Solve the following equation for x and y
`log_(100)|x+y| = 1/2, log_(10)y - log_(10)|x| = log_(100) 4,`

Options

A. `(8/3, 16/3) (-8 , -16)`
B. `(10/3, 20/3), (-10, 20)`
C. `(-10/3, -20/3), (70,20)`
D. none of these

Answer

To solve the given equations for \( x \) and \( y \), we have:

1. \( \log_{100} |x+y| = \frac{1}{2} \)
2. \( \log_{10} y - \log_{10} |x| = \log_{100} 4 \)

### Step 1: Solve the first equation

Starting with the first equation:

\[ \log_{100} |x+y| = \frac{1}{2} \]

We can rewrite this in exponential form:

\[ |x+y| = 100^{\frac{1}{2}} = 10 \]

This gives us two cases to consider:

1. \( x + y = 10 \)
2. \( x + y = -10 \)

### Step 2: Solve the second equation

Now, let's solve the second equation:

\[ \log_{10} y - \log_{10} |x| = \log_{100} 4 \]

We can rewrite \( \log_{100} 4 \) in terms of base 10:

\[ \log_{100} 4 = \frac{\log_{10} 4}{\log_{10} 100} = \frac{\log_{10} 4}{2} \]

Thus, we can rewrite the second equation as:

\[ \log_{10} y - \log_{10} |x| = \frac{\log_{10} 4}{2} \]

Using the properties of logarithms, we can combine the left side:

\[ \log_{10} \left( \frac{y}{|x|} \right) = \frac{\log_{10} 4}{2} \]

Rewriting this in exponential form gives:

\[ \frac{y}{|x|} = 10^{\frac{\log_{10} 4}{2}} = 4^{\frac{1}{2}} = 2 \]

Thus, we have:

\[ y = 2|x| \]

### Step 3: Substitute and solve for both cases

Now we substitute \( y = 2|x| \) into both cases derived from the first equation.

#### Case 1: \( x + y = 10 \)

Substituting \( y \):

\[ x + 2|x| = 10 \]

**Subcase 1.1: \( x \geq 0 \)**

Here, \( |x| = x \):

\[ x + 2x = 10 \implies 3x = 10 \implies x = \frac{10}{3} \]

Then, substituting back to find \( y \):

\[ y = 2|x| = 2 \cdot \frac{10}{3} = \frac{20}{3} \]

**Subcase 1.2: \( x < 0 \)**

Here, \( |x| = -x \):

\[ x + 2(-x) = 10 \implies x - 2x = 10 \implies -x = 10 \implies x = -10 \]

Then, substituting back to find \( y \):

\[ y = 2|x| = 2 \cdot 10 = 20 \]

#### Case 2: \( x + y = -10 \)

Substituting \( y \):

\[ x + 2|x| = -10 \]

**Subcase 2.1: \( x \geq 0 \)**

Here, \( |x| = x \):

\[ x + 2x = -10 \implies 3x = -10 \implies x = -\frac{10}{3} \text{ (not valid since } x \geq 0\text{)} \]

**Subcase 2.2: \( x < 0 \)**

Here, \( |x| = -x \):

\[ x + 2(-x) = -10 \implies x - 2x = -10 \implies -x = -10 \implies x = 10 \text{ (not valid since } x < 0\text{)} \]

### Final Solutions

From the valid cases, we have:

1. \( x = \frac{10}{3}, y = \frac{20}{3} \)
2. \( x = -10, y = 20 \)

### Summary of Solutions

The solutions for \( (x, y) \) are:

1. \( \left( \frac{10}{3}, \frac{20}{3} \right) \)
2. \( (-10, 20) \)