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INSTRUCTORSCarleen EatonGrant Fraser
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Lecture Comments (2)

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Post by Dr Carleen Eaton on July 31, 2014

Darin, I cannot find this part of the lecture. Can you clarify at what time this occurs?

0 answers

Post by Darin Weaver on July 29, 2014

The fourth practice question is incorrect.  Going from step 12 to 13, it states that 5 + 9 = 13 resulting in the answer to the equation 5 + 9 = 2x being x = (13/2).

5 + 9 = 2x should result in x = 7.   (5 + 9 = 14, NOT 13)

Related Articles:

Solving Equations

  • A solution to an equation is a number that makes the equation true.
  • Equality satisfies the reflexive, symmetric, and transitive properties.
  • If both sides of an equation are increased, decreased, multiplied, or divided by the same number, the resulting equation is equivalent to the original one and has the same solutions.
  • Use these properties to solve a formula for a variable.
  • To solve a word problem, define a variable and create an equation. Then solve that equation.

Solving Equations

Write an algebraic expression for:
The product of five times the square of one number and the square root of another number.
  • First, asign variables to the unknown numbers
  • x for the square of one number
  • y for the square root of another number
  • The product means that these numbers will be multiplying
5x2g√y
Write an algebraic expression for
The sum of the square of a number and the cube of another number is equal to the square root of the product of the first and second number.
  • x for the square of a number
  • y for the cube of another number
  • This question is an equation, because it is stated that is equal to
  • Left side translates to: x2 + y3
  • Right side translates to: √{x ×y}
Putting them together we have: x2 + y3 = √{x ×y}
Write a verbal expression for x3 − y3 = (x − y)(x2 + 2xy + y2)
  • Break the equation into three parts, Left Side and First Part Right, Second Part Right
  • Left Side: The difference of the cube of a number and the cube of another number is equal to
  • First Part Right:(x − y) can be written as The quantity of the difference of the first number and the second number.
  • Second Part Right (x2 + 2xy + y2) can be written as: The quantity of the sum of the square of the first number and two times the product of the first and second number plus the second number squared.
  • Putting everything together you get:
The difference of the cube of a number and the cube of another number IS EQUAL to the product of the quantity of the difference of the first number and the second number AND
the quantity of the sum of the square of the first number and two times the product of the first and second number plus the second number squared.
Solve 2(x + 3) + 3(2x + 1) = 5(2x − 1)
  • First, use the distributive property to distribute the (2), (3) and (5)
  • 2(x) + 2(3) + 3(2x) + 3(1) = 5(2x) − 5(1)
  • Multiply whenever possible
  • 2x + 6 + 6x + 3 = 10x − 5
  • Combine like terms on the left side of the equation.
  • (2x + 6x) + (6 + 3) = 8x + 9 = 10x − 5
  • Subtract 8x from both sides of the equation
  • ( − 8x) + (8x) + 9 = 10x − 5 + ( − 8x)
  • Simplify by eliminating the negatives and the positives
  • 9 = 2x − 5
  • Add (5) on both sides of the equation.
  • (5) + 9 = 2x − 5 + (5)
  • 13 = 2x
  • Divide by 2 on both sides
x = [13/2]
The area of a trapezoid is given by A = [h/2](b1 + b2). Solve for h.
  • First, get rid of the 2 by multiplying both sides by 2
  • (2)A = [h/2](b1 + b2)(2)
  • This yields the following result
  • 2A = h ×(b1 + b2)
  • Rather than distributing the h, we'll divide both sides by (b1 + b2)
  • [2A/((b1 + b2))] = [(h ×(b1 + b2))/((b1 + b2))]
  • This leaves h by itself, therefore
h = [2A/((b1 + b2))]
The interior angle of a regular polygon is given by I = [((n − 2) ×180)/n] where n is the number of sides. Solve for n.
  • Multiply both sides by n the n in the numerator and denominator cancel out.
  • n ×I = [((n − 2) ×180)/n] ×n
  • n ×I = (n − 2) ×180
  • Next, distribute the 180.
  • n ×I = n ×180 − 360
  • Next, add 360 to both sides
  • 360 + n ×I = n ×180 − 360 + 360
  • Next, eliminate the positives and negatives.
  • 360 + n ×I = n ×180
  • Next, subtract n ×I from both sides
  • 360 + n ×I − n ×I = n ×180 − n ×I
  • 360 = n ×180 − n ×I
  • Next, factor out an n
  • 360 = n ×(180 − I)
  • Next, divide both sides by (180 − I)
  • [360/(180 − I)] = [(n ×(180 − I))/(180 − I)]
  • Next, cross out (180 − I) from the top and bottom to get the result
The number of sides of a regular polygon given the interior angle I is n = [360/(180 − I)]
Solve: − 8 + 8( − 7 − 5b) = − 8(1 + 5b) + 7b
  • Distribute using the distributive property
  • − 8 − 56 − 40b = − 8 − 40b + 7b
  • Combine like terms
  • − 64 − 40b = − 8 − 33b
  • Move variable to one side of the equation
  • − 56 = 7b
  • Solve
b = − 8
Solve: − 3(8 + x) + 6x = 7( − 4x + 1)
  • Distribute
  • − 24 − 3x + 6x = − 28x + 7
  • Combine like terms
  • − 24 + 3x = − 28x + 7
  • Isolate the variables to one side if the equation
  • − 31 = − 31x
  • Solve
x = 1
Solve: 2(n + 3) = 7(3 + n)
  • Distribute using the distributive property
  • 2n + 6 = 21 + 7n
  • Move variable to one side of the equation
  • − 15 = 5n
  • Solve
n = − 3
Solve: − 7(1 − 4v) = 7(5v + 5)
  • Distribute
  • − 7 + 28v = 35v + 35
  • Isolate the variables to one side if the equation
  • − 42 = 7v
  • Solve
v = − 6

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Solving Equations

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Translations 0:06
    • Verbal Expressions and Algebraic Expressions
    • Example: Sum of Two Numbers
    • Example: Square of a Number
  • Properties of Equality 3:20
    • Reflexive Property
    • Symmetric Property
    • Transitive Property
    • Addition Property
    • Subtraction Property
    • Multiplication Property
    • Division Property
  • Solving Equations 6:58
    • Example: Using Properties
  • Solving for a Variable 8:25
    • Example: Solve for Z
  • Example 1: Write Algebraic Expression 10:15
  • Example 2: Write Verbal Expression 11:31
  • Example 3: Solve the Equation 14:05
  • Example 4: Simplify Using Properties 17:26

Transcription: Solving Equations

Welcome to Educator.com.0000

In today's lesson, we are going to be talking about solving equations.0002

Recall that verbal expressions can be translated into algebraic expressions, and vice versa.0007

So, you can go from verbal expressions to algebraic, and from algebraic back to verbal.0014

For example, if I were given a sentence (a verbal expression) such as, "The sum of two numbers is equal to 7,"0019

I could translate that into an algebraic expression.0041

And the technique is to first assign variables; so I am looking, and I see that I have two numbers.0043

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.0049

This is actually an equation, not just an expression; so if I see that I have an equal sign (like this is equal),0057

I note where that is, because that is going to divide it into the left and right sides of the equation.0064

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."0068

So, sum means adding; and my two numbers I am going to call x and y; "is equal to 7."0075

And then, double-check: the sum of two numbers (x + y--that works out) is equal to (I have that taken care of) 7.0084

Or another example: A number is 5 less than the square of another number.0094

And you need to be careful with this "is less than," because it is easy to get that backwards.0113

So, first I notice that I have two numbers; I have "a number" and "another number."0118

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.0123

Now, it says "a number is," so I have "is"--that tells me where the equal sign is.0132

So, the left side of the equation is just a number.0141

With "5 less than" or "something less than," the temptation is sometimes to say "5 minus."0146

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.0152

So, a number is, and then 5 less than...what is it 5 less than? the square of another number.0161

I am calling my other number y, so y2 - 5, not 5 - y2.0168

So, checking it: A number (x) is (=) 5 less than the square of another number (y2 - 5).0176

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.0188

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).0201

The first property is the reflexive property: the reflexive property states that, for every real number a, a is equal to a.0209

So, a number is equal to itself.0219

The symmetric property states that, for all real numbers, if a equals b, then b equals a.0222

So, I can switch the left and right sides of the equation, and I am not going to change anything fundamental.0233

So, those are the reflexive and symmetric properties.0239

Another property is the transitive property: the transitive property states that, if a equals b, and b equals c, then a equals c.0241

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 5.0257

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 5.0270

So, I go back; then 5 equals 5, or a equals c--it just follows through.0277

OK, that is the first three properties discussed up here.0284

But equality also satisfies the addition, subtraction, multiplication, and division properties.0288

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.0293

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.0299

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.0309

And these 5's cancel out; that will give me x = 15.0325

So, the resulting equation after I add 5, if I add 5 to both sides--this equation is also true.0329

The subtraction principle is the same idea, that if I have an equation (x + 3 = 5), I can subtract 3 from both sides;0338

I can subtract the same number from both sides, and the resulting equation is also true.0357

Multiplication principle: again, this is something you have used before to solve equations.0363

And it states that, if the same number is multiplied by both sides of the equation, the resulting equation is true.0367

So, I might have x/2 = 12; to solve that, I need to multiply both sides by 2; and that is allowable.0376

Division: the same idea--in this case, I may have 4x = 6; to solve that, I am going to divide both sides by 4.0391

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.0406

OK, knowing those properties and the properties of real numbers (the properties of equality and the properties of real numbers), you can solve equations.0417

And we are going to use these properties frequently, even if you don't always know them by name.0427

You know how to do them, and you are going to be applying them throughout the course.0432

For example, if I have 19 = 3x + 4, since usually we have the variable on the left,0437

I can apply the symmetric property and just change this to 3x + 4 = 19.0447

That is the symmetric property that allows me to flip around the two sides of the equation.0451

Next, I need to isolate x, so I am going to subtract 4 from both sides.0459

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 = 15.0465

Well, now I need to isolate x with one last step; and I am going to do that by using the division property.0480

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 = 5.0487

So, by applying those principles, I was able to solve this simple equation.0499

Sometimes, in algebra, you will be asked to solve for a variable.0505

These properties can be used to solve a formula for a variable.0509

I may be given a formula such as 4x - 2y + z = 8.0514

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.0521

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.0532

So again, I am going to apply these same principles.0541

First, I am going to use the subtraction principle; and I can start out by subtracting 4x from both sides.0545

That is going to give me...these 4x's will cancel out; -2y + z = 8 - 4x.0561

Now, I need to add 2y to both sides, because I had -2y; so then, I am going to add 2y to both sides,0570

so that my 2y's cancel out to give me z = 8 - 4x + 2y.0590

And we usually write this so that we have the x's, then the y (in alphabetical order), and then the constant.0600

So, z equals -4x + 2y + 8; this is solving for z in terms of x and y.0607

First example: write an algebraic expression for the sum of three times the cube of one number and twice the square of another number.0617

OK, starting out, I am figuring out how many variables I have.0627

The sum of three times the cube of one number and twice the square of another number...x an y.0632

I am going to assign those as my variables, because I have two variables.0640

And there is no equal sign; this is an expression, not an equation, so I don't have an equal sign in it.0643

And it says "the sum," so I am doing addition.0651

"The sum of three times the cube of one number"--that is 3x3,0656

"and"--so that tells me that that is where the addition goes--"twice the square of another number";0663

so here is the other number: "twice the square of the other number."0670

Checking that: the sum of three times the cube of one number (3x3) and (+) twice--two times the square of another number (2y2).0675

Next, we are asked to do the opposite, which is to write a verbal expression for this algebraic expression.0692

OK, so looking at what I have: here, first, I have the difference of two squares;0699

and I am also looking and seeing that I have two variables.0706

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)...0709

Next, I have an equal sign: so "is equal to"...0737

the sum of the square (we have sum--we are adding), and I want to make it clear0743

that we are talking about the same number here and here, so I am going to say "the square of the first number,"0759

and (I am adding again) 3 times (multiplying three times x times y) the product of0770

(my first number and my second number) the first number and the second number, plus the square of the second number.0787

OK, so this was quite long, but you can check it.0820

"The difference of the square of a number and the square of another number" (I have that right here)0823

"is equal to" (there is my equal sign) "the sum" (it's a sum) "of the square of the first number,0827

and three times the product of the first number and the second number, plus the square of the second number."0833

So, it is pretty long, but we got everything covered clearly.0839

The third example is to solve: and we can use the properties that we discussed today in order to solve this.0845

First, I am going to need to get rid of these parentheses; and I can do that by using the distributive property.0852

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 -9.0858

So, just checking: I have -2x + -2(-8) + -3(9) + -3(-2x) = 4(2x) + 4(-9).0892

Multiplying everything out to simplify: -2x; and then -2 times -8 gives me +16.0909

And then, I have -3 times 9; that is actually a negative, -27.0919

-3 times -2x; that is positive, +6x.0928

That equals...4 times 2x is 8x; 4 times -9 is -36.0934

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.0941

I have constants: I have 16 - 27, which is going to give me -11.0951

I can't really do anything further with that right side.0957

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.0961

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 -25.0969

Next, I am going to subtract 8x from both sides--again, trying to get like terms together and isolate the x.0984

4x - 8x is going to give me -4x; equals...these cancel; that is -25.0994

Now, I am going to divide both sides by -4 to get x = -25/-4.1003

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 1/4.1011

So again, starting out, getting rid of the parentheses by using the distributive property and carefully multiplying out each section got me to here.1020

We're adding like terms to get to this step, then isolating the x through using the addition property,1031

the subtraction property, and the division property to get x = 6 1/4.1039

OK, in this example, we are asked to solve for h.1047

And we are given...this is actually the surface area of a cylinder; the surface area of a cylinder1050

equals 2π, times the radius, times the height, plus 2πr2.1055

So, I need to solve for h in terms of these other variables.1061

And I am going to start out by using the symmetric property to rewrite this with the h on the left.1069

Let me rewrite this as 2πrh + 2πr2 = s.1077

To get the h by itself, I can start out by subtracting 2πr2 from both sides.1083

OK, once I have done that, then this is going to be gone from this side; and I will have 2πrh.1095

2πrh = s - 2πr2.1106

In order to get the h by itself, I need to divide both sides by 2πr.1111

2πr will cancel out, leaving h isolated on the left, and s - 2πr2 all divided by 2πr.1121

And you really can't go any farther with this.1130

I started out by rewriting this with the value, the h, which we are trying to isolate on the left,1132

using the subtraction property and the division property to isolate the h.1140

That concludes this session of Educator.com, and I will see you next lesson.1147