Class 8 Math Linear Equations in One Variable Notes

Linear Equations in One Variable – Class 8

Hi everyone! This chapter is about Linear Equations in One Variable. These are like puzzles where we need to find a missing number.

What are Linear Equations?

A linear equation is an equation where the highest power of the variable (usually ‘x’) is 1. It’s like a straight line when you graph it (that’s why it’s called “linear”). It has an equals sign (=).

Examples: 2x + 3 = 7, 5y – 2 = 13, z/2 + 1 = 4

Non-Examples: x² + 2 = 5 (x has a power of 2), 1/x = 3 (x is in the denominator)

Solving Linear Equations

Solving a linear equation means finding the value of the variable that makes the equation true. We do this by isolating the variable on one side of the equals sign.

Steps to Solve:

Simplify both sides of the equation by combining like terms.
Use inverse operations to move terms to the correct side. Remember:
- To undo addition, subtract.
- To undo subtraction, add.
- To undo multiplication, divide.
- To undo division, multiply.
Check your answer by plugging it back into the original equation.

Solve: 3x – 5 = 10

Add 5 to both sides: 3x – 5 + 5 = 10 + 5 => 3x = 15
Divide both sides by 3: 3x / 3 = 15 / 3 => x = 5
Check: 3(5) – 5 = 15 – 5 = 10 (It works!)

Applications of Linear Equations

Linear equations are used to solve many real-world problems. Here are some examples:

1. Age Problems:

Example: Rohan is 5 years older than his brother. In 3 years, Rohan will be twice as old as his brother. How old are they now?

Solution: Let the brother’s current age be ‘x’. Rohan’s current age is ‘x + 5’. In 3 years, the brother will be ‘x + 3’ and Rohan will be ‘x + 5 + 3’ or ‘x + 8’. The equation is: x + 8 = 2(x + 3). Solving this gives x = 2 (brother’s age). Rohan’s age is 2 + 5 = 7.

2. Word Problems (Numbers):

Example: The sum of two numbers is 25. One number is 7 more than the other. Find the numbers.

Solution: Let one number be ‘x’. The other number is ‘x + 7’. The equation is: x + (x + 7) = 25. Solving this gives x = 9 (one number). The other number is 9 + 7 = 16.

3. Geometry Problems:

Example: The length of a rectangle is 3 times its width. If the perimeter is 48 cm, find the length and width.

Solution: Let the width be ‘w’. The length is ‘3w’. The perimeter is 2(length + width) = 2(3w + w) = 8w. The equation is: 8w = 48. Solving this gives w = 6 (width). The length is 3 * 6 = 18.

Practice solving lots of equations, and you’ll become a pro!

Linear Equations Quiz – Application Problems

1. Age Problem: A father is three times as old as his son. In 12 years, the father will be twice as old as his son. How old are they now?

Father: 36 years, Son: 12 years

Let the son’s current age be ‘x’. The father’s current age is ‘3x’. In 12 years, the son will be ‘x + 12’ and the father will be ‘3x + 12’. The equation is: 3x + 12 = 2(x + 12). Solving this gives x = 12 (son’s age). The father’s age is 3 * 12 = 36.

2. Number Problem: The sum of two consecutive even integers is 78. Find the integers.

38 and 40

Let the first even integer be ‘x’. The next consecutive even integer is ‘x + 2’. The equation is: x + (x + 2) = 78. Solving this gives x = 38. The other integer is 38 + 2 = 40.

3. Geometry Problem: The length of a rectangle is 5 cm more than its width. If the perimeter is 38 cm, find the length and width.

Length: 12 cm, Width: 7 cm

Let the width be ‘w’. The length is ‘w + 5’. The perimeter is 2(length + width) = 2(w + 5 + w) = 4w + 10. The equation is: 4w + 10 = 38. Solving this gives w = 7 (width). The length is 7 + 5 = 12.

4. Money Problem: A person has 20 notes consisting of Rs. 10 and Rs. 50 denominations. If the total amount of money is Rs. 580, how many notes of each type are there?

12 Rs. 10 notes and 8 Rs. 50 notes

Let ‘x’ be the number of Rs. 10 notes. Then the number of Rs. 50 notes is ’20 – x’. The equation is: 10x + 50(20 – x) = 580. Solving this gives x = 12. So, there are 12 Rs. 10 notes and 20 – 12 = 8 Rs. 50 notes.

5. Speed Problem: A car travels at a speed of 60 km/h for a certain time. If it had traveled at a speed of 70 km/h, it would have covered 50 km more in the same time. Find the distance traveled by the car.

300 km

Let ‘t’ be the time. Distance = speed * time. Distance at 60 km/h is 60t. Distance at 70 km/h is 70t. The equation is: 70t = 60t + 50. Solving this gives t = 5 hours. The distance is 60 * 5 = 300 km.

6. Mixture Problem: A shopkeeper mixes two types of rice costing Rs. 30 per kg and Rs. 50 per kg in the ratio 2:3. If the selling price of the mixed variety is Rs. 40 per kg, find the cost price of the mixed variety.

Rs. 42 per kg

Let the quantities of rice be 2x and 3x. Total cost = 30(2x) + 50(3x) = 60x + 150x = 210x. Total quantity = 2x + 3x = 5x. Cost price of mixture = Total cost / Total quantity = 210x / 5x = Rs. 42 per kg.

7. Work Problem: A can do a piece of work in 10 days and B can do it in 15 days. How many days will they take to complete the work together?

6 days

A’s work per day = 1/10. B’s work per day = 1/15. Combined work per day = (1/10) + (1/15) = 1/6. Number of days to complete the work together = 1 / (1/6) = 6 days.

8. Investment Problem: A person invests a sum of Rs. 8000 partly at 5% per annum and the remaining at 6% per annum simple interest. If the total interest earned after 3 years is Rs. 1260, find the amount invested at 6% per annum.

Rs. 6000

Let ‘x’ be the amount invested at 5%. The amount at 6% is ‘8000 – x’. Simple Interest = (Principal * Rate * Time) / 100. The equation is: (x * 5 * 3)/100 + ((8000 – x) * 6 * 3)/100 = 1260. Solving this gives x = 2000. Amount at 6% is 8000 – 2000 = Rs. 6000.

9. Shopping Problem: Rohan bought 3 notebooks and 2 pens for Rs. 80. Each notebook costs Rs. 10 more than a pen. What is the cost of one notebook?

Rs. 20

Let the cost of a pen be ‘p’. The cost of a notebook is ‘p + 10’. The equation is: 3(p + 10) + 2p = 80. Solving this gives p = 10 (cost of pen). Cost of notebook is 10 + 10 = Rs. 20.

10. Distance-Time Problem: A train travels a distance of 300 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 2 hours less for the same journey. Find the speed of the train.

50 km/hr

Let the speed be ‘s’. Time = Distance/Speed. Time taken at speed ‘s’ is 300/s. Time taken at speed ‘s + 5’ is 300/(s + 5). The equation is: 300/s – 300/(s + 5) = 2. Solving this gives s = 50 km/hr.