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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) ?

$2 x+3 y=-8$ $\newline$ $3 y^{2}-8 y=2 x+10$ $\newline$ 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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Write an equation from words

Full solution

Q.

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}

?

Express $x$ in terms of $y$ : Given the system of equations: $\newline$ $1$ ) $2x + 3y = -8$ $\newline$ $2$ ) $3y^2 - 8y = 2x + 10$ $\newline$ We need to find the product of the $x$ -coordinates of the distinct solutions to these equations. $\newline$ First, let's express $x$ from the first equation in terms of $y$ . $\newline$ $2x = -8 - 3y$ $\newline$ $x = \frac{-8 - 3y}{2}$
Substitute $x$ into second equation: Now, substitute the expression for $x$ from the first equation into the second equation. $\newline$ $3y^2 - 8y = 2((-8 - 3y) / 2) + 10$ $\newline$ Simplify the equation. $\newline$ $3y^2 - 8y = -8 - 3y + 10$ $\newline$ $3y^2 - 8y = 2 - 3y$
Set equation to zero: Next, move all terms to one side to set the equation to zero. $\newline$ $3y^2 - 8y + 3y - 2 = 0$ $\newline$ Combine like terms. $\newline$ $3y^2 - 5y - 2 = 0$
Solve 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, $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , 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 = $49$
Calculate discriminant: Since the discriminant is positive, we have two distinct real solutions for $y$ . $\newline$ Now, let's find the two solutions for $y$ using the quadratic formula. $\newline$ $y_{1,2} = \frac{5 \pm \sqrt{49}}{2 \times 3}$ $\newline$ $y_{1,2} = \frac{5 \pm 7}{6}$ $\newline$ We have two solutions for $y$ : $\newline$ $y_{1} = \frac{5 + 7}{6}$ $\newline$ $y_{1} = \frac{12}{6}$ $\newline$ $y_{1} = 2$ $\newline$ $y_{2} = \frac{5 - 7}{6}$ $\newline$ $y_{2} = \frac{-2}{6}$ $\newline$ $y$ $0$
Find 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. $\newline$ For $y_1 = 2$ : $\newline$ $x_1 = (-8 - 3(2)) / 2$ $\newline$ $x_1 = (-8 - 6) / 2$ $\newline$ $x_1 = -14 / 2$ $\newline$ $x_1 = -7$
Find x values: For $y_{2} = -\frac{1}{3}$ : $\newline$ $x_{2} = ( -8 - 3(-\frac{1}{3})) / 2$ $\newline$ $x_{2} = (-8 + 1) / 2$ $\newline$ $x_{2} = -\frac{7}{2}$
Calculate product of x values: Finally, we find the product of $x_{1}$ and $x_{2}$ . $\newline$ Product = $x_{1} \times x_{2}$ $\newline$ Product = $(-7) \times (-7 / 2)$ $\newline$ Product = $\frac{49}{2}$

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