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Foot251
@Foot251
May 2019
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Exercice 39 svp j'ai besoin d'aide merci d'avance
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penalite2002
Je sors x de la premiere equation et je le remets ds la 2eme
2x+3y=1 dc x=(1-3y)/2 ds la 2eme 3x+4y=2 dc 3(1-3y)/2+4y=2 ce qui donne
(3-3y)/2+4y=2 dc 3y/2+4y=2-3/2 DC 11Y/2 =1/2 dc 11y=1 dc y=1/11 ensuite 0n remplace y ds la 1ere equation et on trouve x
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penalite2002
dsl erreur de calcul mais le principe est le bon
foot251
Merci
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2x+3y=1 dc x=(1-3y)/2 ds la 2eme 3x+4y=2 dc 3(1-3y)/2+4y=2 ce qui donne
(3-3y)/2+4y=2 dc 3y/2+4y=2-3/2 DC 11Y/2 =1/2 dc 11y=1 dc y=1/11 ensuite 0n remplace y ds la 1ere equation et on trouve x