Solve the following system of equations: 10 (x + y) = 0 and 2 10 x = 10 (y + 1)

Using the definition of logarithm, if $\log_b a = c$, then $b^c = a$.

From this, we can express $y$ in terms of $x$:

Using the logarithm property $c \log_b a = \log_b (a^c)$:

Since the logarithms have the same base, we can equate their arguments:

Substitute $y = 1 - x$ into $x^2 = y + 1$:

Rearrange the equation into a standard quadratic form:

Factor the quadratic equation:

This gives two possible values for $x$:

For $\log_{10} x$ to be defined, $x$ must be greater than 0.
Therefore, $x = -2$ is not a valid solution because $\log_{10}(-2)$ is undefined.
The only valid solution for $x$ is $x = 1$.

Substitute $x = 1$ into the equation $y = 1 - x$:

Check the solution $(x, y) = (1, 0)$ in the original equations:
For equation 1: $\log_{10} (1 + 0) = \log_{10} (1) = 0$. (Correct)
For equation 2: $2 \log_{10} (1) = 2 \times 0 = 0$. And $\log_{10} (0 + 1) = \log_{10} (1) = 0$. So $0 = 0$. (Correct)
All arguments of the logarithms are positive ($x=1$, $x+y=1$, $y+1=1$), so the solution is valid.

The solution to the simultaneous equations is: