\( \frac { a } { 3 } = \frac { b } { 4 } \quad \frac { a + 3 b } { 4 a - b } = \)
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To solve the equations, we first express \( a \) in terms of \( b \) using the first equation: \[ \frac{a}{3} = \frac{b}{4} \implies a = \frac{3b}{4} \] Next, we substitute \( a \) into the second equation: \[ \frac{a + 3b}{4a - b} = \frac{\frac{3b}{4} + 3b}{4\left(\frac{3b}{4}\right) - b} \] Now simplifying both the numerator and the denominator: **Numerator:** \[ \frac{3b}{4} + 3b = \frac{3b}{4} + \frac{12b}{4} = \frac{15b}{4} \] **Denominator:** \[ 4\left(\frac{3b}{4}\right) - b = 3b - b = 2b \] Now we combine these results: \[ \frac{\frac{15b}{4}}{2b} = \frac{15b}{4} \cdot \frac{1}{2b} \] The \( b \) terms cancel out: \[ = \frac{15}{8} \] So, the final expression is: \[ \frac{a + 3b}{4a - b} = \frac{15}{8} \]