The equation #14y^2-31xy=10x^2# is just like any quadratic equation, as it can be written as
#14y^2-31xy-10x^2=0# and dividing each term by #x^2#
#14y^2/x^2-31(xy)/x^2-10=0#
or #14(y/x)^2-31y/x-10=0# and assuming #y/x=u# this is
#14u^2-31u-10=0#
Note that in such cases we have #f(x,y)# and degree of each term is #2#. Hence, it is called as homogeneous equation.
Let us now factorize it in normal way and for this we have to identify two numbers whose sum is #-31# and product is #14x(-10)=-140#. These are #-35# and #4#.
#14y^2-31xy-10x^2=0#
or #14y^2-35xy+4xy-10x^2=0#
or #7y(2y-5x)+2x(2y-5x)=0#
or #(7y+2x)(2y-5x)=0#
Therefore either #7y+2x=0# i.e. #7y=-2x# i.e. #y=-2/7x#
or #2y-5x=0# i.e. #2y=5x# i.e. #y=5/2x#
