If a + b + c = $$\frac{7}{{12}},$$ 3a - 4b + 5c = $$\frac{3}{4}$$ and 7a - 11b - 13c = $$ - \frac{7}{{12}},$$  then what is the value of a + c?

Solution (By Examveda Team)

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