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A compound inequality with no solution

Sal solves the compound inequality 5x-3<12 AND 4x+1>25, only to realize there's no x-value that makes both inequalities true. Created by Sal Khan and Monterey Institute for Technology and Education.

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  • blobby green style avatar for user coryhunnicutt1
    my question is whats the point of this. when will i use this in the real world lmao
    (6 votes)
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  • starky sapling style avatar for user Gabriel870
    At that point couldn't you bend the number line like you can bend space?
    I know you can't, but still.
    (5 votes)
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  • duskpin seedling style avatar for user Isabella  Braquet
    how do you know when to switch the inequality symbol?
    (3 votes)
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  • blobby green style avatar for user Goldfish
    can there be a no solution for an OR compound inequality or is it just for AND compound inequalities?
    **If YES to no solution for OR compound inequalities can you provide an example Please?
    (2 votes)
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    • starky ultimate style avatar for user KLaudano
      If any of the inequalities in the compound OR inequality have a valid solution, the compound OR inequality will also have a valid solution.

      The inequality below has no solutions because x^2 + 1 is never less than 0 and -x^2 - x - 2 is never greater than 0.
      x^2 + 1 < 0 OR -x^2 - x - 2 > 0
      (2 votes)
  • male robot hal style avatar for user Seifelkhashab
    What does itmean that all numbers are real, also I keep on getting the answer right but I can't find it in the choices?
    (3 votes)
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  • stelly blue style avatar for user chiragaugusta
    The inequalities numbers like ex: 5x-3<12 and 4x+1>25 so that mean you will substitute the number right @Sal Khan
    (2 votes)
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  • piceratops ultimate style avatar for user flamethrower  🔥⚡
    @ sal says that there is no solution to the example equation, but i was wondering if it did have a solution like 1/ 0 as anything by zero gives infinity or negative infinity. really crazy question but just asking
    (2 votes)
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    • leafers ultimate style avatar for user NathanFischer
      Let's assume that when solving for any equation - or "x" in this case - the answer comes out to be "1/0". If this happens, the answer is thus undefined and there is no solution.

      Therefore, to help you clarify, anything divided by zero - as with the case of 1/0 - is NOT infinity or negative infinity. It is simply undefined. More accurately, it would be better to say in your above statement that anything which APPROACHES 1/0 is positive infinity or negative infinity. Numbers that approach 1/0 would be something like "1/0.1", "1/0.001", or "1/0.000001" - where the last example number would equal to 1,000,000. So, here in the example, we are able to show that as the denominator get closer and closer to zero, the fraction as a whole get closer and closer to a really BIG number - or infinity. However, when the denominator becomes zero, it is NOT infinity but an undefined number.

      Here's a khanacademy video that explains this nicely: https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:foundation-algebra/x2f8bb11595b61c86:division-zero/v/why-dividing-by-zero-is-undefined

      However, if you want to get more in-depth, here's an amazing and easy to follow animated TED-Ed video that explains the whole idea in less than five minutes REALLY well: https://www.youtube.com/watch?v=NKmGVE85GUU

      Hope this helps!
      (2 votes)
  • blobby green style avatar for user Tatenda2007
    What do you do if you have a fraction in one of your constraints
    (2 votes)
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    • stelly blue style avatar for user Kim Seidel
      I assume you mean that you have already solved each inequality and one of your inequalities has a fraction as its solution. You use the fraction in the same way that you use integers. Fractions sit in between the integers on a number line. You should be able to graph that inequality starting at the fraction and put the appropriate arrow in place just like for an integer. Then, you can do the union / intersection of the two inequalities. For example:
      x < 3/4 and x > -4 has an intersection of: -4 < x < 3/4

      If this doesn't answer your questions, comment back. If you need help with placing fractions on a number line, use the search bar and search for "fractions on a number line" or "mixed numbers on a number line" or "decimals on a number line". There are videos on KA that given examples for how to do them.
      (2 votes)
  • hopper jumping style avatar for user littlesisiscool
    How do you eliminate options in the problems. What is the difference between AND and OR? I am REALLY struggling with this concept. Its like math block. I feel like I've never struggled more with a concept than this one. AAAH! Please help.
    (2 votes)
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    • stelly blue style avatar for user Kim Seidel
      The word AND tells you to find the intersection of both solution sets. An intersection is the solutions in common, or that overlab.

      The word OR tells you to find the union of the 2 solution sets. A union is 2 sets combine all possible solutions from both sets.

      To learn more about these, search for "intersection and union of sets". There is a video on KA that walks you thru them.
      (3 votes)
  • hopper cool style avatar for user Johnathon Aaron
    Sal states that there is no solution, but what if x was a function of some sorts or a liner equation with multiple places on the number line that fall into the constraints both less then 3 and greater than 6? example, a solution set of (2,7)
    (2 votes)
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Video transcript

Solve for x, 5x - 3 is less than 12 "and" 4x plus 1 is greater than 25. So let's just solve for X in each of these constraints and keep in mind that any x has to satisfy both of them because it's an "and" over here so first we have this 5 x minus 3 is less than 12 so if we want to isolate the x we can get rid of this negative 3 here by adding 3 to both sides so let's add 3 to both sides of this inequality. The left-hand side, we're just left with a 5x, the minus 3 and the plus 3 cancel out. 5x is less than 12 plus 3 is 15. Now we can divide both sides by positive 5, that won't swap the inequality since 5 is positive. So we divide both sides by positive 5 and we are left with just from this constraint that x is less than 15 over 5, which is 3. So that constraint over here. But we have the second constraint as well. We have this one, we have 4x plus 1 is greater than 25. So very similarly we can subtract one from both sides to get rid of that one on the left-hand side. And we get 4x, the ones cancel out. is greater than 25 minus one is 24. Divide both sides by positive 4 Don't have to do anything to the inequality since it's a positive number. And we get x is greater than 24 over 4 is 6. And remember there was that "and" over here. We have this "and". So x has to be less than 3 "and" x has to be greater than 6. So already your brain might be realizing that this is a little bit strange. This first constraint says that x needs to be less than 3 so this is 3 on the number line. We're saying x has to be less than 3 so it has to be in this shaded area right over there. This second constraint says that x has to be greater than 6. So if this is 6 over here, it says that x has to greater than 6. It can't even include 6. And since we have this "and" here. The only x-es that are a solution for this compound inequality are the ones that satisfy both. The ones that are in the overlap of their solution set. But when you look at it right over here it's clear that there is no overlap. There is no x that is both greater than 6 "and" less than 3. So in this situation we have no solution.