To solve the system of equations:
[tex]\[
\left\{\begin{array}{r}
(x+3)^2+(y+2)^2=17 \\
x+y=-2
\end{array}\right.
\][/tex]
we will follow the steps below.
1. Rewrite the second equation for one of the variables:\
[tex]\[x + y = -2\][/tex]
Rewriting for [tex]\(y\)[/tex]:
[tex]\[ y = -2 - x \][/tex]
2. Substitute [tex]\(y\)[/tex] into the first equation:\
[tex]\[ (x + 3)^2 + (y + 2)^2 = 17 \][/tex]
Substitute [tex]\(y = -2 - x\)[/tex]:
[tex]\[ (x + 3)^2 + ((-2 - x) + 2)^2 = 17 \][/tex]
[tex]\[ (x + 3)^2 + (-x)^2 = 17 \][/tex]
[tex]\[ (x + 3)^2 + x^2 = 17 \][/tex]
3. Expand and simplify the equation:
[tex]\[ (x + 3)^2 = x^2 + 6x + 9 \][/tex]
So, the equation becomes:
[tex]\[ x^2 + 6x + 9 + x^2 = 17 \][/tex]
[tex]\[ 2x^2 + 6x + 9 = 17 \][/tex]
4. Solve the quadratic equation:
[tex]\[ 2x^2 + 6x + 9 - 17 = 0 \][/tex]
[tex]\[ 2x^2 + 6x - 8 = 0 \][/tex]
Divide the entire equation by 2 to simplify:
[tex]\[ x^2 + 3x - 4 = 0 \][/tex]
Factorize the quadratic equation:
[tex]\[ (x + 4)(x - 1) = 0 \][/tex]
So, we have two possible values for [tex]\(x\)[/tex]:
[tex]\[ x = -4 \quad \text{or} \quad x = 1 \][/tex]
5. Find the corresponding [tex]\(y\)[/tex] values for each [tex]\(x\)[/tex]:
- For [tex]\(x = -4\)[/tex]:
[tex]\[ y = -2 - (-4) \][/tex]
[tex]\[ y = -2 + 4 \][/tex]
[tex]\[ y = 2 \][/tex]
The first solution is [tex]\((-4, 2)\)[/tex].
- For [tex]\(x = 1\)[/tex]:
[tex]\[ y = -2 - 1 \][/tex]
[tex]\[ y = -3 \][/tex]
The second solution is [tex]\((1, -3)\)[/tex].
6. Order the pairs by the smallest [tex]\(x\)[/tex]-coordinate first:
- [tex]\((-4, 2)\)[/tex]
- [tex]\((1, -3)\)[/tex]
Therefore, the solutions are:
[tex]\[
([-4, 2], [1, -3])
\][/tex]
So, we submit:
(([ -4, 2 ]), ([ 1, -3 ]))
