Finding the Unknown: Solving Equations Like a Balance

An equation is a mathematical statement that two expressions are equal, written with an equals sign between them — for example, 4x + 1 = 21. The expression on the left is called the Left Hand Side (LHS), and the expression on the right is the Right Hand Side (RHS). Solving the equation means finding the value of the unknown (here, x) that makes LHS exactly equal to RHS.

One way to solve an equation is trial and error: try a value, check if LHS equals RHS, and adjust. For 4x + 1 = 21: trying x = 1 gives LHS = 4(1)+1 = 5, far below 21. Trying x = 4 gives LHS = 4(4)+1 = 17, closer. Trying x = 5 gives LHS = 4(5)+1 = 21, which matches the RHS exactly — so x = 5 is the solution.

Trial and error works, but it can take many guesses for harder equations, especially when the answer isn't a small whole number. This chapter develops a faster, systematic method that works every single time, without any guessing at all.

Picture a weighing scale that is perfectly balanced. If you add the same weight to both pans, it stays balanced. If you remove the same weight from both pans, it also stays balanced. An equation behaves in exactly the same way: since LHS and RHS are equal in value, adding, subtracting, multiplying, or dividing both sides by the same number keeps the equation true.

Consider the equation 3x + 5 = 20. To find x, the goal is to get x completely alone on one side. First, subtract 5 from both sides (removing the +5 term from the LHS): 3x + 5 − 5 = 20 − 5, giving 3x = 15. Then, divide both sides by 3 (removing the factor of 3): 3x ÷ 3 = 15 ÷ 3, giving x = 5.

Always check the answer by substituting it back into the original equation: LHS = 3(5) + 5 = 15 + 5 = 20, and RHS = 20. Since LHS = RHS, x = 5 is confirmed correct. This substitution check is quick, and it catches almost every careless mistake before it becomes a wrong final answer.

Consider 7x − 2 = 4x + 13, which has x appearing on both sides. The first move is to collect the x-terms onto one side. Subtract 4x from both sides: 7x − 2 − 4x = 4x + 13 − 4x, giving 3x − 2 = 13. Now add 2 to both sides to remove the constant term from the left: 3x − 2 + 2 = 13 + 2, giving 3x = 15. Finally, divide both sides by 3: x = 5.

Checking: LHS = 7(5) − 2 = 35 − 2 = 33. RHS = 4(5) + 13 = 20 + 13 = 33. Since both sides equal 33, x = 5 is confirmed correct.

Equations with brackets are solved the same way, after first expanding the brackets. Take 2(x + 3) = 16: expanding the left side gives 2x + 6 = 16. Subtracting 6 from both sides gives 2x = 10, and dividing both sides by 2 gives x = 5. Checking: LHS = 2(5+3) = 2(8) = 16, and RHS = 16 — matching exactly.

Consider 3(x − 2) = 2(x + 1), which has brackets on both sides. Expanding each side first: the left becomes 3x − 6, and the right becomes 2x + 2. The equation is now 3x − 6 = 2x + 2. Subtracting 2x from both sides gives x − 6 = 2. Adding 6 to both sides gives x = 8.

Checking: LHS = 3(8 − 2) = 3(6) = 18. RHS = 2(8 + 1) = 2(9) = 18. Both sides match, confirming x = 8 is correct.

A common slip is mishandling a minus sign when expanding or moving terms — for instance, forgetting that subtracting (x + 3) means subtracting both the x and the 3, not just one of them. This is exactly why the substitution check matters so much: plugging the found value back into the original, unexpanded equation (not a simplified version you created along the way) will immediately expose a sign mistake if LHS and RHS don't come out equal.

A taxi charges a fixed fee of ₹800 plus ₹20 for every kilometre travelled, and a particular ride costs ₹2200 in total. Letting k stand for the number of kilometres travelled, the total cost is 800 + 20k, which must equal 2200: 800 + 20k = 2200. Subtracting 800 from both sides gives 20k = 1400, and dividing both sides by 20 gives k = 70. Checking: 800 + 20(70) = 800 + 1400 = 2200, which matches exactly — the ride covered 70 km.

Two friends, Meera and Fatima, start saving money on the same day. Meera starts with ₹4000 and saves ₹650 every month; Fatima starts with ₹5050 and saves ₹500 every month. After how many months will they have saved the same total amount? Letting m stand for the number of months, Meera's savings are 4000 + 650m and Fatima's savings are 5050 + 500m. Setting these equal: 4000 + 650m = 5050 + 500m. Subtracting 500m from both sides gives 4000 + 150m = 5050; subtracting 4000 from both sides gives 150m = 1050; dividing both sides by 150 gives m = 7. Checking: Meera has 4000 + 650(7) = 4000 + 4550 = 8550, and Fatima has 5050 + 500(7) = 5050 + 3500 = 8550 — the same amount after exactly 7 months.

The pattern is always the same: name the unknown, translate every piece of the situation into an algebraic expression using that name, set two matching expressions equal to each other, then solve and check.