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If
(x,y) is a solution to the system of equations shown, what is the product of the
y-coordinates of the solutions?

x^(2)+y^(2)=9

x+y=3

If $(x, y)$ is a solution to the system of equations shown, what is the product of the $y$ -coordinates of the solutions? $\newline$ $x^{2}+y^{2}=9$ $\newline$ $x+y=3$

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Evaluate recursive formulas for sequences

Full solution

Q. If

(x, y)

is a solution to the system of equations shown, what is the product of the

y

-coordinates of the solutions?

\newline

x^{2}+y^{2}=9

\newline

x+y=3

Given Equations: We are given the system of equations: $\newline$ $1$ . $x^2 + y^2 = 9$ $\newline$ $2$ . $x + y = 3$ $\newline$ We need to find the y-coordinates of the solutions to these equations and then calculate their product. Let's start by expressing x in terms of y using the second equation. $\newline$ $x = 3 - y$
Expressing x in terms of y: Now, we substitute $x = 3 - y$ into the first equation to find y. $\newline$ $(3 - y)^2 + y^2 = 9$ $\newline$ Expanding the squared term, we get: $\newline$ $9 - 6y + y^2 + y^2 = 9$ $\newline$ Simplifying, we combine like terms: $\newline$ $2y^2 - 6y = 0$
Substituting x into the first equation: We can factor out a y from the equation: $\newline$ $y(2y - 6) = 0$ $\newline$ This gives us two possible solutions for y: $\newline$ $1$ . $y = 0$ $\newline$ $2$ . $2y - 6 = 0$ which simplifies to $y = 3$
Factoring out y: We have two y-coordinates for the solutions: $y = 0$ and $y = 3$ . To find the product of the y-coordinates, we multiply these two values together: $\newline$ Product of y-coordinates = $y_1 \times y_2 = 0 \times 3 = 0$
Finding the y-coordinates: The product of the y-coordinates of the solutions to the system of equations is $0$ .

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