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Contents

## Inequalities

We've just seen that if we have negative 2 is less than 2x is less than 10, you can divide everything by two. And we'll end up with negative 1 is less than x is less than 5. Let's look further into how we can solve inequalities. But this time we'll just start by dealing with single inequalities. Let's say I have 3x is greater than or equal to 12. Can you write an inequality in this box just for x. So I'd like you to isolate x on one side of the inequality and then have just a number on the other side.

## Inequalities

Dividing both sides by 3 gives us x is greater than or equal to 4. Let's check this on a number line. Here is what this inequality should look like on the number line. The number line is thus divided into two regions. There's one region over here on the left of values that x is not allowed to equal. And then the region on the right of the number line, that's all the values that x is allowed to equal take on. Let's pick one value from each, and see if they verify our solution. First, I'll pick a value of x that should work in the original inequality. Maybe, 6. 3 times 6 is equal to 18. And, yes, 18 is greater than or equal to 12. So that works. Now let's check a value from the left side, the area that's not shaded in. So when I plug in a value from over her to the original inequality, I should not make this true. A value from over here should not satisfy this. Let's pick 0. 3 times 0 is equal to 0, and 0 is not greater than or equal to 12, so this also confirms our solution.

## Dividing by Negatives

So, it works to divide both sides in an inequality by the same number to get a solution, or does it? What happens if we have negative x is greater than or equal to 4? Well, using the method we used before, we would divide both sides by negative 1. And we'd find that x is greater than or equal to negative 4. If this is right, then that means that this inequality up here, the original one we had, is satisfied by any value of x that is greater than or equal to negative 4. Now, last time I went through and I did a sort of sanity check, where I plugged in some test values that fit the solution interval we've come up with. And I plugged them in to the original equation to make sure that my answer made sense. Now, can you do the sanity check yourself this time, and tell me if this solution right here is correct? Just tell me, yes or no.

## Dividing by Negatives

No. If I plug in 0, which is definitely greater than or equal to negative 4, then I get that 0 is greater than or equal to 4, and that is just not right. Let's see what's going on here. We have something to figure out.

## Looking at Numbers

I'm having kind of a hard time when I have all these x's floating around. So, let's just take it easy. Let's write a super simple inequality, maybe 1 is less than 2. Those are two numbers that I think I can handle right now. Let's check this on a number line to make sure that I'm not crazy. Well here's 1 and here's is all of these values to the left of 2. 1 is definitely in this interval. But, I was looking at our earlier inequality, I ran into a problem when I divided both sides of the inequality by negative 1. I'll make some room here, and now, let's see what happens if I divide both sides of the inequality by negative 1. So, what number do we get on either side of our inequality, if we divide both sides by negative 1? Make sure you keep the numbers on the sides that they're on right now for the purposes of this answer. After that, I would like you to consider your answer, and tell me whether or not this makes sense.

## Looking at Numbers

us negative 2. Now let's look at our number line again. Of course what's shown right now is the original inequality. We're just going to use the number ticks for a second. Negative 1 is to the right of negative 2. It's greater than negative 2. So no, this definitely does not make sense at all.

## Fill in the Inequality

Just like with an equation, if we do the same thing to an inequality, we should be able to find some way to keep the statement true as we move down through our manipulation. So let's say, I'm super committed to dividing both sides of this by negative 1. What inequality do we need here then, in between the negative 1 and the negative 2? Please fill in the proper symbol in this box.

## Fill in the Inequality

Negative 1 is greater than negative 2, so we need a greater than sign. Let's look at the two number lines for these two inequalities. Let's say that this time, I want to see if the 2's are correct in relation to the 1's. So, I'm going to draw these number lines from the perspective of the number 1's. So, I want to know if 2 is greater than 1. That means it should fall in the category of all numbers that are greater than 1. Start at 1 I need to shade to the right. Sure enough here's 2, smack dab in the middle of that range, well not in the middle but in it and that's all that matters. Now we want to know if negative 2 is in fact in the range of numbers that are less than negative 1. Let's go over to negative 1 and shade to the left of negative 1. And once again we have negative we have pretty much exactly the same thing, except that the intervals here are fractions of one another across the 0 mark. So if I match the 2 of them together on 1 number line, can see that we have the exact same thing to the right of the dividing by negative 1 is just like flipping everything to the opposite side of the number line. So, when we divided by negative 1, we flipped our point of reference across the 0, the same distance away to the other side of the number line. And that's why we had to switch our sign, so that it faced the opposite direction as well. The two numbers now have the opposite relationship. Since being further away from zero than another number on the positive end of the number line means that you're greater than that other number. Whereas being further from zero than another number on the other end of the number line; the negative side means that you're less than that other number.

## Four Problems

Now that we've talked a little bit more about changing coefficients and sine issues of inequalities. Let's try a few problems. For each one of these solutions, I've already written the x for you. And in the orange box, I'd like you to fill in an inequality sign, so greater than, less than, greater than or equal to, or less than or equal to. And in the teal box, I'd like you to fill in the number that this is relating x to.

## Four Problems

For number one, we need to divide both sides by negative 5. 35 divided by negative 5 is negative 7. And, because we're dividing by a negative number, we need to flip the sign of the inequality. So it goes from being a less than sign to a greater than sign. We can go through this similar process for the rest of the problems. And, here are our solutions.

## Isolate x

What if we have negative x over 8 is greater than 2. What inequality could you write for x then? Remember we want to fully isolate x.

## Isolate x

We can think of negative x over 8 as x divided by negative 8, since the negative sign can go either with the numerator or the denominator. It just belongs to the fraction as a whole. That means to get x by itself, you just need to multiply both sides by negative 8. On the left side, that leaves us with x. On the right side, we have negative 16. But, just like before, we have to flip the sign direction. We get x is less than negative 16. As always, I'm going to check my solution with a number line. X is less than 16 is this region over here. Let's pick one value inside this region, and see if it satisfies this inequality. Negative 20 is over there, so negative 20 over negative 8 is just 5 over 2, or 2 and a half. 2 and a half is greater than 2, so we're good on that. Now let's pick one value to the right, in the region where values that we plug in should not satisfy this inequality. I'll pick zero, since it's always my favorite number to pick. Zero divided by anything, including negative 8, is just equal to zero. And zero is not greater than two, so this also confirms our solution set. So we can see that whether we divide or we multiply both sides of an inequality by a negative number, we have to switch the sign of the inequality. Remember, division and multiplication are very, very intimately linked.

## Flip the Direction

So now we know how to divide and multiply with inequalities. In fact, inequalities seem to be just like equations, except that if we multiply or divide by a negative number, when we're doing an inequality, we need to flip the direction that the inequality faces. So, lets try a slightly tougher one now. We have 3x plus 2 is less then equal to 11. Now 11 happens to be my favorite number in the entire world, so treat it well when you're solving this inequality. Please solve for x and write the inequality that you get in this box.

## Flip the Direction

We treat this just as if we were solving an equation, except that of course, we have an inequality sign running throughout. We're dividing by a positive number, this positive 3, not a negative one, so we have no funny business with the signs. And our final answer is, x is less than or equal to 3. And of course, you should check your answer. But by now, you know how to do that, so I'm not going to take the time to.

## Check the Inequality Sign

Just one more question, and then you'll have a few more practice questions to have some fun with. What if we have 12 is less than 4 x? I'd like you, however, to put the x on the left side of the solution, and then fill-in what inequality symbol belongs here. Input answer belongs on this side. You might want to check on either side of the endpoint of the inequality or the dividing mark between the solutions and the non-solutions to see if you have the inequality sign the right way around.

## Check the Inequality Sign

If we divide both sides by four, we end up with 3 is less than x. However, I had said that I wanted x on the left side of the inequality. So that means we need to flip this whole thing around. The whole thing becomes a mirror image of itself. Then we have x is greater than 3.

## More Problems

Here are a few more problems for you to try. They're definitely going to get a little bit trickier. But you have all of the skills that you need, and the understanding that you need to solve them properly. So just give them a try.

## More Problems

I'm going to momentarily make these problems at the bottom small and just show you the answers to the first three. Then we'll move onto the last two later. So here's the work and the answers for the first three problems. We have x is less than negative 3, x is greater than or equal to negative 2, and x is greater than three. Of course with this last one, we started out with x on the right side of the equation, and some people prefer to put the variable on the left side in the answer. So if you want to follow that convention, then to get x onto the left side, you also need to flip the inequality so that it's facing the opposite direction. Saying that 3 is less than x is also the same as saying, x is greater than 3. And now moving on to our last two problems. And after several steps we end up with x is greater than or equal to 4 and x is greater than 4.

## Deluxe Duo

Grant has decided that he wants to diversify his product base. He's created a new kind of wiper set to add to his glasses family. This is the Deluxe Duo, a set of glasses wipers unlike any other every created. They have superior wiping power and non function curly qs for decoration on the ends of them. With them. Now, as silly as these glasses wipers might seem to you or me, Grant has a reason for creating them. He can charge a lot of money. In fact, of just one set of deluxe duo glasses wipers, he can charge \$75. This is big money in the glasses wiper business. If we let the number of deluxe glasses wipers sold equal m, and Grant's total profit in dollars from them equal p. Then I've created an equation for you that describes his profit based on the number of deluxe glasses wipers he sold. You know \$75 is how much he charges per pair of deluxe glasses wipers, and \$5000 is the amount of money that he had to put down to invest in the materials and everything he needed to start building these. So, we have an equation, and we have a wonderful product.

## Deluxe Duo Profit

Grant as we know is ambitious, he wants to make at least \$20,000 in profit off of these deluxe glasses wiper sales. So knowing that, what inequality can we write combining our earlier equation with this new information about his profit.

## Deluxe Duo Profit

Since we know that Grant wants his profit to be at least \$20,000 and p stands for profit in dollars. We know that p has to be greater than or equal to 20,000. Now we just have a simple case of substitution, p is equal to this entire quantity, 75m minus 5,000. So you replace p here in the inequality with that quantity. So here is our final answer. 75m minus 5,000 is greater than or equal to 20,000.

## Minimum Number of Sales

Let's use our lovely new inequality. Now think critically about this. What is the minimum number of wipers that Grant must sell to reach his desired profit? Please fill in the proper value of m right here.

## Minimum Number of Sales

Manipulating this in the ways that we've practiced gives us m is greater than or equal to 333 and 1 3rd. Let's look at this on a number line. So, m can be anything to the right of, or including, 333 and a third. But wait, what are we talking about anyway? What was m equal to? M is equal to the number of deluxe wipers sold. Now, you can't sell a third of a wiper set, so that means we're going to have to make m into a whole number. Now, if we make it 333, then we won't quite make it to our \$20,000 profit level. So, that mean that we're going to have to round up. M has to be greater than or equal to 334 wipers. Well, I'm going to leave our number line looking just like this, because these would be about in the same place. You can just imagine that our bracket has shifted slightly to the right. What we want is the minimum possible value for m, so the smallest value in this range of possible solutions. That, of course, is going to be just this endpoint solution, 334 exactly. So we want m equals 334.

## Absolute Value

Let's have another look at an inequality involving an absolute value sign. We have the absolute value of x is less than 5. Now, I'd like you to get rid of the absolute value signs around x, in a mathematically correct way, of course, and write the solution as a chained compound inequality. Remember, that's a compound inequality where we get rid of the and sign, and just write everything as one connected mathematical statement.

## Absolute Value

The answer is negative 5 is less than x is less than 5. Let's look at that on a number line. We know that the distance between x and 0 needs to be less than 5, so it can be any value between negative 5 and 5 on the number line.

## Absolute Value to Inequality

We've just seen how we can write an absolute value inequality. Which we see, involves only 2 things related to one another, into a compound inequality. Where we have 3 quantities related to one another. And, of course, I've used G notation. This is because when you take the absolute value of a number. You're creating an upper and a lower bound for its size. Going off of this, what if we had the absolute value of 3x minus 5 is less than 4. How could we write this as a chained inequality? For right now just leave the quantity 3x minus 5 as a single blocked unit.

## Absolute Value to Inequality

We get negative 4 is less than 3x minus 5, which is less than 4.

## Absolutely Valuable

Now that we've rewritten this as a compound inequality, we know how to solve for x. So please isolate x by doing the same thing to all three sections at once. Then write your answer in these three boxes. I have been really nice, and I've already filled in your inequality signs for you. So you just have to fill in the quantities that go in between them.

## Absolutely Valuable

And our final answer, using the manipulation tactics we've learned before, is 1 inequalities. I know this actually pretty difficult stuff so, thank you for sticking with and for thinking deeply about it.

## Fractional Equation

We've been working so much with inequalities that I'm not sure I even remember how to solve an equation. You may recall this is an equal sign and when we isolate x on one side of an equation, we're finding actually a single value that it's equal to on the other on the other side. Or at least that's what we're doing as long as there's one number over there. Anyway, just to get our brains jogging about equations again and to make sure you haven't forgotten everything you've learned in this class, let's solve this equation. Now before you go through all the steps of solving this, I'd like you to start by writing the two fractions on the left hand side of the equation as a single fraction. So find the common denominator, figure out how that changes the numerators and then combine like terms.

## Fractional Equation

The least common multiple of 2 and 3 is 6. So we know that that's still going to need to be the denominator of our combined fraction. To change the fraction x over 2 to have a denominator of 6 instead, we need to multiply it by 1 in the form of 3 over 3, since 2 times 3 is 6. For x over 3, we need to multiply it by fractions again. It's never too late to do some extra practice. Carrying out this multiplication gives us this. And then we just need to combine like terms. We end up with 5 x over 6 equals 4.

## Solving Fractional Equations

Now, let's finally solve for x. What should x equal, if we have 5x over 6 equals

## Solving Fractional Equations

We know that 5x over 6 is just the same as 5 6ths times x. So to get rid of this coefficient of 5 6ths, we just need to multiply each side by its reciprocal, which is 6 5ths. And we end up with x equals 24 over 5. Let's check our answer just to be sure. We plug 24 over 5 back into the original equation. We're going to start by just working with the left hand side of the equation and seeing if what we end up with is equal to the right side. 5 times 24 over 5 is just 24, and we leave the denominator the same, since we didn't do anything to it. 24 over 6 is 4. The left hand side simplifies to 4, which is equal to our right hand side. So we are good.

## Denominator x

Okay, so clearly we haven't totally forgotten how to solve equations. That's awesome. It's going to help us a lot in algebra moving forward. Let's look at a new equation. We have 3 over x minus 1 over 5 is equal to 7. Just like the last example, we have two fractions, but this time there's not an x anywhere in the second fraction. We have two choices as I see it, or at least two choices in terms of how to start the problem. The first choice that I see is the first step we could take, would be to combine the two fractions that are on the left-hand side. That's definitely an option. But there's another choice. We could isolate the terms, or term in this case, involving x on the left-hand side and move everything else over to the right-hand side. I have a feeling that it's going to be a little bit less work if we go with the second option. I'll put a link to how to do the problem in the other way in the lesson notes. But for now, let's go with this method. So our goal right off the bat is to get this term by itself on the left-hand side. So for the equation that I want you to write in this box, I would like you to isolate the term with x on the left-hand side. Get everything else on the right-hand side and simplify it so that it's written as an improper fraction. What do you get then?

## Denominator x

All we need to do is get of this term of negative 1 5th on the left-hand side. To do that you just need to add 1 5th to both sides. 7 and 1 5th is equivalent to 36 5ths. So we have 3 over x equals 36 5ths.

## Solving with x in the Denominator

In the new version of the equation we've come up with, we have an x as a denominator, and we don't really want it there. So, I'd like you to solve for x. And, just as a hint, you could probably start by multiplying both sides of the equation by x. What do you get as your final answer then?

## Solving with x in the Denominator

Going through the pretty typical steps to solving this equation gives us an answer of x equals 5 over 12. Be sure to reduce your fraction. I initially ended up with 15 over 36. But 15 and 36 share a common factor of 3. Dividing both the numerator and the denominator by 3 gave me this final answer.

## New Denominator

Here is another equation. 3 over x minus 1 over the quantity 5x is equal to 7. Again, we have choices about what to do first. I'm going to say that I'd like our first step this time to be to combine the two fractions over here on the left hand side. We can see that x is involved in both these fractions, so it seems like a pretty logical first step to take to me. So if we decide to do that, what should we make the denominator of the single fraction we'll have here on the left side? Make sure that this is the least common denominator we could have.

## New Denominator

The answer is 5x. 5x is the least common multiple of x and 5x. We just need to multiply both the top and the bottom of the first fraction by 5.

## Single Fraction

Now that we know what the denominator of this single fraction over here needs to be, please combine these two terms, and then write the fraction as simplified as you can get it, in this box. We're just going to leave the right side of the equation the same for now. So we're not doing anything that changes both sides of the equation. We're just simplifying within the left side right now.

## Single Fraction

Remember that to get that common denominator in both terms, we need to multiply the first term by 5 over 5, so the denominator becomes 5x. Then we just combine our like terms, and simplify. This gives us a new equation of 14 over 5x equals

## x Equals What

Now, I'd like you to just solve for x.

## x Equals What

Just like we did before, we need to multiply both sides by the denominator of the single fraction on the left-hand side. Status 5x. Isolating x gives us x equals 2 over 5. And of course, you want to substitute this back in to check and make sure that it works.

## Get Rid of Fractions

Let's go back to the equation that we started this whole problem off with, 3 over x minus 1 over 5x equals 7. So we already solved the problem one way and that was to combine our 2 fractions into 1 and then proceed. However there is another method, and it's always good to explore a little bit more. The second thing that we could start off by doing, would be to Get rid of our fractions right off the bat. And we know that the way we get rid of fractions is to multiply fractions by numbers. So, let's think about what number we could use. Since x of the denominator of the first fraction on the left hand side, let's start off trying with that. What equation do we have if we take the single step of multiplying both sides of this equation by x? Fill in the left side that we'll be left with here, and the right side that we'll be left with here. You can of course, combine like terms and simplify on either side of the equation if you want to. I just don't want you to do anything else that modifies both sides of the equation.

## Get Rid of Fractions

We know that we need to multiply both sides by x. But, what does it mean to multiply this entire side of the equation by x? Well, it means that multiplication by x needs to distribute to both terms on this side of the equation, since we're multiplying the entire side by x. 7 times x is, of course, just 7x. Then we can distribute and simplify the left-hand side, x times 3 over x just gives us 3. And x times negative 1 over 5x gives us negative 1 over 5. And in the end, we get 14 over 5 equals 7x, which of course looks like what we had before.

## Multiplying Both Sides

In the last quiz, we started off by multiplying both sides by x, since that's the denominator of this first fraction. But what if instead we picked 5x? That was after all the common denominator that we used the first time around that we tried this problem. So please, in these two boxes, fill in what the left-hand side and the right-hand side of this equation will be if we multiply each side by 5x. Again, you can simplify within either side as much as you want to.

## Multiplying Both Sides

You remember that the 5x needs to multiply the entire side of the equation, and of course, both sides. So you just do multiplication by 5x to both terms. This time we get 14 equals 35x, which is also something that we had earlier. I think that multiplying by the lowest common denominator of our two fractions initially helped us out a lot. Made this whole process much easier in fewer steps. So that's a good tip for moving forward in the future. If you want to get rid of your fractions right off the bat, then the factor that you should multiply everything by should be the least common denominator of all of your fractions that you want to combine.

## First Step

Speaking of fractions, here's another equation, with yet another fraction involved. We've just been talking about how many different ways there are to start problems, especially those involving fractions. So for this problem in particular, if we'd like to solve for z, what do you think our first step should be? Should we start off by simplifying the fraction? Should we multiply both sides by z? Should we multiply both sides by 3? Or should we multiply both sides by 3z plus 6?

## First Step

We should definitely multiply both sides by 3z plus 6. That will make it so that our left hand side doesn't have a fraction at all.

## Solve for z

We've got a little bit of a leg up on this problem, since we've already determined our first step. So I'l like you to take that first step, and then take all of the others, and just right here, give me your answer for z. What should z equal to satisfy this equation?

## Solve for z

After a series of various steps, all of which you're pretty familiar with, we end up with a final answer of z equals negative 3. And of course, it's super important to check our answer. We just plug negative 3 in to the left side of the equation and negative 15 over negative 3 is indeed equal to 5, which is what our right-hand side equals. So we're good.

## Multiplying by Denominator

Since we love fractions so much, let's just maximize our use of them. We have a fraction on the left side, 6x minus 5 over the quantity 7 minus 3x. And we also have one on the right, 4 minus 8x over the quantity 4x plus 2. We just saw how effective it is to try to get rid of our denominators, that at least, leads us to a point where we can start to manipulate this in ways that we can get x by itself. But here, we have two denominators, we have one on the left and one on the right. I'd still like us to get rid of them and we know that we do that through multiplying by them. So in this box, I'd like you to write out the entire equation you'll get at the very first stage, where you no longer have fractions in this equation. So I don't want you to multiply anything out I don't want you to distribute or simplify on either side. I just want you to get this to a place where there are no more fractions.

## Multiplying by Denominator

Let's start by considering the left side. To make this no longer a fraction, we need to multiply both sides to the equation by its denominator. However, doing that alone doesn't help us with the right side. To deal with this fraction, we need to multiply both sides by the quantity 4x plus 2. Now let's see what that leaves us with. On the left side, multiplying by 7 minus 3x gets rid of the denominator, but then we still are multiplying by this other factor. I made a mistake and forgot to write our parentheses around this entire factor we're multiply by. That's super important, and clearly, it's very easy mistake to make. On the right-hand side, since we're multiplying by 4x plus 2, that deals with the denominator. But we still have a factor of 7 minus 3x. Wonderful. This is our final answer.

## Simplify Each Side

The next step I'd like you take is to simplify within each side of the equation. So carry out this multiplication and then combine like terms. Do not however, yet, do anything that changes both sides of the equation. Just simplify within each side.

## Simplify Each Side

Multiplying each term in the first set of parenthesis by each term in the second set of parenthesis. And doing this, of course, for both sides gives us this long messy thing. Thankfully, it will start to look a little bit prettier in a second. Now we need to combine like terms. Our like terms are of course, as we can see, terms that have an x to the first power in them. And for now, we're only going to do that within either side of the equation. Of course, it doesn't matter what order you have the terms in either side of the equation in. So let's say, for example, you have 24x squared as the first term in the right-hand side. That's totally fine.

## Solving for x

And now, finally, I would like you to actually solve for x. What does x need to equal in order to satisfy this equation?

## Solving for x

The first step we need to take, and this is actually a little bit tricky, is to recognize that we have a term of 24x squared on both sides of the equation. That means, if we subtracted from both sides, we'll just get rid of it entirely. That's definitely going to make our lives simpler. Now, I'm going to move all of my x terms, so that they're on the left side of the equation. And all of my constant terms, so that they're on the right. All that's left to do is divide both sides by 60. Recognizing that 38 and 60 are both divisible by 2, we can cancel out that common factor, and get a final answer of x equals 19 over 30. Awesome. I know that was a really, pretty intricate and complicated equation to solve. But look at where we came from. We started off with all of these messy fractions, and we ended up with, well, an answer that is a fraction, but it's an answer nonetheless. I'd like you to plug this answer into the places of x in the original equation, to make sure that it actually works. Please remember to check the left hand side on its own, and the right hand side on its own, and then compare the two.

## Checking LHS and RHS

If we look back at our original equation, 6x minus 5 over 7 minus 3x equals 4 minus 8x over 4x plus 2, what does each side individually simplify to if we plug in the value of x that we just calculated? So substitute in 19 over 30 in each spot of x for the left-hand side first and then enter the answer that, that equals here, and then the right-hand side.

## Checking LHS and RHS

So this is a lot of work. But that's important. It's good to practice your arithmetic, good to practice manipulating fractions, and multiplying things. Lucky for us, both the left-hand side and the right-hand side simplify to negative 4 over 17. So our answer of x equals 19 over 30 is correct.

## Rewrite with Exponent

Now here's something we haven't seen in awhile, a square root sign. We have the square root of the quantity 2x plus 4 is equal to 5. And just to jog your memory of the relationship between exponents and square roots, please rewrite this entire equation with the square root replaced by the proper exponent. Remember to take into consideration what the exponent should apply to, which terms it should apply to.

## Rewrite with Exponent

Remember that square root signs are the same as exponents of 1 half. So, you need to take the entire quantity 2x plus 4 place it in parenthesis and then, just outside the parenthesis put a power of 1 over 2, and we didn't do anything to change the right hand side of the equation at all.

## What is the Next Step

Now that we've reminded ourselves of what a square root sign actually means, what do you think the next step in moving towards solving our equation should be? Remember, we want to get x by itself on one side of the equation eventually. So should we start off by multiplying both sides by the quantity 2x plus 4? Should we multiply both sides by the square root of 2x plus 4? Should we square both sides, or should we multiply both sides by 2? Clearly these are all steps that we could take and would be legal in the context of solving equations. But which one will actually move us as close to our answer as we can get in one step?

## What is the Next Step

We need to square both sides, doing, that's going to get rid of our exponent of give us another number. So this is the best choice.

## To the Half

Now that we've determined what our first step should be, squaring both sides, please take that step, and then continue on until you get a final value for x. What does x need to equal to make this first statement true?

## To the Half

First things first, we square both sides, just like we decided to. Squaring the left-hand side just gets rid of this exponent of one-half, and we get 2 x plus 4. 5 squared is 25. So, 2 x plus 4 equals 25. Now, things are super easy. We keep doing our regular old algebra until we get to a final answer of x equals well, or 10 and one half. Let's check this just to be sure, since this is sort of a new kind of equation for us. Dealing only with the left hand side of the equation, I plug in 21 over 2 in the spot of x. We get to the square of 25 which is, of course, 5. And this works.

## Solve This

So this new equation was super fun to solve. It was great. We knew exactly what we were doing. But what if we change one little tiny thing? What if I make this a negative 5, instead of a positive 5. The question I have for you is, is this an equation that we can solve? In other words, is this an equation at all? Tell me, yes or no.

## Solve This

And the answer unfortunately is no, we can't solve this. Remember that when we used the square root sign without any sign explicitly indicated outside of it. This gives us the positive square root of whatever quantity is inside. Because we know then for sure this quantity on the left is positive, there is no way that it can equal a negative number. Which is what's on the right side. So even thought we now know how to solve a ton of different looking equations, probably a lot of which it didn't seem at first like we did know how to solve, there are some that are just not mathematically possible. That has nothing to do with our math expertise, though. And you've done a really, really awesome job, on this lesson and on the ones before.