document.write( "Question 50939: solve for |x+1|+|x-2|<=2\r
\n" ); document.write( "\n" ); document.write( "absolute value of x+1 and absolute value of x-2 is less than or equal to 2 \n" ); document.write( "
Algebra.Com's Answer #33962 by AnlytcPhil(1806)\"\" \"About 
You can put this solution on YOUR website!
solve for |x+1| + |x-2|< 2 \r\n" );
document.write( "absolute value of x+1 and absolute value \r\n" );
document.write( "of x-2 is less than or equal to 2\r\n" );
document.write( "\r\n" );
document.write( "Use the principle that\r\n" );
document.write( "\r\n" );
document.write( "if A > 0 then |A| = A \r\n" );
document.write( "if A < 0 then |A| = -A\r\n" );
document.write( "if A = 0 then |A| = 0\r\n" );
document.write( "\r\n" );
document.write( "There are four cases to consider:\r\n" );
document.write( "\r\n" );
document.write( "Case 1:  x+1 > 0 AND x-2 > 0\r\n" );
document.write( "           x > -1 AND x > 2\r\n" );
document.write( "\r\n" );
document.write( "This means x > 2\r\n" );
document.write( "\r\n" );
document.write( "|x+1| + |x-2| < 2\r\n" );
document.write( "    x+1 + x-2 < 2\r\n" );
document.write( "       2x - 1 < 2\r\n" );
document.write( "           2x < 3\r\n" );
document.write( "            x < 3/2\r\n" );
document.write( "\r\n" );
document.write( "This contradicts x > 2, so Case 1 is impossible.\r\n" );
document.write( "\r\n" );
document.write( "Case 2:  x+1 > 0 AND x-2 < 0\r\n" );
document.write( "           x > -1 AND x < 2\r\n" );
document.write( "\r\n" );
document.write( "This means -1 < x < 2\r\n" );
document.write( "\r\n" );
document.write( " |x+1| + |x-2| < 2\r\n" );
document.write( "  x+1 + -(x-2) < 2\r\n" );
document.write( " x + 1 - x + 2 < 2\r\n" );
document.write( "             3 < 2\r\n" );
document.write( "  \r\n" );
document.write( "This is never true, so Case 2 is impossible.\r\n" );
document.write( "\r\n" );
document.write( "Case 3:  x+1 < 0 AND x-2 > 0\r\n" );
document.write( "           x < -1 AND x > 2\r\n" );
document.write( "\r\n" );
document.write( "This is impossible so Case 3 is impossible.\r\n" );
document.write( "\r\n" );
document.write( "Case 4:  x+1 < 0 AND x-2 < 0\r\n" );
document.write( "           x < -1 AND x < 2\r\n" );
document.write( "\r\n" );
document.write( "This means x < -1\r\n" );
document.write( "\r\n" );
document.write( "   |x+1| + |x-2| < 2\r\n" );
document.write( " -(x+1) + -(x-2) < 2\r\n" );
document.write( "     -x-1 - x+2  < 2\r\n" );
document.write( "           -2x+1 < 2\r\n" );
document.write( "             -2x < 1\r\n" );
document.write( "               x > -1/2\r\n" );
document.write( "     \r\n" );
document.write( "But this contradicts x < -1\r\n" );
document.write( "\r\n" );
document.write( "So Case 4 is impossible also.\r\n" );
document.write( "\r\n" );
document.write( "There is no solution.\r\n" );
document.write( "\r\n" );
document.write( "Edwin
\n" ); document.write( "
\n" );