Ta có :
\(f\left(x\right)=ax+b\)
+) \(f\left(1\right)=3\)
\(\Leftrightarrow a.1+b=3\)
\(\Leftrightarrow a+b=3\)\(\left(1\right)\)
+) \(f\left(-2\right)=2\)
\(\Leftrightarrow a.\left(-2\right)+b=2\)
\(\Leftrightarrow-2a+b=2\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\left(a-b\right)+\left(-2a+b\right)=3+2\)
\(\Leftrightarrow-1a=5\)
\(\Leftrightarrow a=5\)
\(\Leftrightarrow b=-2\)
Vậy ....