Question

Use elimination to solve each system below:

Answer 1

Answer:

SYSTEM 1

$4x + 3y = 4 - - - eqn(i) \\ - 2x - 3y = - 8 - - - eqn(ii) \\ eqn(i) + eqn(ii) \\ = > 4x - 2x = 4 - 8 \\ 2x = - 4 \\ x = \frac{ - 4}{2} \\ x = - 2 \\ in \: eqn(i) \: \: 4x + 3y = 4 \\ but \: x = - 2 \\ = > 4( - 2) + 3y = 4 \\ = > - 8 + 3y = 4 \\ 3y = 12 \\ y = \frac{12}{3} \\ y = 4$

SYSTEM 2

$3x + 2y = 7 - - - eqn(i) \\ 2x - y = 7 - - - eqn(ii) \\ multiply \: eqn(ii) \: by \: 2 \\ = > 4x - 2y = 14 - - - eqn(iii) \\ eqn(i) + eqn(iii) \\ = > 7x = 21 \\ x = \frac{21}{7} \\ x = 3 \\ in \: eqn(ii) \: 2x - y = 7 \\ but \: x = 3 \\ hence \: \: 2(3) - y = 7 \\ y = 6 - 7 \\ y = - 1$