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Algebra 1

Course: Algebra 1 > Unit 2

Lesson 3: Analyzing the number of solutions to linear equations
Math
Algebra 1
Solving equations & inequalities
Analyzing the number of solutions to linear equations
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Creating an equation with no solutions

CCSS.Math:
8.EE.C.7
,
8.EE.C.7a
Google Classroom
Sal shows how to complete the equation -11x + 4 = __x + __ so that it has no solutions. Created by Sal Khan.

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  • blobby green style avatar for user Alexandra Hamilton
    Posted 5 years ago. Direct link to Alexandra Hamilton's post “How do I find the value o...”
    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)
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    • stelly blue style avatar for user Kim Seidel
      Posted 5 years ago. Direct link to Kim Seidel's post “I think you are saying th...”
      Good Answer
      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.
      (13 votes)
  • leaf green style avatar for user Anonymous
    Posted 5 years ago. Direct link to Anonymous's post “Wait. Equations with no s...”
    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.
    (10 votes)
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    • leafers ultimate style avatar for user Scott R.
      Posted 5 years ago. Direct link to Scott R.'s post “That is actually almost t...”
      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.
      (9 votes)
  • leaf blue style avatar for user Logeswaran Sach
    Posted 2 years ago. Direct link to Logeswaran Sach's post “Is there any simple trick...”
    Is there any simple trick to find the equation which has no solution without even solving it
    (5 votes)
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    • mr pink green style avatar for user David Severin
      Posted 2 years ago. Direct link to David Severin's post “This trick is based on si...”
      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.
      (6 votes)
  • orange juice squid orange style avatar for user Jon Winder
    Posted 7 years ago. Direct link to Jon Winder's post “Is there any real world a...”
    Is there any real world application for making an equation with no solution?
    (6 votes)
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  • aqualine ultimate style avatar for user Lindsay MacVean
    Posted 7 years ago. Direct link to Lindsay MacVean's post “I cant wrap my head aroun...”
    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)
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    • stelly blue style avatar for user Kim Seidel
      Posted 7 years ago. Direct link to Kim Seidel's post “The thing you are missing...”
      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.
      (6 votes)
  • leafers tree style avatar for user PeterM
    Posted 3 months ago. Direct link to PeterM's post “wow i dont get it”
    wow i dont get it
    (3 votes)
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  • duskpin ultimate style avatar for user busyargirl
    Posted 8 months ago. Direct link to busyargirl's post “How do I know which to st...”
    How do I know which to start by cancelling out on a quiz? The constants or the coefficients and their variables? Or should I work the equation twice, each time cancelling out a different part first?
    (2 votes)
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    • stelly blue style avatar for user Kim Seidel
      Posted 8 months ago. Direct link to Kim Seidel's post “The properties of equalit...”
      The properties of equality are vary flexible - there is no strict sequence. Many students find it less confusing it they move the variables (and their coefficients first. You would then have the variable on one side and it is clear that anything other than the variable must be moved to the other side. In an equation with no solution, move the variable to one side will result in the variable adding to zero on both sides. You are then left with constants only on both sides. If the sides are not equal, the equation has no solution. If the constants are equal, then the equation has a solution of all real numbers.

      Hope this helps.
      (6 votes)
  • blobby green style avatar for user shawn.neidlow
    Posted a month ago. Direct link to shawn.neidlow's post “Guys, I think Sal shows h...”
    Guys, I think Sal shows how to complete the equation -11x + 4 = _x + _ so that it has no solutions.
    (4 votes)
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  • duskpin tree style avatar for user Pan
    Posted 9 years ago. Direct link to Pan's post “why would we want to make...”
    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?
    (1 vote)
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    • leaf grey style avatar for user Arzal
      Posted 3 years ago. Direct link to Arzal's post “Wow, this comment is old,...”
      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.
      (4 votes)
  • blobby green style avatar for user Kelli Diaz
    Posted a year ago. Direct link to Kelli Diaz's post “How is this fun at all”
    How is this fun at all
    (4 votes)
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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.