To solve the equation [tex]\( -90x + 84x^2 + 90y - 200 = 120xy - 147y^2 \)[/tex], we'll proceed through the following steps:
1. Rewrite the Equation: We start with the original equation:
[tex]\[
-90x + 84x^2 + 90y - 200 = 120xy - 147y^2
\][/tex]
2. Combine Like Terms and Simplify:
Move all terms to one side of the equation to set it to equal zero:
[tex]\[
84x^2 - 90x + 90y - 200 - 120xy + 147y^2 = 0
\][/tex]
3. Group Terms:
Group the terms involving [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[
84x^2 - 120xy + 147y^2 - 90x + 90y - 200 = 0
\][/tex]
4. Solve the System of Equations:
Given the result for solving this equation:
[tex]\[
(x, y) = \left(\frac{5y}{7} - \frac{\sqrt{-8748y^2 - 2160y + 18825}}{84} + \frac{15}{28}, y\right) \text{ and } \left(\frac{5y}{7} + \frac{\sqrt{-8748y^2 - 2160y + 18825}}{84} + \frac{15}{28}, y\right)
\][/tex]
This means we have two potential solutions for [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex]:
1. [tex]\( x = \frac{5y}{7} - \frac{\sqrt{-8748y^2 - 2160y + 18825}}{84} + \frac{15}{28} \)[/tex]
2. [tex]\( x = \frac{5y}{7} + \frac{\sqrt{-8748y^2 - 2160y + 18825}}{84} + \frac{15}{28} \)[/tex]
Thus, each solution consists of a pair [tex]\((x, y)\)[/tex] as:
[tex]\[
\left(\frac{5y}{7} - \frac{\sqrt{-8748y^2 - 2160y + 18825}}{84} + \frac{15}{28}, y\right)
\][/tex] and
[tex]\[
\left(\frac{5y}{7} + \frac{\sqrt{-8748y^2 - 2160y + 18825}}{84} + \frac{15}{28}, y\right)
\][/tex]
In conclusion, the solutions express [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex]. These solutions suggest a parametric form where [tex]\(y\)[/tex] can theoretically take any value, and the corresponding values of [tex]\(x\)[/tex] are determined as shown. To interpret or contextualize these solutions further, we might consider specific values for [tex]\(y\)[/tex] or analyze the nature of the quadratic expression under the square root.
