Applied Finite Mathematics HW Solutions 5

Applied Finite Mathematics HW Solutions 5 - EXERCISES 1.1...

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Unformatted text preview: EXERCISES 1.1 19- Β§($β€”5)_+4:= %(2β€”32)+1 r) G) (a: β€” 5) + 6(4) = 6 (g) (2 β€” 3.1;) +6(1) 2(3~5)+24=2β€”3$+_6 '23β€”10+24=3β€”32: 23+3az=8β€”14 53:2β€”6 β€”6 $0.23 β€˜5β€˜5 The original equation was linear. 20- 3(1β€”2m)+i=%(32:β€”1)+1 E 1 1 , 1 20(3)(1β€”2:z)+20 Z =20 5 (3.2:β€”-1)+20 2 60(1β€”23)+5=4(32:β€”1)+10 60β€”1202:+5=12:cβ€”4+10 β€”1203: +65 =123+6 β€”1203 β€”- 12.1: = 6 β€”- 65 β€”132:c = β€”59 _ β€”59 _ 59 _ β€”132 _ 132 The original equation was linear. 21. 5 _ 5 1 = 33β€”2 3 . 3 7 + 2 +2: + 42 (-3.) β€” 42 (g) + 42(1) = 42 (333) + 42(2) +_42(3) 142β€”62 +42 =63z+42z +126 ' 32: + 42 = 1052: +126 82 β€” 1052. = 126 4 42 β€”-972=84 z:fiia_E . β€”97 97 The original equation was linear. _ 22. 2(3β€”1)+6=4x+1β€”2(4+3) ' 23*2+6=42:+1-8β€”2:c 23+4=22:β€”7 23β€”29;: β€”-7β€”4 0=-11 The eqnation has 'no solution. The original equation was not linear. 23. . _3x+4_+(1β€”4m)=1+xβ€”2(a:;16) 33+.4+1β€”4m=1+3β€”22:+32 β€”a:+5 : β€”$+33 β€”-:1:β€”lβ€”3: =33β€”5 0:23 The equationhas no solutions. The original equation was not linear. 'lβ€˜hismatedal hasbeenrepmduoed in awordanwwiflicopyrigln law by the University ofMani'toba Bookstore. Furlher remodudim in narl or in a." as drir'llurmnhlhflnrl ...
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