.Which of the following equations has no real solutions? A. 9x - 3 = 9x - 3 B. 9(x - 3) = 9x - 27 C. 9x - 3 = 27x - 3 D. 9(x + 3) = 9(x + 11)
@armyengineer225 can u help me on this one?
First simplify each equations\
@nelsonjedi can u explain how to do that? sorry...
Let's do this one in steps: A) Subtract 9x from both sides, and you would have 0-3 = 0-3. Add 3 to both sides, and it is 0 = 6. B) Divide by 9, and you have x-3 = x-3. Again, 0 = 6. C) Subtract 9x from both sides, and you have -3 = 16x -3. 16x = -6, so x = -6/16. This one has a solution. D) Divide by 9, and x+3 = x+11. x = x+8, 0 = 8... Are you supposed to choose only one answer?
yes isnt there two that i could choose?
It appears that there should be 3 you can choose, A, B, and D, since a singular number cannot equal 6. Perhaps the question is, which one does have a solution? In which case, the answer would be C.
I'm sorry, I meant a singular number cannot equal 0.
@armyengineer225
i dont know why it says that... there is a possibility it could of been a typo
i went ahead and chose c it was wrong
Perhaps, though I may be wrong. You may want to talk to your teacher.
Ah I see, so they multiplied everything out.
oh... okay
Distribute to simplify and what do you have
well can anybody help me with this? @armyengineer225 @luckyted @nelsonjedi
solution to these equations: A) for every x in R, 9x - 3 = 9x -3, left side is exactly equals to right B) just like A. C) simplify to 9x=27x, so x = 0. D) no solution for x, such that 9x-27 = 9x-99. so the answer is D.
oh that makes sense thank you so much @luckyted
but i still need help with the last question i asked
(x/5)+2 = 4. Subtract 2 from 4, you have (x/5) = 2. Multiply by 5 to clear the fraction, you have x = 10.
thank you! @armyengineer225
No problem. Sorry about the first question!
it ok! :)
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