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

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 = ±11.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.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 71 to get y = -380.We have found the two sets of solutions for the system of equations: (11, 522) and (-11, -380). Write the solution as coordinate points.

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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.

\newline

y = 41x + 71

\newline

y = x^2 + 41x - 50

\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 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.