Solve the system of equations. y = x^2 + 40x - 20 y = 40x - 4 Write the coordinates in exact form. Simplify all fractions and radicals. (______,______) (______,______)

Set Equations Equal: We have the system of equations: y = x^2 + 40x - 20 y = 40x - 4 To find the intersection points, set the two equations equal to each other. x^2 + 40x - 20 = 40x - 4Simplify and Rearrange: Simplify the equation by subtracting 40x from both sides and adding 4 to both sides. x^2 + 40x - 20 - 40x + 4 = 40x - 4 - 40x + 4 x^2 - 16 = 0Solve Quadratic Equation: Solve the quadratic equation for x. x^2 - 16 = 0 Factor the left side of the equation. (x - 4)(x + 4) = 0Factor and Solve: Set each factor equal to zero and solve for x. (x - 4) = 0 or (x + 4) = 0 x = 4 or x = -4Find y-Values: Find the corresponding y-values for each x-value by substituting back into either of the original equations. We'll use y = 40x - 4. For x = 4: y = 40(4) - 4 y = 160 - 4 y = 156Coordinate Calculation: For x = -4: y = 40(-4) - 4 y = -160 - 4 y = -164Write the coordinates in exact form. First Coordinate: (4, 156) Second Coordinate: (-4, -164)

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Solve the system of equations. $\newline$ $y = x^2 + 40x - 20$ $\newline$ $y = 40x - 4$ $\newline$ Write the coordinates in exact form. Simplify all fractions and radicals. $\newline$ (,) $\newline$ (,)

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Algebra 1

Solve a system of linear and quadratic equations

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Q. Solve the system of equations.

\newline

y = x^2 + 40x - 20

\newline

y = 40x - 4

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

\newline

(______,______)

\newline

(______,______)

Set Equations Equal: We have the system of equations: $\newline$ $y = x^2 + 40x - 20$ $\newline$ $y = 40x - 4$ $\newline$ To find the intersection points, set the two equations equal to each other. $\newline$ $x^2 + 40x - 20 = 40x - 4$
Simplify and Rearrange: Simplify the equation by subtracting $40x$ from both sides and adding $4$ to both sides. $\newline$ $x^2 + 40x - 20 - 40x + 4 = 40x - 4 - 40x + 4$ $\newline$ $x^2 - 16 = 0$
Solve Quadratic Equation: Solve the quadratic equation for $x$ . $x^2 - 16 = 0$ Factor the left side of the equation. $(x - 4)(x + 4) = 0$
Factor and Solve: Set each factor equal to zero and solve for $x$ . $(x - 4) = 0 \text{ or } (x + 4) = 0$ $x = 4 \text{ or } x = -4$
Find y-Values: Find the corresponding y-values for each x-value by substituting back into either of the original equations. We'll use $y = 40x - 4$ . $\newline$ For $x = 4$ : $\newline$ $y = 40(4) - 4$ $\newline$ $y = 160 - 4$ $\newline$ $y = 156$
Coordinate Calculation: For $x = -4$ : $y = 40(-4) - 4$ $y = -160 - 4$ $y = -164$
Coordinate Calculation: For $x = -4$ : $y = 40(-4) - 4$ $y = -160 - 4$ $y = -164$ Write the coordinates in exact form. $\text{First Coordinate}: (4, 156)$ $\text{Second Coordinate}: (-4, -164)$