{"id":1146,"date":"2013-07-31T04:20:01","date_gmt":"2013-07-31T04:20:01","guid":{"rendered":"?page_id=1146"},"modified":"2018-07-11T14:20:50","modified_gmt":"2018-07-11T14:20:50","slug":"page-156","status":"publish","type":"page","link":"https:\/\/www.collegeprepalgebra.com\/cpa\/content\/math-help\/chapter-3\/section-7\/page-156\/","title":{"rendered":"Page 156"},"content":{"rendered":"
\n

Math Help for Section 3.7, Page 156<\/h2>\n<\/h2>\n

\nStudy
\nTip<\/span>
\nAn absolute
\nvalue inequality of the form $|x|<a$  (or $|x|\\le a$)
\ncan be
\nsolved with a double inequality, but an inequality of the
\nform $|x|>a$  (or $|x|\\ge a$)
\ncannot. Instead, you
\nmust solve two separate
\ninequalities, as demonstrated
\nin Example 7 on page 90.\n<\/p>\n

\nExample
\n7: Tip<\/span>
\nThe solution set in set notation is $\\{x|x\\le
\n-{1\\over3}\\}\\cup\\{x|x\\ge 3\\}$.<\/p>\n

Example
\n7: Check<\/span>
\nThe solution was found to be $x\\le -{1\\over3}$ or $x\\ge 3$, so check
\nthat
\n$x=-{1\\over3}$ and $x=3$
\nsatisfy the inequality and that $x=0$ does not.  <\/p>\n\n\n\n\n\n\n\n\n\n\n
$x=-{1\\over3}$:<\/td>\n<\/tr>\n
  <\/td>\n$\\eqalign{|3x-4|\\ge & 5
\n&{\\small\\color{red}\\quad\\quad \\text{Write original
\ninequality.}} \\cr \\left|3\\left({\\color{red}-{1\\over3}}\\right)-4\\right|
\n\\overset{?}{\\ge}& 5 &{\\small\\color{red}\\quad\\quad
\n\\text{Substitute }-{1\\over3}\\text{ for }x.}
\n\\cr |-1-4|\\overset{?}{\\ge}& 5
\n&{\\small\\color{red}\\quad\\quad \\text{Multiply.}} \\cr 
\n|-5|\\overset{?}{\\ge}& 5
\n&{\\small\\color{red}\\quad\\quad \\text{Subtract.}} \\cr 5\\ge
\n& 5 &{\\small\\color{red}\\quad\\quad
\n\\text{Solution
\nchecks. }\\checkmark} } $<\/td>\n<\/tr>\n
\n<\/td>\n<\/td>\n<\/tr>\n
$x=3$:<\/td>\n<\/tr>\n
<\/td>\n$\\eqalign{|3x-4|\\ge & 5
\n&{\\small\\color{red}\\quad\\quad \\text{Write original
\ninequality.}} \\cr |3({\\color{red}3})-4|
\n\\overset{?}{\\ge}& 5 &{\\small\\color{red}\\quad\\quad
\n\\text{Substitute }3\\text{ for }x.}
\n\\cr |9-4|\\overset{?}{\\ge}& 5
\n&{\\small\\color{red}\\quad\\quad \\text{Multiply.}} \\cr 
\n|5|\\overset{?}{\\ge}& 5
\n&{\\small\\color{red}\\quad\\quad \\text{Subtract.}} \\cr 5\\ge
\n& 5 &{\\small\\color{red}\\quad\\quad
\n\\text{Solution
\nchecks. }\\checkmark} } $<\/td>\n<\/tr>\n
\n<\/td>\n<\/td>\n<\/tr>\n
$x=0$:<\/td>\n<\/tr>\n
<\/td>\n$\\eqalign{|3x-4|\\ge & 5
\n&{\\small\\color{red}\\quad\\quad \\text{Write original
\ninequality.}} \\cr |3({\\color{red}0})-4|
\n\\overset{?}{\\ge}& 5 &{\\small\\color{red}\\quad\\quad
\n\\text{Substitute }0\\text{ for }x.}
\n\\cr |0-4|\\overset{?}{\\ge}& 5
\n&{\\small\\color{red}\\quad\\quad \\text{Multiply.}} \\cr 
\n|-4|\\overset{?}{\\ge}& 5
\n&{\\small\\color{red}\\quad\\quad \\text{Subtract.}} \\cr 4\\not\\ge
\n& 5 &{\\small\\color{red}\\quad\\quad
\n\\text{Solution does not
\ncheck. ✗ }} } $<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\nExample
\n8: Tip<\/span>
\nThe solution set in set notation is $\\{x|\\,5.97\\le x\\le6.03\\}$.<\/p>\n

Example 8: Check<\/span>
\nThe solution was found to be $5.97\\le x\\le6.03$, so check
\nthat
\n$x=6$
\nsatisfies the inequality and that $x=5$ and $x=7$ do not.  <\/p>\n\n\n\n\n\n\n\n\n\n\n
$x=6$:<\/td>\n<\/tr>\n
  <\/td>\n$\\eqalign{\\left|2-{x\\over3}\\right| \\le & 0.01
\n&{\\small\\color{red}\\quad\\quad \\text{Write original
\ninequality.}} \\cr \\left|2-{{\\color{red}6}\\over3}\\right|
\n\\overset{?}{\\le}& 0.01 &{\\small\\color{red}\\quad\\quad
\n\\text{Substitute 6 for }x.}
\n\\cr |2-2|\\overset{?}{\\le}& 0.01
\n&{\\small\\color{red}\\quad\\quad \\text{Divide.}} \\cr 
\n|0|\\overset{?}{\\le}& 0.01
\n&{\\small\\color{red}\\quad\\quad \\text{Subtract.}} \\cr 0\\le
\n& 0.01 &{\\small\\color{red}\\quad\\quad
\n\\text{Solution
\nchecks. }\\checkmark} } $<\/td>\n<\/tr>\n
\n<\/td>\n<\/td>\n<\/tr>\n
$x=5$:<\/td>\n<\/tr>\n
<\/td>\n$\\eqalign{\\left|2-{x\\over3}\\right| \\le & 0.01
\n&{\\small\\color{red}\\quad\\quad \\text{Write original
\ninequality.}} \\cr \\left|2-{{\\color{red}5}\\over3}\\right|
\n\\overset{?}{\\le}& 0.01 &{\\small\\color{red}\\quad\\quad
\n\\text{Substitute 5 for }x.}
\n\\cr |2-1.\\overline{6}|\\overset{?}{\\le}& 0.01
\n&{\\small\\color{red}\\quad\\quad \\text{Write }{5\\over3}\\text{ as }1.\\overline{6}.} \\cr 
\n|0.\\overline{3}|\\overset{?}{\\le}& 0.01
\n&{\\small\\color{red}\\quad\\quad \\text{Subtract.}} \\cr 0.\\overline{3}\\not\\le
\n& 0.01 &{\\small\\color{red}\\quad\\quad
\n\\text{Solution
\ndoes not check. ✗}} } $<\/td>\n<\/tr>\n
\n<\/td>\n<\/td>\n<\/tr>\n
$x=7$:<\/td>\n<\/tr>\n
<\/td>\n$\\eqalign{\\left|2-{x\\over3}\\right| \\le & 0.01
\n&{\\small\\color{red}\\quad\\quad \\text{Write original
\ninequality.}} \\cr \\left|2-{{\\color{red}7}\\over3}\\right|
\n\\overset{?}{\\le}& 0.01 &{\\small\\color{red}\\quad\\quad
\n\\text{Substitute 7 for }x.}
\n\\cr |2-2.\\overline{3}|\\overset{?}{\\le}& 0.01
\n&{\\small\\color{red}\\quad\\quad \\text{Write }{7\\over3}\\text{ as }2.\\overline{3}.} \\cr 
\n|-0.\\overline{3}|\\overset{?}{\\le}& 0.01
\n&{\\small\\color{red}\\quad\\quad \\text{Subtract.}} \\cr 0.\\overline{3}\\not\\le
\n& 0.01 &{\\small\\color{red}\\quad\\quad
\n\\text{Solution
\ndoes not check. ✗}} } $<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n


\n
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Math Help for Section 3.7, Page 156 Study Tip An absolute value inequality of the form $|x|<a$  (or $|x|\\le a$) can be solved with a double inequality, but an inequality of the form $|x|>a$  (or $|x|\\ge a$) cannot. Instead, you must … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"parent":1141,"menu_order":780,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_acf_changed":false,"footnotes":""},"class_list":["post-1146","page","type-page","status-publish","hentry"],"acf":[],"_links":{"self":[{"href":"https:\/\/www.collegeprepalgebra.com\/cpa\/wp-json\/wp\/v2\/pages\/1146","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.collegeprepalgebra.com\/cpa\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.collegeprepalgebra.com\/cpa\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.collegeprepalgebra.com\/cpa\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/www.collegeprepalgebra.com\/cpa\/wp-json\/wp\/v2\/comments?post=1146"}],"version-history":[{"count":2,"href":"https:\/\/www.collegeprepalgebra.com\/cpa\/wp-json\/wp\/v2\/pages\/1146\/revisions"}],"predecessor-version":[{"id":8895,"href":"https:\/\/www.collegeprepalgebra.com\/cpa\/wp-json\/wp\/v2\/pages\/1146\/revisions\/8895"}],"up":[{"embeddable":true,"href":"https:\/\/www.collegeprepalgebra.com\/cpa\/wp-json\/wp\/v2\/pages\/1141"}],"wp:attachment":[{"href":"https:\/\/www.collegeprepalgebra.com\/cpa\/wp-json\/wp\/v2\/media?parent=1146"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}