If a + b + c = 712, 3a - 4b + 5c = 3/4 and 7a - 11b - 13c = - 712, then what is the value of a + c?

Answer: Option B

$$\begin{array}{rl}& \text{We have,}\\ & \Rightarrow a+b+c=\frac{7}{12}......\left(1\right)\\ & \Rightarrow 3a-4b+5c=\frac{3}{4}......\left(2\right)\\ & \Rightarrow 7a-11b-13c=-\frac{7}{12}......\left(3\right)\end{array}$$
Subtracting 3 times of equation (1) from equation (2), we get
$$\begin{array}{rl}& \Rightarrow -7b+2c=\frac{3}{4}-3\times \frac{7}{12}\\ & \Rightarrow -7b+2c=\frac{3}{4}-\frac{7}{4}\\ & \Rightarrow -7b+2c=-1......\left(4\right)\end{array}$$
Subtracting equation (3) from 7 times equation (1), we get
$$\begin{array}{rl}& \Rightarrow 18b+20c=\frac{49}{12}+\frac{7}{12}\\ & \Rightarrow 18b+20c=\frac{56}{12}\\ & \Rightarrow 18b+20c=\frac{14}{3}......\left(\text{5}\right)\end{array}$$
Subtracting 10 times of equation (4) from equation (5), we get
$$\begin{array}{rl}& \Rightarrow 88b=\frac{14}{3}+10\\ & \Rightarrow 88b=\frac{44}{3}\\ & \Rightarrow b=\frac{1}{6}\end{array}$$
Substituting in equation (1), we get
$$\begin{array}{rl}& \Rightarrow a+\frac{1}{6}+c=\frac{7}{12}\\ & \Rightarrow a+c=\frac{7}{12}-\frac{1}{6}=\frac{5}{12}\\ & \therefore \text{The value of }a+c\text{ is }\frac{5}{12}\end{array}$$