# Q. 5.     What should be taken away from $3x^{2}-4y^{2}+5xy+20$  to obtain $-x^{2}-y^{2}+6xy+20 ?$

$\\3x^2 - 4y^2 + 5xy + 20 - a = -x^2 - y^2 + 6xy + 20 \\a = 3x^2 - 4y^2 + 5xy + 20 - ( -x^2 - y^2 + 6xy + 20 ) \\a = 3x^2 - 4y^2 + 5xy + 20 + x^2 + y^2 - 6xy - 20 \\a = ( 3 + 1 )x^2 + ( -4 + 1 )y^2 + ( 5 - 6 )xy + 20 - 20 \\a = 4x^2 - 3y^2 - xy$