Solve the system of equations. $\newline$ $y = 3x^2 + 28x - 3$ $\newline$ $y = 28x + 45$ $\newline$ 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.

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y = 3x^2 + 28x - 3

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y = 28x + 45

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

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(______,______)

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(______,______)

Set Equations Equal: We have the system of equations: $\newline$ $y = 3x^2 + 28x - 3$ $\newline$ $y = 28x + 45$ $\newline$ To find the intersection points, we set the two equations equal to each other. $\newline$ $3x^2 + 28x - 3 = 28x + 45$
Subtract and Simplify: Subtract $28x + 45$ from both sides to move all terms to one side and set the equation to zero. $\newline$ $3x^2 + 28x - 3 - 28x - 45 = 0$ $\newline$ $3x^2 - 48 = 0$
Isolate Quadratic Term: Add $48$ to both sides to isolate the quadratic term. $\newline$ $3x^2 = 48$
Solve for x: Divide both sides by $3$ to solve for $x^2$ . $\newline$ $x^2 = \frac{48}{3}$ $\newline$ $x^2 = 16$
Find y Values: Take the square root of both sides to solve for $x$ . $x = \sqrt{16}$ or $x = -\sqrt{16}$ $x = 4$ or $x = -4$
Coordinates of Intersection Points: Now we have two values for $x$ : $x_1 = 4$ and $x_2 = -4$ . We need to find the corresponding $y$ values for each $x$ by substituting them into either of the original equations. We'll use $y = 28x + 45$ . For $x_1 = 4$ : $y = 28(4) + 45$ $y = 112 + 45$ $y = 157$
Coordinates of Intersection Points: Now we have two values for $x$ : $x_1 = 4$ and $x_2 = -4$ . We need to find the corresponding $y$ values for each $x$ by substituting them into either of the original equations. We'll use $y = 28x + 45$ . For $x_1 = 4$ : $y = 28(4) + 45$ $y = 112 + 45$ $y = 157$ For $x_2 = -4$ : $x_1 = 4$ $1$ $x_1 = 4$ $2$ $x_1 = 4$ $3$
Coordinates of Intersection Points: Now we have two values for $x$ : $x_1 = 4$ and $x_2 = -4$ . We need to find the corresponding $y$ values for each $x$ by substituting them into either of the original equations. We'll use $y = 28x + 45$ . For $x_1 = 4$ : $y = 28(4) + 45$ $y = 112 + 45$ $y = 157$ For $x_2 = -4$ : $x_1 = 4$ $1$ $x_1 = 4$ $2$ $x_1 = 4$ $3$ We have found the $y$ values corresponding to each $x$ . Therefore, the coordinates of the intersection points are: First Coordinate: $x_1 = 4$ $6$ Second Coordinate: $x_1 = 4$ $7$