Thereis x. If you add
30/7
to it, from the sum you get subtract
22/7
, and then subtract that difference from
59/7
, then you will get
19/7
what is x?
\(\frac{59}{7}-[(x+\frac{30}{7})-\frac{22}{7}]=\frac{19}{7} \\~\\ \frac{59}{7}-[x+\frac{30}{7}-\frac{22}{7}]=\frac{19}{7} \\~\\ \frac{59}{7}-x-\frac{30}{7}+\frac{22}{7}=\frac{19}{7} \\~\\ \frac{59}{7}-\frac{30}{7}+\frac{22}{7}=\frac{19}{7}+x \\~\\ \frac{59}{7}-\frac{30}{7}+\frac{22}{7}-\frac{19}{7}=x \\~\\ \frac{32}{7}=x\)
.\(\frac{59}{7}-[(x+\frac{30}{7})-\frac{22}{7}]=\frac{19}{7} \\~\\ \frac{59}{7}-[x+\frac{30}{7}-\frac{22}{7}]=\frac{19}{7} \\~\\ \frac{59}{7}-x-\frac{30}{7}+\frac{22}{7}=\frac{19}{7} \\~\\ \frac{59}{7}-\frac{30}{7}+\frac{22}{7}=\frac{19}{7}+x \\~\\ \frac{59}{7}-\frac{30}{7}+\frac{22}{7}-\frac{19}{7}=x \\~\\ \frac{32}{7}=x\)