# Linear Equation in One Variable Examples

In mathematics, linear equations in one variable contains number of equations with one variable. The general form is ax + b = 0, where a is any non zero number. In linear algebra, we deal with linear, quadratic and cubic equations mostly. Solution of the equation is based on the type of equations.  A linear equation have single solution. If we have a quadratic equation, then it would have exactly two solutions. Similarly for a cubic equation, there would be three solutions. In this section will concentrate more on linear equations with one variable. We have already examined linear equation rules to describe linear relationships found in numerical patterns. The knowledge of which operations preserve equality and which disturb them are important generalizations about the properties of numbers among these operations, mostly distribution property and some basic algebraic properties help to solve such problems. Linear equations in one variable can be represented in various ways. You may solve in one step or two or more steps. Let us assume some of the forms and see how to solve such equations.

## Solving one-step Linear Equations

You can get your result by isolating the variable in one step. Just we have to figure out what action we need to take. for example if variable is added to something. we will subtract and get the answer.
Example 1:  x + 5 = 0
Solution: x + 5 - 5 = -5
x = -5
Example 2: Solve 3x = 9
Solution: In this case variable is multiplied by 3, so we have to divide the equation by 3.
3x = 9
x = 9/3 = 3
Value of x is 3. Answer!

## Solving Two Steps Linear Equations

Use two-step method, if equation Involves more than one operation keeping BODMAS in mind.

Example: Solve 3x - 10 = 40
Solution: 3x - 10 = 41
Step 1: Add 10 both sides
3x - 10 + 10 = 41 + 10
3x = 51
Step 2: To isolate x, divide both the sides by 3
3x/3 = 51/3
x = 27
Value of x is 27.

## Simplifying before Solving Equation

If the equation has variable both the sides (RHS and LHS). Simplification process is required before we proceed to isolating variable. Using distributive property and BODMAS remove the brackets and combine the similar terms.

Example: Solve the equation 13 + 3x = 5 - 2(7x - 10).
Solution: 13 + 3x = 5 - 2(7x - 10)
13 + 3x = 5 - 2(7x) + 2(10)  (using distributive property)
13 + 3x = 5 - 14x + 20
13 + 3x = 25 - 14x
Subtract 13 from both the sides
13 + 3x - 13 = 25 - 14x - 13
3x = 12 - 14x
3x + 14x = 12 - 14x + 14x
17x = 12
Divide each side by 17
17x/17 = 12/17

## Reducing Fraction before Isolating the Variable

Fractions increase the difficultly level. Will resolve the fractions and convert the equation into simpler form by multiplying both sides by the LCD(lowest common denominator).

Example : Solve $\frac{3(x+2)}{5}$ = 3 + $\frac{7x}{15}$
Solution
$\frac{3(x+2)}{5}$ = 3 + $\frac{7x}{15}$
Multiply both the sides by 15 (LCM of 5 and 15 is 15)

15($\frac{3(x+2)}{5}$ = 3 + $\frac{7x}{15}$)
15 $\times$ $\frac{3(x+2)}{5}$ = 15 $\times$ 3 + 15 $\times$ $\frac{7x}{15}$
3(3(x+2) = 45 + 7x
9(x + 2) = 45 + 7x
9x + 18 = 45 + 7x
9x - 7x = 45 - 18 + 7x - 7x
2x = 27
x = 27/2

## Practice Problems

Practice below problems:
Practice Problem 1: Solve the linear equation x + 2 - 3(x + 1) = 7(x - 3)
Practice Problem 2: Solve for m, 12m - 10 = 34
Practice Problem 3: Solve $\frac{7(x-3)}{10}$ = 11 + $\frac{2x}{50}$
Practice Problem 4: Solve for p: 3(p - 2) + 10 = 0