Question 20·25 Super-Hard SAT Advanced Math Questions·Advanced Math

Since $x$ and $y$ are positive, $xy$ isn't zero. Multiply the first equation by $xy$ to clear both fractions:

Set $s=x+y$. The square of the sum links what you need to $xy$. Expand the square of a sum, keeping the $2xy$ term:

Replace $x^2+y^2$ using the first equation:

Combine like terms:

You need to replace the product $xy$ in the second equation. Multiply both sides of $s^2=\frac{9}{2}xy$ by $\frac{2}{9}$:

From $s^2=x^2+2xy+y^2$, replacing $x^2+y^2$ with $s^2-2xy$ makes the quadratic part of the second equation:

Replace $xy$ with $\frac{2}{9}s^2$:

Multiply:

Combine like terms:

The other part of the second equation is twice the sum:

Replace both parts in the equation that equals $51$:

Type $\frac{s_1^2}{3}+2s_1\sim51\{s_1>0\}$ into Desmos. Use $s_1$ and $\sim$ so Desmos solves for a value, and keep $s_1>0$ because $s=x+y$ is positive. Under PARAMETERS, Desmos reports $s_1\approx9.727922$. That's a decimal for the sum, not yet an exact answer choice.

Type each exact choice on its own Desmos line, in the order shown. Desmos prints approximately $15.7279$, $9.3693$, $9.7279$, and $5.4853$. Only the third matches the positive sum $s_1$. The decimals identify the exact expression; they don't replace it. So $x+y=9\sqrt{2}-3$. Choice C.

Start from $\frac{s^2}{3}+2s=51$, found above. Multiply every term by $3$ to clear the denominator:

Half of $6$ is $3$, and $3^2=9$. To make a perfect square on the left, add $9$ to both sides:

The left side is $(s+3)^2$, because $(s+3)(s+3)=s^2+6s+9$. Include the added $9$ on the right:

Since $s=x+y>0$, $s+3$ is positive. Take the positive square root:

Use $162=81\cdot2$ to simplify:

Subtract $3$ to get the requested sum:

Since $s=x+y$, the value of $x+y$ is $9\sqrt{2}-3$. Choice C.