Linear Equations - Video Tutorials & Practice Problems

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Introduction to Solving Linear Equations

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Everyone. In the previous chapter, we worked with linear expressions, some combination of numbers variables and operations like two X plus three. Now, if I take that linear expression and I say that it needs to be equal to something, let's say this two X plus three needed to equal five. I now have a linear equation. Now, equations are gonna be a really important important part of this course because we're gonna have to solve a bunch of different types of A. But don't worry, we're gonna use a lot of the skills that we learned with linear expressions and I'm gonna walk you through solving these step by step. So like I said a linear expression, if I just take it and add an equal sign, it is now a linear equation. Now it's called a linear equation because it is equal to something. If it's equal to something, it's an equation. Now, when we worked with linear expressions, we would always simplify or evaluate it. For some known value of X, we would be given that X is equal to a particular number like four. And we would simply replace X in our expression with that value and get an answer. Now with linear expressions, we don't know what X needs to be and we're tasked with solving for that unknown value of X. We have no idea what it has to be, but we want to find our value of X that makes our statement true. So for this particular equation, I would want to find a value of X that makes two X plus three equal five. Now, if I were to just guess what this X needed to be, let's say that that's how I wanted to solve it. I might guess that X is equal to zero. I could check that by saying two times zero plus three, but two times zero is zero. I would just get three, that's not equal to five. So X equals zero wouldn't be my answer here. Then I could go and say X equals one maybe. So if I plug that in two times one plus three, two times one is two less that three. This actually does give me five, which is great, but we probably don't want to do that for every linear equation that we're getting because that would honestly just become really annoying. So we're gonna need to use some other skills here in order to solve our linear equations. So you're actually gonna need to use all of the different operations in your toolbox, addition, subtract multiplication and division in order to isolate X to get it by itself. Now, these operations should always be done to both sides of the equation. This is super important and it's gonna be important throughout this course, whatever you do to one side of the equation you have to do to the other. So let's get some practice with isolating X. In this example, we want to identify and perform the operation needed to isolate X by applying it again to both sides. So looking at my first example, I have X plus two equals zero. Now this two is being added to the X. So how do I get rid of it in order to get X by itself? Well, the opposite of adding to would be subtracting two. So to get rid of it, I could subtract two and I need to do that to both sides. So on my left side here, it would cancel and I'm just left with X equals zero minus two, gives me negative two. And I have isolated X. Looking at our second example, I have three, X equals 12. Now here my three isn't being added to the X. It's actually multiplying it. So what operation could I do in order to get rid of that three? That's multiplying my X. Well, the opposite of multiplication is division. So I could go ahead and divide by three on both sides to cancel it out. And again, I'm just left with X equals 12, divided by three, gives me four. So I've isolated X here. Now you might have noticed that for these, we were doing opposite operations. If ever, I want to get rid of something in order to isolate X, I'm always gonna do the opposite operation of whatever is happening in the equation. So when we saw something being added, the opposite operation to get rid of that was subtraction. And we, when we saw something being multiplied, the opposite operation was division. Now this of course, also works the other way. So if I see something being subtracted, I could add it in order to get rid of it. And if I see something being divided, I could always multiply to get rid of it. But for these, we only saw one operation needed in order to isolate X. And you're often going to actually have to do multiple operations in order to solve a linear equation. So let's take a look at that. In this example, I want to solve the equation two times X minus three equals zero. So let's go ahead and take a look at our steps. Our very first step is to distribute our constants and looking at my equation here, I have this too that needs to get distributed to both the X and the negative three. So if I do that, I get two X and then two times negative three gives me negative six equals zero. So step one is done. Now step two asks me to combine like terms. So taking a look at my equation, I have two X minus six equals zero. I don't have any like terms here because I just have an X term and I have a constant, I can't combine that any further. So step two is also done. Now, in step three, we want to group terms with X and our constants on opposite sides. Now it doesn't matter which side, I put my X terms and which side I put my constants on as long as they're on opposite sides. So let's go ahead and do that. I have two X minus six equals zero. So I want to pull this six over to get it on the opposite side. So in order to do that, I need to do my opposite operation here, which is gonna be adding six to both sides, it will then cancel there. And I'm left with two X equals zero plus six gives me six. So step three is done. Now, step four is to isolate X. You might also hear this called solve for X. So let's go ahead and isolate X, I have my two multiplying my X here. So that means to get rid of that too. I need to divide. So if I divide both sides by two, my two will cancel here. I am left with X equals six, divided by two, gives me three. And this is actually the answer. So I'm done with step four and three. My solution here or my answer here is called the solution or the root of the equation. You might hear it called either an answer a solution or a root. Any of these are referring to the same thing. So we actually do have one more step here. Step number five. And that is to check our solution by replacing X in our original equation. So I'm gonna take my original equation two X minus three equals zero and make sure that I found the value of X that makes that true. So I take two, the value I got for X was three minus three equals zero. Now this then gives me 23 minus three is zero equals zero. And we know that anything times zero is zero. So I am left with zero equals zero, which is definitely a true statement. So I've completed solving this linear equation. That's all for this one guys. Thanks for watching.

2

Problem

Problem

Solve the Equation. $3\left(2-5x\right)=4x+25$

A

$x=-27$

B

$x=-1$

C

$x=1$

D

$x=16$

3

concept

Solving Linear Equations with Fractions

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4m

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Hey, everyone, as you solve a bunch of different linear equations, you may come across a linear equation that has fractions in it. Now, I know that fractions can be scary and sometimes a little bit challenging to work with. But don't worry, we're gonna get rid of these fractions as quickly as possible and get back to a linear equation that looks exactly like what you've already been solving. So when we see a linear equation with fractions, we need to eliminate those fractions using the least common denominator. So let's look at that in an example. So we want to solve this equation and it says 1/4 times X plus two minus one third X is equal to 112. Now, this equation definitely has fractions in it and I wanna get rid of those as soon as possible. So here we're actually adding a step zero before we do anything. And our step zero is gonna be to multiply by our L CD, our least common denominator in order to eliminate our fractions. So looking at my equation here, my denominators are 43 and 12. So my least common denominator here is going to be 12. So to get rid of those fractions, I need to multiply my entire equation by 12. When we multiply by our least common denominator, we want to make sure to distribute it to every single term in our equation. So let's go ahead and simplify this. I have 12/4 times X plus two minus 12/3 times X equals 12/12. Now, we can even simplify this a little bit further. So if I take 12, divided by four, that gives me three times X plus two minus or X equals one, now all of my fractions are gone and I can continue solving this just as I would any other linear equation. So my step zero is done, I can move on to step one which is to distribute our constants. Now, looking at this, I have this three here that needs to get distributed to both the X and my two. So that gives me three X plus six minus four X. And the right side of my equation is gonna stay the same. I don't have anything to distribute there. So step one is done. Now, step two is to combine like terms. So looking at my equation, I have a couple of like terms I have three X and negative four X and those are gonna combine to give me negative one X, everything else is gonna stay the same. So I still have that plus six equals one. So step two is done, I have combined my like terms. Now looking at step three, I want to group my terms with X on one side and constants on the other to get them on opposite sides. So I'm gonna go ahead and move this six over to the other side in order to get all my constants on one side. So in order to do that, I'm just going to subtract six from both sides And here it will cancel and my negative one X stays there. And then I have one minus six which will give me negative five. So I've completed step three, I have moved my constants moved my uh X terms to be on opposite sides. And now I want to do step four which is to isolate X. Now I just have a negative one multiplying my X. So in order to get rid of that negative one, I need to divide by it on both sides. So we'll cancel over here and I am left with X equals negative five divided by negative one, gives me positive five. And this is my solution. So I've completed step number four. Now step number five again is to check by placing X in our original equation. Now, when we have fractions in our equation, this can get a little bit complicated and sometimes be time consuming. So this step is optional. But remember you can always put it back in your original equation to double check that's all for this one guys. Thanks for watching.

4

Problem

Problem

Solve the equation.

$\frac92+\frac14\left(x+2\right)=\frac34x$

A

$x=10$

B

$x=4$

C

$x=8$

D

$x=-1$

5

concept

Categorizing Linear Equations

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6m

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Hey everyone. So we know how to solve linear equations. But once we solve a linear equation, we can actually then place it in a category. So there are three possible categories that linear equations can be put into based on how many solutions they have. Now these solutions may be written as a solution set which thinking back to set notation, this would just be our solution written in between curly brackets. And that would be our solution set. Let's go ahead and jump into an example. So we want to solve and then categorize these linear equations. So looking at my first example, I have two X plus four equals 10. So my first step here is gonna be to get all of my constants on one side, all of my X terms on the other. So to move that four over, I need to go ahead and subtract it from both sides. So I'll cancel over here and I'm gonna be left with two X equals 10 minus four, which gives me six. My next step here is to isolate that X variable. So, so to do that here, I'm going to divide by two on both sides, remember whatever I do to one side of the equation I have to do to the other. So my two is going to be canceled and I'm left with X equals six divided by two, which is three. So this is my solution. And for this particular linear equation, I only have one solution, the only value that would solve this and I could plug back in to get a true statement is three. So my solution set is simply that number three in curly brackets. Now all of the linear equations we've seen up to this point have had one solution. And so their solution set would just be written as whatever number I get for X inside of those curly brackets. Now when I only have one solution, this is called a conditional linear equation. And the reason it's called conditional is because it is only true on the condition that X equals some number. So for this particular linear equation, it is only true on the condition that X is equal to three. So it's a conditional equation. Now all of our linear equations up until this point, like I said, have only had one solution. So let's take a look at some other possibilities. So looking at my next linear equation, I have X plus five equals X plus two plus three. So let's go ahead and simplify that. So on this side, I just have X plus five and then equals X and then this two and this three are both constant. So I can go ahead and combine them two and three together is going to give me five. So you might see where this is going. But let's take this a step further. So if I want to move all of my constants to one side and all of my X terms to the other, I need to go ahead and subtract my five over here to cancel it out. Whatever I do to one side, I have to do to the other. And then to move my X over, I would need to subtract X from both sides. So what happens here is I then have X minus X which is zero equals five minus five, which is also zero. So I just get zero equals zero, which is an undeniably true statement. Zero is always equal to zero. So this time us that I could plug in any value for X at all and end up with a true statement. So I actually have an infinite number of solutions here because no matter what value I plug in for X, it will always be true. So that means that my solution set is actually all the real numbers. So all real numbers I could plug back into my equation and get a true statement and we denote real numbers or all real numbers with this fancy R just with an extra line in it. So when my, when I have an infinite number of solutions. This is actually called an identity equation. So it's an identity equation because it's always going to be true no matter what infinite solutions identity equation. Let's look at our final possibility. So over here I have X equals X plus four. So if I wanna go ahead and get all of my X terms on one side with all of my constants on the other, I need to subtract X from this side. It will cancel here. Whatever I do to one side, I have to do to the other. So I'm left with X minus X. This gives me zero. Then on the right side of my equation, I'm left with four. So I end up with zero equals four. Now this is definitely not a true statement. So what I'm left with here is just a completely false statement. So what this tells me is that no matter what value I plug in for X, it's just gonna end up giving me something a little outlandish. A little crazy like zero equals four, which we know is not true. So I actually have no solution here. There is no value that I could plug in for X to make this statement true. So my solution set is actually going to be written still in curly brackets with a zero with a slash through it. And this is called the empty set because there is nothing inside that set. There is no number that I could plug into that equation to make it true. Now, when this happens, when I have no solutions, this is going to be called an inconsistent equation. And you might also hear it called as a contradiction because it gives you something crazy like zero equals four or seven equals 10 or something that you know is not true. So this is the last type or last category of our linear equations. Our first one was conditional. And when we have a one solution, that is a conditional linear equation. Now, when we have infinite solutions, we have our second category which is an identity equation. And then my very last one, if I have no solutions, that's my third type. It is an inconsistent equation. That's all for this one. Let me know if you have questions.

6

Problem

Problem

Solve the equation. Then state whether it is an identity, conditional, or inconsistent equation.

$5x+17=8x+12-3\left(x+4\right)$

A

Identity

B

Conditional

C

Inconsistent

7

Problem

Problem

Solve the equation. Then state whether it is an identity, conditional, or inconsistent equation. $\frac{x}{4}+\frac16=\frac{x}{3}$

A

Identity

B

Conditional

C

Inconsistent

8

Problem

Problem

Solve the equation then state whether it is an identity, conditional, or inconsistent equation.