Solve the system of equations. y = 38x + 13 y = x^2 + 28x + 34 Write the coordinates in exact form. Simplify all fractions and radicals. (______,______) (______,______)

Set Equations Equal: Set the two equations equal to each other since they both equal y. y = 38x + 13 y = x^2 + 28x + 34 So, 38x + 13 = x^2 + 28x + 34.Rearrange and Solve: Rearrange the equation to set it to zero and solve for x. x^2 + 28x + 34 - 38x - 13 = 0 x^2 - 10x + 21 = 0Factor Quadratic Equation: Factor the quadratic equation. (x - 7)(x - 3) = 0Solve for x: Solve for x by setting each factor equal to zero. x - 7 = 0 or x - 3 = 0 So, x = 7 or x = 3Substitute x = 7: Substitute x = 7 into one of the original equations to find the corresponding y value. y = 38(7) + 13 y = 266 + 13 y = 279Substitute x = 3: Substitute x = 3 into one of the original equations to find the corresponding y value. y = 38(3) + 13 y = 114 + 13 y = 127

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Solve the system of equations. $\newline$ $y = 38x + 13$ $\newline$ $y = x^2 + 28x + 34$ $\newline$ 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 = 38x + 13

\newline

y = x^2 + 28x + 34

\newline

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

\newline

(______,______)

\newline

(______,______)

Set Equations Equal: Set the two equations equal to each other since they both equal $y$ . $y = 38x + 13$ $y = x^2 + 28x + 34$ So, $38x + 13 = x^2 + 28x + 34$ .
Rearrange and Solve: Rearrange the equation to set it to zero and solve for $x$ . $x^2 + 28x + 34 - 38x - 13 = 0$ $x^2 - 10x + 21 = 0$
Factor Quadratic Equation: Factor the quadratic equation. $(x - 7)(x - 3) = 0$
Solve for x: Solve for x by setting each factor equal to zero. $\newline$ $x - 7 = 0$ or $x - 3 = 0$ $\newline$ So, $x = 7$ or $x = 3$
Substitute $x = 7$ : Substitute $x = 7$ into one of the original equations to find the corresponding $y$ value. $\newline$ $y = 38(7) + 13$ $\newline$ $y = 266 + 13$ $\newline$ $y = 279$
Substitute $x = 3$ : Substitute $x = 3$ into one of the original equations to find the corresponding $y$ value. $\newline$ $y = 38(3) + 13$ $\newline$ $y = 114 + 13$ $\newline$ $y = 127$