{VERSION 5 0 "IBM INTEL NT" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 261 "" 0 1 128 128 128 1 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 262 "" 0 1 128 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE " " -1 263 "" 0 1 128 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 128 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 128 0 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 266 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE " " -1 268 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 128 128 128 1 0 0 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }} {SECT 0 {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 257 49 "High School Modul es > Algebra by Gregory A. Moore" }}{PARA 3 "" 0 "" {TEXT -1 4 " " }{TEXT 256 23 "Absolute Value Problems" }}{PARA 0 "" 0 "" {TEXT -1 54 "\nSolving all types of basic absolute value equations.\n" }}{PARA 0 " " 0 "" {TEXT 258 153 "[Directions : Execute the Code Resource section \+ first. Although there will be no output immediately, these definitions are used later in this worksheet.]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{SECT 0 {PARA 3 "" 0 "" {TEXT -1 5 "Code " }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 21 "restart: with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 817 "PointsPlot := proc( a, b )\n #description \+ \"plot a finite interval on a number line\";\nlocal wid,Axesplot,leftp t,rightpt,leftend,rightend,LT,RT,ZT,textpos,zp;\n \n\nwid := ceil( (b- a)/15) ;\ntextpos := max(wid, 2) ;\nLT := textplot( [a, textpos, a] ): \nRT := textplot( [b, textpos, b] ):\nZT := textplot( [0, textpos, 0] \+ ):\nleftend := min( a - 2*wid, 0);\nrightend := max( b + 2*wid, 0);\n \n\nAxesplot := plot( 0, x = leftend..rightend , \n \+ y = (-wid)..(2*wid), axes = none, thickness = 4, color = blue,\n \+ scaling = constrained, tickmarks = [1,1] ):\nzp := pl ot( [[0,-wid/3],[0,wid/3]], color = black );\n\nleftpt := plottools[di sk]([ a, 0], wid/2, color=red):\nrightpt := plottools[disk]([ b, 0], w id/2, color=red ):\n\ndisplay( [Axesplot, LT, RT, ZT, leftpt, rightpt ,zp ]);\n\nend proc:\n\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1289 "intfin := proc( a, b, leq, req )\n #description \"plot \+ a finite interval on a number line\";\nlocal r, M, v1, v2, wid, bar, A xesplot, leftpt, rightpt, pt_color,\n leftend, rightend, LT, RT, \+ IT, textpos, lt, rt;\n \nr := b-a ;\nM := evalf( 1.2 * max( b-a, abs(a ), abs(b) ) );\nv1 := [a, 0];\nv2 := [b, 0];\nwid := ceil( M/15) ;\n\n bar := arrow( v1, v2,\n shape = double_arrow, color = red , differe nce = true,\n width = wid, head_width = 0, head_length = 0):\n\n\ni f( leq ) then lt := `[`; else lt := `(`; fi;\nif( req ) then rt := `]` ; else rt := `)`; fi;\n\ntextpos := max(wid, 2) ;\nLT := textplot( [a, textpos, a] ):\nRT := textplot( [b, textpos, b] ):\nIT := textplot( [ (a+b)/2, textpos + wid, cat(lt,a,`,`,b,rt)], color = green ):\n\nlefte nd := min( a - 2*wid, 0);\nrightend := max( b + 2*wid, 0);\n\nAxesplo t := plot( 0, x = leftend..rightend , \n y = (-wid )..(2*wid), axes = none, thickness = 2,\n scaling \+ = constrained, tickmarks = [1,1] ):\n\n\nif( leq ) then pt_color := re d; else pt_color := white; fi;\nleftpt := plottools[disk]([ a, 0], wid /2, color=pt_color):\nif( req ) then pt_color := red; else pt_color := white; fi;\nrightpt := plottools[disk]([ b, 0], wid/2, color=pt_color ):\n\ndisplay( leftpt, rightpt, bar, Axesplot, LT, RT, IT );\n\nend \+ proc:\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1201 "#_____________ _____________________________________________\nintinfr := proc(a, leq \+ )\n# description \"plot an infinite interval going to +inf\"; \nlocal bb, r, M, v1, v2, wid, bar, Axesplot, textpos,\n le ftpt, rightpt, pt_color, leftend, rightend, LT, RT, IT,lt;\n\nbb := ma x( 3*abs(a), 5) ; \n\nr := bb-a;\nM := evalf(1.2* max( abs(r), abs(a)) );\nv1 := [a, 0];\nv2 := [bb, 0];\nwid := ceil( M/15);\nbar := arrow( \+ v1, v2,\n shape = double_arrow, color = red , difference = true,\n \+ width = wid, head_width = 1.8*wid, head_length = 1.2*wid):\n\nif( le q ) then lt := `[`; else lt := `(`; fi;\n\ntextpos := max(wid, 3) ;\nL T := textplot( [a, textpos, a] ):\nRT := textplot( [bb, textpos, `Infi nity`] ): \nIT := textplot( [(a+bb)/2, textpos + wid, cat(lt,a,`, infi nty)`)], color = green ):\n\nleftend := min( 0, a*1.5);\nrightend := m ax( 0, bb);\n\nAxesplot := plot( 0, x = leftend..rightend , \n \+ y = (-wid)..(3*wid), axes = none, thickness = 2,\n \+ scaling = constrained, tickmarks = [1,1] ):\n\n\nif( leq ) then pt_color := red; else pt_color := white; fi;\nleftpt := plotto ols[disk]([ a, 0], wid/2, color=pt_color):\n\ndisplay( leftpt, bar, \+ Axesplot, LT, RT, IT );\n\nend proc:\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1211 "#__________________________________________________ ________\nintinfl := proc( b, req )\n# description \"plot a in finite interval going to -inf\";\nlocal aa, bb, r, M, v1, v2, wid, \+ bar, Axesplot,textpos, \n leftpt, rightpt, pt_color, leftend, \+ rightend, LT, RT, IT, rt;\n\naa := min( -3*abs(b), -5); \n\nr := b-aa; \nM := evalf(1.2*max( abs(r), abs(b) ));\nv1 := [aa, 0];\nv2 := [b, 0] ;\nwid := ceil( M/15);\nbar := arrow( v2, v1,\n shape = double_arro w, color = red , difference = true,\n width = wid, head_width = 1.8* wid, head_length = 1.2*wid):\n\ntextpos := max(wid, 3) ;\nif( req ) th en rt := `]`; else rt := `)`; fi;\n\n\nLT := textplot( [aa, textpos, ` -Infinity`] ):\nRT := textplot( [b, textpos, b] ):\nIT := textplot( [( aa+b)/2, textpos + wid, cat(`(-infinity,`,b,rt)], color = green ):\n\n leftend := min( 0, aa);\nrightend := max( 0, b*1.5);\n\nAxesplot := pl ot( 0, x = leftend..rightend , \n y = (-wid)..(3*w id), axes = none, thickness = 2,\n scaling = const rained, tickmarks = [1,1] ):\n\n\nif( req ) then pt_color := red; else pt_color := white; fi;\nrightpt := plottools[disk]([ b, 0], wid/2, co lor=pt_color):\n\ndisplay( rightpt, bar, Axesplot, LT, RT , IT);\n\n end proc:\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 899 "#__________ ________________________________________________\nintall := proc()\n \+ #description \"plot a infinite interval of all real numbers \" ;\nlocal a, b, v1, v2, wid, bar1, bar2, LT, RT, Axesplot ;\na := -2 5; b := 25;\n\nv1 := [a, 0];\nv2 := [b, 0];\nwid := ceil( (b-a)/15);\n bar1 := arrow( v1,\n shape = double_arrow, color = red , difference = false,\n width = wid, head_width = 1.8*wid, head_length = 1.2*wid ):\n\nbar2 := arrow( v2,\n shape = double_arrow, color = red \+ , difference = false,\n width = wid, head_width = 1.8*wid, he ad_length = 1.2*wid):\n\nLT := textplot( [a, max( wid, 6), `Infinity`] ):\nRT := textplot( [b, max( wid, 6), `Infinity`] ): \n\n\nAxesplot : = plot( 0, x = a..b , \n y = (-wid)..(3*wid), axes = none, thickness = 2,\n scaling = constrained, t ickmarks = [1,1] ):\n\ndisplay( bar1, bar2, Axesplot, LT, RT );\nend \+ proc:\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 623 "#______________ ____________________________________________\n#_______________________ ___________________________________\nintv := proc( expr1, expr2 )\n \+ # description \"plot a infinite interval going to -inf\";\nlocal a, b, leq, req;\n\nif type( expr1, `<=`) then leq := true; else le q := false; fi;\nif type( expr2, `<=`) then req := true; else req := f alse; fi;\na := lhs( expr1); b := rhs( expr2);\n\nif( abs(a) <> in finity) \n then if( abs(b) <> infinity) then intfin( a,b, leq, req ); \n else intinfr( a, leq); fi;\n else if( abs(b) <> infinity) t hen intinfl( b, req);\n else intall(); fi;\n\nfi;\n\nend proc:" }}}{PARA 3 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 43 " 1. Equalities " }{TEXT 260 33 "|x| = \+ a | ax+b | = c" }}{PARA 0 "" 0 "" {TEXT -1 170 "\nThere i s a duality to absolute values. When we have a simple equation like |x | = 12, there are two possible answers : +12 because |12| = 12, and -1 2 because |-12| = 12. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "abs(x) = 12;\nx = \{solve(%,x)\};" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "PointsPlot( -12, 12 );" } }}{PARA 0 "" 0 "" {TEXT -1 120 "\nWhen we try to solve these kinds of \+ equations,we break the equation into two linear equations without abso lute values. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "x = 12; solve(%); x = -12; solve(%);" }}}{PARA 0 "" 0 "" {TEXT -1 120 "\n\nWhenever there is an equation |x| = a (a > 0), the solutions will be two points, symmetrically placed about the o rigin." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "abs(x) = 7; x = \+ \{solve(%,x)\};\nPointsPlot( -7,7 );" }}}{PARA 0 "" 0 "" {TEXT -1 203 "\nMore examples - that work the same way. However, since -0 = 0, ther e is only one answer to |x| = 0. And since |x| >= 0, there are no sol utions to having a non-negative value equalling a negative value." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "abs(x) = 100; x = \{solve( %,x)\};" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "abs(x) = 2; x = \{solve(%,x)\};" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "abs(x ) = 0; x = \{solve(%,x)\};" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "abs(x) = -5; x = \{solve(%,x)\};" }}}{PARA 0 "" 0 "" {TEXT -1 124 "\n\nIt's slightly more complicated when we're not just dealing wi th |x| but the absolute value of some linear expression of x." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "abs(x - 3) = 10; x = \{sol ve(%,x)\};" }}}{PARA 0 "" 0 "" {TEXT -1 95 "\nAs before, the way to so lve these by hand is to break them into two separate linear equations. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "(x-3) = 10; solve(%); \+ (x-3) = -10; solve(%);" }}}{PARA 0 "" 0 "" {TEXT -1 97 "\nWhen the a bsolute value expression is more complicated, the solutions are no lon ger symmetrical." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "PointsPl ot( -7, 13);" }}}{PARA 0 "" 0 "" {TEXT -1 25 "\nHere are other example s." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 152 "abs( 8*x + 3 ) = 13; x = \{solve(%,x)\};\n ( 8*x + 3 ) = 13; x = \{solve(%,x)\};\n \+ ( 8*x + 3 ) = -13; x = \{solve(%,x)\};\n PointsPlot( -2, 5/4 ); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 148 "abs( 5*x - 9) = 16; \+ x = \{solve(%,x)\};\n ( 5*x - 9 ) = 16; x = \{solve(%,x)\};\n \+ ( 5*x - 9) = -16; x = \{solve(%,x)\};\nPointsPlot( -7/5, 5 );" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 3 "" 0 " " {TEXT -1 37 " 2. Simple Less Than " }{TEXT 261 28 "| x| < a |x| < a" }}{PARA 0 "" 0 "" {TEXT -1 316 " \nWhen w e consider an inequality of the form | x | < 3, we are looking for all of the positive numbers less than or equal to 3, and their negatives. Lets just look at the integers which satisfy this inequality. (Of cou rse we are looking for all real numbers which satisfy it, but this is \+ just to give us some clues.)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "for k from -8 to 8 do k, eva lb( abs(k) <= 3 ); od;" }}}{PARA 0 "" 0 "" {TEXT -1 302 "\nAs you can see, the only values which satisfy | x | < 3 are -3, -2, -1, 0, 1, 2 , and 3.\n\nTo solve this kind of equation, by hand, we usually make a three-sided inequality : -3 < x < 3. Although we can't do this in Map le, we can do each inequality separately, and take the intersection of the answer.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "abs( x ) \+ <= 3; \n-3, `<`, x, `<`, 3; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "-3 <= x; N := solve(%);\n x <= 3; P := solve(%);\n N intersect P ;\nsolve( abs(x) <= 3, x); " }}}{PARA 0 "" 0 "" {TEXT -1 33 "\nHere i s a graph of the solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "intv( -3 <= x, x <= 3 );" }}}{PARA 0 "" 0 "" {TEXT -1 63 "\n\nHer e is another example where the answer is an open interval." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "abs( x ) < 10;\nsolve(%);\nintv( -1 0 < x, x < 10);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 3 " " 0 "" {TEXT -1 34 " 3. Simple Greater Than " }{TEXT 262 28 " |x| > a |x| < a" }}{PARA 0 "" 0 "" {TEXT -1 204 "\nLets \+ now turn this around and look at the case where the absolute value is \+ greater than or equal to a number. Just to get an idea of what is goin g on, lets examine which integers satisfy this inequality." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "for \+ k from -8 to 8 do k, evalb( abs(k) >= 3 ); od;" }}}{PARA 0 "" 0 "" {TEXT -1 147 "\nAs we can see, this list is quite different than the l ist we got in the last section. It's almost opposite, in fact. All of \+ the numbers that were " }{TEXT 266 4 "true" }{TEXT -1 16 " before are \+ now " }{TEXT 267 5 "false" }{TEXT -1 44 ", and vice-versa, except 3 an d -3 which are " }{TEXT 268 4 "true" }{TEXT -1 398 " in both cases. Wh at we can see from this list is that the solution to this kind of ineq uality consists of two separate groups of numbers - the 'big' positive numbers greater than or equal to 3, and their negatives. These two s ets of numbers are distinct since the numbers in between -3 and 3 do n ot satisfy this inequality. This tells us that the solution will inclu de two separate inequalities.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "abs( x ) >= 3; x >= 3, x <= -3; " }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 26 "abs( x ) >= 3; solve(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "display( intv( 3 <= x, x < infinity ), intv( -i nfinity < x, x <= -3 ) );" }}}{PARA 0 "" 0 "" {TEXT -1 170 "\n\nNote \+ the connection between the inequalities and their solutions :\n \+ |x| < 3 solution is [-3,3]\n |x| \+ > 3 solution is " }{XPPEDIT 18 0 "`union`(-infinity, -3,3, in finity);" "6#-%&unionG6&,$%)infinityG!\"\",$\"\"$F(F*F'" }{TEXT -1 110 "\nTogether these two sets cover the ENTIRE number line. They are \+ like interlocking pieces of a puzzle.\n " }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 110 "display( intv( -3 <= x, x < 3 ) );\ndisplay( intv( 3 <= x, x < infinity ), intv( -infinity < x, x <= -3 ) );" }} }{PARA 0 "" 0 "" {TEXT -1 93 "\n\nHere is another example. Again the i ntervals are symetric for this kind of simpler problem." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 98 "abs(x) > 15 ; solve(%);\ndisplay( \+ intv( 15 < x, x < infinity ), intv( -infinity < x, x < -15 ) );" }}} {PARA 0 "" 0 "" {TEXT -1 1 "\n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 36 " 4. General Less Than " }{TEXT 263 16 "| ax+b | < c \+ " }}{PARA 0 "" 0 "" {TEXT -1 227 "\nWe can also apply these principles to more complicated absolute values. It's important to keep in mind t he essential difference between absolute values greater than or less t han numbers. \n | x | < a or " }{XPPEDIT 18 0 "abs(x) <= a;" "6#1-%$absG6#%\"xG%\"aG" }{TEXT -1 83 " .... solved by a three sided inequality\n | x | > a or " }{XPPEDIT 18 0 "a <= abs(x);" "6#1%\"aG-%$absG6#%\"xG" }{TEXT -1 138 " .... solved by a two inequalities \n\nThese same distinctions appl y to more complicated cases.\n | ax+b | < c or \+ " }{XPPEDIT 18 0 "abs(a*x+b) <= a;" "6#1-%$absG6#,&*&%\"aG\"\"\"%\"xG F*F*%\"bGF*F)" }{TEXT -1 85 " .... solved by a three sided inequal ity\n | ax+b | > c or " }{XPPEDIT 18 0 "a <= ab s(a*x+b);" "6#1%\"aG-%$absG6#,&*&F$\"\"\"%\"xGF*F*%\"bGF*" }{TEXT -1 40 " .... solved by a two inequalities \n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "abs( 5*x + 2) < 17; \n-17, `<`, 5*x + 2, `<`, 17; \n-19, `<`, 5*x , `<`, 15; \+ \n-19/5, `<`, x , `<`, 15/5;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "abs( 5*x + 2) < 17; solve(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "display( intv( 3 < x, x < infinity ), intv( -infinit y < x, x < - 19/5 ) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 " abs( 3*x - 8) <= 7; solve(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "display( intv( 4/3 < x, x < infinity ), intv( -infinity < x, x < -1 ) );" }}}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }}{PARA 0 "" 0 "" {TEXT 259 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 34 " 5. General Greater Than " }{TEXT 264 29 "| ax+b | > c | ax+b | \263 c" }}{PARA 0 "" 0 "" {TEXT -1 160 "\nWe \+ can also apply these principles to more complicated absolute values wh ich are greater than or greater than or equal. \n | x | > a or " }{XPPEDIT 18 0 "a <= abs(x);" "6#1%\"aG-%$absG6#%\" xG" }{TEXT -1 79 " .... solved by a two inequalities \n \+ | ax+b | > c or " }{XPPEDIT 18 0 "a <= abs(a*x+b);" "6#1% \"aG-%$absG6#,&*&F$\"\"\"%\"xGF*F*%\"bGF*" }{TEXT -1 40 " .... sol ved by a two inequalities \n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "abs( 3*x + 2) > 7; solve (%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "display( intv( 5/3 \+ < x, x < infinity ), intv( -infinity < x, x < -3 ) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "abs( 6*x - 1) >= 7; solve(%);" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "display( intv( 4/3 < x, x \+ < infinity ), intv( -infinity < x, x < -1 ) );" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 31 " 6. Graphs of Absolute Values " }{TEXT 265 12 "[ Opti onal ]" }}{PARA 0 "" 0 "" {TEXT -1 94 "\nYou can visualize an absolute vallue equation by looking at the intersection of two graphs.\n." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 147 "abs( x ) = 3; solve( %);\np lot( \{lhs(%%), rhs(%%)\}, x = -5..5, thickness = 2);\n`Note two point s of intersection mean two solutions to the equation`;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 146 "abs( x ) = 0; solve( %);\nplot( \{ lhs(%%), rhs(%%)\}, x = -5..5, thickness = 2);\n`Note ONE points of in tersection mean ONE solution to the equation`;" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 138 "abs(x) = -4; solve( %);\nplot( \{abs(x), -4\} , x = -5..5, thickness = 2);\n`Note NO points of intersection mean NO \+ solutions to the equation`;" }}}{PARA 0 "" 0 "" {TEXT -1 51 "\nHere ar e instances of the more complicated cases. " }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 84 "abs( 2*x + 7) = 12; solve( %);\nplot( \{lhs(%%), rh s(%%)\}, x = -12..5, thickness = 2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "abs( 7*x - 3) = 11; solve( %);\nplot( \{lhs(%%), rhs( %%)\}, x = -5..7, thickness = 2);" }}}{PARA 0 "" 0 "" {TEXT -1 106 "\n \nWhat about inequalities rather than equations? The equalities show w here the points of intersection are." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "abs( 2*x - 7) = 9; solve( %);\nplot( \{lhs(%%), rhs(% %)\}, x = -4..12, thickness = 2);" }}}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }}{PARA 0 "" 0 "" {TEXT -1 117 "\nWhen the absolute value is LESS than the constant, then we are looking for where the V is BELOW the horizo ntal line." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 162 "display( plot ( \{ abs( 2*x - 7), 9 \}, x = -5..12, thickness = 2),\n polygo nplot( \{ [[8,9], [-1,9], [3.5,0]] \} , color = blue ) );\n abs ( 2*x - 7) < 9;" }}}{PARA 0 "" 0 "" {TEXT -1 120 "\nWhen the absolute \+ value is GREATER than the constant, then we are looking for where the \+ V is ABOVE the horizontal line." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 204 "display( plot( \{ abs( 2*x - 7), 9 \}, x = -5..12, thickness \+ = 2),\n polygonplot( \{ [[8,9], [12,9], [12,17]],\n \+ [[-1,9], [-5,9], [-5,17]] \} , color = blue ) );\n abs( 2*x - 7) > 9;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT 269 45 " \n\n \251 2002 Waterloo Maple Inc " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 1" 26 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 } {PAGENUMBERS 0 1 2 33 1 1 }