Solve the system of equations. y = x^2 - 39x - 43 y = -39x + 101 Write the coordinates in exact form. Simplify all fractions and radicals. (______,______) (______,______)

Set Equations Equal: We have the system of equations: y = x^2 - 39x - 43 y = -39x + 101 Set the two equations equal to each other to find the x-values where they intersect. x^2 - 39x - 43 = -39x + 101Simplify and Standardize: Simplify the equation by adding 39x to both sides and subtracting 101 from both sides to get the quadratic equation in standard form. x^2 - 39x - 43 + 39x - 101 = -39x + 101 + 39x - 101 x^2 - 144 = 0Solve Quadratic Equation: Solve the quadratic equation for x. x^2 - 144 = 0 Add 144 to both sides. x^2 = 144 Take the square root of both sides. x = ±12Find y-values: Find the corresponding y-values for each x-value by substituting x back into one of the original equations. We can use the second equation y = -39x + 101 for simplicity. For x = 12: y = -39(12) + 101 y = -468 + 101 y = -367Find Second y-value: Find the y-value for the second x-value. For x = -12: y = -39(-12) + 101 y = 468 + 101 y = 569Write Coordinates: Write the coordinates in exact form. The first coordinate is (12, -367). The second coordinate is (-12, 569).

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Solve the system of equations. $\newline$ $y = x^2 - 39x - 43$ $\newline$ $y = -39x + 101$ $\newline$ Write the coordinates in exact form. Simplify all fractions and radicals. $\newline$ (,) $\newline$ (,)

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Algebra 2

Solve a system of linear and quadratic equations: parabolas

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Q. Solve the system of equations.

\newline

y = x^2 - 39x - 43

\newline

y = -39x + 101

\newline

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

\newline

(______,______)

\newline

(______,______)

Set Equations Equal: We have the system of equations: $\newline$ $y = x^2 - 39x - 43$ $\newline$ $y = -39x + 101$ $\newline$ Set the two equations equal to each other to find the $x$ -values where they intersect. $\newline$ $x^2 - 39x - 43 = -39x + 101$
Simplify and Standardize: Simplify the equation by adding $39x$ to both sides and subtracting $101$ from both sides to get the quadratic equation in standard form. $\newline$ $x^2 - 39x - 43 + 39x - 101 = -39x + 101 + 39x - 101$ $\newline$ $x^2 - 144 = 0$
Solve Quadratic Equation: Solve the quadratic equation for $x$ . $x^2 - 144 = 0$ Add $144$ to both sides. $x^2 = 144$ Take the square root of both sides. $x = \pm12$
Find y-values: Find the corresponding $y$ -values for each $x$ -value by substituting $x$ back into one of the original equations. We can use the second equation $y = -39x + 101$ for simplicity. $\newline$ For $x = 12$ : $\newline$ $y = -39(12) + 101$ $\newline$ $y = -468 + 101$ $\newline$ $y = -367$
Find Second y-value: Find the y-value for the second x-value. $\newline$ For $x = -12$ : $\newline$ $y = -39(-12) + 101$ $\newline$ $y = 468 + 101$ $\newline$ $y = 569$
Write Coordinates: Write the coordinates in exact form. $\newline$ The first coordinate is $(12, -367)$ . $\newline$ The second coordinate is $(-12, 569)$ .