Solve using substitution method

We have the system:

1. $3p + 2e = 190$
2. $2p + 5e = 290$ From equation (1), let's express $e$ in terms of $p$:

$$2e = 190 - 3p$$ $$e = \frac{190 - 3p}{2}$$

Substitute the expression for $e$ from Step 1 into equation (2):

$$2p + 5\left(\frac{190 - 3p}{2}\right) = 290$$

Multiply the entire equation by 2 to eliminate the fraction:

$$2(2p) + 2 \cdot 5\left(\frac{190 - 3p}{2}\right) = 2(290)$$ $$4p + 5(190 - 3p) = 580$$ $$4p + 950 - 15p = 580$$ $$-11p + 950 = 580$$ $$-11p = 580 - 950$$ $$-11p = -370$$ $$p = \frac{-370}{-11}$$ $$p = \frac{370}{11}$$

$$e = \frac{190 - 3p}{2}$$

Substitute $p = \frac{370}{11}$:

$$e = \frac{190 - 3\left(\frac{370}{11}\right)}{2}$$ $$e = \frac{190 - \frac{1110}{11}}{2}$$

Find a common denominator for the terms in the numerator:

$$e = \frac{\frac{190 \times 11}{11} - \frac{1110}{11}}{2}$$ $$e = \frac{\frac{2090}{11} - \frac{1110}{11}}{2}$$ $$e = \frac{\frac{2090 - 1110}{11}}{2}$$ $$e = \frac{\frac{980}{11}}{2}$$ $$e = \frac{980}{11 \times 2}$$ $$e = \frac{980}{22}$$ $$e = \frac{490}{11}$$

The cost of one pencil is $\boxed{\text{KSh } \frac{370}{11}}$ and the cost of one exercise book is $\boxed{\text{KSh } \frac{490}{11}}$.