Different Types of Problems in Linear Equation in One Variable

In previous topics we have learnt a lot a about linear equations in one variable. Under this topic we will learn about different types of questions that we come across in linear equations having one variable.

Mostly there are two types of questions that we come across in this topic, one is solving simple linear equation and the other one is solving word problems using linear equations in one variable. Within these two types only, there are multiple types of problems but there is unique process of step of solving them, i.e., bring all unknown variables on the left hand side and all constants on the right hand side of the equation by using simple addition, subtraction, multiplication and division and then solve the so formed equation using suitable algebraic operation.

Now for having a better understanding of the concept let us solve some problems based of the concept.

Type 1: Variable on one side:

1) Solve 2x + 4 = 17.

2) Solve 3x – 9 =20.

3) Solve 4x - 5 = 15.

4) Solve 6x + 12 = 54.

Solution:

1) 2x + 4 =17.

Separating variables on right hand side and constants on the left hand side:

         2x = 17 – 4

         2x = 13

          x = 13/2.

2) 3x – 9 = 20.

      3x = 20 – 9

      3x = 11

       x = 11/3.

3) 4x – 5 =15.

     4x = 15 + 5

     4x = 20

      x = 20/4 = 5

      x = 5.

4) 6x + 12 = 54

6x = 54 – 12

6x = 48

x = 42/6

x = 7.

Type 2:  When there are variables present on both sides of the equation:

In this case also, variables are taken on the left hand side of the equation and constants on the right hand side of the equation by using simple mathematical operations. The formed equation is then solved.

1) Solve 2x + 10 = 3x – 20.

2) Solve 3x – 12 = 4x + 15.

3) Solve 3x – 2 = 4x +8.

Solutions:

1) 2x + 10 = 3x – 20.

     2x – 3x = 20 – 10

             -x = 10.

Multiply both sides of the equation by negative sign.

                    x = -10.

2) 3x – 12 = 4x + 15.

     3x – 4x = 15 + 12

       -x = 27

Multiply both sides of the equation by negative sign.

x = -27.

3. 3x – 2 = 4x + 8.

    3x – 4x = 8 + 2

    -x = 10

Multiplying both sides of the equation by negative sign.

      x = -10.

Type 3: When equation given is in the form of fractions.

In such cases where equations given are in form of fraction, take the L.C.M. of the fraction on both sides of the equation and then cross multiply the denominator of both L.H.S. and R.H.S. and then the solve the equation formed after cross multiplying the denominators.

Examples: 

1) Solve \(\frac{x}{2}\) + \(\frac{x}{4}\) = \(\frac{3}{8}\)

2) Solve \(\frac{5x}{6}\) - \(\frac{2x}{3}\) = \(\frac{2}{9}\)

Solution:

1) Solve \(\frac{x}{2}\) + \(\frac{x}{4}\) = \(\frac{3}{8}\)

\(\frac{x}{2}\) + \(\frac{x}{4}\) = \(\frac{3}{8}\)

\(\frac{2x+x}{4}\) = \(\frac{3}{8}\)

\(\frac{3x}{4}\) = \(\frac{3}{8}\)

(3x) x 8 = 3 x 4

24x = 12

x = 12/24

x = 1/2.

2) Solve \(\frac{5x}{6}\) - \(\frac{2x}{3}\) = \(\frac{2}{9}\)

\(\frac{5x}{6}\) - \(\frac{2x}{3}\) = \(\frac{2}{9}\)

\(\frac{5x-4x}{6}\) = \(\frac{2}{9}\)

\(\frac{x}{6}\) = \(\frac{2}{9}\)

On cross multiplication:

9x = 12

x = 12/9

x = 4/3.

These were some basic types of problems that might come under solving simple linear equations.

Now let us move on the problems based on word problems in linear equation in one variable:

Word problems come in form of simple English language form rather than coming in mathematical form. So first of all, we need to understand the English language form and then we need to convert that into mathematical language in linear equation form and then solve the equation to get the value of the variable. Now there are innumerable number of problems on the word problems based on the linear equation in one variable. We cannot study them separately but there are some common steps which are involved in all the word problems related to the linear equation in one variable.

Steps involved in solving word problems based on linear equation in one variable are as follows:

Step 1: First of all read the given problem carefully and note down the given and required quantities separately.

Step 2: Denote the unknown quantities as ‘x’, ‘y’, ‘z’, etc. 

Step 3: Then translate the problem into mathematical language or statement.

Step 4: Form the linear equation in one variable using the given conditions in the problem.

Sep 5: solve the equation for the unknown quantity.

Now let us solve some word problems on linear equation in one variable.

1) Sum of two numbers is 50. If one number is 4 times the other, find the numbers.

Solution:

Let one of the numbers be ‘x’. then second number is 4x.

Then, x + 4x = 50

             5x = 50

              x = 50/5

              x = 10.

So 1st number = 10.

      2nd number = 40.

2) Rajeev is 5 times older than his son. After 2 years sum of ages will be 40. Calculate their present ages.

Solution:

Let Rajeev’s present age be 5x years.

His son’s present age = x years.

After 2 years:

Rajeev’s age = 5x + 2 years.

His son’s age = x + 2 years.

Now, 5x + 2 + x + 2 = 40.

         6x + 4 = 40

          6x = 40 – 4

           6x = 36.

           x = 36/6

             x = 6.

Hence Rajeev’s age = 5x = 5 × 6 = 30 years.

His son’s age = x = 6 years.

3) A bag contains some number of white balls, twice number of white balls are blue balls, thrice the number of blue balls are the red balls. If the total number of balls in the bag are 27. Calculate the number of balls of each color present in the bag.

Solution:

Let number of white balls be ‘x’.

Number of blue balls = 2x.

Number of red balls = 3 × (2x)

Total number of balls = 27.

So, x + 2x + 3 × (2x) = 27

 x + 2x + 6x = 27

9x = 27

x = 27/9

x = 3.

So, number of white balls = x = 3.

Number of blue balls = 2x = 2 × 3 = 6.

Number of red balls = 3 × (2x) = 3 × 6 = 18.

All other word problems can be solved by following the above mentioned steps.

9th Grade Math

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