To solve the system of equations algebraically:
[tex]\[
\begin{cases}
x^2 + y^2 = 18 \\
x - 2y = -3
\end{cases}
\][/tex]
1. Isolate [tex]\( x \)[/tex] from the second equation:
[tex]\[
x - 2y = -3 \implies x = 2y - 3
\][/tex]
2. Substitute [tex]\( x \)[/tex] in the first equation:
[tex]\[
(2y - 3)^2 + y^2 = 18
\][/tex]
3. Expand [tex]\((2y - 3)^2\)[/tex]:
[tex]\[
(2y - 3)^2 = 4y^2 - 12y + 9
\][/tex]
4. Substitute back into the first equation:
[tex]\[
4y^2 - 12y + 9 + y^2 = 18
\][/tex]
5. Combine like terms:
[tex]\[
5y^2 - 12y + 9 = 18
\][/tex]
6. Bring all terms to one side of the equation:
[tex]\[
5y^2 - 12y + 9 - 18 = 0 \implies 5y^2 - 12y - 9 = 0
\][/tex]
7. Solve the quadratic equation [tex]\( 5y^2 - 12y - 9 = 0 \)[/tex] using the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 5 \)[/tex], [tex]\( b = -12 \)[/tex], and [tex]\( c = -9 \)[/tex]:
[tex]\[
y = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 5 \cdot (-9)}}{2 \cdot 5}
\][/tex]
[tex]\[
y = \frac{12 \pm \sqrt{144 + 180}}{10}
\][/tex]
[tex]\[
y = \frac{12 \pm \sqrt{324}}{10}
\][/tex]
[tex]\[
y = \frac{12 \pm 18}{10}
\][/tex]
[tex]\[
y = \frac{12 + 18}{10} \quad \text{or} \quad y = \frac{12 - 18}{10}
\][/tex]
[tex]\[
y = 3 \quad \text{or} \quad y = -\frac{6}{5}
\][/tex]
8. Find the corresponding [tex]\( x \)[/tex] values for each [tex]\( y \)[/tex]:
For [tex]\( y = 3 \)[/tex]:
[tex]\[
x = 2(3) - 3 = 6 - 3 = 3
\][/tex]
For [tex]\( y = -\frac{6}{5} \)[/tex]:
[tex]\[
x = 2 \left(-\frac{6}{5}\right) - 3 = -\frac{12}{5} - 3 = -\frac{12}{5} - \frac{15}{5} = -\frac{27}{5}
\][/tex]
9. Combine the solutions:
[tex]\[
(x, y) = (3, 3) \quad \text{and} \quad \left( -\frac{27}{5}, -\frac{6}{5} \right)
\][/tex]
Thus, the solutions to the system of equations are:
[tex]\[
(x, y) = (3, 3) \quad \text{and} \quad \left( -\frac{27}{5}, -\frac{6}{5} \right).
\][/tex]
