Just for fun!-V

We need that these lines intersect the y-axis , so we make x=0 , taking it $y=\frac{-6-2x}{-3}=\frac{18-2x}{3}$ And with this we find the two intesections in y-axis $y_1=\frac{-6-2(0)}{-3}=2$ and $y_2=\frac{18-2(0)}{3}=6$ So, we have the intersections in y-axis of the the lines: (0,2) and (0,6) . Now, to find the intersection of these two lines we need to find the point y (or I needed), and we can do it as this $x=\frac{3y-6}{2}=\frac{18-3y}{2}$ $3y-6=18-3y$ $6y=24$ $y=4$ And now, we know the point of intersection of lines is (x,4) , and to find the value of x , only need to put the value of y in any of equations, so $x=\frac{3y-6}{2}=\frac{3(4)-6}{2}=\frac{6}{2}=3$ And our intersection point of the lines is (3,4) , then we watch in our plane that we have an isosceles triangle with side lenghts a,b,c (I make the plane and watched it), where $a=4, b=\sqrt{5}, c=\sqrt{5}$ And by the same plane i made, i find the base= a = 4 (because i want it to be my base) and the high= 3 . So the area is $\alpha=\frac{3*4}{2}=3*2=6$ And that's all folks! Questions?

Find the area of shaded region, Triangle DBE

Base BD=4 units,

Height(From E to Y-axis)= 3 units

So $\boxed{Area = 6 }$ sq.units