Main content

## Algebra 1

### Course: Algebra 1 > Unit 2

Lesson 5: Multi-step inequalities# Inequalities with variables on both sides (with parentheses)

Sal solves the inequality 5x+7>3(x+1), draws the solution on a number line and checks a few values to verify the solution. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- say you have to graph an inequality,once you solve the equation such as :2r+5<19 would be 2 times 7 +5=19 right

but after that when you graph this on a # line how do you know which # to put the hollow or solid circle above?(24 votes)- If the it's just < or >, then you draw a hollow circle because your not including that point. x>4.You are not including 4 on the number line, but all the points greater than 4. x<3. You are not including 3 on the number line but all the points less than 3.

If the inequality sign is greater than or equal to or less than or equal to, then you shade the dot because your including that point. x is greater than OR equal to 4. X can be greater than 4 OR it can be equal to 4, so since 4 is one of the solutions, you need to use the solid dot. If x is less than or equal to 3, then you shade the dot because three is part of the solution set, x is greater than OR equal to 3.(8 votes)

- Do you also Swap The Symbol if you're ADDING or SUBTRACTING by a negative number?(9 votes)
- No, because adding and subtracting doesn't really make one side bigger than another if the original was the opposite. For example, 1 < 2 times -1 = -1 > -2. but 1+1 < 1 + 2 keeps the sign, because nothing except the numbers changed.(3 votes)

- At3:40couldn't you subtract 3 instead of 7?(7 votes)
- no, because if you subtract three there will still be the seven (which would become four) on the other side, and you want to get rid of that.(2 votes)

- I guess "false" and "no solution" are the very close, if not identical, and close also to "undefined" in meaning. E.g.:

Is "y = x/0" false? Is the system of equations "y = 3x and y = 3x+1" false? (I think so, since they can't both be true (for the same x). "4 < 3" seems to be just false, and for this, "no solution" seems inappropriate.

The system "2y = 2x+2 and 7y = 7x+7" is true for all x. I don't understand how "2 < 3" is true for all x when there is no x in the inequality. To me it's just a true statement about 2 and 3.

Does anyone have any thoughts about these things one way or the other?(3 votes)- Note: The following is from my own thought. You can agree or disagree with me.
`y=x/0`

is not necessarily false. Dividing by 0 is undefined. It cannot be wrong should there be no right.

Now let's talk about`2<3`

. Let's say you have an inequality and you manage to get to this point. Since 2 IS less than 3, that solution (or inequality without a variable) would be true.

For example, we can have`4<6+2`

. Simplify that and you will get`1<2`

, which proves the inequality is true.

I would also note that "no solution" and "false" have similar meanings. I will illustrate this.

In`4<3`

, 4 is obviously not less than 3. I would, however, say it is "false", since there are no variables to make 3 greater than 4 or 4 less than 3.

However, for`y=3x`

and`y=3x+1`

, I would say there is "no solution", since (for the same x) there is no way to make the equations equal to each other.

My conclusion is that "false" and "no solution" have similar but not quite the same meanings. "Undefined" has a completely different meaning from "false" and a rather different meaning compared to "no solution."

Please tell me what you think about my thought.(7 votes)

- When I do my math, my sign comes out flipped. Am I doing something wrong?

5x+7 > 3(x+1)

5x+7 > 3x+3

2x > -4

x < -2(2 votes)- It should not be flipped. You only need to flip the sign when you multiply or divide both sides by a negative number. In that last step, you are dividing by 2 which is a positive number. So your sign should not be flipped. Hope this helps!(8 votes)

- in equation we do things on both side so its true. since inequations < are not equation= why we apply same rules(3 votes)
- The rules are not exactly the same. The rules used maintain the relationship of the 2 sides of the inequality.

1) If we add/subtract the same value to both sides of an inequality, the relationship is unchanged. For example:

2<5 becomes 6<9 if we add 4 to both sides. The left side is still less then the right side.

2) If we multiply or divide both sides by the same**positive**value, the relationship is unchanged. For example:

3<9 becomes 6<18 if we multiply both sides by +2. The left side is still less than the right side.

3) This is the rule that is different. If we multiply or divide both sides by the same**negative**value, the relationship between the numbers reverses. So, we change the direction of the inequality. For example:

-2<7 becomes 4>-14 if we multiply both sides by -2. The left side is now larger than the right side, so we reverse the inequality.

Hope this helps.(5 votes)

- My sign comes out flipped. What am I doing wrong?

5x+7>3x+3 /-3

5x+4>3x /-5x

4>-2x /:-2

-2<x(0 votes)- You did it it correctly. You are right to flip the sign when multiplying or dividing by a negative number.(10 votes)

- So, about the open circle thing, does it only work on negative numbers or just in this case? And the filled in circle are for positive numbers?(2 votes)
- The open circle has to do with inequalities < and > where the value that is circled does not count. The closed circle has to do with inequalities ≥ and ≤ where the point counts. It has nothing to do with the sign of the number.(3 votes)

- honestly i dont like these vids cause they talk too much and this guy repeats himself like 8 times 1/10(3 votes)
- In one of the sets of practice I got to the following result;

P= -q-11/5q-5r+1

Why do you simplify further by multiplying by -1? How does that make it simpler. It seems to just flip the positive and negative values. Also when the denominator has some positive values and some negative values how do you determine when to multiply by -1 to make it positive?(3 votes)- the variable needs to be positive for the answer to be correct. multiplying a negative by a negative makes the variable positive.(2 votes)

## Video transcript

Solve for x. And we have 5x plus 7 is greater than 3 times x plus 1. So let's just try to isolate "x" on one side of this inequality. But before we do that, let's just simplify this righthand side. so we get 5x plus 7 is greater than - let's distribute this 3. So 3 times x plus 1 is the same thing as 3 times x plus 3 times 1 so it's going to be 3x plus 3 times 1 is 3. Now if we want to put our x's on the lefthand side, we can subtract 3x from both sides. That will get rid of this 3x on the righthand side. So let's do that. Let's subtract 3x from both sides, and we get on the lefthand side: 5x minus 3x is 2x plus 7 is greater than - 3x minus 3x - those cancel out. That was the whole point behind subtracting 3x from both sides - is greater than 3. Is greater than 3. No we can subtract 7 from both sides to get rid of this positive 7 right over here. So, let's subtract, let's subtract 7 from both sides. And we get on the lefthand side... 2x plus 7 minus 7 is just 2x. Is greater than 3 minus 7 which is negative 4. And then let's see, we have 2x is greater than negative 4. If we just want an x over here, we can just divide both sides by 2. Since 2 is a positive number, we don't have to swap the inequality. So let's just divide both sides by 2, and we get x is greater than negative 4 divide by 2 is negative 2. So the solution will look like this. Draw the number line. I can draw a straighter number line than that. There we go. Still not that great, but it will serve our purposes. Let's say that's -3, -2, -1, 0, 1, 2, 3. X is greater than negative 2. It does not include negative 2. It is not greater than or equal to negative 2, so we have to exclude negative 2. And we exclude negative 2 by drawing an open circle at negative 2, but all the values greater than that are valid x's that would solve, that would satisfy this inequality. So anything above it - anything above it will work. And let's just try, let's try just try something that should work. and then let's try something that shouldn't work. So 0 should work. It is greater than negative 2. It's right over here. So, let's verify that. 5 times 0 plus 7 should be greater than 3 times 0 plus 1. So this is 7 - 'cause this is just a 0 - 7 should be greater than 3. Right. 3 times 1. So 7 should be greater than 3, and it definitely is. Now let's try something that should not work. Let's try negative 3. So 5 times negative 3... 5 times negative 3 plus 7, let's see if it is greater than 3 times negative 3 plus 1. So this is negative 15 plus 7 is negative 8 That is negative 8. Let's see if that is greater than negative 3 plus 1 is negative 2 times 3 is negative 6. Negative 8 is not - is not greater than negative 6. Negative 8 is more negative than negative 6. It's less than. So, it is good that negative 3 didn't work 'cause we didn't include that in our solution set. So we tried something that is in our solution set and it did work. And something that is not, and it didn't work. So we are feeling pretty good.