Solving Linear Equations in One Variable

 2 x = – 8

  x= – 4

The solution set consists of one number:$\{-4\}$. It is the only solution and, therefore, we have solved a conditional equation.

An inconsistent equation results in a false statement. For example, if we are to solve 5 x−15 = 5(x−4),

We have the following:

5 x−15=5 x−20

5 x−15−5 x=5 x−20−5 x

 −15 ≠−20(Subtract 5 x from both sides. False statement)

Indeed, $-15$≠−20. There is no solution because this is an inconsistent equation.

Solving linear equations in one variable involves the fundamental properties of equality and basic algebraic operations. A brief review of those operations follows.

Note :

A linear equation in one variable can be written in the form  ax+b=0 where a and b are real numbers, $a\ne 0$.

How to : Given a linear equation in one variable use algebra to solve it.

The following steps are used to manipulate an equation and isolate the unknown variable, so that the last line reads x=_________, if x is the unknown. There is no set order, as the steps used depend on what is given:

1. We may add, subtract, multiply, or divide an equation by a number or an expression as long as we do the same thing to both sides of the equal sign. Note that we cannot divide by zero.
2. Apply the distributive property as needed: 
   $a(b+c)=ab+ac .$Isolate the variable on one side of the equation.
3. When the variable is multiplied by a coefficient in the final stage, multiply both sides of the equation by the reciprocal of the coefficient.

Solving Equation in one variable

Problem 1:  Solve the following equation:

$2 x+7=19$

Solution

This equation can be written in the form

$ax+b=0,$

by subtracting $19$ from both sides. However, we may proceed to solve the equation in its original form by performing algebraic operations.

2 x+7=19

2 x=12 Subtract 7 from both sides. x=6

Multiply both sides by 12 or divide by 2.

The solution is

$x= 6$

Problem 2: The sum of two numbers is 25. One of the numbers exceeds the other by 9. Find the numbers.

Solution:

Let the number be x, Then the other number = x + 9

Sum of two numbers = 25

According to question, x + x + 9 = 25

⇒ 2 x + 9 = 25

⇒ 2 x = 25 – 9 (transposing 9 to the R.H.S changes to -9)

⇒ 2 x = 16

⇒ 2 x/2 = 16/2 (divide by 2 on both the sides)

⇒ x = 8

Therefore, x + 9 = 8 + 9 = 17

Therefore, the two numbers are 8 and 17.

Problem 3: The difference between the two numbers is 48. The ratio of the two numbers is 7:3. What are the two numbers?

Solution:

Let the common ratio be x.

Their difference = 48

According to the question,

7 x – 3 x = 48

⇒ 4 x = 48

⇒ x = 48/4

⇒ x = 12

Therefore, 7 x = 7 × 12 = 84

3 x = 3 × 12 = 36

Therefore, the two numbers are 84 and 36.

Problem 4: The length of a rectangle is twice its breadth. If the perimeter is 72 metre, find the length and breadth of the rectangle.

Solution: 

Let the breadth of the rectangle be x, Then the length of the rectangle = 2 x

Perimeter of the rectangle = 72

Therefore, according to the question

2(x + 2 x) = 72

⇒ 2 × 3 x = 72

⇒ 6 x = 72

⇒ x = 72/6

⇒ x = 12

We know, length of the rectangle = 2 x

= 2 × 12 = 24

Therefore, length of the rectangle is 24 m and breadth of the rectangle is 12 m.

Problem 5: The sum of two consecutive multiples of 5 is 55. Find these multiples.

Solution:

Let the first multiple of 5 be x.

Then the other multiple of 5 will be x + 5 and their sum = 55

Therefore, x + x + 5 = 55

⇒ 2 x + 5 = 55

⇒ 2 x = 55 – 5

⇒ 2 x = 50

⇒ x = 50/2

⇒ x = 25

Therefore, the multiples of 5, i.e., x + 5 = 25 + 5 = 30

Therefore, the two consecutive multiples of 5 whose sum is 55 are 25 and 30.