$3 x+y=1$ $\newline$ $y=6 x^{2}-4 x-1$ $\newline$ If $(x, y)$ is a solution to the system of equations shown, which of the following are $x$ -coordinates of the solutions? $\newline$ Choose $1$ answer: $\newline$ (A) $-\frac{1}{2}$ and $\frac{5}{2}$ $\newline$ (B) $-\frac{2}{3}$ and $\frac{1}{2}$ $\newline$ (C) $\frac{2}{3}$ and $-\frac{1}{2}$ $\newline$ (D) $\frac{2}{3}$ and $-1$

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Algebra 2

Domain and range

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3 x+y=1

\newline

y=6 x^{2}-4 x-1

\newline

(x, y)

is a solution to the system of equations shown, which of the following are

x

-coordinates of the solutions?

\newline

Choose

1

answer:

\newline

(A)

-\frac{1}{2}

and

\frac{5}{2}

\newline

(B)

-\frac{2}{3}

and

\frac{1}{2}

\newline

(C)

\frac{2}{3}

and

-\frac{1}{2}

\newline

(D)

\frac{2}{3}

and

-1

Substitute $y$ into first equation: We have a system of two equations: $\newline$ $1$ ) $3x + y = 1$ $\newline$ $2$ ) $y = 6x^2 - 4x - 1$ $\newline$ To find the $x$ -coordinates of the solutions, we can substitute the expression for $y$ from the second equation into the first equation.
Combine and set to zero: Substitute $y = 6x^2 - 4x - 1$ into the first equation: $\newline$ $3x + (6x^2 - 4x - 1) = 1$
Factor the quadratic equation: Combine like terms and move all terms to one side to set the equation to zero: $\newline$ $3x + 6x^2 - 4x - 1 - 1 = 0$ $\newline$ $6x^2 - x - 2 = 0$
Solve for $x$ : Now we have a quadratic equation in the form $ax^2 + bx + c = 0$ . We can solve for $x$ by factoring, completing the square, or using the quadratic formula. Let's try to factor the equation first.
Solve for $x$ : Factor the quadratic equation: $\newline$ $6x^2 - x - 2 = (3x + 2)(2x - 1)$
Final x-coordinates: Set each factor equal to zero and solve for x: $\newline$ $3x + 2 = 0$ or $2x - 1 = 0$
Final x-coordinates: Set each factor equal to zero and solve for $x$ : $3x + 2 = 0$ or $2x - 1 = 0$ Solve the first equation for $x$ : $3x + 2 = 0$ $3x = -2$ $x = -\frac{2}{3}$
Final x-coordinates: Set each factor equal to zero and solve for $x$ : $3x + 2 = 0$ or $2x - 1 = 0$ Solve the first equation for $x$ : $3x + 2 = 0$ $3x = -2$ $x = -\frac{2}{3}$ Solve the second equation for $x$ : $2x - 1 = 0$ $2x = 1$ $3x + 2 = 0$ $0$
Final x-coordinates: Set each factor equal to zero and solve for $x$ : $3x + 2 = 0$ or $2x - 1 = 0$ Solve the first equation for $x$ : $3x + 2 = 0$ $3x = -2$ $x = -\frac{2}{3}$ Solve the second equation for $x$ : $2x - 1 = 0$ $2x = 1$ $3x + 2 = 0$ $0$ We have found two x-coordinates for the solutions to the system of equations: $x = -\frac{2}{3}$ and $3x + 2 = 0$ $0$ . These correspond to choice (B).