Which of the following represents all solutions (x,y) to the system of equations shown? y+x=6 y=x^(2)-2x-6 Choose 1 answer: (A) (-4,10) and (3,3) (B) (4,2) and (-3,9) (c) (4,-3) (D) (-4,3)

Write equations: Write down the system of equations to be solved. The system of equations is: y + x = 6 y = x^2 - 2x - 6Set equations equal: Since both equations equal y, set them equal to each other to find the x-values that satisfy both equations. x^2 - 2x - 6 = x + 6Rearrange and solve: Rearrange the equation to set it to zero and solve for x. x^2 - 2x - 6 - x - 6 = 0 x^2 - 3x - 12 = 0Factor quadratic equation: Factor the quadratic equation to find the values of x. (x - 4)(x + 3) = 0Solve for x: Solve for x by setting each factor equal to zero. x - 4 = 0 or x + 3 = 0 x = 4 or x = -3Find y-values: Plug the x-values back into the original equations to find the corresponding y-values. For x = 4: y + 4 = 6 y = 6 - 4 y = 2 For x = -3: y - 3 = 6 y = 6 + 3 y = 9Write solution pairs: Write down the solution pairs (x,y) for the system of equations. The solution pairs are (4,2) and (-3,9).

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Which of the following represents all solutions
(x,y) to the system of equations shown?

y+x=6

y=x^(2)-2x-6
Choose 1 answer:
(A)
(-4,10) and
(3,3)
(B)
(4,2) and
(-3,9)
(c)
(4,-3)
(D)
(-4,3)

Which of the following represents all solutions $(x, y)$ to the system of equations shown? $\newline$ $y+x=6$ $\newline$ $y=x^{2}-2 x-6$ $\newline$ Choose $1$ answer: $\newline$ (A) $(-4,10)$ and $(3,3)$ $\newline$ (B) $(4,2)$ and $(-3,9)$ $\newline$ (C) $(4,-3)$ $\newline$ (D) $(-4,3)$

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Math Problems

Algebra 1

Domain and range of relations

Full solution

Q. Which of the following represents all solutions

(x, y)

to the system of equations shown?

\newline

y+x=6

\newline

y=x^{2}-2 x-6

\newline

Choose

1

answer:

\newline

(A)

(-4,10)

and

(3,3)

\newline

(B)

(4,2)

and

(-3,9)

\newline

(C)

(4,-3)

\newline

(D)

(-4,3)

Write equations: Write down the system of equations to be solved. $\newline$ The system of equations is: $\newline$ $y + x = 6$ $\newline$ $y = x^2 - 2x - 6$
Set equations equal: Since both equations equal $y$ , set them equal to each other to find the $x$ -values that satisfy both equations. $x^2 - 2x - 6 = x + 6$
Rearrange and solve: Rearrange the equation to set it to zero and solve for $x$ . $x^2 - 2x - 6 - x - 6 = 0$ $x^2 - 3x - 12 = 0$
Factor quadratic equation: Factor the quadratic equation to find the values of $x$ . $(x - 4)(x + 3) = 0$
Solve for x: Solve for x by setting each factor equal to zero. $\newline$ $x - 4 = 0$ or $x + 3 = 0$ $\newline$ $x = 4$ or $x = -3$
Find y-values: Plug the x-values back into the original equations to find the corresponding y-values. $\newline$ For $x = 4$ : $\newline$ $y + 4 = 6$ $\newline$ $y = 6 - 4$ $\newline$ $y = 2$ $\newline$ For $x = -3$ : $\newline$ $y - 3 = 6$ $\newline$ $y = 6 + 3$ $\newline$ $y = 9$
Write solution pairs: Write down the solution pairs $(x,y)$ for the system of equations. $\newline$ The solution pairs are $(4,2)$ and $(-3,9)$ .