\( \frac { a } { 3 } = \frac { b } { 4 } \quad \frac { a + 3 b } { 4 a - b } = \)

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}\frac{a}{3}=\frac{b}{4}\\\frac{a+3b}{4a-b}=x\end{array}\right.\) - step1: Rewrite the expression: \(\left\{ \begin{array}{l}\frac{a}{3}=\frac{b}{4}\\x=\frac{a+3b}{4a-b}\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\frac{a}{3}=\frac{b}{4}\) - step3: Multiply both sides of the equation by \(3:\) \(\frac{a}{3}\times 3=\frac{b}{4}\times 3\) - step4: Multiply the terms: \(a=\frac{b\times 3}{4}\) - step5: Divide the terms: \(a=\frac{3b}{4}\) - step6: Substitute the value of \(a:\) \(x=\frac{\frac{3b}{4}+3b}{4\times \frac{3b}{4}-b}\) - step7: Simplify: \(x=\frac{15}{8}\) - step8: Calculate: \(\left(a,b,x\right) = \left(\frac{3b}{4},b,\frac{15}{8}\right),b \in \mathbb{R}\) - step9: Alternative Form: \(\textrm{Infinitely many solutions}\) The system of equations given is: 1. \( \frac{a}{3} = \frac{b}{4} \) 2. \( \frac{a + 3b}{4a - b} = x \) Upon solving these equations, we find that there are infinitely many solutions. This indicates that the equations are dependent, meaning that one can be expressed in terms of the other. To express \( b \) in terms of \( a \) from the first equation, we can cross-multiply: \[ 4a = 3b \implies b = \frac{4a}{3} \] Now, substituting \( b \) into the second equation: \[ \frac{a + 3\left(\frac{4a}{3}\right)}{4a - \left(\frac{4a}{3}\right)} = x \] Calculating the numerator: \[ a + 3\left(\frac{4a}{3}\right) = a + 4a = 5a \] Calculating the denominator: \[ 4a - \left(\frac{4a}{3}\right) = \frac{12a}{3} - \frac{4a}{3} = \frac{8a}{3} \] Now substituting these back into the equation: \[ \frac{5a}{\frac{8a}{3}} = x \] This simplifies to: \[ x = \frac{5a \cdot 3}{8a} = \frac{15}{8} \] Thus, the value of \( \frac{a + 3b}{4a - b} \) is: \[ \frac{15}{8} \]