Problems on Linear Equations in One Variable

Solved algebra problems on linear equations in one variable are explained below with the detailed explanation.

Let’s once again recall the methods of solving linear equations in one variable. 

 Read the linear problem carefully and note what is given in the question and what is required to find out. 

 Denote the unknown by any variable as x, y, ……. (any variable) 

 Translate the problem to the language of mathematics or mathematical statements. 

 Form the linear equation in one variable using the conditions given in the problems. 

 Solve the equation for the unknown. 

 Verify to be sure whether the answer satisfies the conditions of the problem. 

Worked-out problems on linear equations in one variable:

1. The sum of three consecutive multiples of 4 is 444. Find these multiples. 

Solution: 

If x is a multiple of 4, the next multiple is x + 4, next to this is x + 8. 

Their sum = 444

According to the question, 

x + (x + 4) + (x + 8) = 444 

⇒ x + x + 4 + x + 8 = 444

⇒ x + x + x + 4 + 8 = 444 

⇒ 3x + 12 = 444

⇒ 3x = 444 - 12 

⇒ x = 432/3 

⇒ x = 144

Therefore, x + 4 = 144 + 4 = 148 

Therefore, x + 8 - 144 + 8 – 152

Therefore, the three consecutive multiples of 4 are 144, 148, 152.



2. The denominator of a rational number is greater than its numerator by 3. If the numerator is increased by 7 and the denominator is decreased by 1, the new number becomes 3/2. Find the original number.

Solution:

Let the numerator of a rational number = x

Then the denominator of a rational number = x + 3

When numerator is increased by 7, then new numerator = x + 7

When denominator is decreased by 1, then new denominator = x + 3 - 1

The new number formed = 3/2

According to the question,

(x + 7)/(x + 3 - 1) = 3/2

⇒ (x + 7)/(x + 2) = 3/2

⇒ 2(x + 7) = 3(x + 2)

⇒ 2x + 14 = 3x + 6

⇒ 3x - 2x = 14 - 6

⇒ x = 8

The original number i.e., x/(x + 3) = 8/(8 + 3) = 8/11




3. The sum of the digits of a two digit number is 7. If the number formed by reversing the digits is less than the original number by 27, find the original number.

Solution:

Let the units digit of the original number be x.

Then the tens digit of the original number be 7 - x

Then the number formed = 10(7 - x) + x × 1

                                     = 70 - 10x + x = 70 - 9x

On reversing the digits, the number formed

                  = 10 × x + (7 - x) × 1

                  = 10x + 7 - x = 9x + 7

According to the question,

New number = original number - 27

⇒ 9x + 7 = 70 - 9x - 27

⇒ 9x + 7 = 43 - 9x 

⇒ 9x + 9x = 43 – 7

⇒ 18x = 36 

⇒ x = 36/18 

⇒ x = 2 

Therefore, 7 - x

              = 7 - 2

              = 5

The original number is 52


4. A motorboat goes downstream in river and covers a distance between two coastal towns in 5 hours. It covers this distance upstream in 6 hours. If the speed of the stream is 3 km/hr, find the speed of the boat in still water.

Solution:

Let the speed of the boat in still water = x km/hr.

Speed of the boat downstream = (x + 3) km/hr.

Time taken to cover the distance = 5 hrs

Therefore, distance covered in 5 hrs = (x + 3) × 5   (D = Speed × Time)

Speed of the boat upstream = (x - 3) km/hr

Time taken to cover the distance = 6 hrs.

Therefore, distance covered in 6 hrs = 6(x - 3)

Therefore, the distance between two coastal towns is fixed, i.e., same.

According to the question,

5(x + 3) = 6(x - 3)

⇒ 5x + 15 = 6x - 18

⇒ 5x - 6x = -18 – 15

⇒ -x = -33

⇒ x = 33

Required speed of the boat is 33 km/hr.



5. Divide 28 into two parts in such a way that 6/5 of one part is equal to 2/3 of the other.

Solution:

Let one part be x.

Then other part = 28 - x

It is given 6/5 of one part = 2/3 of the other.

⇒ 6/5x = 2/3(28 - x)

⇒ 3x/5 = 1/3(28 - x)

⇒ 9x = 5(28 - x)

⇒ 9x = 140 - 5x

⇒ 9x + 5x = 140

⇒ 14x = 140

⇒ x = 140/14

⇒ x = 10

Then the two parts are 10 and 28 - 10 = 18.




6. A total of $10000 is distributed among 150 persons as gift. A gift is either of $50 or $100. Find the number of gifts of each type.

Solution:

Total number of gifts = 150

Let the number of $50 is x

Then the number of gifts of $100 is (150 - x)

Amount spent on x gifts of $50 = $ 50x

Amount spent on (150 - x) gifts of $100 = $100(150 - x)

Total amount spent for prizes = $10000

According to the question,

50x + 100 (150 - x) = 10000

⇒ 50x + 15000 - 100x = 10000

⇒ -50x = 10000 - 15000

⇒ -50x = -5000

⇒ x = 5000/50

⇒ x = 100

⇒ 150 - x = 150 - 100 = 50

Therefore, gifts of $50 are 100 and gifts of $100 are 50.


The above step-by-step examples demonstrate the solved problems on linear equations in one variable.


 Equations

What is an Equation?

What is a Linear Equation?

How to Solve Linear Equations?

Solving Linear Equations

Problems on Linear Equations in One Variable

Word Problems on Linear Equations in One Variable

Practice Test on Linear Equations

Practice Test on Word Problems on Linear Equations


 Equations - Worksheets

Worksheet on Linear Equations

Worksheet on Word Problems on Linear Equation








7th Grade Math Problems 

8th Grade Math Practice 

From Problems on Linear Equations in One Variable to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.




Share this page: What’s this?

Recent Articles

  1. 5th Grade Circle Worksheet | Free Worksheet with Answer |Practice Math

    Jul 11, 25 02:14 PM

    Radii of the circRadii, Chords, Diameters, Semi-circles
    In 5th Grade Circle Worksheet you will get different types of questions on parts of a circle, relation between radius and diameter, interior of a circle, exterior of a circle and construction of circl…

    Read More

  2. Construction of a Circle | Working Rules | Step-by-step Explanation |

    Jul 09, 25 01:29 AM

    Parts of a Circle
    Construction of a Circle when the length of its Radius is given. Working Rules | Step I: Open the compass such that its pointer be put on initial point (i.e. O) of ruler / scale and the pencil-end be…

    Read More

  3. Combination of Addition and Subtraction | Mixed Addition & Subtraction

    Jul 08, 25 02:32 PM

    Add and Sub
    We will discuss here about the combination of addition and subtraction. The rules which can be used to solve the sums involving addition (+) and subtraction (-) together are: I: First add

    Read More

  4. Addition & Subtraction Together |Combination of addition & subtraction

    Jul 08, 25 02:23 PM

    Addition and Subtraction Together Problem
    We will solve the different types of problems involving addition and subtraction together. To show the problem involving both addition and subtraction, we first group all the numbers with ‘+’ and…

    Read More

  5. 5th Grade Circle | Radius, Interior and Exterior of a Circle|Worksheet

    Jul 08, 25 09:55 AM

    Semi-circular Region
    A circle is the set of all those point in a plane whose distance from a fixed point remains constant. The fixed point is called the centre of the circle and the constant distance is known

    Read More