Question

помогите решить систему

log 4(x) + log4(y)=1,

y-2x=7

Answer 1

$log_{4}x+lg_{4}y=1\ y-2x=7\\ y-2x=7\ y=7+2x\\ log_{4}x+lg_{4}y=1\ lg_{4}(x*y)=1\ lg_{4}(x(7+2x))=1\ lg_{4}(x(7+2x))=lg_{4}4\ x(7+2x)=4\ 7x+2x^2-4=0\ D=49-4*2*(-4)=81\ x_{1,2}=frac{-7pm 9}{4} = 0,5; -4.\ y_{1}=7+2(0,5)=8\ y_{2}=7+2*(-4)=-1\\ x_{1}=0,5; x_{2}=-4; y_{1}=8; y_{2}=-1$