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that what i got but hold on let me double check
But, how did you get it
yea im not sure whats the answer, but the way i was taught to do this is since after the equal sign there is another equation, you put a line in the middle to break it down, and you solve them both separately until its small enough to combine
you put a line in the middle of less than and equal too
hold on im going to draw it
so i got zero for the answer
its sloppy but do you get that?
im kinda confused im not sure if i should of stopped at -1
but i hope that helps
Maybe I wasn't clear, the inequality is to be solved with intervals.
oh you tight then, idk -1,0
such inequalities should be first broken into the 2 parts , x interval should be computed and then the intersection of the intervals is the answer.
\[ 4x < 2x+1 ≤ 3x+2 \] is short for two different relations: \[ 4x < 2x+1 \\ 2x+1 ≤ 3x+2 \] Except for divide by a negative number (which will flip the relation operator), you solve these the same way you solve an equality. the first relation: 4x < 2x + 1 add -2x both sides 4x -2x < 2x - 2x + 1 simplify 2x < 1 divide both sides by 2 x < 1/2 the second relation 2x+1 ≤ 3x+2 add -2x to both sides 2x -2x + 1 ≤ 3x -2x +2 1 ≤ x + 2 add -2 to both sides 1 -2 ≤ x + 2 -2 -1 ≤ x you have x greater than or equal to -1: -1 ≤ x and x less than 1/2: x < 1/2 you can write that in "short hand" as -1 ≤ x < 1/2 on a numer line it would be graphed like this |dw:1377977048397:dw|
Thanks Phi, I understood exactly where to go once you showed how to separate them into two separate inequalities. I was stuck because I didn't process the fact that you can split them and have it be correct if the signs don't change direction by multiplication or division of -1.