Solve the system of equations. y = 8x + 27 y = 3x^2 + 8x - 21 Write the coordinates in exact form. Simplify all fractions and radicals. (______,______) (______,______)

Set Equations Equal: We have the following system of equations: y = 8x + 27 y = 3x^2 + 8x - 21 To find the solution, we will set the two equations equal to each other since they both equal y. 8x + 27 = 3x^2 + 8x - 21Subtract to Eliminate x Term: Subtract 8x from both sides of the equation to eliminate the x term on one side: 8x + 27 - 8x = 3x^2 + 8x - 21 - 8x This simplifies to: 27 = 3x^2 - 21Add to Isolate Quadratic Term: Add 21 to both sides of the equation to isolate the quadratic term: 27 + 21 = 3x^2 - 21 + 21 This simplifies to: 48 = 3x^2Divide to Solve for x^2: Divide both sides by 3 to solve for x^2: 48 / 3 = 3x^2 / 3 This simplifies to: 16 = x^2Take Square Root for x: Take the square root of both sides to solve for x: sqrt(16) = sqrt(x^2) This gives us two solutions for x: x = 4 and x = -4Substitute x = 4: Substitute x = 4 into the original equation y = 8x + 27 to find the corresponding y-value: y = 8(4) + 27 y = 32 + 27 y = 59 So one set of coordinates is (4, 59).Substitute x = -4: Substitute x = -4 into the original equation y = 8x + 27 to find the corresponding y-value: y = 8(-4) + 27 y = -32 + 27 y = -5 So the second set of coordinates is (-4, -5).

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Solve the system of equations. $\newline$ $y = 8x + 27$ $\newline$ $y = 3x^2 + 8x - 21$ $\newline$ Write the coordinates in exact form. Simplify all fractions and radicals. $\newline$ (,) $\newline$ (,)

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Math Problems

Algebra 1

Solve a system of linear and quadratic equations

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Q. Solve the system of equations.

\newline

y = 8x + 27

\newline

y = 3x^2 + 8x - 21

\newline

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

\newline

(______,______)

\newline

(______,______)

Set Equations Equal: We have the following system of equations: $\newline$ $y = 8x + 27$ $\newline$ $y = 3x^2 + 8x - 21$ $\newline$ To find the solution, we will set the two equations equal to each other since they both equal $y$ . $\newline$ $8x + 27 = 3x^2 + 8x - 21$
Subtract to Eliminate x Term: Subtract $8x$ from both sides of the equation to eliminate the x term on one side: $\newline$ $8x + 27 - 8x = 3x^2 + 8x - 21 - 8x$ $\newline$ This simplifies to: $\newline$ $27 = 3x^2 - 21$
Add to Isolate Quadratic Term: Add $21$ to both sides of the equation to isolate the quadratic term: $\newline$ $27 + 21 = 3x^2 - 21 + 21$ $\newline$ This simplifies to: $\newline$ $48 = 3x^2$
Divide to Solve for $x^2$ : Divide both sides by $3$ to solve for $x^2$ : $\frac{48}{3} = \frac{3x^2}{3}$ This simplifies to: $16 = x^2$
Take Square Root for x: Take the square root of both sides to solve for x: $\newline$ $\sqrt{16} = \sqrt{x^2}$ $\newline$ This gives us two solutions for x: $\newline$ $x = 4$ and $x = -4$
Substitute $x = 4$ : Substitute $x = 4$ into the original equation $y = 8x + 27$ to find the corresponding y-value: $\newline$ $y = 8(4) + 27$ $\newline$ $y = 32 + 27$ $\newline$ $y = 59$ $\newline$ So one set of coordinates is $(4, 59)$ .
Substitute $x = -4$ : Substitute $x = -4$ into the original equation $y = 8x + 27$ to find the corresponding y-value: $\newline$ $y = 8(-4) + 27$ $\newline$ $y = -32 + 27$ $\newline$ $y = -5$ $\newline$ So the second set of coordinates is $(-4, -5)$ .