Hello, and welcome to today's mathematics lesson. We are going to explore linear equations in one variable, along with practical problems based on these equations. By the end of this lesson, you will understand what makes an equation linear, how to solve various types of linear equations, and how to apply this powerful tool to real-world situations.
Let us begin with the fundamentals. An equation is a mathematical statement that declares two algebraic expressions are equal. For instance, 7x² + 8 = x - 3 is not linear because x is squared, while x - 3 = 3x + 8 is linear because x appears only to the first power.
A linear equation in one variable is special. It involves exactly one unknown quantity, and that variable appears only to the first power — meaning its highest power is one. Examples include 3x - 5 = 0, 8 - y = 15, and 7 + 3z = 10. Notice how each variable stands alone, without squares, cubes, or other complications.
To solve an equation means to discover the value of its variable — this value is called the root or solution. A linear equation in one variable always has exactly one solution.
Now, here is something crucial: the rules that keep an equation balanced. An equation remains unchanged when you perform the same operation on both sides. You may add the same number to both sides, subtract the same number from both sides, multiply both sides by the same non-zero number, or divide both sides by the same non-zero number. Think of an equation as a perfectly balanced scale — whatever you do to one side, you must do to the other to maintain that balance.
Let us see how this works in practice. Consider the equation 21 - 3(a - 7) = a + 20. First, we expand the brackets: minus 3 times minus 7 gives plus 21, so we get 21 - 3a + 21 = a + 20, which simplifies to 42 - 3a = a + 20. Now, let us gather like terms. Subtract 20 from both sides to get 22 - 3a = a, then add 3a to both sides: 22 = 4a. Therefore, a = 22/4 = 5½.
What about equations with fractions? Take (y+2)/4 - (y-3)/3 = 1/2. The denominators are 4, 3, and 2. The least common multiple of 4, 3, and 2 is 12. Multiply every term by 12. This gives 3(y + 2) - 4(y - 3) = 6. Expanding: 3y + 6 - 4y + 12 = 6, which becomes -y + 18 = 6, so -y = -12. So -y = -12, and y = 12.
Some equations have variables in the denominator. For 5/x = 7/(x-4), we use cross-multiplication. This gives 7x = 5(x - 4). So 7x = 5x - 20. Subtracting 5x from both sides: 2x = -20, therefore x = -10.
For equations like (a-2)/(a+4) = (a-3)/(a+1), cross-multiply to get (a - 2)(a + 1) = (a - 3)(a + 4). Expanding both sides: a² - a - 2 = a² + a - 12. Subtract a² from both sides, then simplify: -2a = -10, so a = 5.
Now we turn to the exciting part — solving real problems using linear equations. The method is straightforward but requires careful thinking.
First, read the problem thoroughly to identify what is given and what you need to find. Second, represent the unknown quantity with a letter — typically x, though y, z, a, or b work equally well. Third, translate the words into a mathematical relationship, creating your equation. Finally, solve the equation to find your answer.
Let us try this together. Find a number where one-fifth of it is less than its one-fourth by 3. Let the number be x. One-fifth of x is x/5, and one-fourth is x/4. The difference is 3, so x/4 - x/5 = 3. The least common multiple of 4 and 5 is 20. Multiplying both sides by 20: 5x - 4x = 60, which simplifies to x = 60. Therefore, x = 60.
Here is another type: consecutive numbers. The difference of squares of two consecutive even natural numbers is 92. Let the smaller be x, so the larger is x plus 2, since consecutive even numbers differ by 2. The equation becomes (x+2)² - x² = 92. Expanding: x² + 4x + 4 - x² = 92, which simplifies to 4x + 4 = 92. Subtracting 4 from both sides: 4x = 88, so x = 22. The larger number is 24.
Remember these patterns: Consecutive numbers are x, x + 1, x + 2. Consecutive even numbers are x, x + 2, x + 4, where x is even. Consecutive odd numbers are x, x + 2, x + 4, where x is odd. And consecutive multiples of 3 are x, x + 3, x + 6, where x is a multiple of 3.
Geometry problems work beautifully with linear equations. A rectangle is 8 centimetres long and 5 centimetres wide. Its perimeter doubles when each side increases by x centimetres. The original perimeter is 2(8 + 5) = 26 centimetres. That is 26 centimetres. The new perimeter is 2(8 + x + 5 + x) = 26 + 4x centimetres. This equals 52, so 26 + 4x = 52, giving 4x = 26, and x = 26/4 = 6.5 centimetres. The new length is 8 plus 6.5, which is 14.5 centimetres.
Age problems are classic applications. A man is 24 years older than his son. In 2 years, his age will be twice his son's age. Let the son's present age be x, so the father's is x plus 24. In 2 years, the father will be x + 26 and the son will be x + 2, giving x + 26 = 2(x + 2). This gives x + 26 = 2x + 4, so x = 22. The son is 22, the father is 46.
Speed and distance problems require careful unit handling. A boy walks to school at 4 kilometres per hour and arrives 10 minutes late. The next day he runs at 8 kilometres per hour and arrives 5 minutes early. Let the distance be x kilometres. Time at 4 kilometres per hour is x/4 hours; at 8 kilometres per hour is x/8 hours. The time difference is 15 minutes, which is 1/4 hour. So x/4 - x/8 = 1/4. Multiply by 8: 2x - x = 2, therefore x = 2 kilometres.
Finally, consider work and payment problems. A worker earns 150 rupees for each day worked and is fined 30 rupees for each day absent. In 40 days, he earns 3300 rupees. Let x be days worked, so days absent are 40 minus x. The equation is: 150x - 30(40 - x) = 3300. This becomes 150x - 1200 + 30x = 3300, which gives 180x = 4500, and therefore x = 25.
He worked 25 days and was absent 15 days.
Let us recap the essential points. First, a linear equation in one variable contains a single unknown raised only to the first power. Second, you can add, subtract, multiply, or divide both sides by the same non-zero number without changing the equation's balance. Third, for fractional equations, find the least common multiple of denominators and clear fractions by multiplication. Fourth, when variables appear in denominators, cross-multiplication is your efficient tool. Fifth, for word problems, define your variable clearly, translate relationships into equations, solve systematically, and verify your answer makes sense in context.
Linear equations are remarkably powerful. With this one technique, you can solve problems spanning numbers, ages, geometry, speed, work, and countless real situations. Practice translating words into mathematics, and you will find these equations becoming natural and intuitive.
Thank you for joining this lesson. Keep practising, stay curious, and remember — every complex problem becomes manageable when you break it down step by step. Until next time, happy learning!