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${[20x^(2)-3y^(2)+12 y=-16],[20x^(2)+3y^(2)=128]:}$

$4$ ) $\left\{\begin{array}{l}20 x^{2}-3 y^{2}+12 y=-16 \\ 20 x^{2}+3 y^{2}=128\end{array}\right.$

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Interpret measures of center and variability

Full solution

Q.

4

)

\left\{\begin{array}{l}20 x^{2}-3 y^{2}+12 y=-16 \\ 20 x^{2}+3 y^{2}=128\end{array}\right.

Combine equations: Add the two equations to eliminate $y$ . $\$20x^2 - 3y^2 + 12$ + $20x^2 + 3y^2$ = $-16$ + $128$ \) $40x^2 + 12 = 112$
Solve for $x^2$ : Solve for $x^2$ . $40x^2 = 112 - 12$ $40x^2 = 100$ $x^2 = \frac{100}{40}$ $x^2 = 2.5$
Find x value: Find the value of x. $\newline$ $x = \pm\sqrt{2.5}$ $\newline$ $x = \pm1.5811$ (rounded to four decimal places)
Substitute $x$ for $y$ : Substitute $x$ back into one of the original equations to find $y$ . Using the first equation: $20(2.5) - 3y^2 + 12 = -16$ $50 - 3y^2 + 12 = -16$ $-3y^2 = -16 - 50 - 12$ $-3y^2 = -78$ $y^2 = 26$
Solve for y: Solve for y. $\newline$ $y = \pm\sqrt{26}$ $\newline$ $y = \pm5.099$ (rounded to three decimal places)

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