akar dari sisten persamaan p/2+q/4=13/4 dan p/4+q/3=21/4

[tex] \frac{p}{2} + \frac{q}{4} = \frac{13}{4} \\ - - - - - - - ( \times 4) \\ 2p + q = 13...(1) \\ \frac{p}{4} + \frac{q}{3} = \frac{21}{4} \\ - - - - - - - ( \times 12) \\ 3p + 4q = 63...(2) \\ eliminasi \: pers \: (1) \: dan \: (2) \\ 2p + q = 13 \: \: \: | \times 4| \\ 3p + 4q = 63 \: | \times 1| \\ - - - - - - - - - \\ 8p + 4q = 52 \\ 3p + 4q = 63 \\ - - - - - - - ( - ) \\ 5p = - 11 \\ p = \frac{ - 11}{5} \\ subsitusi \: ke \: pers \: (1) \\ 2( \frac{ - 11}{5} ) + q = 13 \\ - - - - - - - - ( \times 5) \\ - 22 + 5q = 65 \\ 5q = 87 \\ q = \frac{87}{5} [/tex]

Maaf kalo salah, semoga membantu