Applied biology professionals in Kenya frequently analyze biological data and model biological processes using algebraic equations. Mastery of solving linear equations and simultaneous equations enables precise quantification of biological variables such as population growth rates, enzyme activities, or nutrient concentrations. This chapter develops skills to solve these equations using algebraic and graphical methods, empowering students to apply mathematical reasoning in biological research and practical applications.

2.1 Solution of linear equations

Linear equations are algebraic expressions where the highest power of the variable is one. In applied biology, linear equations model relationships such as enzyme-substrate interaction rates, nutrient uptake, or linear growth patterns. The general form of a linear equation in one variable \(x\) is:

$$ ax + b = 0 $$

where \(a\) and \(b\) are constants, and \(a \neq 0\).

Worked Examples

Example 1:
A biologist measures the growth of a bacterial culture and finds the population \(P\) (in millions) relates to time \(t\) (in hours) by the equation \(3t + 5 = 0\). Find the time when the population reaches zero.

Given: \(3t + 5 = 0\)$$ 3t + 5 = 0 $$$$ 3t = -5 $$$$ t = \frac{-5}{3} $$$$ t = -1.67 \text{ hours} $$ Answer: \(t = -1.67\) hours (The negative time implies the population was zero 1.67 hours before the start of measurement.)

Example 2:
The concentration \(C\) of a nutrient in a solution decreases linearly with time \(t\) by the equation \(0.4t - 2 = 0\). Find the time when the nutrient concentration becomes zero.

Given: \(0.4t - 2 = 0\)$$ 0.4t - 2 = 0 $$$$ 0.4t = 2 $$$$ t = \frac{2}{0.4} $$$$ t = 5 \text{ hours} $$ Answer: \(t = 5\) hours

Example 3:
In a plant physiology experiment, the rate of photosynthesis \(R\) (in \(\mu mol \, CO_2/m^2/s\)) is modeled by \(5R + 20 = 0\). Find the rate when the equation holds.

Given: \(5R + 20 = 0\)$$ 5R + 20 = 0 $$$$ 5R = -20 $$$$ R = \frac{-20}{5} $$$$ R = -4 \, \mu mol \, CO_2/m^2/s $$ Answer: \(R = -4 \, \mu mol \, CO_2/m^2/s\) (Negative rate may indicate respiration exceeding photosynthesis.)

Example 4:
A technician models the linear relationship between enzyme activity \(E\) and substrate concentration \(S\) as \(2E - 7 = 0\). Calculate \(E\).

Given: \(2E - 7 = 0\)$$ 2E - 7 = 0 $$$$ 2E = 7 $$$$ E = \frac{7}{2} $$$$ E = 3.5 \text{ units} $$ Answer: \(E = 3.5\) units

Example 5:
The linear equation \(0.75x + 4.5 = 0\) represents a biological variable \(x\). Find \(x\).

Given: \(0.75x + 4.5 = 0\)$$ 0.75x + 4.5 = 0 $$$$ 0.75x = -4.5 $$$$ x = \frac{-4.5}{0.75} $$$$ x = -6 $$ Answer: \(x = -6\)

Chapter Summary

This chapter focused on solving algebraic equations and expressions starting with linear equations, where methods to find the value of unknown variables were explored. It then covered simultaneous equations, presenting three approaches: elimination, substitution, and graphical methods, each providing a way to solve two or more equations at once. The chapter also explained how to transpose formulas, which involves rearranging equations to isolate a specific variable. Attention was given to quadratic equations, introducing three techniques for their solution: factorization, completing the square, and using the quadratic formula. Each method was demonstrated with clear steps to enable practical application. The comprehensive treatment ensures a solid foundation in manipulating and solving different types of algebraic equations relevant to scientific and engineering problems.