( f 1 ) y + x = 56 {\displaystyle (f1)y+x=56} ( f 2 ) 2 y + x = 110 , 5 | / 2 {\displaystyle (f2)2y+x=110,5|/2}
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( f 1 ) y + x = 56 | − ( f 2 ′ ) {\displaystyle (f1)y+x=56|-(f2')} ( f 2 ′ ) y + 0.5 x = 55 , 25 {\displaystyle (f2')y+0.5x=55,25}
0.5 x = 0 , 75 | ∗ 2 {\displaystyle 0.5x=0,75|*2} x = 1 , 5 {\displaystyle x=1,5}
x in (f1): y + 1 , 5 = 56 | − 1 , 5 {\displaystyle y+1,5=56|-1,5} y = 54 , 5 {\displaystyle y=54,5}
Probe in (f1) und (f2): ( f 1 ) 57 , 5 − 1 , 5 = 56 {\displaystyle (f1)57,5-1,5=56} 56 = 56 w . A . {\displaystyle 56=56w.A.} ( f 2 ) ( 2 ∗ 54 , 5 ) + 1 , 5 {\displaystyle (f2)(2*54,5)+1,5} 109 + 1 , 5 = 110 , 5 {\displaystyle 109+1,5=110,5} 110 , 5 = 110 , 5 w . A . {\displaystyle 110,5=110,5w.A.}
L = ( 1 , 5 | 54 , 5 ) {\displaystyle L={(1,5|54,5)}} --Oxygenpower 16:17, 23. Apr. 2010 (CEST)Beantworten