- 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
Sal attempts to solve 8(3x + 10) = 28x - 14 - 4x only to find that the equation has no solution. Created by Sal Khan and Monterey Institute for Technology and Education.
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- did i do something wrong i only got the number left
Solve for X
8(3x + 10) = 28x – 14 – 4x
Multiplied the numbers in the brackets
24x + 80 = 28x – 14 – 4x
Compressed the 2 ‘x’ numbers left side
24x + 80 = 24x – 14
got rid of the 14 on the right by add 14 to both sides and Took away the 24 by x on both sides
- 24𝑥 − 24𝑥 + 94 = 24𝑥 − 24𝑥
0 + 94 = 0
94 = 0
This is of course a false statement and means that the equation doesn't have a solution.
No matter what value of 𝑥 we put into the equation we will never get equality.(19 votes)
Concludion: 2=3(2 votes)
- You have a false conclusion.
The inverse property of addition tells us that if you subtract a number from itself, it will always = 0.
It does not tell you that if you apply the property multiple times that the 2 numbers you used will equal each other.
2 will never equal 3.(15 votes)
- If you are solving a problem and you come to one of these special cases, can you just write null set? The symbol is like a 0 with a line crossing diagonally through it. On a test just drawing the symbol might be easier and less time consuming, hence my question.(3 votes)
- I'm not exactly sure if you can use that, some teachers may agree other may not. To be on the safer side I recommend that you say 'No Solution' as an answer. Since we couldn't find what 'x' was exactly there wasn't a solution to the answer.
I hope this helped.(8 votes)
- why are functions so hard for me even though that we have studied them for weeks(5 votes)
- is this actually a equation at all? I'm not sure(5 votes)
- What can help me with the quiz, Re watching the video or redoing it?(4 votes)
- When solving an equation of this specific type is the goal only to see if the beginning set of equalities are actually equal to each other? Would a test ever be looking for the value of x in a question like this? i know the problem is in my order of operations in that i keep combining the like terms of the numbers without a variable before the numbers with a variable. I'm led to believe that when solving for x (or the given variable) it would be correct to go in the order opposite of PEMDAS, whereas when given an equation of this type I'm supposed to go in order of like terms with the variable first, am I wrong in thinking this? I think a more valuable thing to ask is if I was given an equation of any type, what about a given equation would lead me to believe that this route would be the one to take? I hope that made sense to someone; I had trouble finding a good way to word this.(4 votes)
- I think Sal was talking way too fast, I didn't really get the term.(2 votes)
I just want to ask, At1:28what do Sal meant by these guys(2 votes)
Solve for x. We have 8 times the quantity 3x plus 10 is equal to 28x minus 14 minus 4x. So like every equation we've done so far, we just want to isolate all of the x's on one side of this equation. But before we do that, we can actually simplify each of these sides. On the left-hand side, we can multiply the quantity 3x plus 10 times 8. So we're essentially just distributing the 8, the distributive property right here. So this is the same thing as 8 times 3x, which is 24x, plus 8 times 10, which is 80, is equal to-- and over here, we have 28x minus 14 minus 4x. So we can combine the 28x and the minus 4x. If we have 28x minus 4x, that is 24x And then you have the minus 14 right over here. Now, the next thing we could-- and it's already looking a little bit suspicious, but just to confirm that it's as suspicious as it looks, let's try to subtract 24x from both sides of this equation. And if we do that, we see that we actually remove the x's from both sides of the equation because we have a 24x there, and we have a 24x there. You might say, hey, let's put all the x's on the left-hand side. So let's get rid of this 24x. So you subtract 24x right over there, but you have to do it to the left-hand side as well. On the left-hand side, these guys cancel out, and you're left with just 80-- these guys cancel out as well-- is equal to a negative 14. Now, this looks very bizarre. It's making a statement that 80 is equal to negative 14, which we know is not true. This does not happen. 80 is never equal to negative 14. They're just inherently inequal. So this equation right here actually has no solution. This has no solution. There is a no x-value that will make 80 equal to negative 14.