=y (say) ⇒x2y−5xy+9y=x ⇒x2y−(5y+1)x+9y=0 For xεR,D≥0 ⇒b2−4ac≥0 ⇒(5y+1)2−4y(9y)≥0 ⇒25y2+10y−36y2+1≥0 ⇒−11y2+10y+1≥0 ⇒11y2−10y−1≤0 ⇒11y2−11y+y−1≤0 ⇒11y(y−1)+1(y−1) ⇒(y−1)(11y+1)≤0 Critical values of y are 1 and −