carcass wrote:
Is \(xm < ym\) ?
(1) \(x > y\)
(2) \(m < 0\)
this is more simple method
xm<ym
move ym to the left
xm-ym<0
m(x-y)<0
now we can analyse , applying case 1 and 2.
Is \(xm < ym\) ?
(1) \(x > y\)
(2) \(m < 0\)
this is more simple method
xm<ym
move ym to the left
xm-ym<0
m(x-y)<0
now we can analyse , applying case 1 and 2.
![Will Linley – Head First – Single [iTunes Plus M4A]](http://is1-ssl.mzstatic.com/image/thumb/Music211/v4/3a/fe/29/3afe2925-2b4b-c869-f8bf-8a81bcd2ac03/26UMGIM95096.rgb.jpg/316x316bb.webp)
