Question 11: Given that \(f(x)=ax+b\), where \(a\) and \(b\) are constant numbers. If \(f(-2)=3\) and \(f(4)=10\), then find the values of \(a\) and \(b\).
Solution:
\(f(x)=ax+b\)
\(f(-2)=-2a+b\)
\(-2a+b=3\)\(\ \_\_\_\_\_\_\ (1)\)\(\quad \because f(-2)=3\)
\(f(4)=4a+b\)
\(4a+b=10\)\(\ \_\_\_\_\_\_\ (2)\)\(\quad \because f(4)=10\)
Subtracting equation \((1)\) from \((2)\), we have
\(\ \ 4a+b=10\)
\(\underline{\underset{+}{-}2a\underset{-}{+}b=\underset{-}{}3}\)
\(\ \ 6a\ \ \ \ \ \ \ \ =7\)
\(a=\large{\frac{7}{6}}\)
Put value of \(a\) in equation \((1)\), we have
\(-2(\frac{7}{6})+b=3\)
\(-\frac{7}{3}+b=3\)
\(b=3+\frac{7}{3}\)
\(b=\frac{9+7}{3}\)
\(b=\frac{16}{3}\)