Unformatted text preview: EXERCISES 1.1 19 Β§($β5)_+4:= %(2β32)+1 r) G) (a: β 5) + 6(4) = 6 (g) (2 β 3.1;) +6(1) 2(3~5)+24=2β3$+_6
'23β10+24=3β32:
23+3az=8β14 53:2β6
β6 $0.23
β5β5 The original equation was linear. 20 3(1β2m)+i=%(32:β1)+1 E
1 1 , 1
20(3)(1β2:z)+20 Z =20 5 (3.2:β1)+20 2
60(1β23)+5=4(32:β1)+10
60β1202:+5=12:cβ4+10 β1203: +65 =123+6
β1203 β 12.1: = 6 β 65 β132:c = β59
_ β59 _ 59
_ β132 _ 132
The original equation was linear.
21. 5 _ 5 1 = 33β2 3
. 3 7 + 2 +2: + 42 (3.) β 42 (g) + 42(1) = 42 (333) + 42(2) +_42(3)
142β62 +42 =63z+42z +126 '
32: + 42 = 1052: +126
82 β 1052. = 126 4 42 β972=84
z:ο¬ia_E
. β97 97
The original equation was linear. _
22. 2(3β1)+6=4x+1β2(4+3)
' 23*2+6=42:+18β2:c
23+4=22:β7
23β29;: β7β4
0=11
The eqnation has 'no solution. The original equation was not linear.
23. . _3x+4_+(1β4m)=1+xβ2(a:;16)
33+.4+1β4m=1+3β22:+32
βa:+5 : β$+33
β:1:βlβ3: =33β5
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 Fall '11
 Dr.Cornell
 Math

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