Linear equations and simultaneous equations
A linear equation states that two expressions have the same value and that the unknown appears only to the first . In 3x + 5 = 20, the letter x represents a number that makes the statement true. Solving means finding that number, rather than merely moving symbols from one side of the equality sign to the other.
Think of the equality sign as a balance. Adding, subtracting, multiplying or dividing both sides by the same allowed number preserves the balance. To solve 3x + 5 = 20, subtract 5 from both sides to obtain 3x = 15. Divide both sides by 3, so x = 5. Substitution checks the answer: 3(5) + 5 really is 20.
When brackets occur, expand them carefully before collecting like terms. For example, 2(x + 4) = 18 becomes 2x + 8 = 18. A minus sign outside brackets changes every sign inside: −(x − 3) is −x + 3. Writing one operation on each line makes sign errors easier to notice and gives a clear record of the reasoning.
Equations can describe everyday situations. If three identical exercise books and a ₦200 pen cost ₦1,700, let b be the price of one book. The equation is 3b + 200 = 1,700, giving. Define the unknown and its unit before building the equation. A useful final sentence answers the actual question, rather than leaving an unexplained value of x.
Two unknowns usually require two independent equations. If x + y = 10 and x − y = 2, adding the equations eliminates y and gives 2x = 12. Thus x = 6 and y = 4. Elimination works because the same quantities are added to equal quantities. Substitution is another valid method: express one unknown in terms of the other and replace it in the second equation.
Before submitting, check that every division is by a nonzero number and that each algebraic step preserves equality. Distinguish an equation from an expression: 3x + 5 has no equality sign and cannot be solved without more information. If a word problem asks for a length, a negative algebraic result may signal a setup error or a value that the context excludes.
Choose a method that preserves the equation
A common shortcut is to say that a term changes sign when it crosses the equality sign. That description hides the operation. In, subtract 3x from both sides to get 2x + 8 = 20, then subtract 8. The result is 2x = 12 and x = 6. Writing the underlying operation matters when several negative terms or fractions occur, because it gives you a rule you can check instead of a movement to memorise.
For an equation with fractional terms, first identify forbidden denominator values. In x/3 + 2 = 5, multiplying both sides by 3 gives x + 6 = 15, so x = 9. In, multiplying the whole equation by 4 gives 2(x + 1) = x − 3 and hence x = −5. Multiply every term, including constants outside fractions. In an equation containing 1/(x − 2), x cannot equal 2 even if a later simplification appears to permit it.
Understand simultaneous equations, not just elimination
Each linear equation in two variables describes a straight line. A solution common to two independent equations is the point where their lines intersect. Parallel distinct lines have no common solution. Two equations that describe the same line have infinitely many common solutions. Thus x + y = 7 and 2x + 2y = 14 do not determine a unique pair: the second equation is only a multiple of the first. This helps you recognise why elimination can produce 0 = 0 or a false statement such as 0 = 3.
To eliminate a variable, arrange both equations in the same order and make one pair of coefficients equal in magnitude. For 2x + 3y = 19 and 3x − 2y = 9, multiply the first equation by 2 and the second by 3. The equations become 4x + 6y = 38 and 9x − 6y = 27. Adding gives 13x = 65, so x = 5. Substitute into either original equation to obtain y = 3. Check both originals, since a single successful substitution does not establish a simultaneous solution.
Build equations from a situation
Translate the relationships, rather than converting words one at a time. If an adult ticket costs ₦1,000 and a child ticket costs ₦600, the total number of tickets gives one equation and the total money gives another. For eight tickets costing ₦6,400, let a be adult tickets and c child tickets. Then a + c = 8 and 1,000a + 600c = 6,400. Dividing the money equation by 200 simplifies it to 5a + 3c = 32. Solving gives four of each, and both the count and the cost can be checked.
Changing the subject of a formula uses the same balance idea. Starting with, subtract u to obtain v − u = at and divide by a, where a ≠ 0. Brackets are essential because the whole difference is divided by a. When the subject appears more than once, collect its terms before dividing. From, obtain (a − c)x = d − b. A unique solution requires a − c ≠ 0. Otherwise inspect the remaining statement separately.
Check meaning and avoid common mistakes
The expression 2x + 3x is 5x, while 2x × 3x is 6x². Combining unlike terms, dividing only one term of a sum, or dropping brackets changes the problem. A negative answer is not automatically wrong: a temperature or a bank balance can be negative. A negative number of people is impossible in an ordinary counting question. State units, inspect whether a count must be an integer, and use a rough estimate to catch an answer that is algebraically tidy but unreasonable in context.
Self-check: solve 3(x − 2) = 15, then solve x + y = 11 together with x − y = 3. The first answer is x = 7. Adding the simultaneous equations gives 2x = 14, so x = 7 and y = 4. Explain why adding the equations eliminated y, and verify both answers by substitution. For further practice, decide whether x + y = 5 and 2x + 2y = 12 can be satisfied together. They cannot: multiplying the first equation by 2 would require the same left side to equal 10, not 12.
Test whether a proposed answer is complete
If 0x = 0 remains after simplification, every allowed x satisfies that statement. If 0x = 5 remains, no x satisfies it. These cases explain why dividing by a variable without checking zero can erase essential information. For equations with variables in denominators, list the exclusions before clearing fractions, then check every candidate in the original form.