已知P=3xy-8x+1,Q=x-2xy-2,当x≠0时,3P-2Q=7恒成立,则y的值为_.
问题描述:
已知P=3xy-8x+1,Q=x-2xy-2,当x≠0时,3P-2Q=7恒成立,则y的值为______.
答
∵P=3xy-8x+1,Q=x-2xy-2,
∴3P-2Q=3(3xy-8x+1)-2(x-2xy-2)=7恒成立,
∴9xy-24x+3-2x+4xy+4=7,
13xy-26x=0,
13x(y-2)=0,
∵x≠0,
∴y-2=0,
∴y=2;
故答案为:2.