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{:[2x+3y=-8],[3y^(2)-8y=2x+10]:} If (x_(1),y_(1)) and (x_(2),y_(2)) are distinct solutions to the system of equation shown, what is the product of the x_(1) and x_(2) ? ◻

2x+3y=−83y2−8y=2x+10 \begin{array}{l} 2 x+3 y=-8 \\ 3 y^{2}-8 y=2 x+10 \end{array} \newlineIf (x1,y1) \left(x_{1}, y_{1}\right) and (x2,y2) \left(x_{2}, y_{2}\right) are distinct solutions to the system of equation shown, what is the product of the x1 x_{1} and x2 x_{2} ?\newline□ \square

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Q. 2x+3y=−83y2−8y=2x+10 \begin{array}{l} 2 x+3 y=-8 \\ 3 y^{2}-8 y=2 x+10 \end{array} \newlineIf (x1,y1) \left(x_{1}, y_{1}\right) and (x2,y2) \left(x_{2}, y_{2}\right) are distinct solutions to the system of equation shown, what is the product of the x1 x_{1} and x2 x_{2} ?\newline□ \square
  1. Isolate x in first equation: First, let's isolate x in the first equation: 2x+3y=−82x + 3y = -8.\newline2x=−8−3y2x = -8 - 3y\newlinex=−8−3y2x = \frac{-8 - 3y}{2}
  2. Substitute xx in second equation: Now, substitute xx in the second equation: 3y2−8y=2x+103y^2 - 8y = 2x + 10.\newline3y2−8y=2((−8−3y)/2)+103y^2 - 8y = 2((-8 - 3y) / 2) + 10
  3. Simplify the equation: Simplify the equation: 3y2−8y=−8−3y+103y^2 - 8y = -8 - 3y + 10. 3y2−8y=−3y+23y^2 - 8y = -3y + 2
  4. Bring all terms together: Bring all terms to one side: 3y2−8y+3y−2=03y^2 - 8y + 3y - 2 = 0.\newline3y2−5y−2=03y^2 - 5y - 2 = 0
  5. Factor the quadratic equation: Factor the quadratic equation: 3y + 1)(y - 2) = 0\. So, \$y = -\frac{1}{3} or y=2y = 2.
  6. Substitute y values for x: Substitute y values back into x=−8−3y2x = \frac{-8 - 3y}{2} to find x1x_{1} and x2x_{2}. For y=−13y = -\frac{1}{3}: x=−8−3(−13)2=−8+12=−72x = \frac{-8 - 3(-\frac{1}{3})}{2} = \frac{-8 + 1}{2} = -\frac{7}{2} For y=2y = 2: x=−8−3(2)2=−8−62=−142=−7x = \frac{-8 - 3(2)}{2} = \frac{-8 - 6}{2} = -\frac{14}{2} = -7

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