We come across many situations in everyday life where we need to find an unknown quantity. For example, we may want to know the price of a pen when we know the total cost of a few pens, or the number of marbles in a bag when the total count and the shared number are known. In mathematics, such unknown quantities are represented by variables (usually letters like x, y or z), and the relationship between the known and unknown quantities is written as an equation. Solving that equation gives us the value of the unknown.
An algebraic equation is a statement of equality between two algebraic expressions. A linear equation in one variable is an equation of the form ax + b = 0, where a and b are constants and a is not zero. For example, 2x + 3 = 7 is a linear equation in one variable x. In this chapter we will learn how to solve such equations, how to use transposition and cross multiplication, how to reduce equations to simpler forms, and how to apply linear equations to solve real-life word problems.
An equation has two sides: the left-hand side (LHS) and the right-hand side (RHS), connected by the equality sign. The value of the variable which satisfies the equation, that is, which makes LHS equal to RHS, is called the solution of the equation.
For example, in the equation 2x + 3 = 7, putting x = 2 gives LHS = 2(2) + 3 = 7 = RHS. So x = 2 is the solution of the equation.
An equation is like a weighing balance which is balanced. Whatever operation we perform on one side must be performed on the other side as well, so that the balance is not disturbed.
The process of moving a term from one side of the equation to the other side with its sign changed is called transposition.
Example: Solve 2x + 3 = 7. Transpose 3 to RHS: 2x = 7 - 3 = 4. Transpose 2 to RHS: x = 4/2 = 2.
When variables appear on both sides, first bring all variable terms to one side and all constant terms to the other side, then solve.
Example: Solve 2x - 3 = x + 5. 2x - x = 5 + 3, so x = 8.
To solve equations containing fractions, we multiply both sides by the LCM of the denominators, or use cross multiplication.
For example, to solve (x + 1)/2 = (x - 3)/4, we cross multiply: 4(x + 1) = 2(x - 3). This gives 4x + 4 = 2x - 6, so 4x - 2x = -6 - 4, that is, 2x = -10, hence x = -5.
Some equations do not look linear but can be reduced to linear form by cross multiplication. For example, (2x + 1)/(3x - 2) = 5/9 can be cross multiplied to get 9(2x + 1) = 5(3x - 2), which is a linear equation.
To solve word problems we follow these steps: 1. Read the problem carefully and identify the unknown quantity. Let the unknown be x. 2. Form a linear equation using the given conditions. 3. Solve the equation. 4. Verify the answer by substituting it back in the problem.
The sum of three consecutive integers is 51. Find the integers. Let the integers be x, x + 1 and x + 2. x + (x + 1) + (x + 2) = 51 3x + 3 = 51, so 3x = 48 and x = 16. The integers are 16, 17 and 18.
| Operation on LHS | Operation on RHS after transposition |
|---|---|
| Addition (+a) | Subtraction (-a) |
| Subtraction (-a) | Addition (+a) |
| Multiplication (x a) | Division (divided by a) |
| Division (divided by a) | Multiplication (x a) |
| Step | Action |
|---|---|
| Step 1 | Read the problem and identify the unknown, let it be x |
| Step 2 | Form the equation using given conditions |
| Step 3 | Solve the equation using transposition |
| Step 4 | Check the solution by substitution |
| Step 5 | Write the answer with proper units |
Linear equations in one variable are one of the most important tools of algebra. In this chapter we learnt what an equation is, how to find its solution, and how to use transposition and cross multiplication to solve even those equations which appear complicated or fractional. We also studied how to convert real-life situations into equations and solve them, which is a skill used throughout mathematics and science. The concepts of this chapter are the foundation for simultaneous equations, quadratic equations and inequalities that will be studied in higher classes. Mastering the balance idea and practising many word problems is the key to scoring well in examinations.