The number of pairs (x, y) which will satisfy the  equation x2−xy+y2=4(x+y−4) is

we have, x2−xy+y2=4(x+y−4)⇒x2−x(y+4)+y2−4y+16=0∵x∈R∵(−(y+4))2−4⋅1⋅(y2−4y+16)≥0⇒y2+8y+16−4y2+16y−64≥0⇒3y2−24y+48≤0⇒y2−8y+16≤0⇒(y−4)2≤0∴(y−4)2=0∴y=4x2−4x+16=4(x+4−4)x2−8x+16=0(x−4)2=0x=4Number of pairs is 1 i.e., (4, 4).