Question 87·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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The real numbers $x$ and $y$ satisfy the system

\begin{cases} x + y = 12 \\[3pt] x^2 + y^2 = 2k \end{cases}

where $k$ is a constant.

If the system has exactly one real solution pair $(x,y)$ , what is the value of $k$ ?

Your answer

In a symmetric system, swapping $x$ and $y$ leaves both equations unchanged. If the system has exactly one solution pair, that pair must have equal coordinates; otherwise, swapping them would produce a second pair. Use the fixed sum to find those coordinates, then check that no unequal pair can give the same sum of squares. Watch the $2$ in $2k$ .

Hints

Hint 1
Both equations are unchanged if you swap $x$ and $y$ . That symmetry means a solution with different coordinates would have a second, swapped solution. What must be true of the coordinates when there is exactly one pair?
Hint 2
Once the coordinates are equal, use their sum of $12$ to find each one. To check that this pair is the only one, imagine raising one coordinate by $t$ and lowering the other by $t$ .
Hint 3
Compare the sum of squares for the equal pair and an unequal pair. An unequal pair adds a positive squared term, so it cannot have the same sum of squares. Then use $2k$ , not $k$ , as the value of that sum.

Step-by-step

Use symmetry, then check uniqueness
Step 1
Use the one-pair condition
No Desmos needed. Symmetry gives an exact way to identify the only possible pair. Swapping $x$ and $y$ changes neither equation: $x+y$ and $x^2+y^2$ stay the same. So if $(x,y)$ works, $(y,x)$ works too. Exactly one pair forces $x=y$ ; otherwise, those would be two different pairs.
Step 2
Find the equal coordinates
Put $y=x$ into the sum equation:
$x+x=12.$
Combine the equal terms:
$2x=12.$
Divide by $2$ :
$x=6.$
Since $y=x$ , the only possible pair is $(6,6)$ .
Step 3
Show no unequal pair has the same sum of squares
Every pair whose sum is $12$ can be written as $x=6+t$ and $y=6-t$ : one coordinate goes up by $t$ while the other goes down by $t$ . Substitute those forms into the sum of squares:
$x^2+y^2=(6+t)^2+(6-t)^2.$
Expand both squares:
$x^2+y^2=(36+12t+t^2)+(36-12t+t^2).$
Combine terms:
$x^2+y^2=72+2t^2.$
For an unequal pair, $t\ne0$ , so $2t^2>0$ . Only $(6,6)$ gives a sum of squares of $72$ .
Step 4
Find the constant
The second equation says the sum of squares is $2k$ , not $k$ . At the one-pair value, set it equal to $72$ :
$2k=72.$
Divide by $2$ :
$k=36.$
So the value that gives exactly one real solution pair is $36$ . Grid in 36.

Question 87·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

Lessons that teach this