II gimnazjum

Matematyka z plusem 2

Skoro rozwiązaniem ma być para liczb x=3 i y=-2, to podstawiay te wartości do układu równań:&nbsp; {a⋅3+b⋅(−2)=4b⋅3−a⋅(−2)=7​ &nbsp; {3a−2b=4      ∣⋅(−2)2a+3b=7  ∣⋅3​

a) {x2=4x+y=2​ &nbsp;&nbsp; &nbsp; &nbsp; &nbsp; {x=22+y=2​ &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; lub &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; {x=−2−2+y=2​ &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; {x=2y=2−2=0​ &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;lub &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; {x=−2y=2+2=4​ &nbsp; b) {(x−1)7=−1y2−x=9​ &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; {x−1=−1y2−x=9​,      bo −1=(−1)7

a) +{ax+y=−1x−y=2​​ &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; ax+x=1 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; x(a+1)=1 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; x=a+11​ &nbsp; Aby była to liczba całkowita, w mianowniku musi być 1 lub -1, więc a+1=1 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;lub &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; a+1=−1

{5p−3r=−6   ∣⋅23p−2r=2          ∣⋅(−3)​

{3x−4y=7  ∣⋅44x−3y=14     ∣⋅(−3)​

{4x+y=3y+45x−2y=x+y​     {4x−2y=44x−3y=0   ∣⋅(−1)​

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