x²+y²=61 x-y+10=9

To solve the system of equations:

1. \( x^2 + y^2 = 61 \)
2. \( x - y + 10 = 9 \)

Let's start with the second equation to express \(y\) in terms of \(x\):

From the equation \( x - y + 10 = 9 \):

\[ x - y = -1 \]

This simplifies to:

\[ y = x + 1 \]

Now, we can substitute \(y = x + 1\) into the first equation:

\[ x^2 + (x + 1)^2 = 61 \]

Expanding the second term:

\[ x^2 + (x^2 + 2x + 1) = 61 \]

Combining like terms:

\[ 2x^2 + 2x + 1 = 61 \]

Next, we subtract 61 from both sides:

\[ 2x^2 + 2x + 1 - 61 = 0 \]

This simplifies to:

\[ 2x^2 + 2x - 60 = 0 \]

We can divide the entire equation by 2:

\[ x^2 + x - 30 = 0 \]

Now, we can factor this quadratic equation:

\[ (x + 6)(x - 5) = 0 \]

Setting each factor equal to zero gives:

\[ x + 6 = 0 \quad \Rightarrow \quad x = -6 \] \[ x - 5 = 0 \quad \Rightarrow \quad x = 5 \]

Now, we can find the corresponding \(y\) values for each \(x\):

1. For \(x = -6\): \[ y = -6 + 1 = -5 \]

2. For \(x = 5\): \[ y = 5 + 1 = 6 \]

Thus, the solution pairs \((x, y)\) are:

\[ (-6, -5) \quad \text{and} \quad (5, 6) \]

So the final solutions to the system of equations are:

\[ (x, y) = (-6, -5) \quad \text{and} \quad (5, 6) \]