By the end of this chapter, you will be able to:
Mastering these skills will help you confidently handle mathematical problems that are essential in many technical and scientific tasks in your trade.
Algebraic equations and expressions are fundamental tools in Science Laboratory Technology, enabling professionals to analyze data, calculate concentrations, and solve quantitative problems encountered in laboratory experiments. Mastery of linear equations and simultaneous equations allows laboratory technologists to interpret measurements accurately, prepare solutions, and troubleshoot instrument calibration. This chapter focuses on applying algebraic methods to solve equations relevant to laboratory contexts, enhancing problem-solving skills essential for effective laboratory practice.
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 power of one. In laboratory technology, solving linear equations is crucial for determining unknown quantities such as reagent volumes, concentrations, or dilutions based on given data. The general form of a linear equation is
$$ ax + b = 0 $$
where \(a\) and \(b\) are constants, and \(x\) is the unknown variable.
This subtopic covers solving basic linear equations with one variable by isolating the variable on one side of the equation.
$$ ax + b = 0 $$
Example 1: A laboratory technician needs to find the volume \(x\) (in ml) of a stock solution to dilute with water such that the total volume is 100 ml. The equation is \(2x + 10 = 0\).
Given: \(a = 2\), \(b = 10\)
$$ 2x + 10 = 0 $$
$$ 2x = -10 $$
$$ x = \frac{-10}{2} $$
Answer: \(-5\) ml (Note: Negative volume is not physically meaningful, indicating an error in problem setup.)
Example 2: Calculate the concentration \(x\) (in mol/L) when the equation \(5x - 15 = 0\) represents the relationship between concentration and reaction rate.
Given: \(a = 5\), \(b = -15\)
$$ 5x - 15 = 0 $$
$$ 5x = 15 $$
$$ x = \frac{15}{5} $$
Answer: 3 mol/L
Example 3: Determine the amount of solute \(x\) (in grams) when \(0.4x + 2 = 6\).
Given: \(a = 0.4\), \(b = 2\), constant on right side is 6.
Rearranged:
$$ 0.4x + 2 = 6 $$
$$ 0.4x = 6 - 2 $$
$$ 0.4x = 4 $$
$$ x = \frac{4}{0.4} $$
Answer: 10 g
This subtopic explains how to solve linear equations involving fractional coefficients or constants, common when working with ratios or concentrations.
$$ \frac{a}{b}x + c = d $$
Example 1: Find \(x\) in the equation \(\frac{1}{3}x + 4 = 7\), where \(x\) is the volume in ml.
Given: \(a = \frac{1}{3}\), \(c = 4\), \(d = 7\)
$$ \frac{1}{3}x + 4 = 7 $$
$$ \frac{1}{3}x = 7 - 4 $$
$$ \frac{1}{3}x = 3 $$
$$ x = 3 \times 3 $$
Answer: 9 ml
Example 2: Solve for \(x\) in \(\frac{2}{5}x - \frac{1}{2} = 3\).
Given: \(a = \frac{2}{5}\), \(b = -\frac{1}{2}\), \(d = 3\)
$$ \frac{2}{5}x - \frac{1}{2} = 3 $$
$$ \frac{2}{5}x = 3 + \frac{1}{2} $$
$$ \frac{2}{5}x = \frac{6}{2} + \frac{1}{2} = \frac{7}{2} $$
$$ x = \frac{7}{2} \times \frac{5}{2} $$
$$ x = \frac{35}{4} = 8.75 $$
Answer: 8.75 units
Example 3: A lab solution concentration \(x\) satisfies \(\frac{3}{4}x + \frac{5}{6} = 2\). Find \(x\).
Given: \(a = \frac{3}{4}\), \(c = \frac{5}{6}\), \(d = 2\)
$$ \frac{3}{4}x + \frac{5}{6} = 2 $$
$$ \frac{3}{4}x = 2 - \frac{5}{6} $$
To subtract fractions:
$$ 2 = \frac{12}{6} $$
$$ \frac{3}{4}x = \frac{12}{6} - \frac{5}{6} = \frac{7}{6} $$
$$ x = \frac{7}{6} \times \frac{4}{3} = \frac{28}{18} = \frac{14}{9} \approx 1.56 $$
Answer: 1.56 units
This subtopic deals with equations where the variable appears on both sides, requiring rearrangement before solving.
$$ ax + b = cx + d $$
Example 1: Solve for \(x\) in \(3x + 5 = 2x + 10\).
Given: \(a=3\), \(b=5\), \(c=2\), \(d=10\)
$$ 3x + 5 = 2x + 10 $$
Subtract \(2x\) from both sides:
$$ 3x - 2x + 5 = 10 $$
$$ x + 5 = 10 $$
Subtract 5:
$$ x = 10 - 5 $$
Answer: 5 units
Example 2: Find \(x\) in \(0.5x + 3 = 1.2x + 1\).
Given: \(a=0.5\), \(b=3\), \(c=1.2\), \(d=1\)
$$ 0.5x + 3 = 1.2x + 1 $$
Subtract \(0.5x\) from both sides:
$$ 3 = 1.2x - 0.5x + 1 $$
$$ 3 = 0.7x + 1 $$
Subtract 1:
$$ 3 - 1 = 0.7x $$
$$ 2 = 0.7x $$
Divide by 0.7:
$$ x = \frac{2}{0.7} \approx 2.857 $$
Answer: 2.857 units
Example 3: Solve \(4x - 6 = 2x + 10\).
Given: \(a=4\), \(b=-6\), \(c=2\), \(d=10\)
$$ 4x - 6 = 2x + 10 $$
Subtract \(2x\):
$$ 4x - 2x - 6 = 10 $$
$$ 2x - 6 = 10 $$
Add 6:
$$ 2x = 16 $$
Divide by 2:
$$ x = 8 $$
Answer: 8 units
This subtopic applies linear equations to typical laboratory problems such as calculating reagent volumes, concentrations, and dilutions.
Example 1: A technician mixes \(x\) ml of a 5 M solution with 10 ml of water to prepare 15 ml of a diluted solution. Find \(x\).
Given total volume: 15 ml, water volume: 10 ml
The equation:
$$ x + 10 = 15 $$
$$ x = 15 - 10 $$
Answer: 5 ml
Example 2: Calculate the concentration \(x\) (M) of a solution after diluting 20 ml of 8 M stock solution to 100 ml.
Using dilution formula:
$$ C_1 V_1 = C_2 V_2 $$
Given: \(C_1=8\), \(V_1=20\), \(V_2=100\), \(C_2=x\)
$$ 8 \times 20 = x \times 100 $$
$$ 160 = 100x $$
$$ x = \frac{160}{100} = 1.6 $$
Answer: 1.6 M
Example 3: A reagent costs Ksh 500 for 2 kg. Find the cost \(x\) per gram.
Given: total cost = 500 Ksh, total mass = 2000 g
$$ x \times 2000 = 500 $$
$$ 2000x = 500 $$
$$ x = \frac{500}{2000} = 0.25 $$
Answer: Ksh 0.25 per gram
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Create a free accountThis chapter focused on solving algebraic equations and expressions starting with methods to solve linear equations. It then explored solving simultaneous equations using three main techniques: elimination, substitution, and graphical methods, each providing a different approach to find common solutions. The chapter also introduced linear graphs, covering the concepts of coordinates, plotting points on a plane, and drawing graphs of straight lines to visually represent equations. Moving on to quadratic equations, the chapter explained how to solve them using factorization, completing the square, and the quadratic formula, each method offering a systematic way to find the roots of quadratic expressions. Together, these topics build a solid foundation for understanding and manipulating algebraic equations in various scientific contexts.
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