**1. SIMPLE LINEAR EQUATIONS IN ONE UNKNOWN**

In our previous classes, we have learnt some number games like

where we have to fill in the circle with any possible number to make the sentence true.

Suppose we replace the circle by the letter x.

x + 3 = 8

x + 3 = 8 is called an equation in one unknown x.

A solution or root of an equation is a value of the unknown that will make the equation true. For

example, x = 5 is a solution of the above equation but x = 1 is not. To solve an equation means to find the solution(s) to the equation.

A simple equation the form ax + b = c, where a, b and c are constants and $a\ne 0$ is called a linear

equation. We can relate the idea of balance to a linear equation that help us solve it. Let us illustrate how this can be done by using the equation 2x + 3 = 9.

$\begin{array}{l}2x+3=9\\ 2x=9-3\\ \Rightarrow 2x=6\\ x=\frac{6}{2}\\ \Rightarrow x=3\end{array}$

Hence x = 3 is a solution of the equation 2x + 3 = 9.

In general, an equation remains unchanged when both sides are added, subtracted, multiplied or divided by the same number.

**2. EQUATIONS INVOLVING BRACKETS**

We apply the distributive law of multiplication over addition to help us solve equations involving brackets.**Example: **Solve the equation 9(x + 1) = 2(3x + 8)

**Sol.** 9(x + 1) = 2(3x + 8)

9x + 9 = 6x + 16

9x + 9 – 6x = 16

3x + 9 = 16

3x = 16 – 9

3x = 7

$x=\frac{7}{3}$

**3. SIMPLE FRACTIONAL EQUATIONS**

When the unknown of an equation is in the denominator of a term, the equation is called a fractional equation.

Examples of fractional equations are: $\frac{6}{x-2}=3\text{and}\frac{1}{x+3}=\frac{2}{x}$

We can use multiplication to transform simple fractional equations into linear equations.

In solving fractional equations, it is important to check the solutions. They cannot be those values that make a denominator of the original equation zero.**Example: **Solve the equation $\frac{6}{x-2}=3$

**Sol. $\frac{6}{x-2}=3$**

Multiplying both sides by x – 2, we have: $(x-2)\left(\frac{6}{x-2}\right)=(x-2)\left(3\right)$

6 = 3 (x – 2)

Notice that the denominator (x – 2)

6 = 3x – 6 does not appear in this equation

3x = 12

$x=\frac{12}{3}\phantom{\rule{0ex}{0ex}}\therefore x=4$

**4. FORMING LINEAR EQUATIONS TO SOLVE PROBLEMS**

Sometimes we can use linear equations to represent real-life situations. Then by solving the linear equations, we can actually provide solutions to real-life problems. It is therefore useful to learn how to form linear equations from given information.

Here, we shall learn how we can use a linear equation to solve for any unknown quantity.**Example: **The sum of three consecutive integers is 111. Find the integers.**Sol. **

**Step 1:** We are going to find the three integers.**Step 2:** Let x be the smallest integer.**Step 3:** Middle integer = x + 1

Largest integer = x + 2**Step 4:** Sum of 3 integers = 111

$\therefore $ x + (x + 1) + (x + 2) = 111

3x + 3 = 111**Step 5:** 3x = 111 – 3

3x = 108

$x=\frac{108}{3}$

$\therefore $ x = 36**Step 6:** The three integers are 36, 37 and 38