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Mathematics 19 Online
OpenStudy (ubie01):

PLEASE HELP ILL MEDAL!!! Select the graph of the solution. Click until the correct graph appears. |x| + 1 > 2 https://wca.sooschools.com/media/g_alg01_ccss_2015/4/t2_3a.gif https://wca.sooschools.com/media/g_alg01_ccss_2015/4/t2_3b.gif https://wca.sooschools.com/media/g_alg01_ccss_2015/4/t2_3c.gif

OpenStudy (ubie01):

@Hero

OpenStudy (ubie01):

@Michele_Laino @Mehek14

OpenStudy (plainntall):

What does |x| mean? What does its graph look like?

OpenStudy (ubie01):

The graph options r the links

OpenStudy (plainntall):

I see them. |x| means the absolute value of x right?

OpenStudy (plainntall):

What would |-1|+1 equal?

OpenStudy (plainntall):

What would |1|+1 equal?

OpenStudy (plainntall):

2 right?

OpenStudy (michele_laino):

by definition we have this: \[\begin{gathered} \left| x \right| = x,\quad x \geqslant 0 \hfill \\ \hfill \\ \left| x \right| = - x,\quad x < 0 \hfill \\ \end{gathered} \]

OpenStudy (plainntall):

Doesn't that mean we want all the x values to the left of -1 and to the right of 1, but not -1 or 1 or the numbers in between them? So what graph represents that?

OpenStudy (michele_laino):

hint: if \(x \geqslant 0\), then after a substitution, we can rewrite you inequality as follows: \(x+1>2\) please solve

OpenStudy (ubie01):

So wouldn't that be graph #3?

OpenStudy (plainntall):

Yes, https://wca.sooschools.com/media/g_alg01_ccss_2015/4/t2_3a.gif :) Michele_Laino, how do you get people to see that they have to solve for the negative and positive values of x? x>1 works, but how do they get x<-1?

OpenStudy (michele_laino):

whereas if \(x<0\) then I can rewrite the starting inequality as follows: \(-x+1>2\) again, please solve

OpenStudy (plainntall):

Doesn't that solve to x>-1?

OpenStudy (plainntall):

We need x<-1 don't we?

OpenStudy (michele_laino):

hint: the solutions of first inequality, are: \[x > 1\] whereas the solutions of the second inequality are \[x<-1\] now, please make the union set of those solutions above @Ubie01

OpenStudy (plainntall):

I see, when we multiply or divide each side of the inequality we change the direction of the inequality so when x greater than or equal to 0 x+1 >2 we subtract both sides by 1 and our inequality stays the same x>1 But when it is negative or x<0 -x+1>2 subtract both sides by 1 -x>1 divide by -1 causes us to change the equality x<-1

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