Question 205363
because the 1st equation and the 2nd equation actually is the same (the 2nd equation is twice the 1st equation),
let x = any real number k
4x + 5y = 10
4k + 5y = 10
5y = 10 - 4k
y = {{{(10-4k)/5}}}
y = {{{2 - (4/5)k}}}


so, the solution is:
x = k
y = {{{2 - (4/5)k}}}
which k is any real number