Solve the following system of equations: 1) 2(3 (x−1)) = 3 x 2) y = sqrt(x+1)

Using the logarithm property $a \log b = \log b^a$, we can rewrite the first equation:

Since the bases of the logarithms are the same, we can equate their arguments:

Expand the left side of the equation:

Rearrange the terms to form a quadratic equation:

Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to solve for $x$. Here, $a=1$, $b=-3$, $c=1$.

We have two potential solutions for $x$: $x_1 = \frac{3 + \sqrt{5}}{2}$ and $x_2 = \frac{3 - \sqrt{5}}{2}$.

For $\log_3 (x-1)$ to be defined, $x-1 > 0 \implies x > 1$.
For $\log_3 x$ to be defined, $x > 0$.
Both conditions require $x > 1$.
Let's evaluate the two potential solutions:
For $x_1 = \frac{3 + \sqrt{5}}{2}$: Since $\sqrt{5} \approx 2.236$, $x_1 \approx \frac{3 + 2.236}{2} = \frac{5.236}{2} = 2.618$. This value is greater than 1, so it is a valid solution.
For $x_2 = \frac{3 - \sqrt{5}}{2}$: Since $\sqrt{5} \approx 2.236$, $x_2 \approx \frac{3 - 2.236}{2} = \frac{0.764}{2} = 0.382$. This value is not greater than 1, so it is not a valid solution.
Therefore, the only valid value for $x$ is $x = \frac{3 + \sqrt{5}}{2}$.

Substitute $x = \frac{3 + \sqrt{5}}{2}$:

Combine the terms under the square root:

The simultaneous solution is: