document.write( "Question 1199343: Find a and b.\r
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\n" ); document.write( "\n" ); document.write( "If -3 < x < 0, then a < 1/(x + 4) < b
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Algebra.Com's Answer #833190 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "Add four to all sides of the 1st inequality
\n" ); document.write( "-3 < x < 0
\n" ); document.write( "-3+4 < x+4 < 0+4
\n" ); document.write( "1 < x+4 < 4
\n" ); document.write( "The inequality signs stay the same for now.\r
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\n" ); document.write( "\n" ); document.write( "Then apply the reciprocal to all three sides.
\n" ); document.write( "This flips the direction of the inequality signs.
\n" ); document.write( "It's like saying how 2 > 1/3 flips to 1/2 < 3.\r
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\n" ); document.write( "\n" ); document.write( "So,
\n" ); document.write( "1 < x+4 < 4
\n" ); document.write( "1/1 > 1/(x+4) > 1/4
\n" ); document.write( "1 > 1/(x+4) > 1/4\r
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\n" ); document.write( "\n" ); document.write( "Now let's go from the form p > q > r to r < q < p
\n" ); document.write( "1 > 1/(x+4) > 1/4
\n" ); document.write( "1/4 < 1/(x+4) < 1
\n" ); document.write( "In other words, swap the 1st and last sides so that the \"greater than\" signs become \"less than\" signs.\r
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\n" ); document.write( "\n" ); document.write( "Compare that to the form
\n" ); document.write( "a < 1/(x+4) < b
\n" ); document.write( "to find that
\n" ); document.write( "a = 1/4
\n" ); document.write( "b = 1
\n" ); document.write( "which are the final answers.
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