Linear Equations - Online Tutor, Practice Problems & Exam Prep

Linear equations, formed by adding an equal sign to linear expressions, can be solved using operations like addition, subtraction, multiplication, and division to isolate the variable. Equations can be categorized into three types: conditional equations with one solution, identity equations with infinite solutions, and inconsistent equations with no solutions. For example, solving \(2x + 4 = 10\) yields \(x = 3\), while \(x + 5 = x + 2 + 3\) results in \(0 = 0\), indicating infinite solutions. Understanding these concepts is crucial for mastering algebraic problem-solving.

Introduction to Solving Linear Equtions

Hey, everyone. In the previous chapter, we worked with linear expressions, some combination of numbers, variables, and operations like 2x+3. Now if I take that linear expression and I say that it needs to be equal to something, let's say this 2x+3 needed to equal 5, I now have a linear equation. Now equations are going to be a really important part of this course because we're going to have to solve a bunch of different types of equations. But don't worry, we're going to use a lot of the skills that we learned from linear expressions, and I'm going to 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. 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 4, 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 2x+3 equal 5. 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 0. I could check that by saying 2 times 0 plus 3. But 2 times 0 is 0. I would just get 3. That's not equal to 5. So x equals 0 wouldn't be my answer here. Then I could go and say x equals 1, maybe. So if I plug that in, 2 times 1 plus 3, 2 times 1 is 2, plus that 3, this actually does give me 5, 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 going to need to use some other skills here in order to solve our linear equations. So, you're actually going to need to use all of the different operations in your toolbox, addition, subtraction, 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 going to 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 2 equals 0. Now this 2 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 2 would be subtracting 2. So to get rid of it, I could subtract 2, and I need to do that to both sides. So my left side here would cancel, and I'm just left with x equals 0 minus 2 gives me a negative 2, and I have isolated x. Looking at our second example, I have 3x equals 12. Now here, my 3 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 3 that's multiplying my x? Well, the opposite of multiplication is division, so I could go ahead and divide by 3 on both sides to cancel it out. And, again, I'm just left with x equals 12 divided by 3 gives me 4. 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 going to 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 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 2 times x minus 3 equals 0. 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 2 that needs to get distributed to both the x and the negative 3. So if I do that, I get 2x, and then 2∗−3 gives me negative 6 equals 0. So step 1 is done. Now step 2 asked me to combine like terms. So taking a look at my equation, I have 2x−6 equals 0. 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 2 is also done. Now in step 3, 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 2x−6 equals 0. So I want to pull this 6 over to get it on the opposite side. So in order to do that, I need to do my opposite operation here, which is going to be adding 6 to both sides. It will then cancel there, and I'm left with 2x equals 0 +6 gives me 6. So step 3 is done. Now step 4 is to isolate x. You might also hear this called solve for x. So let's go ahead and isolate x. I have my 2 multiplying my x here. So that means to get rid of that 2, I need to divide. So if I divide both sides by 2, my 2 will cancel here. I'm left with x equals 6 divided by 2 gives me 3. And this is actually the answer. So I'm done with step 4. And 3, 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 5, and that is to check our solution by replacing x in our original equation. So I'm going to take my original equation, 2x−3 equals 0, and make sure that I found the value of x that makes that true. So I take 2, the value I got for x was 3 minus 3 equals 0. Now this then gives me 2, 3 minutes 3 is 0, equals 0. And we know that anything times 0 is 0, so I am left with 0 equals 0, which is definitely a true statement. So I've completed solving this linear equation. That's all for this one, guys. Thanks for watching.

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

• A. $x=-27$
• B. $x=-1$
• C. $x=1$
• D. $x=16$

Solving Linear Equations with Fractions

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 going to 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 14 ∗ x + 2 - 13 ∗ x = 112. Now this equation definitely has fractions in it, and I want to get rid of those as soon as possible. So here, we're actually adding a step 0 before we do anything else. And our step 0 is going to be to multiply by our LCD, our least common denominator, in order to eliminate our fractions. So looking at my equation here, my denominators are 4, 3, 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 124 ∗ ( x + 2 ) - 123 ∗ x = 1212. Now we can even simplify this a little bit further. So if I take 12 divided by 4, that gives me 3 times (x+2) minus 4x equals 1. Now all of my fractions are gone and I can continue solving this just as I would any other linear equation. So my step 0 is done.

I can move on to step 1 which is to distribute our constants. Now looking at this, I have this 3 here that needs to get distributed to both the x and my 2. So that gives me 3x+6. We don't have anything to distribute there. So step 1 is done. Now step 2 is to combine like terms. So looking at my equation, I have a couple of like terms. I have 3x and negative 4x, and those are going to combine to give me negative one x. Everything else is going to stay the same. So I still have that plus 6 equals 1. So step 2 is done.

I have combined my like terms. Now looking at step 3, 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 going to go ahead and move this 6 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 6 from both sides. And here, it will cancel, and my negative one x stays there. And then I have 1 minus 6, which will give me negative 5. So I've completed step 3. I have moved my constants, moved my x terms to be on opposite sides. Now I want to do step 4, 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 it will cancel over here, and I am left with x equals negative 5 divided by negative 1 gives me positive 5, and this is my solution. So I've completed step number 4. Now step number 5, again, is to check by replacing 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.

Solve the equation.

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

• A. $x=10$
• B. $x=4$
• C. $x=8$
• D. $x=-1$

Categorizing Linear Equations

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. There are 3 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 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.

Looking at my first example, I have 2x+4=10. So my first step here is going to be to get all of my constants on one side and all of my x terms on the other. To move that 4 over, I need to go ahead and subtract it from both sides. So it'll cancel over here and I'm going to be left with 2x=10-4, which gives me 6. My next step here is to isolate that x variable. So, to do that here, I'm going to divide by 2 on both sides. Remember, whatever I do to one side of the equation, I have to do to the other. So, my 2 is going to be canceled, and I'm left with x=62, which is 3. This is my solution. For this particular linear equation, I only have one solution. The only value that would solve this that I could plug back in to get a true statement is 3. So my solution set is simply that number 3 in curly brackets.

Now all of the linear equations we've seen up to this point have had one solution, and their solution set would just be written as whatever number I get for x inside of those curly brackets. When I only have one solution, this is called a conditional linear equation. The reason it's called conditional is that 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 3.

Looking at my next linear equation, I have x+5=x+2+3. Let's simplify that. On this side, I just have x+5 and then equals x. The 2 and the 3 are both constants, so I can combine them. 2 and 3 together are going to give me 5. So you might see where this is going but let's take this a step further. 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 5 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. What happens here is I then have x minus x, which is 0, equals 5 minus 5, which is also 0. I just get 0 equals 0, which is an undeniably true statement. 0 is always equal to 0. This tells 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 all real numbers. 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. When I have an infinite number of solutions, this is called an identity equation.

Let's look at our final possibility. I have x=x+4. To 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 0. Then on the right side of my equation, I'm left with 4 so I end up with 0 equals 4. Now this is definitely not a true statement. What this tells me is that no matter what value I plug in for x, it's just going to end up giving me something a little outlandish, like 0 equals 4, 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 0 with a slash through it, and this is called the empty set because there is nothing inside that set. When this happens, when I have no solutions, this is going to be called an inconsistent equation.

So we have three types of linear equations. Our first one was conditional, when we have one solution. 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.

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

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

• A. Identity
• B. Conditional
• C. Inconsistent

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 

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

$-2\left(5-3x\right)+x=7x-10$

• A. Identity 
• B. Conditional 
• C. Inconsistent