$\begin{array}{l} 2 x-1=y \\ 3 x-1=y \end{array}$ $\newline$ Consider the given system of equations. Which of the following statements about this system is true? $\newline$ Choose $1$ answer: $\newline$ (A) There is only one $(x, y)$ solution and $y$ is positive. $\newline$ (B) There is only one $(x, y)$ solution and $y$ is negative. $\newline$ (C) There are infinitely many $(x, y)$ solutions. $\newline$ (D) There are no $(x, y)$ solutions.

Full solution

\begin{array}{l} 2 x-1=y \\ 3 x-1=y \end{array}

\newline

Consider the given system of equations. Which of the following statements about this system is true?

\newline

Choose

1

answer:

\newline

(A) There is only one

(x, y)

solution and

y

is positive.

\newline

(B) There is only one

(x, y)

solution and

y

is negative.

\newline

(x, y)

solutions.

\newline

(D) There are no

(x, y)

solutions.

Analyze System of Equations: Analyze the given system of equations. $\newline$ We have the system: $\newline$ $2x - 1 = y$ $\newline$ $3x - 1 = y$ $\newline$ We need to determine the nature of the solutions to this system.
Compare Equations: Compare the two equations. $\newline$ Both equations are equal to $y$ , so we can set them equal to each other: $\newline$ $2x - 1 = 3x - 1$
Solve for x: Solve for x. $\newline$ Subtract $2x$ from both sides to get: $\newline$ $-1 = x - 1$ $\newline$ Now, add $1$ to both sides to find the value of $x$ : $\newline$ $0 = x$
Substitute for y: Substitute $x$ back into one of the original equations to find $y$ . Using the first equation: $2x - 1 = y$ $2(0) - 1 = y$ $y = -1$
Conclude Solution: Conclude the nature of the solution. $\newline$ Since we found a single solution for $x$ and $y$ , there is only one $(x,y)$ solution. The value of $y$ is negative.