Solve for x: |x| − 8 = −5 x = −13 and x = −3 x = 3 and x = −3 x = 3 and x = 13 No solution
|x| - 8 = -5 Add 8 to both sides. -5 + 8 = 3 x=3 Since we know that the absolute value bars also mean the same thing as the distance from the number to 0, the absolute value of a number is always positive. We can say |-3| becuase that will equal to positive 3 (which is x). Your answer can be both |-3| and/or |3|.
So B?
Yes
ok Thank you can u help with a few more?
Sure
I'd be more comfortable (a lot more) NOT using those absolute value symbols in stating your solutions. I'd write "the solutions are 3 and -3."
Solve for x: -3|x + 7| = -12 x = 5 over 3, x = −19 over 3 x = −3, x = −11 x = −3, x = 11 No solution
The original equation is true if you substitute either -3 or 3 for x.
ok.
Do you know how to solve for x in an equation becuase all of these equations are really the same concept?
Wouldnt you distribute the -3?
Yes towards x and 7.
so -3x + -21
Yes
So then would i add 12?
Then you can solve for x from there becuase you know the solution is -12.
Shade: Here's an alternative approach: Divide the original equation by -3. Then we have the simpler |x+7| = 4. Please verify that the two solutions to this are -3 and -11.
I got that so its right -3 and -11?
No. Since you want to isolate the variable to get the original amount. To do that, you want to take out any integer that there are before or after it to only have the coefficient by itself. That means that you have to instead subtract (the inverse of addition) -21 to 12. Using that sum you can then divide by 3 (since multiplication is the inverse of division).
|x| − 8 = −5 + 8 + 8 |x| = 3 ^ That would be one solution -
(1) What do you propose is the correct response to this question? (2) Are -3 and -11 solutions to the original absolute value problem? (3) Are -3 and -11 solutions to the simplified absolute value problem |x+7| = 4?
Oh wait - Wrong problem LOL
Shade, yes, I believe -3 and -11 are the solutions we were looking for. Unforunately, calculusxy disagrees. See my questions (2) and (3) (above). Best wishes to you.
So what is the correct answer yall have me soo confused
OK so i have one more i need to ask
Solve for x: |2x + 6| − 4 = 20 x = 9 and x = 11 x = −9 and x = 15 x = 9 and x = −15 No solution I believe the answer is no solution! Please tell me if i am incorrect
Would be x = -11 and x = -3 ---> http://www.wolframalpha.com/input/?i=+x%3A+-3%7Cx+%2B+7%7C+%3D+-12+solutions+ <--- As @mathmale Said :)
@mathmale I answered ShadePorter's question of "So then would I add -12?"
Shade, please check the proposed solutions yourself. Substitute x = -3 into the original absolute value equation. Is the result true or false? Substitute -11 and do the same. Is the result true or false?
@ShadePorter Looks like you made a mistake - There are two solutions there :)
Charlotte, thanks! Made my day. Good night, all. I need my beauty sleep. Solve absolute value problems in your dreams.
Take out the absolute value bars, since the positive numbers are already positive. Then solve for x.
@mathmale Really?! HUZZAH! Glad to have made your day - You just made mines! xD Goodnight and take care!
please tell me if i am correct on the "no solution problem"
I believe that there is only one solution which is is supposed to be 9. I don't find any other solutions to this other than 9.
@ShadePorter Not correct I'm afraid. You can do it! Try solving it again :)
All i got was 9 as well but all the multiple choice answers also have 11 and i have only got 9? Im soo lost
Or 15
I'll go with no solutions becuase none of the options above make sense other than 9.
Thank you soo much!
No problem :) if you like my answer, you can be my fan :)
Thanks :) have a good night
Let's revisit Shade's second question.| |2x+6| = 24 Dividing through by 2, |x+3| = 12 Case 1: Assume x+3 is already positive. then x+3=12 and x = 9. Case 2: Assume that x+3 is negative. Then the equation becomes -(x+3) = 12, or x+3 = -12, or x=-15. Verify both solutions by substituting them back into the original equation separately.
Join our real-time social learning platform and learn together with your friends!