Ask your own question, for FREE!
Mathematics 17 Online
OpenStudy (anonymous):

[02.05] Solve for x: 4 − (x + 2) < −3(x + 4)

OpenStudy (sleepyhead314):

Hello @peanutball2001 and Welcome to OpenStudy! :) to solve for x, first use the distributive property to get rid of the parenthesis can you do that? :3

OpenStudy (anonymous):

i can on the right side but not the left

OpenStudy (sleepyhead314):

that's ok let's pretend the - (x + 2) is just a + (-1)(x + 2) so on the left we have 4 + (-1)(x + 2) can you distribute the (-1) into the (x + 2) ? :)

OpenStudy (anonymous):

-x and -2?

OpenStudy (sleepyhead314):

right! so on the left we now have 4 - x - 2

OpenStudy (anonymous):

okay so would you end up simplifying that to 2x?

OpenStudy (sleepyhead314):

hmm? 4 - x - 2 < -3x - 12 then add like terms then add 3x to both sides so yes kinda, you'll get 2x on the left so what did you get on the right? :)

OpenStudy (anonymous):

5x< -12??

OpenStudy (sleepyhead314):

no, the 2x was correct for the left side 4 - x - 2 < -3x - 12 2 - x < -3x - 12 +3x +3x 2 + 2x < -12 now add 2 to both sides :)

OpenStudy (anonymous):

4+2x<-10

OpenStudy (sleepyhead314):

*scratches head* hmmm OH whoops >,< sorry I meant to subtract 2 from both sides 2 + 2x < -12 -2 -2

OpenStudy (anonymous):

x<-7

OpenStudy (sleepyhead314):

correct! :D

OpenStudy (anonymous):

okay thanks you the 4-,,,,,, kinda trew me off

OpenStudy (sleepyhead314):

sorry about that >,< I'z kinda sleepy :P

OpenStudy (anonymous):

NO YOUR FINE! i ment in the problem.

OpenStudy (sleepyhead314):

oh that part :P yeah, that was just a subtraction ^_^

OpenStudy (anonymous):

i understand now.

OpenStudy (sleepyhead314):

here's a tutorial on how to use this website better http://openstudy.com/study#/updates/543de42fe4b0b3c6e146b5e8 and if you ever want to tag me to a question just type @sleepyhead314 :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!