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$\begin{aligned} y^2+25&=x \ 10y&=x \end{aligned}$

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Csc, sec, and cot of special angles

Full solution

Q.

\begin{aligned} y^2+25&=x \ 10y&=x \end{aligned}

Start with the second equation: Start with the second equation $10y = x$ .
Substitute into first equation: Substitute $x = 10y$ into the first equation $y^2 + 25 = x$ . $\newline$ $y^2 + 25 = 10y$
Rearrange to form quadratic equation: Rearrange the equation to form a quadratic equation. $\newline$ $y^2 - 10y + 25 = 0$
Factor the quadratic equation: Factor the quadratic equation. $\newline$ $(y - 5)^2 = 0$
Solve for y: Solve for $y$ . $\newline$ $y - 5 = 0$ $\newline$ $y = 5$
Substitute back to find x: Substitute $y = 5$ back into $x = 10y$ to find $x$ . $\newline$ $x = 10 \times 5$ $\newline$ $x = 50$

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