\( \frac { \frac { 2 } { x } - 4 } { \frac { 2 } { x } + 3 } = \square , x \neq 0 \)

To solve the equation \( \frac { \frac { 2 } { x } - 4 } { \frac { 2 } { x } + 3 } = \square, x \neq 0 \), we can multiply both sides by the denominator of the left side to eliminate the fraction. Let's set \( y = \frac{2}{x} \). Then we have: \[ \frac{y - 4}{y + 3} = \square \] Now, multiply both sides by \( y + 3 \): \[ y - 4 = \square(y + 3) \] Expanding the right side, we get: \[ y - 4 = \square y + 3\square \] Rearranging gives us: \[ y - \square y = 3\square + 4 \] Factoring out \( y \) leads us to: \[ y(1 - \square) = 3\square + 4 \] Finally, we can express \( y \): \[ y = \frac{3\square + 4}{1 - \square} \] Now, substituting back \( y = \frac{2}{x} \): \[ \frac{2}{x} = \frac{3\square + 4}{1 - \square} \] And solving for \( x \) involves cross-multiplying and rearranging: \[ 2(1 - \square) = x(3\square + 4) \] Thus, \( x \) can be expressed as: \[ x = \frac{2(1 - \square)}{3\square + 4} \] Remember, \( \square \) is a placeholder for whatever value we want to substitute into the equation, and \( x \neq 0 \). In summary, \( x \) can take on various values depending on \( \square \), as long as \( \square \) does not make \( 3\square + 4 = 0 \), which would lead to an invalid solution since it would make the denominator zero. Keep in mind: no division by zero allowed!