JAMB - Mathematics (1982 - No. 32)
The solution to the simultaneous equations 3x + 5y = 4, 4x + 3y = 5 is
(\(\frac{-13}{11}, \frac{1}{11}\))
(\(\frac{13}{11}, \frac{1}{11}\))
(\(\frac{13}{11}, \frac{-1}{11}\))
(\(\frac{11}{13}, \frac{1}{11}\))
(13, 11)
Explanation
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 = \(\frac{13}{11}\)
= \(\frac{13}{11}, \frac{1}{11}\)
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 = \(\frac{13}{11}\)
= \(\frac{13}{11}, \frac{1}{11}\)
Comments (0)
