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${:[-27 x+54 y=9x^(2)-19],[3x-6y=-5]:} If (x_(1),y_(1)) and (x_(2),y_(2)) are distinct solutions to the system of equations shown, what is the sum of the y-values y_(1) and y_(2) ? ◻$

$\begin{aligned} -27 x+54 y & =9 x^{2}-19 \\ 3 x-6 y & =-5 \end{aligned}$ $\newline$ 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 sum of the $y$ -values $y_{1}$ and $y_{2}$ ? $\newline$ $\square$

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Full solution

\begin{aligned} -27 x+54 y & =9 x^{2}-19 \\ 3 x-6 y & =-5 \end{aligned}

\newline

\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 sum of the

y

-values

y_{1}

and

y_{2}

\newline

\square

Write Equations: First, let's write down the system of equations: $\newline$ $1$ ) $-27x + 54y = 9x^2 - 19$ $\newline$ $2$ ) $3x - 6y = -5$
Simplify Second Equation: We can simplify the second equation by dividing by $3$ to make it easier to work with: $\newline$ $2$ ) $x - 2y = -\frac{5}{3}$
Solve for x: Now, let's solve the second equation for x: $\newline$ $x = 2y - \frac{5}{3}$
Substitute $x$ : Substitute $x$ from the second equation into the first equation: $\newline$ $-27(2y - \frac{5}{3}) + 54y = 9(2y - \frac{5}{3})^2 - 19$
Simplify First Equation: Simplify the equation: $-54y + 45 + 54y = 9(4y^2 - \frac{10y}{3} + \frac{25}{9}) - 19$
Cancel Terms: The $-54y$ and $+54y$ cancel each other out, so we're left with: $\newline$ $45 = 36y^2 - 30y + 25 - 19$
Simplify Right Side: Simplify the right side of the equation: $45 = 36y^2 - 30y + 6$
Set Equation to Zero: Subtract $45$ from both sides to set the equation to zero: $\newline$ $0 = 36y^2 - 30y - 39$
Correct Simplification: Now we need to solve this quadratic equation for $y$ . However, I made a mistake in the previous step. The correct simplification should be: $45 = 36y^2 - 30y + \frac{25}{9} - 19$