Determine the unique pair of real numbers \((x,y)\) satisfying \((3x^2 - 12x + 16)(2y^2 + 6y + 9) = 18.\)
Enter your answer as an ordered pair in the format \((x,y)\), where \(x\) and \(y\) are replaced by appropriate numbers.
Thank you very much for your help!
+23245 The unique answer is (2, -1.5).
One way to determine this answer: (I'm going to put it into graphing form.)
2y2 + 6y + 9 = 18 / (3x2 - 12x + 16)
2y2 + 6y = 18 / (3x2 - 12x + 16) - 9
2(y2 + 3y) = 18 / (3x2 - 12x + 16) - 9
(y2 + 3y) = 9 / (3x2 - 12x + 16) - 9/2 (divide by 2)
y2 + 3y + 9/4 = 9 / (3x2 - 12x + 16) - 9/2 + 9/4 (complete the square)
(y + 3/2)2 = 9 / (3x2 - 12x + 16) - 9/4 (factor; simplify)
(y + 3/2)2 = [ 36 - 9(3x2 - 12x + 16) ] / [ 4(3x2 - 12x + 16) ] (common denominator)
(y + 3/2)2 = [ 36 - 273x2 + 108x - 144) ] / [ 4(3x2 - 12x + 16) ]
(y + 3/2)2 = [ - 27x2 + 108x - 108) ] / [ 4(3x2 - 12x + 16) ]
(y + 3/2)2 = [ - 27(x2 + 4x - 4) ] / [ 4(3x2 - 12x + 16) ]
(y + 3/2)2 = [ - 27(x - 2)2 ] / [ 4(3x2 - 12x + 16) ]
y + 3/2 = +/- sqrt( [ - 27(x - 2)2 ] / [ 4(3x2 - 12x + 16) ] )
y = +/- sqrt( [ - 27(x - 2)2 ] / [ 4(3x2 - 12x + 16) ] ) - 3/2
The graph of y = 3x2 - 12x + 16 has only positive values for y.
The graph of y = -27(x - 2)2 has only negative values for y except when x = 2; then its value is zero.
Therefore, the only real value for the expression under the square root sign occurs when x = 2.
And, when x = 2, the y-value in the original equation is -1.5.
I wonder who thought up this problem!
