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## 8th grade (Illustrative Mathematics)

### Unit 4: Lesson 6

Lesson 7: All, some, or no solutions# Creating an equation with no solutions

Sal shows how to complete the equation -11x + 4 = __x + __ so that it has no solutions. Created by Sal Khan.

## Want to join the conversation?

- How do I find the value of a constant, such as (k) where there are no solutions? How would I solve it if the equation 4(80 + n) = (3k)n ?(17 votes)
- I think you are saying that you need to find a value of "k" so that the equation will have no solution.

For this to happen...

1) the coefficient of "n" must match on both sides of the equation

2) the constant on each side must be different.

Start by simplifying your equation -- distribute the 4: 320 + 4n = 3kn

The constants on each side are different: 320 on left, and 0 on right. So, one condition is met.

We now know that the coefficient of "n" must = 4. You can find "k" by setting 3k = 4 and solving for "k".

Hope this helps.(2 votes)

- Is there any simple trick to find the equation which has no solution without even solving it(5 votes)
- This trick is based on simplifying and as soon as you see the same coefficients of the variable on both sides and any different numbers on the two sides, you know that there are no solutions.

Example: 2(2x+7)= 5x +12 -x

Distribute on left to get 4x +14

Combine like terms on right to get 4x + 12

Since the coefficients of x are both 4, but the constants are different, you know there are no solutions because if you took it to the end, you would get 2=0 which can never be true.(1 vote)

- how many solution does 4(2x – 3) – 5 = 4x have(4 votes)
- The overall coefficient on x on the left-hand side is 4*2=8.

The overall coefficient on x on the right-hand side is 4.

These coefficients are unequal, so this linear equation has exactly one solution.(0 votes)

- why would we want to make a linear equation that has no solution, isn't the point of an equation to solve for the variable?(0 votes)
- Wow, this comment is old, where are you now? anyway, my guess is that by creating a linear equation with no solution at all we could instantly distinguish equations with no solution with just 1 step or less and thus saving time.(5 votes)

- Is there any real world application for making an equation with no solution?(1 vote)
- No, there can't be, because it wouldn't exist. If there is no solution, there can't be an existance.(5 votes)

- Wait. Equations with no solution cannot apply to something in real life because of the laws of thermodynamics, so if these equations have no real life use why are we learning about them at all. Or do they have a real life use.(2 votes)
- That is actually almost true, but the reason we learn them is to show that there are equations with no answer, but yes there is no real life application since you will never in real life with real problems ever really experience something with no solution.(2 votes)

- how do we find the equations with one solution(2 votes)
- The linear equations with exactly one solution are precisely those that have different overall coefficients on x on the two sides.(2 votes)

- For example in the inequality 75x+57= -75x+57 you end up with 150x=0. When it's equal to zero how many solutions does it have??(1 vote)
- 75𝑥 + 57 = −75𝑥 + 57 is an
*equation*that simplifies to 150𝑥 = 0

Dividing both sides by 150, we get 𝑥 = 0, which is the only solution.(3 votes)

- What is the website that the guy used for this video? I think that would be really good to practice on. Thanks!(2 votes)
- It is the old khan academy, if you used khan academy a few years ago it was like this :)(1 vote)

- If Sal put 4x + 4 instead of -11x + -11, would he still get the question right?(2 votes)
- No... if he used 4x+4, the solution would become x = 0, which is not the same as having no solution.(1 vote)

## Video transcript

We're asked to use
the drop-downs to form a linear equation
with no solutions. So a linear equation
with no solutions is going to be one where I don't
care how you manipulate it, the thing on the
left can never be equal to the thing on the right. And so let's see what
options they give us. One, they want us to-- we
can pick the coefficient on the x term and then
we can pick the constant. So if we made this
negative 11x, so now we have a negative
11x on both sides. Here on the left hand side,
we have negative 11x plus 4. If we do something other
than 4 here, so if we did say negative 11x minus
11, then here we're not going to have any solutions. And you say, hey, Sal how
did you come up with that? Well think about
it right over here. We have a negative 11x here,
we have a negative 11x there. If you wanted to solve
it algebraically you could add 11x to both sides
and both of these terms will cancel out with each other
and all you would be left with is a 4 is equal to
a negative 11, which is not possible for
any x that you pick. Another way that you
think about it is here we have negative 11
times some number and we're adding 4
to it, and here we're taking negative 11 times
that same number and we're subtracting 11 from it. So if you take a negative
11 times some number and on one side you add four,
and on the other side you subtract 11, there's no way, it
doesn't matter what x you pick. There's no x for which
that is going to be true. But let's check our
answer right over here.