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${:[y-3=3x],[y+11=50-10 x+x^(2)]:} If (x_(1),y_(1)) and (x_(2),y_(2)) are distinct solutions to the system of equations shown, what is the product of the y_(1) and y_(2) ?$

$y-3=3 x$ $\newline$ $y+11=50-10 x+x^{2}$ $\newline$ 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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Math Problems

Algebra 1

Compare linear, exponential, and quadratic growth

Full solution

y-3=3 x

\newline

y+11=50-10 x+x^{2}

\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 product of the

y_{1}

and

y_{2}

Write Equations: Write down the system of equations. $\newline$ We have the following system of equations: $\newline$ $y - 3 = 3x$ $\newline$ $y + 11 = 50 - 10x + x^2$ $\newline$ We need to find the solutions $(x_1, y_1)$ and $(x_2, y_2)$ and then calculate the product of $y_1$ and $y_2$ .
Express $y$ in terms of $x$ : Express $y$ from the first equation in terms of $x$ . $y = 3x + 3$ Now we have $y$ in terms of $x$ , which we can substitute into the second equation.
Substitute $y$ into second equation: Substitute $y = 3x + 3$ into the second equation. $\newline$ $(3x + 3) + 11 = 50 - 10x + x^2$ $\newline$ Simplify the equation by combining like terms. $\newline$ $3x + 3 + 11 = 50 - 10x + x^2$ $\newline$ $3x + 14 = 50 - 10x + x^2$
Rearrange to quadratic equation: Rearrange the equation to form a quadratic equation. $\newline$ $x^2 + 10x + 3x - 50 + 14 = 0$ $\newline$ Combine like terms. $\newline$ $x^2 + 13x - 36 = 0$ $\newline$ Now we have a quadratic equation in standard form.
Factor quadratic equation: Factor the quadratic equation. $\newline$ We need to find two numbers that multiply to $-36$ and add up to $13$ . $\newline$ The numbers are $16$ and $-9$ . $\newline$ So we can factor the equation as: $\newline$ $(x + 16)(x - 9) = 0$
Solve for x: Solve for x. $\newline$ Set each factor equal to zero and solve for x. $\newline$ $x + 16 = 0$ or $x - 9 = 0$ $\newline$ $x = -16$ or $x = 9$ $\newline$ These are the x-values for our solutions $(x_1, y_1)$ and $(x_2, y_2)$ .
Find corresponding y-values: Find the corresponding y-values. $\newline$ Substitute $x = -16$ and $x = 9$ into $y = 3x + 3$ to find $y_1$ and $y_2$ . $\newline$ For $x = -16$ : $\newline$ $y_1 = 3(-16) + 3 = -48 + 3 = -45$ $\newline$ For $x = 9$ : $\newline$ $y_2 = 3(9) + 3 = 27 + 3 = 30$ $\newline$ Now we have our solutions: $(-16, -45)$ and $x = 9$ $0$ .
Calculate $y_1 \times y_2$ : Calculate the product of $y_1$ and $y_2$ . $y_1 \times y_2 = -45 \times 30$ $y_1 \times y_2 = -1350$ The product of $y_1$ and $y_2$ is $-1350$ .