Linear Equations

What is an equation?

In arithmetic, we usually come across statements of the following type –

  1. 16 + 6 = 22
  2. 7 x ( 4 + 6 ) = 7 x 4 + 7  x 6

Such a statement involving the symbol “=” is called a statement of equality or simply an equality.

None of the above statements involves a variable.

A statement of equality that involves one or more variables is called an equation.

Consider the following statements:

  1. A number x increased by 7 is 15.
  2. 9 exceeds a number x by 3
  3. 4 times a number x is 24
  4. A number y divided by 5 is 7
  5. The sum of a number x and twice the number y is 12

We can write the above statements as under:

  1. x + 7 = 15
  2. 9 – x = 3
  3. 4x = 24
  4. $\frac{y}{5}$ = 7
  5. x + 2y = 12

Each one of the above statements is a statement of equality, containing one or more variables. Thus each one of them is an equation.

What is a Linear equation?

An equation in which the highest power of the variables involved is 1 is called a linear equation. In other words, a linear equation is a mathematical equation that defines a line. While each linear equation corresponds to exactly one line, each line corresponds to infinitely many equations. These equations will have a variable whose highest power is 1.

The sign of equality divides an equation into two sides, namely the left-hand side and the right-hand side, written as L.H.S and R.H.S respectively.

Types of Linear Equations

A linear equation will have the powers of its variables as 1. There are three types of linear equations in general:

  1. The Point Slope Form
  2. Slope Intercept Form
  3. Standard Form

Let us understand these in detail.

The Point Slope Form

As the name implies, the point-slope form gives one point in a line and its slope. This form is not commonly given to help graph a line. It is, however, more commonly used to get from a verbal description or a graphical depiction of a line to slope-intercept or standard form.

If the given point is (x1, y1), the slope is m, the equation of the line in point-slope form is:

y – y1 = m (x – x1)

Slope Intercept Form

Slope-intercept form conveys the slope and y-intercept of a line. It is technically a special case of point-slope form.

If a line has slope m and y-intercept (0, b), the slope-intercept form is:

Y = mx + b

If this point was written in point-slope form, we would have:

y – b = m ( x – 0 )

Simplifying it we get

y = mx – 0 + b

y = mx + b

Standard Form

The standard form of an equation is:

Ax + By = C

Where A, B, and C are all whole numbers, and A is not negative.

Linear Equations Based on the Number of Variables

We have different linear equations based on the number of variables they have. For example,

Equation with one Variable

 An equation may have only one variable such as:

12x – 10 = 0

18x = 12

Equation with two Variables

An equation may have two variables, such as:

12x +10y – 10 = 0

12x +23y = 20

Equation with three Variables

An equation may have three variables such as:

12x +10y -3z – 10 = 0

12x +23y – 12z = 20

Similarly, there can be equations that have any number of variables.  There are different methods to solve linear equations depending on the number of variables they have. Now let us see how to solve linear equations in one variable.

Solution of a Linear Equation

A number that makes L.H.S = R.H.S when it is substituted for the variable in an equation is said to satisfy the equation and is called the solution or root of an equation.

Solving Linear Equations

Before learning the methods of solving the equations, it is important to know that there are certain rules to solve equations. Let us check what these rules are.

Rules for Solving Linear Equations

The following are the rules for solving linear equations:

  1. We can add the same number to both sides of the equation. This means that if x + 5 = 7, then x + 5 + 2 = 7 + 2
  2. We can subtract the same number from both sides of the equation. This means that if x + 5 = 7 then x + 5 – 2 = 7 – 2
  3. We can multiply both sides of the equation by the same non-zero number. This means that if x  + 5 = 4, then 6 ( x + 5 ) = 6 x 4
  4. We can divide both sides of the equation by the same non-zero number. This means that if 3x = 10, then $\frac{3x}{3}=\frac{10}{3}$

Let us now learn what the methods to solve linear equations are. 

There are two methods of solving linear equations:

  • By trial and error method
  • Transposition method

Trial and Error method

In this method, we often make a guess of the root of the equation. We find the values of the left-hand side, L.H.S and the right-hand side, R.H.S of the given equation for different values of the variable. The value of the variable for which L.H.S = R.H.S is the root of the equation.

Let us understand the method through some examples.

Example

Solve the following equations by the trial and error method:

  1. x + 7 = 10
  2. 5x = 30

Solution

As defined in the trial and error method, we will check the values of the L.H.S and the R.H.S of the equation for some values of x and continue to do so until the L.H.S becomes equal to the R.H.S

1. The given equation is x + 7 = 10

We have,

L.H.S  = x + 7 and R.H.S =10

We will substitute different values of x in the L.H.S until we get the result as 10 which is our R.H.S.

The following tables shows our assumptions of trial and error:

Value of “x”L.H.SR.H.SIs L.H.S = R.H.S
11 + 7 = 810No
22 + 7 = 910No
33 + 7 = 1010Yes

From the above table, we have that when x is substituted by 3, we get L.H.S = R.H.S

Hence, x = 3 is the solution for the equation, x + 7 = 10

2. The given equation is 5x = 30

We have L.H.S = 5x and R.H.S = 30

We will substitute different values of x in the L.H.S until we get the result as 30 which is our R.H.S.

Value of “x”L.H.SR.H.SIs L.H.S = R.H.S
15 x 1 = 530No
25 x 2 = 1030No
35 x 3 = 1530No
45 x 4 = 2030No
55 x 5 = 2530No
65 x 6 = 3030Yes

From the above table, we have that when x is substituted by 6, we get L.H.S = R.H.S

Hence, x = 6 is the solution for the equation, 5x = 30

Transposition Method

By transposing a term of an equation with simply means changing its fine and carrying it to the other side of the equation any term of an equation may be taken to the other side with its sign change without affecting the equality this process is called transposition when we carry a term of an equation from L.H.S to R.H.S and R.H.S to L H.S the plus sign of the term changes into the minus sign on the other side and vice versa

The transposition method involves the following steps:

  1. Obtain the linear equation
  2. Identify the unknown quantity (variable)
  3. Simplify the L.H.S and R.H.S by removing grouping symbols (if any)
  4. Transfer all terms containing the variable on the L.H.S and constant terms on the R.H.S of the equation. Note that the size of the terms will change in carrying them from L.H.S to R.H.S and vice versa
  5. Simplify L.H.S and R.H.S in the simplest form so that each side contains just one term
  6. Solve the equation thus obtained according to the rules.

Solved Examples

Example 1   

Solve the following equation:  2x + 7 = 19

Solution:

We have been given that:

2x + 7 = 19

⇒ 2x = 19 – 7

⇒ 2x = 12

⇒ x = 6

Hence x = 6 is the solution of the equation 2x + 7 = 19

Example 2    

Solve the equation x -3 = 5 and check the result

Solution:

We have x -3 = 5

To solve this equation we have to get checked by itself on the left-hand side (L.H.S). To get x by itself on the L.H.S we need to shift -3. This can be done by adding three to both sides of the equation.

We have,

x– 3 = 5

⇒ x – 3 + 3 = 5 + 3

⇒ x + 0 = 8

⇒ x = 8

Hence, x = 8 is the solution to the equation x – 3 = 5

Now let us check the result.

Substituting x = 8 in the given equation, we have

L.H.S = x – 3 = 8 – 3 = 5

We have R.H.S = 5

Since L.H.S = R.H.S, we can say that our answer is correct.

Example 3    

Solve 0.3x + 0.4 = 0.28x + 1.16

Solution:

We have:

0.3x + 0.4 = 0.28x + 1.16

⇒ 0.3x – 0.28x = 1.16 – 0.4

⇒ ( 0.3 – 0.28 )x + 1.16 – 0.4

⇒ 0.02x = 0.76

⇒ x = $\frac{0.76}{0.02}$

⇒ x = 38

Hence, x = 38 is the solution to the problem 0.3x + 0.4 = 0.28x + 1.16

Applications of Linear equations in Real Life

Why is there a need to learn linear equations? Just like addition and subtraction do we use them in our everyday lives? Let us find out.

Here are some of the areas where we use linear equations in real life:

  1. Linear equations are an important tool in science and many everyday applications.
  2. They allow scientists to describe relationships between two variables in the physical world, make predictions, calculate rates, and make conversions, among other things.
  3. Graphing linear equations help make trends visible.
  4. Gaining a strong understanding of linear equations both helps in scientific problem solving and lays a foundation for exploring other, more mathematically complex relationships in science.

Solving Real-Life Problems using Linear Equations

We shall now learn about the formulation and solution of some practical problems. These problems involve relations among unknown quantities which we call variables and known quantities which are known as numbers and are often stated in words. That is why we often refer to this problem that word problem. A word problem is first translated in the form of an equation containing unknown quantities, i.e. variable and known quantities i.e. numbers or constants and then we solve it by using any of the methods we have learnt. The procedure to translate a word problem in the form of an equation is known as the formulation of the problem. Thus the process of solving a word problem consists of two parts namely, formulation and solution

Following steps should be followed to solve a word problem:

  1. Read the problem carefully and note what is given and what is required
  2. Denote the unknown quantity by some letters, say x, y, z, etc.
  3. Translate the statements of the problem into a mathematical statement
  4. Using the condition given in the problem, form the equation
  5. Solve the equation for the unknown
  6. Check whether the solution satisfies the equation

Let us now understand the above steps through some examples.

Example 1    

The sum of two consecutive numbers is 53, find the numbers.

Solution:

We have been given that:

The sum of two consecutive numbers is 53. We need to find the two numbers. So, going by the steps defined above, we first give a name to the unknown value which in this case is the numbers. Therefore,

Let the first number be x.

Then the next consecutive number will be x + 1

It is given that the sum of two consecutive numbers is 53, therefore, we can form an equation in the following manner:

x + ( x + 1 ) = 53

Now that we have the equation, we will solve it for x.

x + ( x + 1 ) = 53

⇒ 2x + 1 = 53

⇒ 2x = 53 – 1

⇒ 2x = 52

⇒ x= 26

Hence the first number is 26. Now since the next number is a consecutive number, the second number will be x + 1 = 27.

Therefore, the two numbers are 26 and 27

Example 2    

Sara’s mother is three times as old as Sara and four times as old as Sara’s sister; Ann. Ann is three years younger than Sara. How old are Sara, Ann and their mother?

Solution:

We have been given that:

Sara’s mother is three times as old as Sara and four times as old as Sara’s sister; Ann. Ann is three years younger than Sara.

So, going by the steps defined above, we first give a name to the unknown value which in this case is the ages of Sara and Ann. Therefore,

Let Sara’s age be x years

It is given that Ann is three years younger than Sara.

Therefore,

Ann’s age = (x – 3) years

Also, it is given that Sara’s mother is three times as old as Sara

Therefore,

Sara’s mother’s age = 3x years………………………….. (1)

Next, Sara’s mother is four times as old as Ann

Therefore,

Sara’s mother’s age = 4 ( x  – 3 ) years ……………………… (2)

From (1) and (2) we have

3x = 4 ( x  – 3 )

⇒ 3x = 4x – 12

⇒ 3x – 4x = -12

⇒ -x = -12

⇒ x = 12

Hence, Sara’s age is 12 years.

Now

Ann’s age = ( x – 3 ) years

Hence, Ann’s age = 12 – 3 = 9 years

Also,

Sara’s mother’s age = 3x years

Therefore,

Sara’s mother’s age = = 3 x 12 = 36 years.

Hence, Sara’s age = 12 years, Ann’s age = 9 years and Sara’s mother’s age = 36 years

Example 3    

Find two numbers such that one of them exceeds the other by 9 and their sum is 81.

Solution:

We have been given that,

There are two numbers such that one of them exceeds the other by 9 and their sum is 81

So, going by the steps defined above, we first give a name to the unknown value which in the case is the numbers.

Let one number be x.

Then the other number will be x + 9

Also, the sum of the two numbers is 81.

Therefore,

x + ( x + 9 ) = 81

⇒ x + x + 9 = 81

⇒ 2x + 9 = 81

⇒ 2x = 81 – 9

⇒ 2x = 72

⇒ x = 36

Hence the first number is 36

The second number will be x + 9 = 36 + 9 = 45

Hence, the two numbers are 36 and 45

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