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Can You Solve This High School Math Challenge Algebra Problem?

1.8K views · Sep 22, 2026 · Education

Comments · 6

  • @NuisMaz · 9 days ago

    The three equations are defined for x ≥ 0, y ≥ 0, and z ≥ 0. Let a = √x, b = √y, and c = √z. We then get: a² - bc = 42, b² - ac = 6, and c² - ab = -30. Subtracting the first two equations, we get (a - b)(a + b + c) = 36. Subtracting the second and third equations, we get (b - c)(a + b + c) = 36. Therefore, a - b = b - c, and a + b + c = 3b. So, (b - c) * 3b = 36, and consequently, c = b - (12/b), and therefore a = b + (12/b).  Adding the three equations, we find: a + b + c - ab - ac - bc = 18. Therefore, (a - b)² + (b - c)² + (c - a)² = 36, i.e., (12/b)² + (12/b)² + 24/b)² = 36. Hence, b = 2√6, and therefore a = 3√6 and c = √6. Hence, x = 54, y = 24, z = 6.

    3

  • @tutorhkdass · 8 days ago

    Good one boss

  • @МиколаДзядук · 9 days ago

    (1)-(2)=&gt;(√X+√Y+√Z)(√X-√Y)=36<br>(2)-(3)=&gt;(√X+√Y+√Z)(√Y-Z)=36=&gt;<br>2√Y=√X+√Z; X-Z=48; (X; Y; Z) діляться на 6, тому X=54; Z=6; Y=24

    1

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