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Solve the system of equations. $\newline$ $y = 27x + 46$ $\newline$ $y = x^2 + 27x + 10$ $\newline$ Write the coordinates in exact form. Simplify all fractions and radicals. $\newline$ (,) $\newline$ (,)

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Solve a nonlinear system of equations

Full solution

Q. Solve the system of equations.

\newline

y = 27x + 46

\newline

y = x^2 + 27x + 10

\newline

Write the coordinates in exact form. Simplify all fractions and radicals.

\newline

(______,______)

\newline

(______,______)

Substitute $y$ into second equation: Substitute $y$ from the first equation into the second equation: $y = x^2 + 27x + 10$ becomes $27x + 46 = x^2 + 27x + 10$ .
Simplify the equation: Subtract $27x$ from both sides to get $46 = x^2 + 10$ .
Find $x$ : Subtract $10$ from both sides to get $36 = x^2$ .
Calculate $y$ for $x=6$ : Take the square root of both sides to find $x$ . This gives us $x = \pm 6$ .
Calculate $y$ for $x=-6$ : Substitute $x$ back into the first equation $y = 27x + 46$ to find $y$ when $x = 6$ . This gives us $y = 27(6) + 46$ .
Calculate $y$ for $x=-6$ : Substitute $x$ back into the first equation $y = 27x + 46$ to find $y$ when $x = 6$ . This gives us $y = 27(6) + 46$ .Calculate $y$ for $x = 6$ . This gives us $y = 162 + 46$ , so $x=-6$ $0$ .
Calculate $y$ for $x=-6$ : Substitute $x$ back into the first equation $y = 27x + 46$ to find $y$ when $x = 6$ . This gives us $y = 27(6) + 46$ .Calculate $y$ for $x = 6$ . This gives us $y = 162 + 46$ , so $x=-6$ $0$ .Substitute $x$ back into the first equation $y = 27x + 46$ to find $y$ when $x=-6$ $4$ . This gives us $x=-6$ $5$ .
Calculate $y$ for $x=-6$ : Substitute $x$ back into the first equation $y = 27x + 46$ to find $y$ when $x = 6$ . This gives us $y = 27(6) + 46$ .Calculate $y$ for $x = 6$ . This gives us $y = 162 + 46$ , so $x=-6$ $0$ .Substitute $x$ back into the first equation $y = 27x + 46$ to find $y$ when $x=-6$ $4$ . This gives us $x=-6$ $5$ .Calculate $y$ for $x=-6$ $4$ . This gives us $x=-6$ $8$ , so $x=-6$ $9$ .

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