Main content

## Algebra 1 (Eureka Math/EngageNY)

### Unit 1: Lesson 9

Topic C: Lesson 14: Solving inequalities- Testing solutions to inequalities
- Testing solutions to inequalities
- Plotting inequalities
- Plotting an inequality example
- Plotting inequalities
- One-step inequalities examples
- One-step inequalities: -5c ≤ 15
- One-step inequalities
- Two-step inequalities
- Two-step inequalities
- Inequalities with variables on both sides
- Inequalities with variables on both sides (with parentheses)
- Multi-step inequalities
- Multi-step linear inequalities

© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# One-step inequalities: -5c ≤ 15

In addition to solving the inequality, we'll graph the solution. Remember to swap if you mutiply both sides of the inequality by a negative number. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- Why does the inequality get flipped when you multiply both sides by a negative?(65 votes)
- It is the rule.

When you mult by - the value goes opposite so the sign that says what is opposite must change

x < 5......... lets say x is 4

-x > -5 so that means -4 is greater than -5

Remember the number line and see that as you go to the right things get bigger.

That is the point of using the number line.(94 votes)

- couldn't you just divide both by -5 and get the same answer?(26 votes)
- yeah i see that now, i think u could! but he flipped the sign and i think that means something(0 votes)

- How could this be used in everyday life?(20 votes)
- When you are trying to find the max profit or minimum scores you need to get in order to get an A. Something like that.(5 votes)

- So why exactly must we flip the signs?(6 votes)
- Think of it this way: if I have -3 x< 9, suppose I do not like to flip inequality signs by dividing by a negative Starting with the equation -3x<9

My other choice is to move my x by opposites

-3x + 3x < 9 + 3x

0< 9 +3x I still do not have x isolated, so I subtract 9

0-9<9-9+3x

-9 < 3x divide by 3

-9/3 < 3/3x

-3 < x if -3 is less than x, then when I flip this equation

x > -3 x must be greater than -3

Starting from beginning -3x < 9 divide by -3 and flip inequality sign

-3/-3x > 9/-3

x > -3

It is easier to remember to flip the inequality sign than to go through the whole process of moving my x to make it positive which makes me move everything else also - the difference between one step and multiple steps

Hope that helps and you understand the inequality flip now(16 votes)

- how did sal get 15 times -1/5 = -3 im confused(5 votes)
- Multiplying by a fraction is the same as dividing by the reciprocal of the fraction. (A reciprocal is the number you get if you switch the numerator and denominator. So, the reciprocal of 2/3 is 3/2 and the reciprocal of 1/5 is 5.)

So, 15 × (-⅕) = 15 ÷ (-5) = -3(15 votes)

- can't you just divide instead of using the inverse I've tested this on multiple problems and it seems to work every time? just wondering now instead of figuring out I can't do this down the road once problems become more complicated. Thanks.(3 votes)
- Dividing by -5 and multiplying by -1/5 are exactly the same operation and create the same result.

For example: -5c * -1/5 = -5c/(-5) which is division by -5.

So, use the one you are comfortable with.(9 votes)

- can't we divide both sides by 5(2 votes)
- You can, but you're left with "-c", not "c". You're solving for "c", not "-c".

Dividing both sides of "-5c ≤ 15" by 5 gives "-c ≤ 3". Perhaps you're uncomfortable with "flipping" the equality.

If you're still uncomfortable with flipping the side then a way to see why.

-c ≤ 3 [ The "-c" expression ]

0 ≤ 3 + c [ Add "c" to both sides ]

-3 ≤ c [ Subtract "-3" from both sides ]

Now the "c" is on the greater than side of the expression.

Flipping the sign is the correct procedure.

I hope this helps.(5 votes)

- What if you were to add a negative number to both sides? Would you flip the sign then?

Like in x+(-8)>6(3 votes)- No. You only switch the sign when multiplying or dividing by a negative.(3 votes)

- in0:32where does the 1/5 comes from(3 votes)
- The -1/5 works because you have to divide -5c do get c on its own. You can do it 2 different ways. You can either multiply by -1/5 , or you can divide it by -5. Both ways work, so you can decide which way you would want to do it.(2 votes)

- what if you multiply a negative number? does the sign flip(2 votes)
- Yes, adding to Allison's answer, because you would technically have to divide by a negative in a case like that. For example in -x/8>10 (I'm going to go step by step -not going straight to multiply each side by -8) Multiply each side by 8 to get -x>80, to isolate x from here, you have to divide by a negative 1, so the sign does switch. Hope this clears up any confusion you have on this(3 votes)

## Video transcript

Solve for c and
graph the solution. We have negative 5c is
less than or equal to 15. So negative 5c is less
than or equal to 15. I just rewrote it a
little bit bigger. So if we want to
solve for c, we just want to isolate the c
right over here, maybe on the left-hand side. It's right now being
multiplied by negative 5. So the best way to just have
a c on the left-hand side is we can multiply both
sides of this inequality by the inverse of negative
5, or by negative 1/5. So we want to multiply
negative 1/5 times negative 5c. And we also want to multiply
15 times negative 1/5. I'm just multiplying both
sides of the inequality by the inverse of negative 5,
because this will cancel out with the negative 5 and
leave me just with c. Now I didn't draw
the inequality here, because we have to remember,
if we multiply or divide both sides of an inequality
by a negative number, you have to flip the inequality. And we are doing that. We are multiplying both
sides by negative 1/5, which is the equivalent of dividing
both sides by negative 5. So we need to turn
this from a less than or equal to a
greater than or equal. And now we can
proceed solving for c. So negative 1/5 times
negative 5 is 1. So the left-hand
side is just going to be c is greater than or
equal to 15 times negative 1/5. That's the same thing as
15 divided by negative 5. And so that is negative 3. So our solution is c is greater
than or equal to negative 3. And let's graph it. So that is my number line. Let's say that is 0, negative
1, negative 2, negative 3. And then I could go above, 1, 2. And so c is greater than
or equal to negative 3. So it can be equal
to negative 3. So I'll fill that
in right over there. Let me do it in a
different color. So I'll fill it in
right over there. And then it's
greater than as well. So it's all of these values
I am filling in in green. And you can verify that it works
in the original inequality. Pick something that should work. Well, 0 should work. 0 is one of the numbers
that we filled in. Negative 5 times 0 is 0, which
is less than or equal to 15. It's less than 15. Now let's try a number
that's outside of it. And I haven't drawn it here. I could continue with the
number line in this direction. We would have a negative 4 here. Negative 4 should
not be included. And let's verify that
negative 4 doesn't work. Negative 4 times negative
5 is positive 20. And positive 20 is
not less than 15, so it's good that we did
not include negative 4. So this is our solution. And this is that
solution graphed. And I wanted to do that
in that other green color. Here you go. That's what it looks like.