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 equations shown, what is the product of the x_(1) and x_(2) ?

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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 equations shown, what is the product of the  x_(1) and  x_(2) ?

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Express x in terms of y: Given the system of equations: 1) 2x + 3y = -8 2) 3y^2 - 8y = 2x + 10 We need to find the product of the x-coordinates of the distinct solutions to these equations. First, let's express x from the first equation in terms of y. 2x = -8 - 3y x = (-8 - 3y) / 2Substitute x into second equation: Now, substitute the expression for x from the first equation into the second equation. 3y^2 - 8y = 2((-8 - 3y) / 2) + 10 Simplify the equation. 3y^2 - 8y = -8 - 3y + 10 3y^2 - 8y = 2 - 3ySet equation to zero: Next, move all terms to one side to set the equation to zero. 3y^2 - 8y + 3y - 2 = 0 Combine like terms. 3y^2 - 5y - 2 = 0Solve quadratic equation for y: Now, we need to solve the quadratic equation for y. To find the roots of the quadratic equation, we can use the quadratic formula, y = (-b ± √(b^2 - 4ac)) / (2a), where a = 3, b = -5, and c = -2. Let's calculate the discriminant (b^2 - 4ac) first. Discriminant = (-5)^2 - 4(3)(-2) Discriminant = 25 + 24 Discriminant = 49Calculate discriminant: Since the discriminant is positive, we have two distinct real solutions for y. Now, let's find the two solutions for y using the quadratic formula. y_(1,2) = (5 ± √49) / (2 * 3) y_(1,2) = (5 ± 7) / 6 We have two solutions for y: y_(1) = (5 + 7) / 6 y_(1) = 12 / 6 y_(1) = 2 y_(2) = (5 - 7) / 6 y_(2) = -2 / 6 y_(2) = -1/3Find solutions for y: Now that we have the values for y_(1) and y_(2), we can find the corresponding x-values using the expression for x we found earlier. For y_(1) = 2: x_(1) = (-8 - 3(2)) / 2 x_(1) = (-8 - 6) / 2 x_(1) = -14 / 2 x_(1) = -7Find x values: For y_(2) = -1/3: x_(2) = (-8 - 3(-1/3)) / 2 x_(2) = (-8 + 1) / 2 x_(2) = -7 / 2Calculate product of x values: Finally, we find the product of x_(1) and x_(2). Product = x_(1) * x_(2) Product = (-7) * (-7 / 2) Product = 49 / 2

$2 x+3 y=-8$ $\newline$ $3 y^{2}-8 y=2 x+10$ $\newline$ If $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ are distinct solutions to the system of equations shown, what is the product of the $x_{1}$ and $x_{2}$ ?

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