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7y^(2)=50 x-150

y=(3-x)/(2)
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
y_(1) and
y_(2) ?

$7 y^{2}=50 x-150$ $\newline$ $y=\frac{3-x}{2}$ $\newline$ 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 $y_{1}$ and $y_{2}$ ?

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Solve polynomial equations

Full solution

Q.

7 y^{2}=50 x-150

\newline

y=\frac{3-x}{2}

\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

y_{1}

and

y_{2}

?

Equations: We have two equations: $\newline$ $1$ ) $7y^{2} = 50x - 150$ $\newline$ $2$ ) $y = \frac{3 - x}{2}$ $\newline$ To find the solutions to the system, we can substitute the second equation into the first one to eliminate $y$ and solve for $x$ .
Substitution: Substitute $y$ from the second equation into the first equation: $7\left(\frac{3 - x}{2}\right)^{2} = 50x - 150$
Expand and Simplify: Expand and simplify the equation: $\newline$ $\frac{7(9 - 6x + x^2)}{4} = 50x - 150$ $\newline$ Multiply both sides by $4$ to get rid of the denominator: $\newline$ $7(9 - 6x + x^2) = 200x - 600$
Rearrange the Equation: Distribute the $7$ on the left side of the equation: $63 - 42x + 7x^2 = 200x - 600$
Quadratic Formula: Rearrange the equation to form a quadratic equation: $\newline$ $7x^2 - 42x - 200x + 63 + 600 = 0$ $\newline$ $7x^2 - 242x + 663 = 0$
Quadratic Formula to find roots: We need to solve the quadratic equation for $x$ . However, the quadratic does not factor easily, so we will use the quadratic formula: $\newline$ $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $\newline$ where $a = 7$ , $b = -242$ , and $c = 663$ .
Calculate Discriminant: Calculate the discriminant $b^2 - 4ac$ : $\newline$ Discriminant = $(-242)^2 - 4(7)(663)$ $\newline$ Discriminant = $58564 - 18564$ $\newline$ Discriminant = $40000$
Calculate the values of $x_{1}$ and $x_{2}$ Calculate Product of y: Since the discriminant is positive, we have two distinct real solutions for $x$ . Now we calculate the solutions using the quadratic formula: $\newline$ $x_{1,2} = \frac{242 \pm \sqrt{40000}}{14}$ $\newline$ $x_{1} = \frac{242 + \sqrt{40000}}{14} = \frac{242 + 200}{14} = \frac{442}{14} = \frac{221}{7}$ $\newline$ $x_{2} = \frac{242 - \sqrt{40000}}{14} = \frac{242 - 200}{14} = \frac{42}{14} = 3$ $\newline$ Therefore, the values of $x_1 = \frac{221}{7}$ and $x_2 = 3$ .
Calculate the values of $y_{1}$ and $y_{2}$ : Substitute the values of $x_{1}$ and $x_{2}$ in $y = \frac{3 - x}{2}$ to find the values of $y_{1}$ and $y_{2}$ respectively. $\newline$ $y_1 = \frac{3 - x_1}{2} = \frac{3 - \frac{221}{7}}{2} = \frac{\frac{21-221}{7}}{2} = -\frac{200}{14} = -\frac{100}{7}$ $\newline$ $y_2 = \frac{3 - x_2}{2} = \frac{3 - 3}{2} = \frac{0}{2} = 0$ $\newline$ Therefore, the values of $y_1 = -\frac{100}{7}$ and $y_2 = 0$ .
Final Calculation: Calculate the product of $y_{1}$ and $y_{2}$ : $\newline$ $y_{1} \times y_{2} = -\frac{100}{7} \times 0 = 0$ $\newline$ Therefore, the product of the $y_{1}$ and $y_{2}$ is $0$ .

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