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### Course: Algebra 1 > Unit 2

Lesson 3: Analyzing the number of solutions to linear equations- Number of solutions to equations
- Worked example: number of solutions to equations
- Number of solutions to equations
- Creating an equation with no solutions
- Creating an equation with infinitely many solutions
- Number of solutions to equations challenge

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# 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 ?(25 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.(23 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.(16 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.(15 votes)

- Is there any simple trick to find the equation which has no solution without even solving it(10 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.(10 votes)

- Am I aloud to just use photomath this is confusing.(10 votes)
- That's if your teacher is okay with it.But if you need help, your teacher can probably help.(4 votes)

- What are the energy points used for?(6 votes)
- You use energy points to upgrade your avatar and achieve some badges. But, even though you may think the points are useless, there are incoming updates to Khan Academy, so stay tuned!(12 votes)

- Is there any real world application for making an equation with no solution?(6 votes)
- No, there can't be, because it wouldn't exist. If there is no solution, there can't be an existance.(5 votes)

- I cant wrap my head around this. In the form ax+b=cx+d ... if a!=c then there is apparently only one solution. That means both if ( a!=c AND b=d ) Or ( a!=c AND b!=d) in either case the equation is supposed to have just one result. But for the first option I can rearrange to cancel out b and d so

ax = cx

Now if if I solve for any value of x I get two different values on both sides of the equality, and in my head this would surely indicate that there is no solution.

I can do the same for the second option with the added step of moving b and d around, but still ending up with different values on either side of the equality.

So what have I missed. Is it to do with visualising the equation on a graph or is it some other more obvious fallacy. Im just missing something important about how to think about these equations?(3 votes)- The thing you are missing is if you get to ax = cx when b=d, you can solve for x.

Subtract either cx or ax from both sides.

ax - cx = 0

"a" and "c" are different values, so when subtracted, they would create some new number (but not zero). Let's call this new number "n" where a - c = n.

Then, ax - cx would = nx.

nx = 0

Divide by n

x = 0/n = 0. (one solution)

Try it. Plus in different values for "a" and "b" and work thru the steps. x = 0 would be solution for when a!=c AND b=d.(8 votes)

- What do you not get?(6 votes)

- whos gonna make a equation with no solution why would u use that?(5 votes)
- an equation with no solutions is a wrong equation(1 vote)

- 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?(2 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)

## 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.