document.write( "Question 567162: If you have a problem like |x-2| + 3 = 5 you would solve it with a positive and negative equation changing the signs of both the 5 and the 3. If you had an inequality |x-2| + 3 > 5 your positive and negative equations would not include changing the sign of the 3. Why is that?
\n" ); document.write( "|x-2| + 3 = 5
\n" ); document.write( "x + 1 = 5
\n" ); document.write( "x=4
\n" ); document.write( "|x-2| - 3 = -5
\n" ); document.write( "x - 5 = -5
\n" ); document.write( "x=0\r
\n" ); document.write( "\n" ); document.write( "|x-2| + 3 > 5
\n" ); document.write( "x + 1 > 5
\n" ); document.write( "x>4
\n" ); document.write( "|x-2| + 3 < -5
\n" ); document.write( "x + 1 < -5
\n" ); document.write( "x<-6 \n" ); document.write( "

Algebra.Com's Answer #366455 by richard1234(7193) $\"\"$ $\"About$

You can put this solution on YOUR website!
I don't know, your solution is a bit flawed. You can't really say that |x-2| + 3 is always equal to x+1, since there are negative solutions involved too.\r
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\n" ); document.write( "\n" ); document.write( "You should move the 3 to the RHS first:\r
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\r
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\n" ); document.write( "\n" ); document.write( "Now we can take positive and negative solutions:\r
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, this yields x=4 and x=0. Here, you would change the sign of the 3, but I wouldn't do that; you're more likely to make a mistake this way.\r
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\n" ); document.write( "\n" ); document.write( "For the inequality, we have\r
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2\">, this can either be

2\"> or

(since we can have a negative solution -x-2 > 2, multiplying by -1 reverses the direction of the inequality). The solutions are x > 4 and x < 0. \n" ); document.write( "

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