Question 263407
Here is our original system
(i) {{{2x-4y=8}}}
(ii) {{{1x-2y=4}}}
step 1 - multiply (ii) by -2 to get
(iii) {{{-2x + 4y = -8}}}
add (ii) and (iii) to get
(iv) {{{0 + 0 = 0}}}
this mans that we have the same equations but in disguise.
These equations are "dependent"