Hello, and welcome to today's mathematics lesson! Today, we are going to explore Chapter Fourteen: Simple Linear Equations. By the end of this lesson, you will understand what equations are, how to solve them step by step, and how to tackle word problems using equations. Let us begin this exciting journey into algebra!
Let us start with the fundamental question: what exactly is an equation? An equation is a mathematical statement that declares two expressions are equal. Think of it as a balance scale where both sides weigh exactly the same.
For example, if we have the expressions 3x + 2 and x − 7, and we state they are equal, we write 3x + 2 = x − 7. This is now an equation. Similarly, 8y − 15 = y/2 is also an equation, stating that 8y − 15 equals y/2.
Now, what does it mean to solve an equation? To solve an equation means to find the value of the unknown variable that makes the equation true. Suppose we have x + 2 = 6. We need to find what number x represents. Subtracting 2 from both sides, we get x = 6 − 2, so x = 4. We call x = 4 the solution, or root, of the equation.
Here are four crucial rules that keep an equation balanced. These are your tools for solving any equation.
First: you may add the same number to both sides of an equation. If x − 5 = 6, adding 5 to both sides gives x − 5 + 5 = 6 + 5, so x = 11.
Second: you may subtract the same number from both sides. If x + 5 = 6, subtracting 5 from both sides yields x = 1.
Third: you may multiply both sides by the same number. If x/3 = 2, multiplying both sides by 3 gives x = 6.
Fourth: you may divide both sides by the same non-zero number. If 4x = 12, dividing both sides by 4 gives x = 3.
Let us apply these rules together. Take x + 3 = 9. We subtract 3 from both sides: x + 3 − 3 = 9 − 3, giving x = 6. Simple and elegant!
Some equations require two operations to solve. Consider 3x + 8 = 14. First, we isolate the term with x by subtracting 8 from both sides: 3x = 6. Then we divide by 3: x = 2. The second step depends on the first — this is what we call a contextual problem.
Another example: 2x + 5 = 11. Subtract 5: 2x = 6. Divide by 2: x = 3.
Or try x/3 − 8 = 10. Add 8 to both sides: x/3 = 18. Multiply by 3: x = 54. Always work step by step, and you cannot go wrong!
Now let me introduce you to a powerful shortcut: transposition. Instead of writing out every operation on both sides, we can move terms from one side to another by changing their signs.
Here is how it works. When you transpose a positive term, it becomes negative. From x + 3 = 6, moving +3 to the right gives x = 6 − 3, so x = 3.
When you transpose a negative term, it becomes positive. From x − 3 = 6, moving −3 to the right gives x = 6 + 3, so x = 9.
When a term is multiplying the variable, it moves as division. From 3x = 6, moving 3 gives x = 6/3, so x = 2.
When a term is dividing the variable, it moves as multiplication. From x/3 = 6, moving 3 gives x = 6 × 3, so x = 18.
By convention, we prefer to keep the variable on the left side of the equation.
Let us practice with fractions. Take 2/3 x = 16. First, transpose the 3 in the denominator: 2x = 16 × 3, so 2x = 48. Then transpose the 2: x = 48/2, giving x = 24.
Or try 3/4 x + 5 = 8. Transpose +5: 3/4 x = 3. Transpose the 4: 3x = 12. Transpose the 3: x = 4.
What happens when the variable appears on both sides of the equation? Do not worry — we simply gather all variable terms on one side and all constants on the other.
Consider 8x − 3 = 5x + 9. Transpose 5x to the left and −3 to the right: 8x − 5x = 9 + 3. This simplifies to 3x = 12, so x = 4.
Another example: 7 + 4x = 9x − 13. We can transpose 4x to the right and −13 to the left: 7 + 13 = 9x − 4x. This gives 20 = 5x, so x = 4. Notice we get the same answer regardless of which side we choose for the variable!
For equations with brackets, first expand them. Take 2(x − 5) + 3(x − 2) = 8 + 7(x − 4). Expanding: 2x − 10 + 3x − 6 = 8 + 7x − 28. Simplifying both sides: 5x − 16 = 7x − 20. Transposing: 5x − 7x = −20 + 16, so −2x = −4, giving x = 2.
When fractions appear in equations, we use the lowest common multiple, or LCM, to clear them.
Consider x/5 + x = 12. Write x as x/1. The LCM of 5 and 1 is 5, so we combine: (x + 5x)/5 = 12, giving 6x/5 = 12. Multiply both sides by 5: 6x = 60, so x = 10.
A more complex example: (x − 1)/3 − (2x − 3)/5 = 1. The LCM of 3 and 5 is 15. Multiply each term appropriately: (5(x − 1) − 3(2x − 3))/15 = 1. Expanding the numerator: (5x − 5 − 6x + 9)/15 = 1, which simplifies to (−x + 4)/15 = 1. Multiply by 15: −x + 4 = 15, so −x = 11, giving x = −11.
Now comes the most exciting part: using equations to solve real-world problems! The key is to translate words into mathematical expressions.
Example one: five added to twice a whole number gives thirty-five. Let the number be x. Twice the number is 2x, and five added gives 2x + 5. So 2x + 5 = 35. Solving: 2x = 30, so x = 15.
Example two: one-fifth of a number is four. Let the number be x. Then x/5 = 4, so x = 20.
Example three: divide eighty into two parts where the greater part is four times the smaller. Let the smaller part be x. Then the greater part is 4x. Together: x + 4x = 80, so 5x = 80, giving x = 16. The parts are sixteen and sixty-four.
Example four: a number multiplied by nine, then eleven added, gives fifty-six. Let the number be x: 9x + 11 = 56. So 9x = 45, and x = 5.
Example five: two tables and three chairs cost six hundred eighty rupees. A table costs forty rupees more than a chair. Let a chair cost x rupees. Then a table costs x + 40 rupees. The equation: 2(x + 40) + 3x = 680. Expanding: 2x + 80 + 3x = 680, so 5x = 600, giving x = 120. A chair costs one hundred twenty rupees, and a table costs one hundred sixty rupees.
Example six: age problems. A father is three times as old as his son. In ten years, the father will be twice as old as his son. Let the son's present age be x years. The father's present age is 3x years. In ten years: son will be x + 10, father will be 3x + 10. The condition: 3x + 10 = 2(x + 10). Expanding: 3x + 10 = 2x + 20, so x = 10. The son is ten years old, and the father is thirty.
Example seven: a rectangle's perimeter is fifty centimetres. The length exceeds the breadth by five centimetres. Let breadth be x centimetres. Length is x + 5 centimetres. Perimeter formula: 2(length + breadth). So 2(x + 5 + x) = 50, giving 2(2x + 5) = 50, so 4x + 10 = 50. Thus 4x = 40, and x = 10. The breadth is ten centimetres, and the length is fifteen centimetres.
Example eight: a purse contains sixty notes of ten and fifty rupee denominations, totalling one thousand four hundred rupees. Let ten-rupee notes be x. Then fifty-rupee notes are 60 − x. Value equation: 10x + 50(60 − x) = 1400. Expanding: 10x + 3000 − 50x = 1400, so −40x = −1600, giving x = 40. There are forty ten-rupee notes and twenty fifty-rupee notes.
Let us now recap the key takeaways from this lesson.
First, an equation states that two expressions are equal, and solving means finding the variable's value.
Second, you can add, subtract, multiply, or divide both sides by the same number to maintain balance.
Third, transposition is a shortcut: move terms across the equals sign by reversing their operations — plus becomes minus, minus becomes plus, multiplication becomes division, and division becomes multiplication.
Fourth, when variables appear on both sides, gather them on one side and constants on the other.
Fifth, for fractions, use the LCM to clear denominators before solving.
Sixth, for word problems, define your variable clearly, translate the given information into an equation, solve step by step, and verify your answer makes sense in the original problem.
Excellent work today! You have learned to wield one of mathematics' most powerful tools — the equation. With practice, you will solve them with confidence and ease. Keep practicing, stay curious, and remember: every problem has a solution waiting to be found. Until next time, happy solving!