Question

If x2 + y2 = 38 and x - y = 2, then what is x? If there are two possible answers, then enter the larger of the two.

Answer 1

If x-y=2 then x=2+y

And if you insert that in the first equation

$(2 + y) {}^{2} + y {}^{2} = 38$
$4 + 4y + y {}^{2} + {y}^{2} = 38$
$2y {}^{2} + 4y - 34 = 0$
$y = - 1 + \sqrt{17}$
Or
$y = - 1 - \sqrt{17}$


The first one is bigger so the answer is

$y = - 1 - \sqrt{17}$

Answer 2

Answer:

$y= 3\sqrt{2} -1 \\x = 1 +3\sqrt{2}$

Step-by-step explanation:

First we can find x from the linear equation. So:

If x-y=2 then x=2+y

After that. We can inster x value in the cuadratic equation:

$(2 + y) {}^{2} + y {}^{2} = 38$

$4 + 4y + y {}^{2} + {y}^{2} = 38$

$2y {}^{2} + 4y + 4 - 38 = 0$

$2y {}^{2} + 4y -34 = 0$

After reorganizing terms and using factorization methods:

$2 (y+1)^2 - 36 = 0\\(y+1)^2 = 18\\y + 1 = +-\sqrt{18} \\y1 = 3\sqrt{2} -1 \\x1 = 2 -1 + 3\sqrt{2}=1 +3\sqrt{2}\\y2 = -1 - 3\sqrt{2}\\x2 = 2 -1 - 3\sqrt{2}=1 -3\sqrt{2}$

Finally we choose the largest value for x.

$y = 3\sqrt{2} -1 \\x = 1 +3\sqrt{2}$