplify, and our solution is x = 7.

 Now Check Your Answer  2x – 5 = 9  Rewrite the original equation  2 (7) – 5 = 9  Substitute 7 in for x.  14 – 5 = 9  Distribute (multiply the 7 and 2).  9 = 9  The Solution is Correct!

 Don’t forget! We have to write a summary sentence for this problem!  Our Summary Sentence is: Therefore, Amber is 9 years old and Dan is 7 years old.

 Now Try This problem: Mike sells cakes at a local bakery. He charges 10 dollars for each cake. At the end of the day, there was 180 dollars worth of cake sold. How many cakes did Mike sell? Mike Sold 18 CakesMike Sold 10 Cakes Mike Sold 170 CakesMike Sold 8 Cakes

Back to Content Back to Question

Please Continue

 Solving Equations with Fractions and Decimals in them is just the same as solving using addition, subtraction, multiplication, division or a combination, except instead of integers, fractions and decimals are used.

 What is a fraction? A fraction is a number that represents a part of a whole. These numbers are written with a forward slash or horizontal line.  Ex: ½, 2/3, 5/8,  NUMERATOR (the number on top)  DENOMENATOR (the number on bottom)

 What is a decimal? A decimal is also a number that represents part of a whole number. This time, it is represented with by a number after a period.  Ex:  1.4, 5.9,.08,.01763, etc  What is an integer? An integer is a whole number. It is every natural number, (0, 1, 2, 3,...) and every negative natural number (-1, -2, -3,...)

 Since you will solve these equations in the same way that you solved the other ones, we will proceed by example.  So, let’s try this example: Solve 0.2x + 5 = 15

 0.2x + 5 = 15  Rewrite the original equation  0.2x + 5 – 5 = 15 – 5  Subtract 5 from each side.  0.2x = 10  Simplify.  Keep GOING!!!

 0.2x =  Divide each side by 0.2  You Can Use Your Calculator if you need it!  x = 50.  Simplify once more, and our solution is x = 50.

 -1.2x - 3 = 9 x = 1 x = -1 x = 10 x = -10

Please Continue

Back to Content Back to Question

 Let’s try this example:  Solve ¼x + 6 = 7  ¼x + 6 = 7  Rewrite the original equation  ¼x + 6 – 6 = 7 – 6  Subtract 6 from each side.  ¼x = 1  Simplify.

 ¼x = 1  Rewrite the equation again  ¼x (4) = 1 (4)  Multiply each side by 4.  Note that 4 is the RECIPROCAL of ¼.  And remember two numbers that are reciprocals of one another when multiplied together leaves only 1, thus x is left here.  x = 4.  Simplify, and our solution is x = 4.

 ½x + 6 = 9 x = 1 x = 3 x = 6 x = 9

Please Continue

Back to Content Back to Question