The solution to the simultaneous equations 3x + 5y = 4, 4x + 3y = 5 is

A.

(\(\frac{-13}{11}, \frac{1}{11}\))

B.

(\(\frac{13}{11}, \frac{1}{11}\))

C.

(\(\frac{13}{11}, \frac{-1}{11}\))

D.

(\(\frac{11}{13}, \frac{1}{11}\))

E.

(13, 11)

Correct answer is B

3x + 5y = 4, 4x + 3y = 5

3x + 5y = 4 x 4

4x + 3y = 5 x 3

12x + 20y = 16.....(i)

12x + 9y = 15.......(ii)

subtract eqn.(ii) from eqn.(i)

11y = 1

y = \(\frac{1}{11}\)

12x + 20 x \(\frac{1}{11}\) = 16

12x = \(\frac{156}{11}\)

x = \(\frac{13}{11}\)

= \(\frac{13}{11}, \frac{1}{11}\)