can someone tell me how to solve this:

pretend square brackets are absolute value signs:

[3x+1] +x greater than or equal to [2x-1]

squaring both sides doesn't work ): and i've never leant how to do the "plus x" bit ):

physics, massive fail. i promise myself never to study the night before. this will never again happen. my results will reflect my poor effort. i kept on telling my: key words, key words, key words and then i forgot about "Gravity" and red-shift and doppler effect. all those easy concepts to write about.

and i am bad at bs-ing so they are going to take marks of me for incorrect info ):

the test itself wasn't hard, and time wasn't an issue, if ONLY I MADE A BIGGER EFFORT APART FROM THE NIGHT BEFORE then i think i would have been fine, but obviously i'm not a cramming genius, like some people are ):

and i have a really bad memory ):

mc was hard ):

and i was talking to a girl and she said she's done 90% of the paper before D= those curry tutoring places *shakes fist

but i defs need phys tutoring next year.

and english, i also promise myself to have essays with less than 900 words so that i can add more stuff to it during the exam. i didn't even have time to write my last word: vulnerability, so i ended up writing a v and then a squiggle.

the question wasn't as hard as i thought it would be, but i think i didn't answer the question properly ):

i hope tomorrow is going to be productive, unlike today after phys which i was like ceebs, i have tomorrow....

and i was just watching the yr12 finale, variety night, and i have to agree, 3:54 is very amusing, well, i couldn't stop laughing.

I also failed phys (again) because of night-before study but that's okay because I'm probably dropping phys.

ReplyDeleteTo solve equations with multiple absolute values the cleanest way is to take each absolute value as a conditional statement.

From the definition of absolute value, |x| is:

x for x>=0

-x for x < 0

so |3x+1| is the same as

3x+1 for 3x+1>=0,

-3x-1 for 3x+1<0

So basically you have two critical points, -1/3 and 1/2 where the graphs change. Below -1/3 |3x+1| = -3x-1 because 3x+1 is less than zero, and 2x-1 = -2x+1 because 2x-1 is less than zero.

Then solve each section of the equation (x<-1/3, -1/3<=x<=1/2, x>1/2) with the absolute value, apply the condition to the result you get, and then get all the conditions together.

So for example:

If -1/3 <= x <= 1/2:

3x+1 + x >= -2x +1 (since |3x+1| = 3x+1 when 3x+1>0, and |2x-1| = -2x+1 when 2x-1<0)

4x + 1 >= -2x + 1

6x >= 0

x >= 0 FOR -1/3 <= x <= 1/2

0<= x <= 1/2

Solve in the same way for other two conditions.

Note that if you solve into equation and get no solution, you can just ignore the solution.

Squaring both sides *should* work...

ReplyDeleteBut failing that, there's a good, Icy method, right there.

thanks (:

ReplyDelete