Solve the system of equations. $\newline$ $y = 41x + 71$ $\newline$ $y = x^2 + 41x - 50$ $\newline$ Write the coordinates in exact form. Simplify all fractions and radicals. $\newline$ (,) $\newline$ (,)

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Algebra 2

Solve a nonlinear system of equations

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Q. Solve the system of equations.

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y = 41x + 71

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y = x^2 + 41x - 50

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

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(______,______)

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(______,______)

Substitute $y$ into second equation: Substitute $y$ from the first equation into the second equation since they are both equal to $y$ . This gives us the equation $41x + 71 = x^2 + 41x - 50$ .
Simplify the equation: Simplify the equation by subtracting $41x$ from both sides and subtracting $71$ from both sides to get $0 = x^2 - 50 - 71$ .
Combine like terms: Combine like terms to get the simplified equation $0 = x^2 - 121$ .
Solve for x: Solve for x by finding the square roots of $121$ . Since the equation is $x^2 = 121$ , we take the square root of both sides to get $x = \pm11$ .
Substitute $x=11$ into first equation: Now that we have the two possible values for $x$ , we can substitute them back into the first equation $y = 41x + 71$ to find the corresponding $y$ values. First, let's substitute $x = 11$ into the equation.
Calculate $y$ for $x=11$ : Substituting $x = 11$ into $y = 41x + 71$ gives us $y = 41(11) + 71$ .
Substitute $x=-11$ into first equation: Calculate the value of $y$ when $x = 11$ . This gives us $y = 451 + 71$ .
Calculate $y$ for $x=-11$ : Add $451$ and $71$ to get $y = 522$ .
Find two sets of solutions: Now let's substitute $x = -11$ into the equation $y = 41x + 71$ to find the corresponding $y$ value for the second solution.
Find two sets of solutions: Now let's substitute $x = -11$ into the equation $y = 41x + 71$ to find the corresponding $y$ value for the second solution.Substituting $x = -11$ into $y = 41x + 71$ gives us $y = 41(-11) + 71$ .
Find two sets of solutions: Now let's substitute $x = -11$ into the equation $y = 41x + 71$ to find the corresponding $y$ value for the second solution.Substituting $x = -11$ into $y = 41x + 71$ gives us $y = 41(-11) + 71$ .Calculate the value of $y$ when $x = -11$ . This gives us $y = -451 + 71$ .
Find two sets of solutions: Now let's substitute $x = -11$ into the equation $y = 41x + 71$ to find the corresponding $y$ value for the second solution.Substituting $x = -11$ into $y = 41x + 71$ gives us $y = 41(-11) + 71$ .Calculate the value of $y$ when $x = -11$ . This gives us $y = -451 + 71$ .Add $-451$ and $y = 41x + 71$ $0$ to get $y = 41x + 71$ $1$ .
Find two sets of solutions: Now let's substitute $x = -11$ into the equation $y = 41x + 71$ to find the corresponding $y$ value for the second solution.Substituting $x = -11$ into $y = 41x + 71$ gives us $y = 41(-11) + 71$ .Calculate the value of $y$ when $x = -11$ . This gives us $y = -451 + 71$ .Add $-451$ and $y = 41x + 71$ $0$ to get $y = 41x + 71$ $1$ .We have found the two sets of solutions for the system of equations: $y = 41x + 71$ $2$ and $y = 41x + 71$ $3$ . Write the solution as coordinate points.