Question

Помогите, пожалуйста!!

Answer 1

$\left \{ {{xz+x=18} \atop {xz+z=14}} \right.$

$xz+x-xz-z=18-14,\\x-z=4,\\x=4+z.$

Тогда

$(z+4)*z+z=14,\\z^{2} +4z+z=14,\\z^{2} +5z-14=0.$

$D=5^{2} -4*1*(-14)=25+56=81;\\\\z_{1}=\frac{-5+9}{2*1} =\frac{4}{2}=2,\\\\ z_{2}=\frac{-5-9}{2*1} =\frac{-14}{2}=-7.$

$x=4+z$

$z_{1}=2,\\x_{1} =4+z=4+2=6.\\\\z_{2}=-7,\\x_{2}=4+z=4+(-7)=4-7=-3.$