Is (4, 2) a solution to the system y-x=2 and -3x-2y=-8?

It is only a solution to y = x - 2, not -3x - 2y = -8

Step-by-step explanation:

I) Check y = x - 2:

y = x - 2

2 = 4 - 2

As you can see, (4, 2) is a solution to this system.

II) Check -3x - 2y = -8:

-3x - 2y = -8

(-3)(4) - (2)(2) = -8

-12 - 4 =/= 8

-12 - 4 = -16, not 8. So (4, 2) is not a solution to this system.

Make a table with the angle theta as independent variable and the radius r as dependent variable:

theta  radius = 4+2cos theta    radius

   0      4+2                                6
  pi/6   4+2cos pi/6 =                 4+2(sqrt(3)/2

Perhaps you have already plotted this using webassign (but remember that you have not shared an illustration here).  (Please don't type "webassign plot" repeatedly, as it accomplishes nothing.)

Generally, when one wishes to find the area of a region defined by polar functions (as is the case here), one first determines suitable limits of integration from the finished curve and checks them through actual integration.  

Which formula should you use to find the area:  Look up "areas in polar coordinates," as I did.  The formula is as follows:

Enclosed area = Integral from alpha to beta of  (1/2)r^2 d(theta).  Note that the initial radius here is 6 (since r = 4 plus 2 cos theta is 4+2 when theta = 0).