To solve the given system of simultaneous equations:
[tex]\[
\begin{cases}
e^{2y} - x + 2 = 0 \\
\ln(x + 3) - 2y - 1 = 0
\end{cases}
\][/tex]
we need to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations.
1. Rewrite the first equation:
[tex]\[
e^{2y} - x + 2 = 0 \implies e^{2y} = x - 2
\][/tex]
2. Rewrite the second equation:
[tex]\[
\ln(x + 3) - 2y - 1 = 0 \implies \ln(x + 3) = 2y + 1
\][/tex]
3. Express [tex]\( y \)[/tex] from the logarithm equation:
[tex]\[
\ln(x + 3) = 2y + 1 \implies 2y = \ln(x + 3) - 1 \implies y = \frac{\ln(x + 3) - 1}{2}
\][/tex]
4. Replace [tex]\( y \)[/tex] in the exponential equation:
Substitute [tex]\( y = \frac{\ln(x + 3) - 1}{2} \)[/tex] into [tex]\( e^{2y} = x - 2 \)[/tex]:
[tex]\[
e^{2\left(\frac{\ln(x + 3) - 1}{2}\right)} = x - 2
\][/tex]
5. Simplify the exponential expression:
[tex]\[
e^{\ln(x + 3) - 1} = x - 2
\][/tex]
Using properties of exponents, we can write:
[tex]\[
e^{\ln(x + 3)} \cdot e^{-1} = x - 2 \implies \frac{x + 3}{e} = x - 2
\][/tex]
6. Clear the fraction:
[tex]\[
x + 3 = e (x - 2)
\][/tex]
7. Distribute [tex]\( e \)[/tex]:
[tex]\[
x + 3 = ex - 2e
\][/tex]
8. Isolate [tex]\( x \)[/tex]:
[tex]\[
x - ex = -2e - 3
\][/tex]
[tex]\[
x(1 - e) = -2e - 3
\][/tex]
Therefore,
[tex]\[
x = \frac{-2e - 3}{1 - e}
\][/tex]
9. Calculate the numerical value of [tex]\( x \)[/tex]:
Using the known value:
[tex]\[
x = 4.91 \quad (\text{to 2 decimal places})
\][/tex]
10. Substitute [tex]\( x \)[/tex] back to find [tex]\( y \)[/tex]:
[tex]\[
y = \frac{\ln(4.91 + 3) - 1}{2}
\][/tex]
11. Evaluate [tex]\( y \)[/tex]:
Also using the known value:
[tex]\[
y = 0.53 \quad (\text{to 2 decimal places})
\][/tex]
Thus, the solutions to the simultaneous equations are:
[tex]\[
x = 4.91, \quad y = 0.53
\][/tex]
These values satisfy both equations within the given constraints.
