Question

Solve the system of linear equations 2x + 3y = 7 and x - 2y = -3. Provide step-by-step explanations and identify the correct solution from the options (1, 1), (-1, 2), (2, 1), or (0, -1).

Answer 1

To solve the system of linear equations \(2x + 3y = 7\) and \(x - 2y = -3\), we will use the substitution or elimination method. Here, we'll use substitution for clarity.

### Step 1: Solve one equation for one variable
Let's solve the second equation for \(x\):
\[
x - 2y = -3
\]
Adding \(2y\) to both sides:
\[
x = -3 + 2y
\]

### Step 2: Substitute this expression into the first equation
Now substitute \(x\) in \(2x + 3y = 7\):
\[
2(-3 + 2y) + 3y = 7
\]
Distributing the \(2\):
\[
-6 + 4y + 3y = 7
\]
Combining like terms:
\[
-6 + 7y = 7
\]

### Step 3: Isolate \(y\)
Now, add \(6\) to both sides:
\[
7y = 13
\]
Now divide by \(7\):
\[
y = \frac{13}{7}
\]

### Step 4: Substitute \(y\) back to find \(x\)
Now substitute \(y = \frac{13}{7}\) back into \(x = -3 + 2y\):
\[
x = -3 + 2\left(\frac{13}{7}\right)
\]
Calculate \(2 \times \frac{13}{7}\):
\[
x = -3 + \frac{26}{7}
\]
Now, convert \(-3\) into a fraction:
\[
-3 = -\frac{21}{7}
\]
Then:
\[
x = -\frac{21}{7} + \frac{26}{7} = \frac{5}{7}
\]

### Final Step: Write the solution as an ordered pair
Thus, our solution is:
\[
\left( \frac{5}{7}, \frac{13}{7} \right)
\]

### Work through the options
Now, we check against the provided options:
1. \( (1, 1) \): \(2(1) + 3(1) = 5 \neq 7\) and \(1 - 2(1) = -1 \neq -3\)
2. \( (-1, 2) \): \(2(-1) + 3(2) = 4 \neq 7\) and \(-1 - 2(2) = -5 \neq -3\)
3. \( (2, 1) \): \(2(2) + 3(1) = 7\) and \(2 - 2(1) = 0 \neq -3\)
4. \( (0, -1) \): \(2(0) + 3(-1) = -3 \neq 7\) and \(0 - 2(-1) = 2 \neq -3\)

None of the provided options \((1, 1)\), \((-1, 2)\), \( (2, 1)\), or \( (0, -1)\) satisfy both equations.

### Conclusion
The solution to the system \( (x, y) = \left( \frac{5}{7}, \frac{13}{7} \right) \) does not match any of the provided options. Please review or provide options that may include the correct solution.