In this video, we're going to focus on graphing inequalities on number lines and also solving inequalities. We're going to solve inequalities with fractions, variables on both sides, absolute value. And we're also going to write the answer in interval notation. So let's begin.
Let's say if x is greater than 2. How can you represent this inequality on a number line? Do we need an open circle or a closed circle? So 2 is at this position.
Notice that there's no underline next to the inequality sign. So therefore, we need an open circle. circle and because X is greater than 2 whenever it's greater than shade to the right if it's less than shade to the left so since X is greater than 2 we're going to shade to the right side and so that's how you can plot it on a number line. All the way to the left is negative infinity, and all the way to the right, positive infinity. So, to write this answer in interval notation, it's going to be 2 to infinity.
Infinity will always have a parenthesis next to it. But notice that we have an open circle at 2. Therefore, we're going to have a parenthesis next to 2. If it was a closed circle, it would be a bracket. Now what about this one?
Let's say if x is greater than or equal to negative 1. Feel free to pause the video, plot it on a number line, and write the answer using interval notation. So here's negative 1. Do we have an open circle or a closed circle at negative 1? Since we have the underline, it's going to be a closed circle. And because x is greater than or equal to negative 1, we need to shade to the right if it's greater than. So that's how you can plot it on a number line.
And so here we have negative infinity all the way to the left side, and positive infinity to the right. So to write it in interval notation, the green part starts at negative 1, and it includes negative 1, and it goes all the way to positive infinity. So because we have a closed circle, we need a bracket at negative 1. Now what about this one?
Let's say if x is less than 4. What's the answer in this case? So here's 4, and let's say 0 is right there. Do we have an open circle or a closed circle?
So we don't have the underline, so we have an open circle at 4. And because it's less than this time, we need to shade to the left side, since x is less than 4. So now how can we write the answer using interval notation? In interval notation, you want to view it from left to right. The lowest value on the number line is negative infinity.
The highest value is 4. That's where the blue line stops. And it doesn't include 4, so we're going to use parentheses. So it's from negative infinity to 4. It starts here, and it stops here.
Now what about this one? Let's say if x is less than or equal to negative 2. Go ahead and plot it on a number line. and write the answer using interval notation.
So this time, we have a closed circle at negative 2, and we're going to shade to the left since it's less than. So the answer in interval notation is going to start from negative infinity and it's going to stop at negative 2. So you can write it like this. And since it includes negative 2, we need to write a bracket. Now what if we have an inequality that looks like this?
X is greater than 1 but less than or equal to 4. How would you plot it on a number line? So here's 1 and here's 4. So we have an open circle at 1, but a closed circle at 4. X is greater than 1, so we need to shade to the right of 1, but it's less than or equal to 4, so we gotta shade to the left of 4. Therefore, we have to shade in between. So, to write the answer in interval notation, X is between 1 and 4, but we have a bracket at 4 since we have a closed circle at 4. Try this one.
Let's say if x is less than negative 2 or x is greater than and equal to 3. How would you plot it on a number line? So the points of interest... are negative 2 and 3. So at negative 2 we have an open circle and since x is less than negative 2 we need to shade to the left side.
So it looks like this and x is equal to or greater than 3 so we have a closed circle at 3 and we need to shade to the right side. Now how can we write this in interval notation? So there's two shaded regions.
Here's the first part, and here's the second part. And we need to combine them using union. So for the first part, it's negative infinity to negative 2, viewing it from left to right.
And then union, using a bracket since we have a closed circle, 3 to infinity. So that's how you can represent this answer using interval notation. So now at this point, we're going to focus on solving inequalities as opposed to graphing the inequality on a number line. So let's say if we have this expression, x plus 4 is greater than 5. Let's solve for x. So, whenever you want to solve for x, you need to get x by itself.
Since 4 is added to x, to move the 4 to the other side, we need to perform the opposite of addition, which is subtraction. So let's subtract 4 from both sides. 4 minus 4 is 0, so what we have is x is greater than 5 minus 4, which is 1. So x is greater than 1. And the interval notation is 1 to infinity.
And you know how to graph it using a number line at this point. Now what about this one? How would you solve for x if you're given this equation?
What's the first thing that you would do? The first thing that we should do is add 5 to both sides. By the way, feel free to pause the video as you work out these examples, and then unpause it to see the solution.
Negative 8 plus 5 is negative 3. So now what's our next step? How can we separate the 3 from the x? Since 3 is multiplied by x to separate, you need to perform the opposite of multiplication, which is division.
So we need to divide both sides by 3. 3 divided by 3 is 1, so on the left side we have 1x or simply x. Negative 3 divided by 3 is negative 1. So this is our answer, x is less than 1. In interval notation we can write it as negative infinity to negative 1 in parentheses. Try this problem. 7 minus 2x is less than or equal to 12. So the first thing that we need to do is we need to subtract 7 from both sides. So what we have left over, negative 2x is less than or equal to 12 minus 7, which is 5. Now to get x by itself, we need to divide both sides by negative 2. Now whenever you multiply or divide by a negative number, the inequality is going to change direction.
So we have a less than symbol, but now it's going to be a greater than symbol. So x is equal to or greater than negative 5 over 2, which is the same as negative 2.5. So in interval notation, the answer is from negative 5 over 2 to infinity. If you graph it on a number line, Let's say this is 0. Negative 2.5 will be somewhere to the left.
So we would have a closed circle, and we would shade to the right side. Now what if we have variables on both sides? 4 minus 2x is equal to or greater than 3x plus 19. How can we solve for x in this particular example?
So there's many ways you can approach this problem, but first let's subtract both sides by 4. So 19 minus 4 is 15. Now what we could do is subtract both sides by 3x. We need to get x on one side of the equation. Negative 2x minus 3x is negative 5x. Adding or subtracting by a negative number will not reverse the sign of the inequality.
You need to multiply or divide by a negative number. So at this point, if we divide both sides by negative 5, the direction of the inequality is going to change. So instead of being greater than or equal to, it's now less than or equal to negative 3. So this is the answer.
So what is the answer using interval notation? So x is less than or equal to negative 3. So it's from negative infinity, and it's going to stop at negative 3. If you plot it, let's say this is 0. Here's negative 3, and we have a closed circle at this point. And since x is less than or equal to negative 3, we need to shade to the left side. So it's from negative infinity to negative 3. Here's another one for you.
This time, this equation is going to have parentheses. So go ahead and try this one. The first thing I would recommend doing is distributing the number in front of the parentheses. So in the first example on the left, let's distribute the 2. Negative 2 times x is negative 2x.
Negative 2 times negative 3. that's positive 6 and if we distribute the 4, 4 times 2x is 8x and 4 times negative 1 is negative 4 and then let's add like terms 5 plus 6 is 11 and 3 minus 4 is negative 1. So now at this point let's add 1 to both sides and simultaneously let's add 2x to both sides. So, negative 1 and 1 will cancel, and the same with 2x. 11 plus 1 is 12. 8 plus 2, that's 10. So, 12 is greater than 10x.
So at this point, to get x by itself, we need to divide both sides by 10. So 12 divided by 10 is greater than x. Now we can reduce the fraction if we divide by 2. Let me write it the other way. 12 divided by 2 is 6, half of 10 is 5, so 6 over 5 is greater than x, or we can write it as x is less than 6 over 5. If you switch the left side and the right side, you need to reverse the direction of the inequality. Consider this problem, 3 over 2x is less than negative 9. What is the answer for this?
What's the first thing that you would do to solve this particular inequality, if you have a fraction? The first thing I would do is try to clear away the fraction. Notice that we have a denominator of 2. So I'm going to multiply the left side and the right side by 2. 2 so looking at the left side 2 times 3 over 2 is 3 the 2s will cancel you can also say 2 times 3 is 6 6 divided by 2 is 3 so you still end up with 3. On the right side, negative 9 times 2 is negative 18. So now, all we've got to do is divide both sides by 3. So, x is less than negative 6. And as an interval notation, that means x is between negative infinity to negative 6, but it does include negative 6. Let's try another example with fractions. 1 third x plus 4 is greater than or equal to 8. So in this example, let's subtract 4 from both sides. So 1 third x is greater than or equal to 8 minus 4, which is 4. Now let's multiply both sides by 4. I mean, not by 4 actually, by 3. So the 3's will cancel.
3 divided by 3 is 1, so we have 1x on the left side. On the right side, 4 times 3 is 12. So x is equal to or greater than 12. So it's from 12 to infinity. Try this problem.
5 over 4x minus 2 is less than or equal to 3 over 2x plus 1 third. So now what would you do if you have many fractions? For a problem like this, you want to clear away all the fractions. Consider the denominators of 4, 2, and 3. What is the least common multiple of 2, 3, and 4? Multiples of two are 2 4 6 8 10 12 14 16 and so forth multiples of three are 3 6 9 12 15 18 21 multiples of four are four eight 12, 16, 20, the least common multiple is 12. 2, 3, and 4 can go into 12. So therefore, we want to multiply both sides by 12. So you want to multiply everything in this equation by 12. So what is 12 times 5 over 4x?
If you multiply 5 over 4 by 12, what do you get? So you can do 5 times 12, which is... is 60 and then divide by 4 which is 15 or you could divide first and then multiply.
12 divided by 4 is 3, 3 times 5 is 15. So 5 over 4x times 12 is 15. 15x. So, we no longer have a fraction at that point. Now we can multiply 12 by a negative 2, which is negative 24. And then 12 times 3 over 2x. 12 divided by 2 is 6. 6 times 3 is 18, so we have 18x. And for the last one, 12 times 1 third is the same as 12 divided by 3, which is 4. So now we can add 24 to both sides.
And at the same time, we can subtract 18x from both sides. So these will cancel. On the left, 15 minus 18, that's negative 3. And on the right side, 4 plus 24 is 28. So now we've just got to divide both sides by negative 3, which will change the sign of the inequality.
So our answer is x is greater than or equal to negative 28 over 3. Let's try another example like that. Try this one. 2 over 3x plus 5 is greater than 1 over 5x plus 3 fourths.
So what should we multiply both sides so that we can clear away every fraction? So what is the least common multiple of 3, 4, and 5? If you're not sure, you can always multiply the three numbers.
3 times 5 is 15. 15 times 4 is 60. Or you can do 4 times 5, which is 20. 20 times 3 is also 60. So let's multiply everything by 60. So what is 2 over 3x times 60? 60 divided by 3 is 20. 20 times 2 is 40, so we have 40x. And then 60 times 5. If 6 times 5 is 30, 60 times 5 is 300. and then 1 fifth x times 60 or 60 divided by 5 that's going to be 12 so 12x and then what is 3 fourths times 60 60 divided by 4 is 15 15 times 3 is 45 so 3 fourths of 60 is 45 now at this point let's go ahead and let's subtract 40x from both sides And let's subtract 45 from both sides.
So these will cancel. 300 minus 45, that's 255. 12 minus 40 is negative 28. So let's divide both sides by negative 28. So the inequality will change direction. So 255 divided by negative 28 is less than x.
Since we divide by a negative number, don't forget to change the sign of the inequality. Now typically, x is usually written on the left side, so we can also say x is greater than negative 255 over 28. Now let's move on into problems containing absolute value. If the absolute value of x is less than 4, what is the answer? How would you solve for this particular inequality?
To get rid of the absolute value, you need to write two equations. First you need to write the original equation, x is less than 4. And then you need to write the negative version of this equation, x is greater than negative 4. You need to flip the inequality as well as change the sign from positive to negative. So we have two answers, x is less than 4 or greater than negative 4. So on the number line, let's say x is less than 4. So to shade that.
This would go in this direction. And the other one, x is greater than negative 4. So we have an open circle going in this direction. So x is between negative 4 and 4. Now we can see if it's true. Let's say if we plug in a number between negative 4 and 4. Let's say 3. 3 is less than 4, so that works.
Now let's pick a negative number. Let's say negative 2. The absolute value of negative 2 is positive 2. 2 is less than 4. So any number in between negative 4 and 4 works. If we choose a number outside of it, it won't be true.
For example, let's choose a number to the right of 4. Let's say 7. The absolute value of 7 is 7. 7 is not less than 4. So 7 doesn't work. If we try negative 5, the absolute value of negative 5 is 5. And 5 is not less than 4. So this equation... The equation is only true when x is between negative 4 and 4, so therefore the answer in interval notation is negative 4 to 4. Now what if you're given a problem that looks like this? How would you solve for x? What's the first thing you need to do?
So the first thing we need to do is we need to get rid of the absolute value and write two equations. So we're going to write the original equation. And the second equation on the left side, let's keep it the same, but the right side, we're going to change the inequality and change the sign from positive to negative.
So now let's solve for x in both equations separately. In the first equation, let's add 3 to both sides. So, 8 plus 3 is 11, and then we need to divide both sides by 2. So, our first answer is x is greater than or equal to 11 over 2, which is the same as 5.5.
Now for this side, let's add 3 to both sides. So negative 8 plus 3 is negative 5. And then let's divide both sides by 2. So x is less than or equal to negative 5 over 2, which is negative 2.5. So now let's plot the solution on a number line. So here's negative 2.5 and here's 5.5 and 0 is somewhere in the middle. So x is less than or equal to negative 2.5.
So we have a closed circle but shaded. to the left and for the other one X is equal to or greater than 5.5 so it's shaded to the right so we have an or statement since they're going in separate directions all the way to the left is negative infinity and to the right positive infinity so the answer in interval notation is basically what we see here negative infinity to negative 2.5 union 5.5 to infinity Let's try another one. 5 minus 3 times the absolute value of 4x plus 1 is equal to or greater than negative 9. Now, at this point, you do not want...
to write two equations yet. You need to get the absolute value expression by itself before you separate it into two equations, meaning that the 5 and the negative 3, you need to move it to the other side of the equation before writing two equations. So let's subtract both sides by 5 first.
So what we now have is negative 3 times the absolute value of 4x plus 1 is greater than or equal to negative 9 minus 5, which is negative 14. So now, let's divide by negative 3. Now keep in mind, whenever you divide both sides by a negative number, the inequality must change sign. So now it's less than or equal to 14 over 3. Now that we have the absolute value expression by itself on the left side, we can now write two equations. So for the first equation, it's going to be 4x plus 1 is less than or equal to 14 over 3. And for the second one, 4x plus 1 is greater than or equal to negative 14 over 3. So, we gotta change the sign and flip the inequality.
Now at this point, let's get rid of the fraction. Let's multiply everything by 3. So 3 times 4x is equal to... X is 12 X 3 times 1 is 3 and 3 times 14 over 3 the 3s cancel and we're just going to get 14 so now let's subtract both sides by 3 so 12 X is less than or equal to 11 so now we can divide both sides by 12 so our first answer is X is equal to or less than 11 over 12, which is just under 1. Now for the next one, let's do the same thing.
Let's multiply everything by 3. So 3 times 4x is 12x. 3 times 1 is 3. And 3 times negative 14 over 3 is negative 14. The 3's disappear. So let's subtract both sides by 3. So 12x is equal to or greater than negative 17. And now let's divide both sides by 12. So x is equal to or greater than negative 17 over 12. So at this point, let's plot our solution on a number line.
So let's put 0 in the middle. 11 over 12 is to the right of 0, and negative 17 over 12 is to the left. So x is greater than or equal to negative 17 over 12. So we have a closed circle, and we need to shade to the right side.
x is also less than and equal to... 11 over 12 so we got a shade to the left since it's less than or equal to this number so we have everything in the middle therefore the answer is between negative 17 over 12 to 11 over 12 using brackets now you can also write your answer this way You can put x in the middle and say it's less than or equal to 11 over 12, but it's greater than or equal to negative 17 over 12. You can write your answer like this as well. Now here's the last question for the day.
Let's say if you receive a problem that looks like this. How would you solve it? Now instead of separating it into two equations, you can solve this system all at once. So you can subtract all three sides by 5. Negative 3 minus 5 is negative 8. And 19 minus 5 is 14. And these 5s, they cancel. So now you can divide all three sides by 2. So negative 8 divided by 2 is...
negative 4 in the middle we have X 14 divided by 2 is 7 so to plot the solution on a number line you can see that X is between negative 4 and 7 so let's put 0 on this side. So we have an open circle at 7 but a closed circle at negative 4 and X is between these two points. In interval notation it's negative 4 to 7 but brackets for negative 4. So that's what you can do if you have an equation like that. You can simultaneously solve it. So that is it for this video.
Thanks for watching and have a great day.