{"id":162629,"date":"2025-02-15T05:16:25","date_gmt":"2025-02-15T05:16:25","guid":{"rendered":"https:\/\/news.gyankatta.org\/?p=162629"},"modified":"2025-02-20T07:16:39","modified_gmt":"2025-02-20T07:16:39","slug":"class-8-math-linear-equations-in-one-variable","status":"publish","type":"post","link":"https:\/\/news.gyankatta.org\/?p=162629","title":{"rendered":"Class 8 Math Linear Equations in One Variable Notes"},"content":{"rendered":"

Linear Equations in One Variable – Class 8<\/h1>\n

Hi everyone! This chapter is about Linear Equations in One Variable<\/span>. These are like puzzles where we need to find a missing number.<\/p>\n

\"Linear<\/p>\n

What are Linear Equations?<\/h2>\n

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 (=).<\/p>\n

\n

Examples: 2x + 3 = 7, 5y – 2 = 13, z\/2 + 1 = 4<\/p>\n

Non-Examples: x\u00b2 + 2 = 5 (x has a power of 2), 1\/x = 3 (x is in the denominator)<\/p>\n<\/div>\n

Solving Linear Equations<\/h2>\n

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.<\/p>\n

Steps to Solve:<\/h3>\n
    \n
  1. Simplify both sides<\/span> of the equation by combining like terms.<\/li>\n
  2. Use inverse operations<\/span> to move terms to the correct side. Remember:\n
      \n
    • To undo addition, subtract.<\/li>\n
    • To undo subtraction, add.<\/li>\n
    • To undo multiplication, divide.<\/li>\n
    • To undo division, multiply.<\/li>\n<\/ul>\n<\/li>\n
    • Check your answer<\/span> by plugging it back into the original equation.<\/li>\n<\/ol>\n
      \n

      Solve: 3x – 5 = 10<\/p>\n

        \n
      1. Add 5 to both sides: 3x – 5 + 5 = 10 + 5 => 3x = 15<\/li>\n
      2. Divide both sides by 3: 3x \/ 3 = 15 \/ 3 => x = 5<\/li>\n
      3. Check: 3(5) – 5 = 15 – 5 = 10 (It works!)<\/li>\n<\/ol>\n<\/div>\n

        Applications of Linear Equations<\/h2>\n

        Linear equations are used to solve many real-world problems. Here are some examples:<\/p>\n

        \n

        1. Age Problems:<\/h3>\n

        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?<\/p>\n

        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.<\/p>\n<\/div>\n

        \n

        2. Word Problems (Numbers):<\/h3>\n

        Example: The sum of two numbers is 25. One number is 7 more than the other. Find the numbers.<\/p>\n

        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.<\/p>\n<\/div>\n

        \n

        3. Geometry Problems:<\/h3>\n

        Example: The length of a rectangle is 3 times its width. If the perimeter is 48 cm, find the length and width.<\/p>\n

        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.<\/p>\n<\/div>\n

        Practice solving lots of equations, and you’ll become a pro!<\/p>\n

        Linear Equations Quiz – Application Problems<\/h1>\n
        \n

        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?<\/strong><\/p>\n

        Father: 36 years, Son: 12 years<\/div>\n
        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.<\/p>\n<\/div>\n<\/div>\n
        \n

        2. Number Problem: The sum of two consecutive even integers is 78. Find the integers.<\/p>\n

        38 and 40<\/div>\n
        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.<\/p>\n<\/div>\n<\/div>\n
        \n

        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.<\/p>\n

        Length: 12 cm, Width: 7 cm<\/div>\n
        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.<\/p>\n<\/div>\n<\/div>\n
        \n

        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?<\/p>\n

        12 Rs. 10 notes and 8 Rs. 50 notes<\/div>\n
        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.<\/p>\n<\/div>\n<\/div>\n
        \n

        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.<\/p>\n

        300 km<\/div>\n
        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.<\/p>\n<\/div>\n<\/div>\n
        \n

        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.<\/p>\n

        Rs. 42 per kg<\/div>\n
        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.<\/p>\n<\/div>\n<\/div>\n
        \n

        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?<\/p>\n

        6 days<\/div>\n
        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.<\/p>\n<\/div>\n<\/div>\n
        \n

        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.<\/p>\n

        Rs. 6000<\/div>\n
        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.<\/p>\n<\/div>\n<\/div>\n
        \n

        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?<\/p>\n

        Rs. 20<\/div>\n
        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.<\/p>\n<\/div>\n<\/div>\n
        \n

        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.<\/p>\n

        50 km\/hr<\/div>\n
        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.<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

        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 […]<\/p>\n","protected":false},"author":1,"featured_media":162670,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"https:\/\/manishchandra.org\/p6\/LinearequationssolutionsEducationInfographic.png","fifu_image_alt":"","footnotes":""},"categories":[3,23],"tags":[],"class_list":["post-162629","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-education","category-math","cat-3-id","cat-23-id","has_thumb"],"yoast_head":"\nClass 8 Math Linear Equations in One Variable Notes - Gyankatta<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/news.gyankatta.org\/?p=162629\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Class 8 Math Linear Equations in One Variable Notes - Gyankatta\" \/>\n<meta property=\"og:description\" content=\"Linear Equations in One Variable – Class 8 Hi everyone! 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