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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 $y_{2}$ ?

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Algebra 1

Compare linear, exponential, and quadratic growth

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}

Simplify equations: We have the system of equations: $\newline$ $1$ ) $-27x + 54y = 9x^2 - 19$ $\newline$ $2$ ) $3x - 6y = -5$ $\newline$ First, we simplify both equations. For the second equation, we can divide by $3$ to simplify it.
Express $y$ in terms: The simplified second equation is: $\newline$ $x - 2y = -\frac{5}{3}$ $\newline$ We can rewrite this as: $\newline$ $2y = x + \frac{5}{3}$
Substitute $y$ into first: Now, let's express $y$ in terms of $x$ from the simplified second equation: $\newline$ $y = \frac{x + \frac{5}{3}}{2}$ $\newline$ $y = \frac{x}{2} + \frac{5}{6}$
Solve for x: Next, we substitute $y = \frac{x}{2} + \frac{5}{6}$ into the first equation to solve for x: $\newline$ \(-27x + $54$ \left(\frac{x}{ $2$ } + \frac{ $5$ }{ $6$ }\right) = $9$ x^ $2$ - $19$
Find x values: We distribute and simplify the equation: $\newline$ $-27x + 27x + 45 = 9x^2 - 19$ $\newline$ The $-27x$ and $+27x$ cancel each other out, so we have: $\newline$ $45 = 9x^2 - 19$
Calculate y values: Now, we solve for $x^2$ : $\newline$ $9x^2 = 45 + 19$ $\newline$ $9x^2 = 64$ $\newline$ $x^2 = \frac{64}{9}$
Sum of y values: Since $x^2 = \frac{64}{9}$ , we find the square root of both sides to solve for $x$ : $x = \pm\sqrt{\frac{64}{9}}$ $x = \pm\frac{8}{3}$
Sum of y values: Since $x^2 = \frac{64}{9}$ , we find the square root of both sides to solve for $x$ : $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $\newline$ $x = \pm\frac{8}{3}$ We have two values for $x$ , which are $x_1 = \frac{8}{3}$ and $x_2 = -\frac{8}{3}$ . Now we will find the corresponding $y$ -values using the equation $y = \frac{x}{2} + \frac{5}{6}$ . $\newline$ For $x_1 = \frac{8}{3}$ : $\newline$ $y_1 = \left(\frac{8}{3}\right)/2 + \frac{5}{6}$
Sum of y values: Since $x^2 = \frac{64}{9}$ , we find the square root of both sides to solve for x: $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $\newline$ $x = \pm\frac{8}{3}$ We have two values for x, which are $x_1 = \frac{8}{3}$ and $x_2 = -\frac{8}{3}$ . Now we will find the corresponding y-values using the equation $y = \frac{x}{2} + \frac{5}{6}$ . $\newline$ For $x_1 = \frac{8}{3}$ : $\newline$ $y_1 = (\frac{8}{3})/2 + \frac{5}{6}$ Calculating $y_1$ : $\newline$ $y_1 = \frac{4}{3} + \frac{5}{6}$ $\newline$ To add these fractions, we need a common denominator, which is $x = \pm\sqrt{\frac{64}{9}}$ $0$ : $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $1$ $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $2$
Sum of y values: Since $x^2 = \frac{64}{9}$ , we find the square root of both sides to solve for $x$ : $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $\newline$ $x = \pm\frac{8}{3}$ We have two values for $x$ , which are $x_1 = \frac{8}{3}$ and $x_2 = -\frac{8}{3}$ . Now we will find the corresponding $y$ -values using the equation $y = \frac{x}{2} + \frac{5}{6}$ . $\newline$ For $x_1 = \frac{8}{3}$ : $\newline$ $x$ $0$ Calculating $x$ $1$ : $\newline$ $x$ $2$ $\newline$ To add these fractions, we need a common denominator, which is $x$ $3$ : $\newline$ $x$ $4$ $\newline$ $x$ $5$ For $x_2 = -\frac{8}{3}$ : $\newline$ $x$ $7$
Sum of y values: Since $x^2 = \frac{64}{9}$ , we find the square root of both sides to solve for x: $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $\newline$ $x = \pm\frac{8}{3}$ We have two values for x, which are $x_1 = \frac{8}{3}$ and $x_2 = -\frac{8}{3}$ . Now we will find the corresponding y-values using the equation $y = \frac{x}{2} + \frac{5}{6}$ . $\newline$ For $x_1 = \frac{8}{3}$ : $\newline$ $y_1 = \left(\frac{8}{3}\right)/2 + \frac{5}{6}$ Calculating $y_1$ : $\newline$ $y_1 = \frac{4}{3} + \frac{5}{6}$ $\newline$ To add these fractions, we need a common denominator, which is $x = \pm\sqrt{\frac{64}{9}}$ $0$ : $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $1$ $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $2$ For $x_2 = -\frac{8}{3}$ : $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $4$ Calculating $x = \pm\sqrt{\frac{64}{9}}$ $5$ : $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $6$ $\newline$ Again, we need a common denominator, which is $x = \pm\sqrt{\frac{64}{9}}$ $0$ : $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $8$ $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $9$ $\newline$ $x = \pm\frac{8}{3}$ $0$
Sum of y values: Since $x^2 = \frac{64}{9}$ , we find the square root of both sides to solve for x: $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $\newline$ $x = \pm\frac{8}{3}$ We have two values for x, which are $x_1 = \frac{8}{3}$ and $x_2 = -\frac{8}{3}$ . Now we will find the corresponding y-values using the equation $y = \frac{x}{2} + \frac{5}{6}$ . $\newline$ For $x_1 = \frac{8}{3}$ : $\newline$ $y_1 = (\frac{8}{3})/2 + \frac{5}{6}$ Calculating $y_1$ : $\newline$ $y_1 = \frac{4}{3} + \frac{5}{6}$ $\newline$ To add these fractions, we need a common denominator, which is $x = \pm\sqrt{\frac{64}{9}}$ $0$ : $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $1$ $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $2$ For $x_2 = -\frac{8}{3}$ : $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $4$ Calculating $x = \pm\sqrt{\frac{64}{9}}$ $5$ : $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $6$ $\newline$ Again, we need a common denominator, which is $x = \pm\sqrt{\frac{64}{9}}$ $0$ : $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $8$ $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $9$ $\newline$ $x = \pm\frac{8}{3}$ $0$ Now we have both y-values: $x = \pm\sqrt{\frac{64}{9}}$ $2$ and $x = \pm\frac{8}{3}$ $0$ . We find the sum of $y_1$ and $x = \pm\sqrt{\frac{64}{9}}$ $5$ : $\newline$ $x = \pm\frac{8}{3}$ $5$
Sum of y values: Since $x^2 = \frac{64}{9}$ , we find the square root of both sides to solve for x: $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $\newline$ $x = \pm\frac{8}{3}$ We have two values for x, which are $x_1 = \frac{8}{3}$ and $x_2 = -\frac{8}{3}$ . Now we will find the corresponding y-values using the equation $y = \frac{x}{2} + \frac{5}{6}$ . $\newline$ For $x_1 = \frac{8}{3}$ : $\newline$ $y_1 = \left(\frac{8}{3}\right)/2 + \frac{5}{6}$ Calculating $y_1$ : $\newline$ $y_1 = \frac{4}{3} + \frac{5}{6}$ $\newline$ To add these fractions, we need a common denominator, which is $x = \pm\sqrt{\frac{64}{9}}$ $0$ : $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $1$ $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $2$ For $x_2 = -\frac{8}{3}$ : $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $4$ Calculating $x = \pm\sqrt{\frac{64}{9}}$ $5$ : $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $6$ $\newline$ Again, we need a common denominator, which is $x = \pm\sqrt{\frac{64}{9}}$ $0$ : $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $8$ $\newline$ $x = \pm\sqrt{\frac{64}{9}}$ $9$ $\newline$ $x = \pm\frac{8}{3}$ $0$ Now we have both y-values: $x = \pm\sqrt{\frac{64}{9}}$ $2$ and $x = \pm\frac{8}{3}$ $0$ . We find the sum of $y_1$ and $x = \pm\sqrt{\frac{64}{9}}$ $5$ : $\newline$ $x = \pm\frac{8}{3}$ $5$ To add these fractions, we need a common denominator, which is $x = \pm\sqrt{\frac{64}{9}}$ $0$ : $\newline$ $x = \pm\frac{8}{3}$ $7$ $\newline$ $x = \pm\frac{8}{3}$ $8$ $\newline$ $x = \pm\frac{8}{3}$ $9$