Question

3 ori acolada 10+2 ori paranteza patrata 40 minus 2 ori paranteza rotunda 28 minus 3 ori 5 inchidem toate parantezele minus 7x=51 va rog!

Answer 1

3·{10+2·[40-2·(28-3·5)]}-7x=51
3·{10+2·[40-2·(28-15)]}-7x=51
3·[10+2·(40-2·13)]-7x=51
3·[10+2·(40-26)]-7x=51
3·(10+2·14)-7x=51
3·(10+28)-7x=51
3·38-7x=51
114-7x=51 | -114
-7x=-63 |×(-1)
7x=63 |:7
x=9