document.write( "Question 1235: Show me how to work this.\r
\n" ); document.write( "\n" ); document.write( "|3x-4| > 11 \n" ); document.write( "
Algebra.Com's Answer #373 by khwang(438)\"\" \"About 
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Case (i) If 3x - 4 >= 0,
\n" ); document.write( " then |3x-4| > 11 converts to
\n" ); document.write( " 3x - 4 > 11
\n" ); document.write( " or 3x > 11 + 4 = 15,
\n" ); document.write( " So, x > 5...(*)
\n" ); document.write( " But,by 3x - 4 >= 0, or 3x >= 4 or x >= 4/3...(1)
\n" ); document.write( " Hence, the numbers x > 5 satisfying the given condition (1)\r
\n" ); document.write( "\n" ); document.write( " Case (ii) If 3x - 4 < 0,
\n" ); document.write( " then |3x-4| > 11 converts to
\n" ); document.write( " -(3x - 4) > 11 , or -3x + 4 > 11,
\n" ); document.write( " or -3x > 11 - 4 = 7,
\n" ); document.write( " Dividing both sides by -3(have to change the direction of the inequality)
\n" ); document.write( " We have x < -7/3 ...(**)
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\n" ); document.write( " But,by 3x - 4 < 0, or 3x < 4 or x < 4/3...(2)
\n" ); document.write( " We see that the numbers x < -7/3 satisfying the given condition (2)\r
\n" ); document.write( "\n" ); document.write( " Answer: x > 5 or x < -7/3
\n" ); document.write( " The solution set is (-oo, -7/3) U (5,+oo)\r
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\n" ); document.write( "\n" ); document.write( " Another way: USe |ax -b| > c (note a > 0,c >=0) is equivalent to
\n" ); document.write( " x > (b+c)/a or x < (c -b)/a\r
\n" ); document.write( "\n" ); document.write( " Now a = 3, b = 4, c = 11 and we have
\n" ); document.write( " x > 5 or x < -7/3\r
\n" ); document.write( "\n" ); document.write( " Moreover, if c < 0, |ax -b| > c is true for all x and the solution
\n" ); document.write( " set is R(all real numbers)\r
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