Algebraic equations and expressions are fundamental tools for Science Laboratory Technology professionals when analyzing experimental data, calibrating instruments, and solving quantitative problems in the laboratory. Mastery of linear equations and simultaneous equations equips technologists to accurately determine unknown quantities and interpret relationships between variables encountered in chemical assays, biological tests, and physical measurements. This chapter focuses on methods to solve linear equations and systems of simultaneous equations relevant to laboratory contexts in Kenya.

2.1 Solution of linear equations

Linear equations express a relationship between variables where each term is either a constant or the product of a constant and a single variable raised to the first power. In laboratory technology, such equations arise when calculating concentrations, dilutions, or instrument calibrations. The general form of a linear equation in one variable \(x\) is

$$ax + b = 0$$

where \(a\) and \(b\) are constants and \(a eq 0\).

2.1.1 Solving simple linear equations

To solve for \(x\), isolate the variable on one side by performing inverse operations on both sides of the equation.

$$ax + b = 0$$

$$ax = -b$$

$$x = \frac{-b}{a}$$

Worked Examples

Example 1: A laboratory technician measures the absorbance \(A\) of a solution and finds that it relates to concentration \(C\) by the equation \(2C + 3 = 11\). Find \(C\).

Given: \(2C + 3 = 11\)

$$2C = 11 - 3$$

$$2C = 8$$

$$C = \frac{8}{2}$$

$$C = 4 \text{ mg/L}$$

Answer: 4 mg/L

Example 2: The volume \(V\) of a reagent added satisfies \(5V - 10 = 20\). Calculate \(V\).

Given: \(5V - 10 = 20\)

$$5V = 20 + 10$$

$$5V = 30$$

$$V = \frac{30}{5}$$

$$V = 6 \text{ mL}$$

Answer: 6 mL

Example 3: The calibration of a pH meter follows the equation \(0.8x + 2 = 10\). Find \(x\).

Given: \(0.8x + 2 = 10\)

$$0.8x = 10 - 2$$

$$0.8x = 8$$

$$x = \frac{8}{0.8}$$

$$x = 10$$

Answer: 10 (pH units)

2.1.2 Solving linear equations with fractions

Equations may include fractional coefficients, common in dilution calculations or reagent proportions. Multiply both sides by the least common denominator (LCD) to clear fractions before isolating the variable.

$$\frac{a}{m}x + \frac{b}{n} = c$$

Multiply both sides by \(lcm(m, n)\) to eliminate denominators.

Worked Examples

Example 1: Solve \(\frac{1}{3}x + \frac{1}{4} = 2\).

Given: \(\frac{1}{3}x + \frac{1}{4} = 2\)

Multiply both sides by 12 (LCD of 3 and 4):

$$12 \times \frac{1}{3}x + 12 \times \frac{1}{4} = 12 \times 2$$

$$4x + 3 = 24$$

$$4x = 24 - 3$$

$$4x = 21$$

$$x = \frac{21}{4}$$

$$x = 5.25$$

Answer: 5.25

Example 2: Solve \(\frac{2}{5}x - \frac{3}{10} = 1\).

Given: \(\frac{2}{5}x - \frac{3}{10} = 1\)

Multiply both sides by 10 (LCD of 5 and 10):

$$10 \times \frac{2}{5}x - 10 \times \frac{3}{10} = 10 \times 1$$

$$4x - 3 = 10$$

$$4x = 10 + 3$$

$$4x = 13$$

$$x = \frac{13}{4}$$

$$x = 3.25$$

Answer: 3.25

Example 3: Solve \(\frac{3}{7}x + \frac{2}{7} = \frac{5}{7}\).

Given: \(\frac{3}{7}x + \frac{2}{7} = \frac{5}{7}\)

Multiply both sides by 7:

$$7 \times \frac{3}{7}x + 7 \times \frac{2}{7} = 7 \times \frac{5}{7}$$

$$3x + 2 = 5$$

$$3x = 5 - 2$$

$$3x = 3$$

$$x = \frac{3}{3}$$

$$x = 1$$

Answer: 1

2.1.3 Solving linear equations with variables on both sides

When variables appear on both sides, rearrange terms to collect variables on one side and constants on the other before solving.

$$ax + b = cx + d$$

Rearranged to

$$(a, c)x = d, b$$

$$x = \frac{d, b}{a, c}$$

Worked Examples

Example 1: Solve \(3x + 5 = 2x + 9\).

Given: \(3x + 5 = 2x + 9\)

$$3x - 2x = 9 - 5$$

$$x = 4$$

Answer: 4

Example 2: Solve \(4x - 7 = 6x + 1\).

Given: \(4x - 7 = 6x + 1\)

$$4x - 6x = 1 + 7$$

$$-2x = 8$$

$$x = \frac{8}{-2}$$

$$x = -4$$

Answer: -4

Example 3: Solve \(5x + 2 = 3x + 10\).

Given: \(5x + 2 = 3x + 10\)

$$5x - 3x = 10 - 2$$

$$2x = 8$$

$$x = \frac{8}{2}$$

$$x = 4$$

Answer: 4

2.1.4 Solving applied linear equations in laboratory context

Applied problems involve formulating linear equations from laboratory data or experimental conditions and solving them to find unknown quantities.

Worked Examples

Example 1: A technician mixes two solutions, one with concentration 5 mg/L and the other 10 mg/L, to get 4 L of a 7 mg/L solution. Find the volume \(x\) of the 5 mg/L solution used.

Given: Let \(x\) = volume of 5 mg/L solution, then \((4 - x)\) = volume of 10 mg/L solution.

Concentration equation:

$$5x + 10(4 - x) = 7 \times 4$$

$$5x + 40 - 10x = 28$$

$$-5x = 28 - 40$$

$$-5x = -12$$

$$x = \frac{-12}{-5}$$

$$x = 2.4 \text{ L}$$

Answer: 2.4 L

Example 2: A reagent costs Ksh 120 per litre and a buffer solution costs Ksh 80 per litre. A mixture of 10 litres costs Ksh 104 per litre. Find the volume of reagent used.

Given: Let \(x\) = volume of reagent; \(10 - x\) = volume of buffer.

Cost equation:

$$120x + 80(10 - x) = 104 \times 10$$

$$120x + 800 - 80x = 1040$$

$$40x = 1040 - 800$$

$$40x = 240$$

$$x = \frac{240}{40}$$

$$x = 6 \text{ L}$$

Answer: 6 L

Example 3: A lab test requires a dilution such that \(x\) mL of stock solution is mixed with 50 mL of water to get a 2% solution. If the stock is 8%, find \(x\).

Given:

$$8x = 2(x + 50)$$

$$8x = 2x + 100$$

$$8x - 2x = 100$$

$$6x = 100$$

$$x = \frac{100}{6}$$

$$x = 16.67 \text{ mL}$$

Answer: 16.67 mL

Chapter Summary

This chapter focused on solving algebraic equations and expressions relevant to science and engineering contexts. It began with methods for finding solutions to linear equations, providing a foundation for more complex problems. The study then extended to simultaneous equations, demonstrating three key approaches: elimination, substitution, and graphical methods, each allowing the determination of variable values in systems of equations. The chapter proceeded to explore linear graphs, covering coordinate systems, plotting points accurately, and sketching graphs of straight lines to visualize linear relationships. Finally, the focus shifted to quadratic equations, where solutions were obtained through factorization, completing the square, and applying the quadratic formula. These techniques equip students with essential algebraic tools for practical problem solving in technical fields.