In the adjoining figure, $$ABC$$ is an equilateral triangle and $$CDE$$ is a right-angled triangle. $$BCD$$, $$ACE$$ and $$GEF$$ are straight lines. $$ACE$$ is perpendicular to $$GEF$$. $$\angle DEF = x + y$$, $$\angle GEA = 5x - y$$. Then the numerical value of $$4x + 3y$$ is

Since $$GEF$$ is perpendicular to the line $$ACE$$ at $$E$$, $$\angle GEA = 90^\circ$$, so $$5x - y = 90$$. As $$BCD$$ and $$ACE$$ are straight lines, $$\angle DCE$$ is vertically opposite $$\angle ACB = 60^\circ$$, and the right angle of $$\triangle CDE$$ at $$D$$ gives $$\angle DEC = 30^\circ$$. Because $$EF$$ is perpendicular to $$EC$$, $$\angle DEF = 90^\circ - 30^\circ = 60^\circ$$, so $$x + y = 60$$. Solving gives $$x = 25$$ and $$y = 35$$, hence $$4x + 3y = 100 + 105 = 205$$.