sat-dog

Choice C is correct. It’s given that the graphs of the equations in the given system ( ) ( ) intersect at exactly one point, x, y, in the xy-plane. Therefore, x, y is the only solution to the given system of equations. The given system of equations can be = + solved by subtracting the second equation, y3 x , farom the first equation, =2 - + - (= 2 - + ) (- +) 2y x21 6x4. This yields y y 2 x 21 x64 3 x ,a or =2 - + - 0 2 24x 64x . Sincae the given system has only one solution, this equation 2 + + = has only one solution. A quadratic equation in the form rx sx t 0, where r, 2- s, and t are constants, has one solution if and only if the discriminant, s 4rt, is - - + equal to zero. Substituting 2 for r, 24 for s, and a 64 for t in the 2- ( )-2 ( )-( )( )- expression s 4rt yields 24 4 2 64 a . Setting this expression equal to ( -)2 (-)( )( -) = + = zero yields 24 4 2 64 a 0, or 8a 64 0. Subtracting 64 from both =- sides of this equation yields 8a 64. Dividing both sides of this equation by 8 =- - =2 - + - yields a 8. Substituting 8 for a in the equation 0 2 24x 64x yieldas =2 - + + =2 - + 0 2 24x 64x 8, or 0 2 24x 72x. Factoring 2 from the right-hand (=2 - +) side of this equation yields 0 2 12x 36x. Dividing both sides of this =2 - + = (- )(- ) equation by 2 yields 0 12x 36x, which is equivalent to 0 x 6 x 6 , or = -( )2 = - x 0 6 . Taking the square root of both sides of this equation yieldsx 0 6. = Adding 6 to both sides of this equation yields x 6. Choice A is incorrect. This is the value of a, not x. Choice B is incorrect and may result from conceptual or calculation errors. Choice D is incorrect and may result from conceptual or calculation errors.