Solve the system of equations. y = 2x^2 + 21x - 18 y = 21x + 14 Write the coordinates in exact form. Simplify all fractions and radicals. (______,______) (______,______)

Set Equations Equal: We have the system of equations: y = 2x^2 + 21x - 18 y = 21x + 14 To find the intersection points, we need to set the two equations equal to each other. 2x^2 + 21x - 18 = 21x + 14Simplify by Subtracting: Subtract 21x from both sides to start simplifying the equation: 2x^2 + 21x - 18 - 21x = 21x + 14 - 21x 2x^2 - 18 = 14Isolate Quadratic Term: Add 18 to both sides to isolate the quadratic term: 2x^2 - 18 + 18 = 14 + 18 2x^2 = 32Solve for x^2: Divide both sides by 2 to solve for x^2: 2x^2 / 2 = 32 / 2 x^2 = 16Solve for x: Take the square root of both sides to solve for x: sqrt(x^2) = sqrt(16) x = 4 or x = -4Substitute x = 4: Now we have two values for x, we need to find the corresponding y values for each. First, let's substitute x = 4 into the second equation y = 21x + 14: y = 21(4) + 14 y = 84 + 14 y = 98Substitute x = -4: Next, let's substitute x = -4 into the second equation y = 21x + 14: y = 21(-4) + 14 y = -84 + 14 y = -70Find Coordinates: We now have the coordinates of the intersection points in exact form: First Coordinate: (4, 98) Second Coordinate: (-4, -70)

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Solve the system of equations. $\newline$ $y = 2x^2 + 21x - 18$ $\newline$ $y = 21x + 14$ $\newline$ Write the coordinates in exact form. Simplify all fractions and radicals. $\newline$ (,) $\newline$ (,)

View step-by-step help

Home

Math Problems

Algebra 1

Solve a system of linear and quadratic equations

Full solution

Q. Solve the system of equations.

\newline

y = 2x^2 + 21x - 18

\newline

y = 21x + 14

\newline

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

\newline

(______,______)

\newline

(______,______)

Set Equations Equal: We have the system of equations: $\newline$ $y = 2x^2 + 21x - 18$ $\newline$ $y = 21x + 14$ $\newline$ To find the intersection points, we need to set the two equations equal to each other. $\newline$ $2x^2 + 21x - 18 = 21x + 14$
Simplify by Subtracting: Subtract $21x$ from both sides to start simplifying the equation: $\newline$ $2x^2 + 21x - 18 - 21x = 21x + 14 - 21x$ $\newline$ $2x^2 - 18 = 14$
Isolate Quadratic Term: Add $18$ to both sides to isolate the quadratic term: $\newline$ $2x^2 - 18 + 18 = 14 + 18$ $\newline$ $2x^2 = 32$
Solve for $x^2$ : Divide both sides by $2$ to solve for $x^2$ : $\newline$ $\frac{2x^2}{2} = \frac{32}{2}$ $\newline$ $x^2 = 16$
Solve for x: Take the square root of both sides to solve for x: $\newline$ $\sqrt{x^2} = \sqrt{16}$ $\newline$ $x = 4$ or $x = -4$
Substitute $x = 4$ : Now we have two values for $x$ , we need to find the corresponding $y$ values for each. First, let's substitute $x = 4$ into the second equation $y = 21x + 14$ : $\newline$ $y = 21(4) + 14$ $\newline$ $y = 84 + 14$ $\newline$ $y = 98$
Substitute $x = -4$ : Next, let's substitute $x = -4$ into the second equation $y = 21x + 14$ : $\newline$ $y = 21(-4) + 14$ $\newline$ $y = -84 + 14$ $\newline$ $y = -70$
Find Coordinates: We now have the coordinates of the intersection points in exact form: $\newline$ First Coordinate: $(4, 98)$ $\newline$ Second Coordinate: $(-4, -70)$