\(a+b=132\)\(\left(1\right)\)
\(a-b=4\) \(\left(2\right)\)
lấy \(\left(1\right)-\left(2\right)\)ta có
\(a+b-a+b=132-4\)
<=> \(2b=128\)
<=> \(b=64\)
=> \(a=4+b=4+64=68\)
\(a+b=132\)
\(a-b=4\)
\(\Rightarrow a=\left(132+4\right)\text{ : }2\)
\(\Rightarrow a=136\text{ : }2\)
\(\Rightarrow a=68\)
\(\Rightarrow b=68-4\)
\(\Rightarrow b=64\)
\(\text{Vậy : }a=68\)
\(b=64\)
a+b=132
a−b=4
⇒a=(132+4) : 2
⇒a=136 : 2
⇒a=68
⇒b=68−4
⇒b=64
Vậy : a=68
b=64