WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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In today's lesson, we are going to be talking about solving equations.
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Recall that verbal expressions can be translated into algebraic expressions, and vice versa.
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So, you can go from verbal expressions to algebraic, and from algebraic back to verbal.
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For example, if I were given a sentence (a verbal expression) such as, "The sum of two numbers is equal to 7,"
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I could translate that into an algebraic expression.
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And the technique is to first assign variables; so I am looking, and I see that I have two numbers.
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And they don't tell me what the numbers are, so I need to assign variables; and I am going to assign x and y.
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This is actually an equation, not just an expression; so if I see that I have an equal sign (like this is equal),
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I note where that is, because that is going to divide it into the left and right sides of the equation.
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So, I have an equal sign; I am going to first deal with one side of the equation, and it says "the sum of two numbers."
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So, sum means adding; and my two numbers I am going to call x and y; "is equal to 7."
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And then, double-check: the sum of two numbers (x + y--that works out) is equal to (I have that taken care of) 7.
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Or another example: A number is 5 less than the square of another number.
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And you need to be careful with this "is less than," because it is easy to get that backwards.
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So, first I notice that I have two numbers; I have "a number" and "another number."
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Once again, I am going to use x for one of the numbers; and for the other number, I am going to use y; so x and y are my variables.
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Now, it says "a number is," so I have "is"--that tells me where the equal sign is.
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So, the left side of the equation is just a number.
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With "5 less than" or "something less than," the temptation is sometimes to say "5 minus."
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But that is not correct; what this is saying is to look at what comes after (the square of another number) and subtract 5 from that.
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So, a number is, and then 5 less than...what is it 5 less than? the square of another number.
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I am calling my other number y, so y² - 5, *not* 5 - y².
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So, checking it: A number (x) is (=) 5 less than the square of another number (y² - 5).
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And later on in the examples, we will also see a situation where we are given an algebraic expression and asked to translate it into a verbal expression.
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As you are solving equations, you need to keep in mind the properties of equality that you have used previously (but it is good to review).
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The first property is the **reflexive property**: the reflexive property states that, for every real number a, a is equal to a.
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So, a number is equal to itself.
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The **symmetric property** states that, for all real numbers, if a equals b, then b equals a.
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So, I can switch the left and right sides of the equation, and I am not going to change anything fundamental.
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So, those are the reflexive and symmetric properties.
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Another property is the **transitive property**: the transitive property states that, if a equals b, and b equals c, then a equals c.
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Think of it this way: if a actually was 5 (let a equal 5), if I told you that a equals b, then the only way it is going to equal b is if b is also 5.
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So, b must be 5; and then, if I went on and said, "OK, well, b equals c," if that is 5, then c must also be 5.
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So, I go back; then 5 equals 5, or a equals c--it just follows through.
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OK, that is the first three properties discussed up here.
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But equality also satisfies the addition, subtraction, multiplication, and division properties.
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And these are properties that you have used before in Algebra I; and again, you can review these in detail in the Algebra I lessons.
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To refresh them here: the **addition principle** says that, if the same number is added to both sides of the equation, the resulting equation is true.
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For example, if I have x - 5 = 10, and I want to solve for x, the way I am going to do that is: I am going to add 5 to both sides.
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And these 5's cancel out; that will give me x = 15.
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So, the resulting equation after I add 5, if I add 5 to both sides--this equation is also true.
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The subtraction principle is the same idea, that if I have an equation (x + 3 = 5), I can subtract 3 from both sides;
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I can subtract the same number from both sides, and the resulting equation is also true.
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Multiplication principle: again, this is something you have used before to solve equations.
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And it states that, if the same number is multiplied by both sides of the equation, the resulting equation is true.
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So, I might have x/2 = 12; to solve that, I need to multiply both sides by 2; and that is allowable.
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Division: the same idea--in this case, I may have 4x = 6; to solve that, I am going to divide both sides by 4.
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So, the important thing is: if you do something to one side, you need to do the same thing to the other side of the equation.
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OK, knowing those properties and the properties of real numbers (the properties of equality and the properties of real numbers), you can solve equations.
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And we are going to use these properties frequently, even if you don't always know them by name.
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You know how to do them, and you are going to be applying them throughout the course.
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For example, if I have 19 = 3x + 4, since usually we have the variable on the left,
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I can apply the symmetric property and just change this to 3x + 4 = 19.
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That is the symmetric property that allows me to flip around the two sides of the equation.
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Next, I need to isolate x, so I am going to subtract 4 from both sides.
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And that utilizes the subtraction property, which is going to give me...subtracting 4 from both sides, I get 3x; the 4's cancel out, and I get 3x = 15.
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Well, now I need to isolate x with one last step; and I am going to do that by using the division property.
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If I divide both sides by 3, I am using the division property; and that is going to give me...the 3's will cancel out, and I will get x = 5.
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So, by applying those principles, I was able to solve this simple equation.
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Sometimes, in algebra, you will be asked to solve for a variable.
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These properties can be used to solve a formula for a variable.
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I may be given a formula such as 4x - 2y + z = 8.
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And I might be asked to solve for z; so what I am going to do is solve for z in terms of x and y.
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While I may not figure out the actual numerical value of z, I can solve for it; I can isolate it and solve for it in terms of the other variables.
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So again, I am going to apply these same principles.
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First, I am going to use the subtraction principle; and I can start out by subtracting 4x from both sides.
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That is going to give me...these 4x's will cancel out; -2y + z = 8 - 4x.
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Now, I need to add 2y to both sides, because I had -2y; so then, I am going to add 2y to both sides,
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so that my 2y's cancel out to give me z = 8 - 4x + 2y.
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And we usually write this so that we have the x's, then the y (in alphabetical order), and then the constant.
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So, z equals -4x + 2y + 8; this is solving for z in terms of x and y.
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First example: write an algebraic expression for the sum of three times the cube of one number and twice the square of another number.
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OK, starting out, I am figuring out how many variables I have.
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The sum of three times the cube of one number and twice the square of another number...x an y.
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I am going to assign those as my variables, because I have two variables.
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And there is no equal sign; this is an expression, not an equation, so I don't have an equal sign in it.
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And it says "the sum," so I am doing addition.
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"The sum of three times the cube of one number"--that is 3x³,
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"and"--so that tells me that that is where the addition goes--"twice the square of another number";
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so here is the other number: "twice the square of the other number."
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Checking that: the sum of three times the cube of one number (3x³) and (+) twice--two times the square of another number (2y²).
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Next, we are asked to do the opposite, which is to write a verbal expression for this algebraic expression.
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OK, so looking at what I have: here, first, I have the difference of two squares;
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and I am also looking and seeing that I have two variables.
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So, the difference of the square of a number (x--I am just calling it "a number") and the square of another number (that is my y²)...
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Next, I have an equal sign: so "is equal to"...
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the sum of the square (we have sum--we are adding), and I want to make it clear
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that we are talking about the same number here and here, so I am going to say "the square of the first number,"
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and (I am adding again) 3 times (multiplying three times x times y) the product of
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(my first number and my second number) the first number and the second number, plus the square of the second number.
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OK, so this was quite long, but you can check it.
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"The difference of the square of a number and the square of another number" (I have that right here)
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"is equal to" (there is my equal sign) "the sum" (it's a sum) "of the square of the first number,
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and three times the product of the first number and the second number, plus the square of the second number."
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So, it is pretty long, but we got everything covered clearly.
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The third example is to solve: and we can use the properties that we discussed today in order to solve this.
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First, I am going to need to get rid of these parentheses; and I can do that by using the distributive property.
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So, -2 times x gives me -2x; plus -2, times -8, minus...actually, let's do plus -3 times 9, plus -3 times -2x, equals 4 times 2x plus 4 times -9.
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So, just checking: I have -2x + -2(-8) + -3(9) + -3(-2x) = 4(2x) + 4(-9).
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Multiplying everything out to simplify: -2x; and then -2 times -8 gives me +16.
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And then, I have -3 times 9; that is actually a negative, -27.
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-3 times -2x; that is positive, +6x.
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That equals...4 times 2x is 8x; 4 times -9 is -36.
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The next step is to combine like terms; and I have -2x and 6x--those can be combined; that is going to give me 4x.
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I have constants: I have 16 - 27, which is going to give me -11.
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I can't really do anything further with that right side.
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Now that I am this far, what I want to do is isolate the x (isolate the variable); I can do that by using my addition property.
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I am going to add 11 to both sides; the 11's cancel out--that is going to leave me with 4x = 8x, and -36 + 11 is -25.
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Next, I am going to subtract 8x from both sides--again, trying to get like terms together and isolate the x.
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4x - 8x is going to give me -4x; equals...these cancel; that is -25.
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Now, I am going to divide both sides by -4 to get x = -25/-4.
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And the negative and the negative is a positive, so x = 25/4; I can either leave it like that, or I can go on and say this equals 6 and 1/4.
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So again, starting out, getting rid of the parentheses by using the distributive property and carefully multiplying out each section got me to here.
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We're adding like terms to get to this step, then isolating the x through using the addition property,
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the subtraction property, and the division property to get x = 6 1/4.
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OK, in this example, we are asked to solve for h.
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And we are given...this is actually the surface area of a cylinder; the surface area of a cylinder
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equals 2π, times the radius, times the height, plus 2πr².
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So, I need to solve for h in terms of these other variables.
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And I am going to start out by using the symmetric property to rewrite this with the h on the left.
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Let me rewrite this as 2πrh + 2πr² = s.
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To get the h by itself, I can start out by subtracting 2πr² from both sides.
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OK, once I have done that, then this is going to be gone from this side; and I will have 2πrh.
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2πrh = s - 2πr².
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In order to get the h by itself, I need to divide both sides by 2πr.
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2πr will cancel out, leaving h isolated on the left, and s - 2πr² all divided by 2πr.
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And you really can't go any farther with this.
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I started out by rewriting this with the value, the h, which we are trying to isolate on the left,
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using the subtraction property and the division property to isolate the h.
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That concludes this session of Educator.com, and I will see you next lesson.