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

Equations Given: We have a system of two equations: 1. 3x + y = 1 2. y = 6x^2 - 4x - 1 To solve the system, we can substitute the expression for y from the second equation into the first equation. This will give us an equation with only one variable, x, which we can then solve.Substitute y into first equation: Substitute y = 6x^2 - 4x - 1 into the first equation: 3x + (6x^2 - 4x - 1) = 1Combine and simplify: Combine like terms and simplify the equation: 3x + 6x^2 - 4x - 1 = 1 6x^2 - x - 1 = 1Quadratic equation: Subtract 1 from both sides to set the equation to zero: 6x^2 - x - 2 = 0Factor the quadratic: Now we have a quadratic equation. We can solve for x by factoring, completing the square, or using the quadratic formula. The equation looks like it might be factorable, so let's try factoring first.Set factors to zero: We look for two numbers that multiply to (6)(-2) = -12 and add up to -1. These numbers are -4 and 3. So we can write the equation as: (2x - 1)(3x + 2) = 0Solve for x: Set each factor equal to zero and solve for x: 2x - 1 = 0 or 3x + 2 = 0Final x-coordinates: Solve the first equation for x: 2x = 1 x = 1/2Solve the second equation for x: 3x = -2 x = -2/3We have found two x-coordinates where the system of equations has solutions: x = 1/2 and x = -2/3. These correspond to answer choice (B).

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

$3x+y=1$ $\newline$ $y=$ $6x^{2}-4x-1$ $\newline$ If $(x,y)$ is a solution to the system of equations shown, which of the following are $\newline$ 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 1

Write two-variable inequalities: word problems

Full solution

3x+y=1

\newline

y=

6x^{2}-4x-1

\newline

(x,y)

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

\newline

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

}

Equations Given: We have a system of two equations: $\newline$ $1$ . $3x + y = 1$ $\newline$ $2$ . $y = 6x^2 - 4x - 1$ $\newline$ To solve the system, we can substitute the expression for $y$ from the second equation into the first equation. This will give us an equation with only one variable, $x$ , which we can then solve.
Substitute $y$ into first equation: Substitute $y = 6x^2 - 4x - 1$ into the first equation: $\newline$ $3x + (6x^2 - 4x - 1) = 1$
Combine and simplify: Combine like terms and simplify the equation: $\newline$ $3x + 6x^2 - 4x - 1 = 1$ $\newline$ $6x^2 - x - 1 = 1$
Quadratic equation: Subtract $1$ from both sides to set the equation to zero: $\newline$ $6x^2 - x - 2 = 0$
Factor the quadratic: Now we have a quadratic equation. We can solve for $x$ by factoring, completing the square, or using the quadratic formula. The equation looks like it might be factorable, so let's try factoring first.
Set factors to zero: We look for two numbers that multiply to $(6)(-2) = -12$ and add up to $-1$ . These numbers are $-4$ and $3$ . So we can write the equation as: $(2x - 1)(3x + 2) = 0$
Solve for x: Set each factor equal to zero and solve for x: $\newline$ $2x - 1 = 0$ or $3x + 2 = 0$
Final x-coordinates: Solve the first equation for x: $\newline$ $2x = 1$ $\newline$ $x = \frac{1}{2}$
Final x-coordinates: Solve the first equation for x: $\newline$ $2x = 1$ $\newline$ $x = \frac{1}{2}$ Solve the second equation for x: $\newline$ $3x = -2$ $\newline$ $x = -\frac{2}{3}$
Final x-coordinates: Solve the first equation for x: $\newline$ $2x = 1$ $\newline$ $x = \frac{1}{2}$ Solve the second equation for x: $\newline$ $3x = -2$ $\newline$ $x = -\frac{2}{3}$ We have found two x-coordinates where the system of equations has solutions: $x = \frac{1}{2}$ and $x = -\frac{2}{3}$ . These correspond to answer choice (B).