AMC 10 · 2013 · #11
Grade 8 algebraReal numbers x and y satisfy the equation x2+y2=10x−6y−34. What is x+y?
Pick an answer.
AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Real numbers x and y make the equation x^2 + y^2 = 10x - 6y - 34 true. Find the value of x + y.
Givens: x and y are real numbers; x^2 + y^2 = 10x - 6y - 34
Unknowns: The value of x + y
Understand
Restated: Real numbers x and y make the equation x^2 + y^2 = 10x - 6y - 34 true. Find the value of x + y.
Givens: x and y are real numbers; x^2 + y^2 = 10x - 6y - 34
Plan
Primary tool: #15 Organize Information in More Ways
Secondary: #14 Extreme Principle, #1 Draw a Diagram
One equation with two unknowns usually cannot pin down each value, so a plain solve will not work. Instead, rearrange the equation and complete the square to rewrite it as a sum of squares. That new form is more useful: a sum of squares can only equal 0 when every square is 0, which forces exact values for x and y.
Execute — Answer: B
6.EE.A.3 Step 1 Move all terms to one side
- Bring every term to the left so the right side is 0.
- Subtract 10x, add 6y, and add 34 to both sides.
💡 Getting 0 on one side sets up a form you can rewrite and read off.
6.EE.A.3 Step 2 Complete the square for each variable
- Group the x-terms and the y-terms.
- Half of 10 is 5, and half of 6 is 3, so x^2 - 10x is (x - 5)^2 minus 25, and y^2 + 6y is (y + 3)^2 minus 9.
💡 Completing the square turns a scattered quadratic into a clean square plus a leftover number.
7.NS.A.1 Step 3 Combine the leftover numbers
- Substitute both rewritten pieces back in.
- The stray constants are -25, -9, and +34, and they add up to 0, so only the two squares remain.
💡 -25 - 9 + 34 cancels to 0, leaving a bare sum of two squares.
8.EE.A.2 Step 4 Force each square to 0
- Each square is greater than or equal to 0 for real numbers, so the only way two of them add to 0 is if both are 0.
- Then (x - 5)^2 = 0 gives x = 5, and (y + 3)^2 = 0 gives y = -3.
- So x + y = 5 + (-3) = 2, which is answer (B).
💡 Two things that can never be negative can only sum to 0 by both being 0.
6.EE.A.3 Bring every term to the left so the right side is 0. Subtract 10x, add 6y, and a 6.EE.A.3 Group the x-terms and the y-terms. Half of 10 is 5, and half of 6 is 3, so x^2 - 7.NS.A.1 Substitute both rewritten pieces back in. The stray constants are -25, -9, and + 8.EE.A.2 Each square is greater than or equal to 0 for real numbers, so the only way two Review
Reasonableness: Put x = 5 and y = -3 back into the original equation. The left side is 25 + 9 = 34. The right side is 10(5) - 6(-3) - 34 = 50 + 18 - 34 = 34. Both sides match, so the point (5, -3) is correct and x + y = 2, which is choice (B).
Alternative: Read the squared form as geometry. The equation (x - 5)^2 + (y + 3)^2 = 0 is a circle centered at (5, -3) with radius 0, so the circle is just the single point (5, -3). Reading its coordinates gives x + y = 2.
CCSS standards used (min grade 8)
6.EE.A.3Apply the properties of operations to generate equivalent expressions (Rearranging the equation to one side and completing the square in x and y.)7.NS.A.1Apply and extend understanding of addition and subtraction to rational numbers (Adding the leftover constants -25, -9, and +34 to get 0.)8.EE.A.2Use square root and cube root symbols to represent solutions (Concluding that (x-5)^2 = 0 gives x = 5 and (y+3)^2 = 0 gives y = -3.)
⭐ Complete the square to turn the equation into a sum of squares equal to 0, then remember a sum of squares is 0 only when every square is 0.
⭐ Complete the square to turn the equation into a sum of squares equal to 0, then remember a sum of squares is 0 only when every square is 0.
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