Solution
We are given the system of equations:
$$ e^{2y} - x + 2 = 0 \tag{1} $$
$$ \ln(x + 3) - 2y - 1 = 0 \tag{2} $$
Step 1: Solve equation (1) for $x$
From equation (1):
$$ e^{2y} - x + 2 = 0 $$
Rearranging for $x$:
$$ x = e^{2y} + 2 \tag{3} $$
Step 2: Substitute equation (3) into equation (2)
Substitute the expression for $x$ from equation (3) into equation (2):
$$ \ln(e^{2y} + 5) - 2y - 1 = 0 $$
Simplifying:
$$ \ln(e^{2y} + 5) = 2y + 1 $$
$$ e^{2y} + 5 = e^{2y + 1} = e^{2y}e $$
$$ e^{2y}(e - 1) = 5 \tag{4} $$
Step 3: Solve equation (4) numerically
Numerically solving equation (4) gives:
$$ 2y = \ln\left(\frac{5}{e - 1}\right) \rightarrow y \approx 0.53 $$
Step 4: Solve for $x$
Substitute the value of $y$ into equation (3):
$$ x = e^{2(0.53)} + 2 \approx 4.91 $$
Thus, the solutions are:
$$ x \approx 4.91, \quad y \approx 0.53 $$
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