If \( \frac{x}{3}=\frac{y}{4}=\frac{z}{5} \), prove that \( : \frac{2 y-z}{3 x-2 y+z}=\frac{1}{2} \)

To prove that \[ \frac{2y - z}{3x - 2y + z} = \frac{1}{2} \] given that \[ \frac{x}{3} = \frac{y}{4} = \frac{z}{5}, \] we can follow these steps: ### Step 1: Introduce a Common Parameter Since \(\frac{x}{3} = \frac{y}{4} = \frac{z}{5}\), let's introduce a common parameter \(k\) such that: \[ \frac{x}{3} = \frac{y}{4} = \frac{z}{5} = k \] This implies: \[ x = 3k,\quad y = 4k,\quad z = 5k \] ### Step 2: Substitute \(x\), \(y\), and \(z\) in the Expression Now, substitute \(x = 3k\), \(y = 4k\), and \(z = 5k\) into the expression \(\frac{2y - z}{3x - 2y + z}\): #### Numerator: \[ 2y - z = 2(4k) - 5k = 8k - 5k = 3k \] #### Denominator: \[ 3x - 2y + z = 3(3k) - 2(4k) + 5k = 9k - 8k + 5k = 6k \] ### Step 3: Simplify the Fraction Now, substitute the simplified numerator and denominator back into the fraction: \[ \frac{2y - z}{3x - 2y + z} = \frac{3k}{6k} = \frac{3}{6} = \frac{1}{2} \] ### Conclusion Thus, we have proven that: \[ \frac{2y - z}{3x - 2y + z} = \frac{1}{2} \]