We are tasked with solving the following system of equations:

1. $x^2 + (y + 1)^2 = 65$
2. $x - y = 10$

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Step 1: Solve the second equation for $x$

From $x - y = 10$, we can express $x$ in terms of $y$: $x = y + 10$

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Step 2: Substitute $x = y + 10$ into the first equation

Substitute $x = y + 10$ into $x^2 + (y + 1)^2 = 65$: $(y + 10)^2 + (y + 1)^2 = 65$

Expand both terms: $(y + 10)^2 = y^2 + 20y + 100$ $(y + 1)^2 = y^2 + 2y + 1$

Combine these into the equation: $(y^2 + 20y + 100) + (y^2 + 2y + 1) = 65$

Simplify: $2y^2 + 22y + 101 = 65$

Subtract 65 from both sides: $2y^2 + 22y + 36 = 0$

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Step 3: Simplify the quadratic equation

Factor out 2: $y^2 + 11y + 18 = 0$

Factorize: $(y + 9)(y + 2) = 0$

Thus, the solutions for $y$ are: $y = -9 \quad \text{or} \quad y = -2$

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Step 4: Find corresponding values for $x$

For each $y$, substitute back into $x = y + 10$:

1. If $y = -9$: $x = -9 + 10 = 1$

2. If $y = -2$: $x = -2 + 10 = 8$

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Final Solutions

The solutions are: $(x, y) = (1, -9) \quad \text{and} \quad (x, y) = (8, -2)$

Would you like more details or have any questions?

5 Related Questions:

1. How can we check whether the solutions satisfy both equations?
2. What if the quadratic equation in Step 3 couldn't be factored? How would we solve it?
3. What are the geometric interpretations of the equations?
4. How would the solution change if the constant 65 were different?
5. Can you solve the system using a graphing method?

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Tip:

Always check your solutions by substituting them back into the original equations to confirm their validity.