(Х-1)2+(у+2)2=25 Х-У=-1

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To solve this system of equations, we can use the substitution method.

We start with the first equation:

(Х-1)² + (у+2)² = 25

Expanding the terms, we get:

Х² - 2Х + 1 + у² + 4у + 4 = 25

Х² - 2Х + у² + 4у - 20 = 0

Next, we use the second equation:

Х -y = -1

We can rearrange this equation to be:

Х = y - 1

Now we substitute this expression for X into the first equation:

(y - 1)² - 2(y - 1) + у² + 4у - 20 = 0

Expanding the terms, we get:

у² - 2у + 1 - 2у + 2 + у² + 4у - 20 = 0

2у² + 15у - 17 = 0

Now we can solve this quadratic equation for y using the quadratic formula:

y = (-15 ± √(15² - 4 2 -17)) / 2 * 2
y = (-15 ± √(225 + 136)) / 4
y = (-15 ± √361) / 4
y = (-15 ± 19) / 4

y = 4/2 = 2
or
y = -34/2 = -17

Now that we have found possible values for y, we can plug them back into the equation X = y - 1 to find corresponding values for X:

If y = 2, then X = 2 - 1 =1
If y = -17, then X = -17 - 1 = -18

So the solutions to the system of equations are:

X = 1, Y = 2
or
X = -18, Y = -17