- 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
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.
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- 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)
- 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.(11 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.(10 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)
- 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.(4 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?(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)
- 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.(1 vote)
- 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.(2 votes)
- The equation -x+1 = x+1 have one solution, that is 0.
I solve this by first subtracting x from both sides of the equation and so on as Sal did in the videos.
However, if I first subtract 1 from both sides of the equation, I get -x=x.
I divided x into both sides, then I got -1=1 .
That means no solution because -1=1 is no true, -1 can never equal 1.
Can anyone explain this?(1 vote)
- Subtracting 1 from both sides is valid, but it would yield -x=x and that doesn't help. (Although if you think about it, 0 is the only real number where -0 and 0 are equal to one another so you could assume that x is 0 and you would be correct; however this is not mathematically sound)
Also, you don't divide anything when there's no multiplication in the equation (division is the inverse of multiplication).
One correct way to do this would be to add x to both sides then solve.
-x+1 = x+1
1 = 2x+1
0 = 2x
0 = x
The answer is x = 0.(4 votes)
- So we are solving one side to be inequal to another? There's infinite solutions for something with no solution, that is surely weird.(1 vote)
- Your question does not make sense, you are not solving one side to be inequal to the other. You are solving an equatio9n and finding that the variables cancel each other out leaving an equation can never be true. So there are not an infinite number of solutions, there are an infinite number of ways to show an equation with no solutions.(4 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)
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.