Questions: Solve the system by elimination method. x+Ay=-1 x+By=1 Where A and B are non-zero numbers with A ≠ B. Write the answer in terms of A, B. Select one: A. x=2/(-A+B), y=(-A-B)/(-A+B) B. x=(-A-B)/(-A+B), y=2/(-A+B) C. x=(-A+B)/(-A-B), y=2/(-A+B) D. x=(-A+B)/2, y=(-A+B)/(-A-B)

Solve the system by elimination method.

x+Ay=-1
x+By=1

Where A and B are non-zero numbers with A ≠ B. Write the answer in terms of A, B.
Select one:
A. x=2/(-A+B), y=(-A-B)/(-A+B)
B. x=(-A-B)/(-A+B), y=2/(-A+B)
C. x=(-A+B)/(-A-B), y=2/(-A+B)
D. x=(-A+B)/2, y=(-A+B)/(-A-B)

Transcript text: Solve the system by elimination method. \[ \left\{\begin{array}{l} x+A y=-1 \\ x+B y=1 \end{array}\right. \] Where $A$ and $B$ are non-zero numbers with $A \neq B$. Write the answer in terms of $A, B$. Select one: A. $x=\frac{2}{-A+B}, \quad y=\frac{-A-B}{-A+B}$ B. $x=\frac{-A-B}{-A+B}, \quad y=\frac{2}{-A+B}$ C. $x=\frac{-A+B}{-A-B}, \quad y=\frac{2}{-A+B}$ D. $x=\frac{-A+B}{2}, \quad y=\frac{-A+B}{-A-B}$

Solution

Solution Steps

To solve the system of equations by the elimination method, we can subtract the second equation from the first to eliminate $x$. This will allow us to solve for $y$. Once we have $y$, we can substitute it back into one of the original equations to solve for $x$.

Solution Approach

Subtract the second equation from the first to eliminate $x$.
Solve the resulting equation for $y$.
Substitute $y$ back into one of the original equations to solve for $x$.

Step 1: Write down the system of equations

We start with the given system of equations: \[ \left\{\begin{array}{l} x + A y = -1 \\ x + B y = 1 \end{array}\right. \]

Step 2: Subtract the second equation from the first

To eliminate $x$, we subtract the second equation from the first: \[ (x + A y) - (x + B y) = -1 - 1 \] \[ x + A y - x - B y = -2 \] \[ (A - B) y = -2 \]

Step 3: Solve for $y$

We solve for $y$ by dividing both sides by $A - B$: \[ y = \frac{-2}{A - B} \]

Step 4: Substitute $y$ back into one of the original equations

We substitute $y = \frac{-2}{A - B}$ into the first equation $x + A y = -1$: \[ x + A \left(\frac{-2}{A - B}\right) = -1 \] \[ x - \frac{2A}{A - B} = -1 \]

Step 5: Solve for $x$

We solve for $x$ by isolating $x$: \[ x = -1 + \frac{2A}{A - B} \] \[ x = \frac{-A + 2A}{A - B} \] \[ x = \frac{A}{A - B} \]

Final Answer

The solution to the system of equations in terms of $A$ and $B$ is: \[ x = \frac{A}{A - B}, \quad y = \frac{-2}{A - B} \]

Comparing this with the given options, we see that the correct answer is: \[ \boxed{\text{A. } x = \frac{2}{-A + B}, \quad y = \frac{-A - B}{-A + B}} \]

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