Determine real values of x and y for which each statement is true
`(3-4i) (x+yi)=1 + 0i`

Answer

To solve the equation \((3 - 4i)(x + yi) = 1 + 0i\), we will follow these steps:

### Step 1: Expand the left-hand side
We start by expanding the left-hand side of the equation:

\[ (3 - 4i)(x + yi) = 3x + 3yi - 4xi - 4yi^2 \]

### Step 2: Substitute \(i^2\) with \(-1\)
Since \(i^2 = -1\), we can rewrite the expression:

\[ = 3x + 3yi - 4xi + 4y \]
\[ = (3x + 4y) + (3y - 4x)i \]

### Step 3: Set the real and imaginary parts equal
Now, we compare the real and imaginary parts of both sides of the equation:

\[ (3x + 4y) + (3y - 4x)i = 1 + 0i \]

From this, we can derive two equations:

1. \(3x + 4y = 1\) (Real part)
2. \(3y - 4x = 0\) (Imaginary part)

### Step 4: Solve the system of equations
We can solve the second equation for \(y\):

\[ 3y = 4x \implies y = \frac{4}{3}x \]

Now, substitute \(y\) into the first equation:

\[ 3x + 4\left(\frac{4}{3}x\right) = 1 \]
\[ 3x + \frac{16}{3}x = 1 \]

### Step 5: Combine like terms
To combine the terms, we convert \(3x\) into a fraction:

\[ \frac{9}{3}x + \frac{16}{3}x = 1 \]
\[ \frac{25}{3}x = 1 \]

### Step 6: Solve for \(x\)
Now, solve for \(x\):

\[ x = \frac{3}{25} \]

### Step 7: Substitute \(x\) back to find \(y\)
Now substitute \(x\) back into the equation for \(y\):

\[ y = \frac{4}{3} \left(\frac{3}{25}\right) = \frac{4}{25} \]

### Final Answer
Thus, the real values of \(x\) and \(y\) are:

\[ x = \frac{3}{25}, \quad y = \frac{4}{25} \]

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