Algebra: turn a situation into an expression and an equation

1. Give the quantity a precise name

A variable is a symbol representing a number. Let n mean the number of notebooks, not “notebooks” in general. If each notebook costs ₹12, where ₹ denotes rupees, the total cost is 12n rupees. Writing 12 next to n means multiplication, so 12n means 12 × n. A variable need not be called x; a useful letter can remind you what you are counting.

The allowed values come from the situation. A notebook count is a nonnegative whole number; a length can be a positive decimal. An algebraic calculation may produce a value that the context rejects. Define the unit too: if t means minutes, using t = 2 for a two-hour activity is incorrect unless you first convert hours into minutes.

Sources: OpenStax: Prealgebra 2e: Use the Language of Algebra ↗

2. Expressions describe; equations assert equality

The expression 12n + 30 describes the notebook cost plus a fixed ₹30 charge. It has no single numerical value until n is specified. The equation 12n + 30 = 126 says this total equals ₹126. The equals sign means both sides have the same value; it does not simply mean “write the next calculation”. Solving the equation finds which notebook count fits the stated total.

Translate relationships rather than hunting for keywords. “Five less than twice a number” becomes 2x − 5, with x the number and − the subtraction sign. “Twice five less than a number” can be ambiguous in ordinary speech; rewrite the sentence or use brackets. For the unambiguous phrase “twice the quantity x minus five”, write 2(x − 5). Brackets specify what is treated as one group.

Sources: OpenStax: Prealgebra 2e: Use the Language of Algebra ↗

3. Substitute carefully and preserve the order

To evaluate 3x + 2 when x = 4, replace x with 4: 3 × 4 + 2 = 14. Multiplication is performed before addition unless brackets change the order. Thus 3(x + 2) at x = 4 gives 3 × 6 = 18. These expressions are different rules. Try another value if you need to test a proposed simplification; one matching value alone does not prove two expressions always agree.

When the replacement is negative, brackets prevent sign mistakes. For x = −2, the expression 5 − 3x becomes 5 − 3(−2) = 5 − (−6) = 11. A negative quantity being subtracted increases the result. Write enough intermediate steps that you can see whether a minus sign belongs to a number or describes a subtraction operation.

Sources: OpenStax: Prealgebra 2e: Use the Language of Algebra ↗

4. Combine like terms and distribute multiplication

Terms are pieces joined by addition or subtraction. In 4x + 3 + 2x, the terms 4x and 2x count the same kind of quantity, so they combine to 6x; the result is 6x + 3. The numbers multiplying x are coefficients. You cannot turn 6x + 3 into 9x because the 3 is a constant, not three copies of x.

Distribution follows from counting equal groups: 3(x + 4) contains three lots of x and three lots of 4, so it equals 3x + 12. With subtraction, 5(x − 2) = 5x − 10. A minus before brackets affects every term: 9 − (x + 2) = 9 − x − 2 = 7 − x. Check at x = 3: both original and simplified expressions equal 4.

Sources: OpenStax: Prealgebra 2e: Distributive Property ↗

5. Solved example: undo operations on both sides

Return to 12n + 30 = 126. Subtract 30 from both sides, leaving 12n = 96. The same subtraction preserves equality, like removing equal masses from balanced pans. Divide both sides by 12 to obtain n = 8. Division is valid because 12 is not zero. Check the original equation: 12 × 8 + 30 = 96 + 30 = 126.

The answer is eight notebooks, not simply “8”. Do not divide only the term 12n while leaving the other terms untouched. You could divide the entire equation by 12 first, but then both 30 and 126 must also be divided. The familiar instruction “move it to the other side” hides the actual operation; naming the operation helps prevent sign errors.

The same equality rules work with decimal coefficients. For an illustrative length calculation, let x be a length in metres and solve 0.5x + 1.2 = 4.7, with every term measured in metres. Subtracting 1.2 gives 0.5x = 3.5; dividing by 0.5 gives x = 7 metres. Dividing by one-half doubles the number, which is sensible because 3.5 metres is only half of x. Check 0.5 × 7 + 1.2 = 4.7. Do not assume that division must always make an answer smaller.

Sources: NCERT Class VIII Mathematics Exemplar: Linear Equation in One Variable ↗ · OpenStax: Prealgebra 2e: Solve Equations Using Subtraction and Addition ↗ · OpenStax: Prealgebra 2e: Solve Equations Using Division and Multiplication ↗

6. Solved example: the unknown appears on both sides

Two illustrative printing plans charge 4p + 20 and 6p + 8 rupees for p pages. To find when their totals agree, set 4p + 20 = 6p + 8. Subtract 4p from both sides: 20 = 2p + 8. Subtract 8: 12 = 2p. Divide by 2: p = 6. Both plans then cost ₹44, since 4 × 6 + 20 = 6 × 6 + 8 = 44.

This solves the equality question only. For 3 pages the plans cost ₹32 and ₹26; for 10 pages they cost ₹60 and ₹68. The cheaper plan changes because one starts higher but adds less per page. Checking values on either side of the solution explains why the equality point is useful without pretending it answers every possible purchase question.

Sources: OpenStax: Prealgebra 2e: Equations with Variables and Constants on Both Sides ↗

7. A solution can be negative, absent or unrestricted

The equation x + 7 = 3 gives x = −4, which is valid for a number or a temperature change but not a notebook count. If simplification gives 2x + 3 = 2x + 8, subtracting 2x produces 3 = 8, so no value works. If it gives 2(x + 3) = 2x + 6, both sides agree for every allowed x. An equation does not automatically have exactly one solution; its relationship determines the result.

Sources: OpenStax: Prealgebra 2e: Equations with Variables and Constants on Both Sides ↗

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