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(3x-y)^2+(x-5)^2=0

2 min read Jun 16, 2024
(3x-y)^2+(x-5)^2=0

Solving the Equation (3x-y)^2 + (x-5)^2 = 0

This equation presents a unique situation where we need to find the values of x and y that satisfy the given condition. Let's break down the process:

Understanding the Equation

The equation (3x-y)^2 + (x-5)^2 = 0 represents the sum of two squared terms. Remember:

  • The square of any real number is always non-negative (greater than or equal to zero).

This implies that both (3x-y)^2 and (x-5)^2 must be equal to zero for the entire equation to hold true.

Solving for x and y

  1. (3x-y)^2 = 0: For this term to be zero, the expression inside the parentheses must be zero. Therefore, 3x - y = 0.

  2. (x-5)^2 = 0: Similarly, for this term to be zero, x - 5 = 0.

  3. Solving the System: We now have two linear equations:

    • 3x - y = 0
    • x - 5 = 0

    Solving this system of equations, we get:

    • x = 5
    • y = 15

Conclusion

The only solution to the equation (3x-y)^2 + (x-5)^2 = 0 is x = 5 and y = 15. This demonstrates that the equation only holds true for this specific pair of values.

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