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anonymous
 one year ago
will give medals but pleaseeee help.
Solve for x: 4 − (x + 2) < −3(x + 4)
x < −7
x > −7
x < −9
x > −9
anonymous
 one year ago
will give medals but pleaseeee help. Solve for x: 4 − (x + 2) < −3(x + 4) x < −7 x > −7 x < −9 x > −9

This Question is Closed

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0This is basically solving for x in a regular equation. So what you would just do is instead of having the less than symbol, you would place the equal sign.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@calculusxy wait ill do the work and show you where i am stuck at

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i don't know how to do with the other side do i take away the brackets and flip the sign of 4 to + and add the like terms ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0after the negative sign there is a 1 so multiply negative one by x and 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it would be 1x 2 ? @joquez14

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0Just like what @joquez14 said. what we can think is that before the parentheses, there is a 1 (negative because of the minus sign). \[41(x+2) <3x + 12\] that means that you would have to do \[1 \times x = x\]and \[1 \times 2 = 2\]

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0Now we can simplify it even further wit those two terms. \[4 x2<3x12\]

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0Sorry i meant 12 not +12

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so we add together 4 and 2 @calculusxy

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0yes to simplify that side

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0combine the like terms to make solving this easier

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so we add the x's

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0so this is where you need to really cancel out an x from a side.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it would be 6<2x

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0\[4x2=3x  12\] \[6  x=3x  12\] \[6 =2x  12\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but then the answer would 4 and none of my answer choices are 4 ? did i do an error ?

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0what are your answer choices then?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its not supposed to ve  2x i thought i think its supposed to be  4x

calculusxy
 one year ago
Best ResponseYou've already chosen the best response.0@joquez14 if we did 3 + 1 it would equal to 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yup thats what I'm getting 2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why you add one x was negative so u souldnt have added an positive 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0aren't you suppose to do the inverse

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im not to sure since were combining like terms

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I wish I could give you all medals. You all have the right idea, it just looks like you got mixed up with the combining 4 − (x + 2) < −3(x + 4) Start by distributing 4  x  2 < 3x  12 Combine like terms on the left 2  x < 3x  12 Add 3x to both sides 2 + 2x < 12 Subtract 2 from both sides 2x < 14 Divide both sides by 2 x < 7

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you so much guys literally all of you have been such a tremendous helppppppppp @joquez14 @calculusxy and thanks for clearing it up @peachpi
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