Question

acolada,paranteza mare (7*x-5)+2*6se inchide paranteza mare/4+5inchidem acolada*10-deschidem paranteza mare(4+6)*5-42inchidem paranteza mare=112.va rog din suflet sa ma ajutati,multumesc anticipat

Answer 1

{[(7×x-5)+2×6]:4+5}×10-[(4+6)×5-42]=112
{[(7×x-5)+12]:4+5}×10-(10×5-42)=112
{[(7×x-5)+12]:4+5}×10-(50-42)=112
{[(7×x-5)+12]:4+5}×10-8=112
{[(7×x-5)+12]:4+5}×10=112+8
{[(7×x-5)+12]:4+5}×10=120
{[(7×x-5)+12]:4+5}=120:10
{[(7×x-5)+12]:4+5}=12
[(7×x-5)+12]:4=12-5
[(7×x-5)+12]:4=7
[(7×x-5)+12]=7×4
[(7×x-5)+12]=28
(7×x-5)=28-12
(7×x-5)=16
7×x=16+5
7×x=21
x=21:7
x=3