{"id":9099,"date":"2026-06-01T21:33:48","date_gmt":"2026-06-01T21:33:48","guid":{"rendered":"https:\/\/kapdec.com\/help\/?p=9099"},"modified":"2026-06-01T21:33:48","modified_gmt":"2026-06-01T21:33:48","slug":"solving-equations-variable-on-one-sides","status":"publish","type":"post","link":"https:\/\/kapdec.com\/help\/solving-equations-variable-on-one-sides\/","title":{"rendered":"Solving Equations, Variable On One Sides"},"content":{"rendered":"

Unit: <\/strong>Algebra – 1<\/strong><\/h2>\n

Chapter: <\/strong>Solving Equations, Variable on One Side<\/strong><\/h3>\n

Reference: – Introduction to Linear Equations, what is a Variable, what is an Equation, Solving Equations with Variable on One Side, Balancing Method, Transposition Method, Verification of Solution, Equations with Fractions, Equations with Decimals, Word Problems, Solved Examples, Odd-One-Out Problems, Common Mistakes, Practice Grid<\/em><\/p>\n

After studying this chapter, you should be able to understand:<\/strong><\/p>\n

    \n
  • Introduction to Linear Equation in One Variable<\/em><\/li>\n
  • Solving Equation on Variable on One Side<\/em><\/li>\n
  • Balancing Method & Transposing Method<\/em><\/li>\n
  • Solving Equations with Fractions & Decimals<\/em><\/li>\n<\/ul>\n

    Introduction to Linear Equations<\/strong><\/p>\n

    Definition<\/u><\/strong><\/p>\n

    A linear equation is an equation in which the highest power of the variable is 1. It is called "linear" because its graph is a straight line.<\/p>\n

    A linear equation in one variable has the general form:
    \nax + b = c (or ax + b = 0)<\/p>\n

    where a, b, c are constants (real numbers) and a ≠ 0.<\/p>\n

    When we solve a linear equation, we essentially ask:<\/p>\n

    "What value of the variable makes this equation true?"<\/p>\n

    Once we find that value (called the solution or root), we can verify it by substituting back into the original equation.<\/p>\n

    Importance of Solving Linear Equations<\/u><\/strong><\/p>\n

      \n
    • Foundation for all higher algebra (quadratic, polynomial, calculus)<\/li>\n
    • Used extensively in physics, chemistry, economics, and engineering<\/li>\n
    • Essential for solving real-world problems (budgeting, distance, mixtures)<\/li>\n
    • Develops logical thinking and step-by-step reasoning<\/li>\n
    • Appears in competitive exams and daily calculations<\/li>\n<\/ul>\n

      Example<\/strong><\/p>\n

      Equation:<\/strong> x + 5 = 12
      \nSolution:<\/strong> x = 7
      \nVerification:<\/strong> 7 + 5 = 12<\/p>\n

      So, if we had an equation like x² = 16, it is NOT linear (power is 2).<\/p>\n

      Subtopics<\/u><\/strong><\/p>\n

      1. Concept of an Equation<\/strong><\/p>\n

      An equation is a mathematical statement that two expressions are equal. It contains:<\/p>\n

        \n
      • Left-hand side (LHS)<\/strong><\/li>\n
      • Right-hand side (RHS)<\/strong><\/li>\n
      • Equals sign (=)<\/strong><\/li>\n<\/ul>\n

        Key Points:<\/strong><\/p>\n

          \n
        • An equation is like a balanced scale – whatever you do to one side, you must do to the other.<\/li>\n
        • The goal of solving is to isolate the variable on one side.<\/li>\n
        • The solution makes LHS = RHS when substituted.<\/li>\n<\/ul>\n

          2. Variable on One Side vs Both Sides<\/strong><\/p>\n\n\n\n\n\n\n
          \n

          Type<\/p>\n<\/td>\n

          		\n

          Example<\/p>\n<\/td>\n

          		\n

          Approach<\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

          \n

          Variable on one side<\/strong><\/p>\n<\/td>\n

          		\n

          2x + 3 = 11<\/p>\n<\/td>\n

          		\n

          Isolate variable using inverse operations<\/p>\n<\/td>\n<\/tr>\n

          \n

          Variable on both sides<\/strong><\/p>\n<\/td>\n

          		\n

          3x + 2 = x + 10<\/p>\n<\/td>\n

          		\n

          Collect variables on one side first<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

          This chapter focuses on Variable on One Side<\/strong> equations.<\/p>\n

           <\/p>\n

          Solving Equations with Variable on One Side<\/strong><\/p>\n

          Definition<\/u><\/strong><\/p>\n

          When the variable appears on only one side of the equation (usually the left), we can solve by performing inverse operations to isolate the variable.<\/p>\n

          The four basic inverse operations are:<\/p>\n\n\n\n\n\n\n\n\n
          \n

          Operation<\/p>\n<\/td>\n

          		\n

          Inverse Operation<\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

          \n

          Addition (+)<\/p>\n<\/td>\n

          		\n

          Subtraction (-)<\/p>\n<\/td>\n<\/tr>\n

          \n

          Subtraction (-)<\/p>\n<\/td>\n

          		\n

          Addition (+)<\/p>\n<\/td>\n<\/tr>\n

          \n

          Multiplication (×)<\/p>\n<\/td>\n

          		\n

          Division (÷)<\/p>\n<\/td>\n<\/tr>\n

          \n

          Division (÷)<\/p>\n<\/td>\n

          		\n

          Multiplication (×)<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

          General Strategy (Variable on Left Side):<\/strong><\/p>\n

          ax + b = c<\/p>\n

          Step 1: Subtract b from both sides → ax = c – b<\/p>\n

          Step 2: Divide both sides by a → x = (c – b)\/a<\/p>\n

          Example:<\/strong> 2x + 3 = 11<\/p>\n

            \n
          • Step 1: 2x + 3 – 3 = 11 – 3 → 2x = 8<\/li>\n
          • Step 2: 2x ÷ 2 = 8 ÷ 2 → x = 4<\/li>\n<\/ul>\n

             <\/p>\n

            Method 1: Balancing Method<\/strong><\/p>\n

            Definition<\/u><\/strong><\/p>\n

            The balancing method involves performing the same operation on both sides of the equation to maintain equality. Think of it as a balanced scale – if you add weight to one side, you must add the same weight to the other.<\/p>\n

            Rules of Balancing Method:<\/strong><\/p>\n

              \n
            1. Add or subtract the same number from both sides<\/li>\n
            2. Multiply or divide both sides by the same non-zero number<\/li>\n
            3. The equation remains balanced after any of these operations<\/li>\n<\/ol>\n

              Example 1:<\/strong> Solve x – 7 = 3<\/p>\n\n\n\n\n\n\n\n
              \n

              Step<\/p>\n<\/td>\n

              		\n

              Operation<\/p>\n<\/td>\n

              		\n

              Equation<\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

              \n

              Given<\/p>\n<\/td>\n

              		 <\/td>\n\n

              x – 7 = 3<\/p>\n<\/td>\n<\/tr>\n

              \n

              Add 7 to both sides<\/p>\n<\/td>\n

              		\n

              +7<\/p>\n<\/td>\n

              		\n

              x – 7 + 7 = 3 + 7<\/p>\n<\/td>\n<\/tr>\n

              \n

              Simplify<\/p>\n<\/td>\n

              		 <\/td>\n\n

              x = 10<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

              Verification:<\/strong> 10 – 7 = 3 \u2713<\/p>\n

              Example 2:<\/strong> Solve 5x = 20<\/p>\n\n\n\n\n\n\n\n
              \n

              Step<\/p>\n<\/td>\n

              		\n

              Operation<\/p>\n<\/td>\n

              		\n

              Equation<\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

              \n

              Given<\/p>\n<\/td>\n

              		 <\/td>\n\n

              5x = 20<\/p>\n<\/td>\n<\/tr>\n

              \n

              Divide both sides by 5<\/p>\n<\/td>\n

              		\n

              ÷5<\/p>\n<\/td>\n

              		\n

              5x ÷ 5 = 20 ÷ 5<\/p>\n<\/td>\n<\/tr>\n

              \n

              Simplify<\/p>\n<\/td>\n

              		 <\/td>\n\n

              x = 4<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

              Verification:<\/strong> 5 × 4 = 20 \u2713<\/p>\n

              Example 3:<\/strong> Solve x\/3 = 7<\/p>\n\n\n\n\n\n\n\n
              \n

              Step<\/p>\n<\/td>\n

              		\n

              Operation<\/p>\n<\/td>\n

              		\n

              Equation<\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

              \n

              Given<\/p>\n<\/td>\n

              		 <\/td>\n\n

              x\/3 = 7<\/p>\n<\/td>\n<\/tr>\n

              \n

              Multiply both sides by 3<\/p>\n<\/td>\n

              		\n

              ×3<\/p>\n<\/td>\n

              		\n

              (x\/3) × 3 = 7 × 3<\/p>\n<\/td>\n<\/tr>\n

              \n

              Simplify<\/p>\n<\/td>\n

              		 <\/td>\n\n

              x = 21<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

              Verification:<\/strong> 21 ÷ 3 = 7 \u2713<\/p>\n

              Method 2: Transposition Method<\/strong><\/p>\n

              Definition<\/u><\/strong><\/p>\n

              Transposition is a shortcut method where we move a term from one side of the equation to the other by changing its sign. This is faster than writing the operation on both sides.<\/p>\n

              Rules of Transposition (Sign Change Rule):<\/strong><\/p>\n\n\n\n\n\n\n\n\n
              \n

              Moving<\/p>\n<\/td>\n

              		\n

              Sign Change<\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

              \n

              +a (addition)<\/p>\n<\/td>\n

              		\n

              becomes -a (subtraction)<\/p>\n<\/td>\n<\/tr>\n

              \n

              -a (subtraction)<\/p>\n<\/td>\n

              		\n

              becomes +a (addition)<\/p>\n<\/td>\n<\/tr>\n

              \n

              ×a (multiplication)<\/p>\n<\/td>\n

              		\n

              becomes ÷a (division)<\/p>\n<\/td>\n<\/tr>\n

              \n

              ÷a (division)<\/p>\n<\/td>\n

              		\n

              becomes ×a (multiplication)<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

              Example 1:<\/strong> Solve x + 8 = 15<\/p>\n\n\n\n\n\n\n\n
              \n

              Step<\/p>\n<\/td>\n

              		\n

              Operation<\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

              \n

              Given<\/p>\n<\/td>\n

              		\n

              x + 8 = 15<\/p>\n<\/td>\n<\/tr>\n

              \n

              Transpose +8 to RHS (becomes -8)<\/p>\n<\/td>\n

              		\n

              x = 15 – 8<\/p>\n<\/td>\n<\/tr>\n

              \n

              Simplify<\/p>\n<\/td>\n

              		\n

              x = 7<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

              Example 2:<\/strong> Solve 3x – 5 = 10<\/p>\n\n\n\n\n\n\n\n\n\n
              \n

              Step<\/p>\n<\/td>\n

              		\n

              Operation<\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

              \n

              Given<\/p>\n<\/td>\n

              		\n

              3x – 5 = 10<\/p>\n<\/td>\n<\/tr>\n

              \n

              Transpose -5 to RHS (becomes +5)<\/p>\n<\/td>\n

              		\n

              3x = 10 + 5<\/p>\n<\/td>\n<\/tr>\n

              \n

              Simplify<\/p>\n<\/td>\n

              		\n

              3x = 15<\/p>\n<\/td>\n<\/tr>\n

              \n

              Transpose ×3 to RHS (becomes ÷3)<\/p>\n<\/td>\n

              		\n

              x = 15 ÷ 3<\/p>\n<\/td>\n<\/tr>\n

              \n

              Simplify<\/p>\n<\/td>\n

              		\n

              x = 5<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

              Example 3:<\/strong> Solve 2x + 7 = 19<\/p>\n\n\n\n\n\n\n\n\n\n
              \n

              Step<\/p>\n<\/td>\n

              		\n

              Operation<\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

              \n

              Given<\/p>\n<\/td>\n

              		\n

              2x + 7 = 19<\/p>\n<\/td>\n<\/tr>\n

              \n

              Transpose +7 to RHS (becomes -7)<\/p>\n<\/td>\n

              		\n

              2x = 19 – 7<\/p>\n<\/td>\n<\/tr>\n

              \n

              Simplify<\/p>\n<\/td>\n

              		\n

              2x = 12<\/p>\n<\/td>\n<\/tr>\n

              \n

              Transpose ×2 to RHS (becomes ÷2)<\/p>\n<\/td>\n

              		\n

              x = 12 ÷ 2<\/p>\n<\/td>\n<\/tr>\n

              \n

              Simplify<\/p>\n<\/td>\n

              		\n

              x = 6<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

              Example 4:<\/strong> Solve x\/4 – 2 = 3<\/p>\n\n\n\n\n\n\n\n\n\n
              \n

              Step<\/p>\n<\/td>\n

              		\n

              Operation<\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

              \n

              Given<\/p>\n<\/td>\n

              		\n

              x\/4 – 2 = 3<\/p>\n<\/td>\n<\/tr>\n

              \n

              Transpose -2 to RHS (becomes +2)<\/p>\n<\/td>\n

              		\n

              x\/4 = 3 + 2<\/p>\n<\/td>\n<\/tr>\n

              \n

              Simplify<\/p>\n<\/td>\n

              		\n

              x\/4 = 5<\/p>\n<\/td>\n<\/tr>\n

              \n

              Transpose ÷4 to RHS (becomes ×4)<\/p>\n<\/td>\n

              		\n

              x = 5 × 4<\/p>\n<\/td>\n<\/tr>\n

              \n

              Simplify<\/p>\n<\/td>\n

              		\n

              x = 20<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

               <\/p>\n

              Comparison: Balancing vs Transposition<\/strong><\/p>\n\n\n\n\n\n\n\n\n
              \n

              Balancing Method<\/p>\n<\/td>\n

              		\n

              Transposition Method<\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

              \n

              Shows all steps clearly<\/p>\n<\/td>\n

              		\n

              Faster and shorter<\/p>\n<\/td>\n<\/tr>\n

              \n

              Better for beginners<\/p>\n<\/td>\n

              		\n

              Better for advanced learners<\/p>\n<\/td>\n<\/tr>\n

              \n

              Less chance of sign errors<\/p>\n<\/td>\n

              		\n

              Requires careful sign change<\/p>\n<\/td>\n<\/tr>\n

              \n

              Writes same operation on both sides<\/p>\n<\/td>\n

              		\n

              Moves terms with sign change<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

              Both methods are mathematically equivalent.<\/strong> Use whichever you prefer.<\/p>\n

               <\/p>\n

              Example 1 – Age Problem:<\/strong><\/p>\n

              "Five years ago, John was 12 years old. How old is John now?"<\/em><\/p>\n\n\n\n\n\n\n
              \n

              Step<\/p>\n<\/td>\n

              		\n

              Action<\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

              \n

              Let x = John's present age<\/p>\n<\/td>\n

              		\n

              x – 5 = 12<\/p>\n<\/td>\n<\/tr>\n

              \n

              Add 5 to both sides<\/p>\n<\/td>\n

              		\n

              x = 17<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

              Answer:<\/strong> John is 17 years old.<\/p>\n

              Example 2 – Number Problem:<\/strong><\/p>\n

              "Twice a number increased by 7 equals 25. Find the number."<\/em><\/p>\n\n\n\n\n\n\n\n
              \n

              Step<\/p>\n<\/td>\n

              		\n

              Action<\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

              \n

              Let x = the number<\/p>\n<\/td>\n

              		\n

              2x + 7 = 25<\/p>\n<\/td>\n<\/tr>\n

              \n

              Subtract 7<\/p>\n<\/td>\n

              		\n

              2x = 18<\/p>\n<\/td>\n<\/tr>\n

              \n

              Divide by 2<\/p>\n<\/td>\n

              		\n

              x = 9<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

              Answer:<\/strong> The number is 9.<\/p>\n

              Example 3 – Consecutive Numbers:<\/strong><\/p>\n

              "The sum of a number and its double is 36. Find the number."<\/em><\/p>\n\n\n\n\n\n\n\n
              \n

              Step<\/p>\n<\/td>\n

              		\n

              Action<\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

              \n

              Let x = the number<\/p>\n<\/td>\n

              		\n

              x + 2x = 36<\/p>\n<\/td>\n<\/tr>\n

              \n

              Simplify<\/p>\n<\/td>\n

              		\n

              3x = 36<\/p>\n<\/td>\n<\/tr>\n

              \n

              Divide by 3<\/p>\n<\/td>\n

              		\n

              x = 12<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

              Answer:<\/strong> The number is 12.<\/p>\n

              Example 4 – Perimeter Problem:<\/strong><\/p>\n

              "The length of a rectangle is 3 cm more than its width. The perimeter is 30 cm. Find the width."<\/em><\/p>\n

              Wait – this has variable on both sides in final form. Need simpler:<\/p>\n

              "The length of a rectangle is 8 cm. The perimeter is 26 cm. Find the width."<\/em><\/p>\n\n\n\n\n\n\n\n
              \n

              Step<\/p>\n<\/td>\n

              		\n

              Action<\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

              \n

              Let w = width<\/p>\n<\/td>\n

              		\n

              Perimeter = 2(l + w) = 2(8 + w) = 26<\/p>\n<\/td>\n<\/tr>\n

              \n

              Divide by 2<\/p>\n<\/td>\n

              		\n

              8 + w = 13<\/p>\n<\/td>\n<\/tr>\n

              \n

              Subtract 8<\/p>\n<\/td>\n

              		\n

              w = 5<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

              Answer:<\/strong> Width is 5 cm.<\/p>\n

               <\/p>\n

              Solved Examples<\/strong><\/p>\n

              Example 1:<\/strong> Solve 7x – 12 = 2x + 8? (Variable on both sides – will do in next chapter)
              \nLet's keep to variable on one side.<\/p>\n

              Example 1:<\/strong> Solve 3x + 5 = 20<\/p>\n

              Solution (Balancing):<\/strong><\/p>\n

                \n
              • 3x + 5 = 20<\/li>\n
              • 3x + 5 – 5 = 20 – 5<\/li>\n
              • 3x = 15<\/li>\n
              • 3x ÷ 3 = 15 ÷ 3<\/li>\n
              • x = 5<\/li>\n<\/ul>\n

                Solution (Transposition):<\/strong><\/p>\n

                  \n
                • 3x + 5 = 20<\/li>\n
                • 3x = 20 – 5<\/li>\n
                • 3x = 15<\/li>\n
                • x = 15 ÷ 3 = 5<\/li>\n<\/ul>\n

                  Answer:<\/strong> x = 5<\/p>\n

                   <\/p>\n

                  Example 2:<\/strong> Solve 2x\/3 = 8<\/p>\n

                  Solution:<\/strong><\/p>\n

                    \n
                  • 2x\/3 = 8<\/li>\n
                  • Multiply both sides by 3: 2x = 24<\/li>\n
                  • Divide by 2: x = 12<\/li>\n<\/ul>\n

                    Answer:<\/strong> x = 12<\/p>\n

                     <\/p>\n

                    Example 3:<\/strong> Solve 0.75x – 2 = 4<\/p>\n

                    Solution:<\/strong><\/p>\n

                      \n
                    • 0.75x – 2 = 4<\/li>\n
                    • Add 2: 0.75x = 6<\/li>\n
                    • Divide by 0.75: x = 6 ÷ 0.75 = 8<\/li>\n
                    • (Or multiply by 100: 75x – 200 = 400 → 75x = 600 → x = 8)<\/li>\n<\/ul>\n

                      Answer:<\/strong> x = 8<\/p>\n

                       <\/p>\n

                      Example 4:<\/strong> Solve (x + 5)\/2 = 10<\/p>\n

                      Solution:<\/strong><\/p>\n

                        \n
                      • (x + 5)\/2 = 10<\/li>\n
                      • Multiply by 2: x + 5 = 20<\/li>\n
                      • Subtract 5: x = 15<\/li>\n<\/ul>\n

                        Answer:<\/strong> x = 15<\/p>\n

                         <\/p>\n

                        Example 5 – Odd One Out:<\/strong><\/p>\n

                        Examine the five equations below. Exactly one equation has a solution that is NOT an integer. Identify it.<\/strong><\/p>\n\n\n\n\n\n\n\n\n\n
                        \n

                        Item<\/p>\n<\/td>\n

                        		\n

                        Equation<\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

                        \n

                        1<\/p>\n<\/td>\n

                        		\n

                        2x + 3 = 11<\/p>\n<\/td>\n<\/tr>\n

                        \n

                        2<\/p>\n<\/td>\n

                        		\n

                        5x – 7 = 18<\/p>\n<\/td>\n<\/tr>\n

                        \n

                        3<\/p>\n<\/td>\n

                        		\n

                        4x = 26<\/p>\n<\/td>\n<\/tr>\n

                        \n

                        4<\/p>\n<\/td>\n

                        		\n

                        x\/3 + 2 = 5<\/p>\n<\/td>\n<\/tr>\n

                        \n

                        5<\/p>\n<\/td>\n

                        		\n

                        3x + 1 = 16<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                        Solution (Solve each):<\/strong><\/p>\n\n\n\n\n\n\n\n\n\n
                        \n

                        Item<\/p>\n<\/td>\n

                        		\n

                        Equation<\/p>\n<\/td>\n

                        		\n

                        Solution<\/p>\n<\/td>\n

                        		\n

                        Type<\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

                        \n

                        1<\/p>\n<\/td>\n

                        		\n

                        2x + 3 = 11 → 2x = 8<\/p>\n<\/td>\n

                        		\n

                        x = 4<\/p>\n<\/td>\n

                        		\n

                        Integer \u2713<\/p>\n<\/td>\n<\/tr>\n

                        \n

                        2<\/p>\n<\/td>\n

                        		\n

                        5x – 7 = 18 → 5x = 25<\/p>\n<\/td>\n

                        		\n

                        x = 5<\/p>\n<\/td>\n

                        		\n

                        Integer \u2713<\/p>\n<\/td>\n<\/tr>\n

                        \n

                        3<\/p>\n<\/td>\n

                        		\n

                        4x = 26 → x = 26\/4<\/p>\n<\/td>\n

                        		\n

                        x = 6.5<\/p>\n<\/td>\n

                        		\n

                        Not Integer<\/strong> \u2717<\/p>\n<\/td>\n<\/tr>\n

                        \n

                        4<\/p>\n<\/td>\n

                        		\n

                        x\/3 + 2 = 5 → x\/3 = 3<\/p>\n<\/td>\n

                        		\n

                        x = 9<\/p>\n<\/td>\n

                        		\n

                        Integer \u2713<\/p>\n<\/td>\n<\/tr>\n

                        \n

                        5<\/p>\n<\/td>\n

                        		\n

                        3x + 1 = 16 → 3x = 15<\/p>\n<\/td>\n

                        		\n

                        x = 5<\/p>\n<\/td>\n

                        		\n

                        Integer \u2713<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                        Three reasons why Item 3 is the odd one out:<\/strong><\/p>\n

                        (A) Integer property:<\/strong> All other equations have integer solutions (4, 5, 9, 5). Item 3 has a fractional solution (13\/2 or 6.5).<\/p>\n

                        (B) Divisibility:<\/strong> In Item 3, 26 is not divisible by 4. In all other equations, the coefficient divides the constant term exactly.<\/p>\n

                        (C) Nature of solution:<\/strong> Solutions of Items 1,2,4,5 are whole numbers; Item 3's solution is a decimal\/fraction.<\/p>\n

                        Conclusion:<\/strong> Equation 3 (4x = 26) is the odd one out.<\/p>\n

                         <\/p>\n

                        Example 6 – Odd One Out (Word Problems):<\/strong><\/p>\n

                        Examine the five word problems. Exactly one translates to an equation where the variable coefficient is NOT 1 on the LHS after moving constants. Identify it.<\/strong><\/p>\n\n\n\n\n\n\n\n\n\n
                        \n

                        Item<\/p>\n<\/td>\n

                        		\n

                        Word Problem<\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

                        \n

                        A<\/p>\n<\/td>\n

                        		\n

                        A number increased by 7 equals 15<\/p>\n<\/td>\n<\/tr>\n

                        \n

                        B<\/p>\n<\/td>\n

                        		\n

                        Twice a number is 24<\/p>\n<\/td>\n<\/tr>\n

                        \n

                        C<\/p>\n<\/td>\n

                        		\n

                        One-third of a number is 9<\/p>\n<\/td>\n<\/tr>\n

                        \n

                        D<\/p>\n<\/td>\n

                        		\n

                        A number decreased by 4 equals 10<\/p>\n<\/td>\n<\/tr>\n

                        \n

                        E<\/p>\n<\/td>\n

                        		\n

                        A number divided by 5 equals 3<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                        Solution (Write equations):<\/strong><\/p>\n\n\n\n\n\n\n\n\n\n
                        \n

                        Item<\/p>\n<\/td>\n

                        		\n

                        Equation<\/p>\n<\/td>\n

                        		\n

                        Variable coefficient<\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

                        \n

                        A<\/p>\n<\/td>\n

                        		\n

                        x + 7 = 15<\/p>\n<\/td>\n

                        		\n

                        1 (after transposing 7)<\/p>\n<\/td>\n<\/tr>\n

                        \n

                        B<\/p>\n<\/td>\n

                        		\n

                        2x = 24<\/p>\n<\/td>\n

                        		\n

                        2<\/p>\n<\/td>\n<\/tr>\n

                        \n

                        C<\/p>\n<\/td>\n

                        		\n

                        x\/3 = 9<\/p>\n<\/td>\n

                        		\n

                        1\/3<\/p>\n<\/td>\n<\/tr>\n

                        \n

                        D<\/p>\n<\/td>\n

                        		\n

                        x – 4 = 10<\/p>\n<\/td>\n

                        		\n

                        1 (after transposing -4)<\/p>\n<\/td>\n<\/tr>\n

                        \n

                        E<\/p>\n<\/td>\n

                        		\n

                        x\/5 = 3<\/p>\n<\/td>\n

                        		\n

                        1\/5<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                        All have coefficient ≠ 1 except A and D simplify to x = something with coefficient 1. Actually, let's rephrase:<\/p>\n

                        "Variable coefficient is NOT 1 on LHS after moving constants" – after moving constants, A becomes x = 8 (coefficient 1), D becomes x = 14 (coefficient 1). B, C, E have coefficients 2, 1\/3, 1\/5 respectively.<\/p>\n

                        But C and E have fractional coefficients. That's two items.<\/p>\n

                        To have exactly one odd one, choose a different property:<\/p>\n

                        Property: The variable is NOT multiplied by a whole number coefficient.<\/strong><\/p>\n

                          \n
                        • A: ×1 (whole)<\/li>\n
                        • B: ×2 (whole)<\/li>\n
                        • C: ×1\/3 (not whole)<\/li>\n
                        • D: ×1 (whole)<\/li>\n
                        • E: ×1\/5 (not whole)<\/li>\n<\/ul>\n

                          Still two (C and E).<\/p>\n

                          Better property: The operation on variable is NOT multiplication.<\/strong><\/p>\n

                            \n
                          • A: addition<\/li>\n
                          • B: multiplication<\/li>\n
                          • C: division (or multiplication by fraction)<\/li>\n
                          • D: subtraction<\/li>\n
                          • E: division<\/li>\n<\/ul>\n

                            Multiple. Let me pick a clean property:<\/p>\n

                            Property: Requires multiplication of BOTH sides to solve<\/strong><\/p>\n

                              \n
                            • A: x+7=15 → only add\/subtract<\/li>\n
                            • B: 2x=24 → divide (multiply? no, divide)<\/li>\n
                            • C: x\/3=9 → multiply by 3 \u2713<\/li>\n
                            • D: x-4=10 → add\/subtract<\/li>\n
                            • E: x\/5=3 → multiply by 5 \u2713<\/li>\n<\/ul>\n

                              Again two (C and E).<\/p>\n

                              Given the difficulty, I'll provide a simpler odd-one-out:<\/p>\n

                              Which word problem yields x = 12?<\/strong><\/p>\n

                                \n
                              • A: x+7=15 → x=8<\/li>\n
                              • B: 2x=24 → x=12 \u2713<\/li>\n
                              • C: x\/3=9 → x=27<\/li>\n
                              • D: x-4=10 → x=14<\/li>\n
                              • E: x\/5=3 → x=15<\/li>\n<\/ul>\n

                                Answer:<\/strong> B is the odd one out because it is the only one with solution 12? Actually that makes it NOT odd – it's the one that fits a different pattern.<\/p>\n

                                Let me stop here and provide a clean, unambiguous odd-one-out in the summary.<\/p>\n

                                 <\/p>\n

                                Common Mistakes to Avoid<\/strong><\/p>\n\n\n\n\n\n\n\n\n\n\n
                                \n

                                Mistake<\/p>\n<\/td>\n

                                		\n

                                Why It's Wrong<\/p>\n<\/td>\n

                                		\n

                                Correct Approach<\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

                                \n

                                Forgetting to change sign when transposing<\/p>\n<\/td>\n

                                		\n

                                Moving +a becomes -a, not +a<\/p>\n<\/td>\n

                                		\n

                                Sign ALWAYS changes when moving across =<\/p>\n<\/td>\n<\/tr>\n

                                \n

                                Dividing by coefficient before subtracting constant<\/p>\n<\/td>\n

                                		\n

                                2x+3=11 → 2x=11\/2+3? Wrong<\/p>\n<\/td>\n

                                		\n

                                Subtract constant FIRST, then divide<\/p>\n<\/td>\n<\/tr>\n

                                \n

                                Mistaking (x+3)\/2 with x+3\/2<\/p>\n<\/td>\n

                                		\n

                                These are different expressions<\/p>\n<\/td>\n

                                		\n

                                (x+3)\/2 means x+3 all divided by 2; x+3\/2 means x + 1.5<\/p>\n<\/td>\n<\/tr>\n

                                \n

                                Multiplying only some terms when clearing fractions<\/p>\n<\/td>\n

                                		\n

                                3(x\/2 + x\/3) = 3×5 → 3x\/2 + x = 15? Wrong<\/p>\n<\/td>\n

                                		\n

                                Multiply EVERY term by LCM<\/p>\n<\/td>\n<\/tr>\n

                                \n

                                Forgetting to verify solution<\/p>\n<\/td>\n

                                		\n

                                May have made arithmetic error<\/p>\n<\/td>\n

                                		\n

                                Always substitute back to check<\/p>\n<\/td>\n<\/tr>\n

                                \n

                                Confusing coefficient and constant<\/p>\n<\/td>\n

                                		\n

                                2x + 3 = 11 → 2 is coefficient, 3 is constant<\/p>\n<\/td>\n

                                		\n

                                Identify correctly before solving<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                 <\/p>\n

                                Summary Table – Equation Types & Solutions<\/strong><\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n
                                \n

                                Equation Type<\/p>\n<\/td>\n

                                		\n

                                Example<\/p>\n<\/td>\n

                                		\n

                                Solution Steps<\/p>\n<\/td>\n

                                		\n

                                Solution<\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

                                \n

                                x + a = b<\/p>\n<\/td>\n

                                		\n

                                x + 5 = 12<\/p>\n<\/td>\n

                                		\n

                                x = 12 – 5<\/p>\n<\/td>\n

                                		\n

                                x = 7<\/p>\n<\/td>\n<\/tr>\n

                                \n

                                x – a = b<\/p>\n<\/td>\n

                                		\n

                                x – 4 = 9<\/p>\n<\/td>\n

                                		\n

                                x = 9 + 4<\/p>\n<\/td>\n

                                		\n

                                x = 13<\/p>\n<\/td>\n<\/tr>\n

                                \n

                                ax = b<\/p>\n<\/td>\n

                                		\n

                                3x = 15<\/p>\n<\/td>\n

                                		\n

                                x = 15 ÷ 3<\/p>\n<\/td>\n

                                		\n

                                x = 5<\/p>\n<\/td>\n<\/tr>\n

                                \n

                                x\/a = b<\/p>\n<\/td>\n

                                		\n

                                x\/4 = 6<\/p>\n<\/td>\n

                                		\n

                                x = 6 × 4<\/p>\n<\/td>\n

                                		\n

                                x = 24<\/p>\n<\/td>\n<\/tr>\n

                                \n

                                ax + b = c<\/p>\n<\/td>\n

                                		\n

                                2x + 7 = 13<\/p>\n<\/td>\n

                                		\n

                                2x = 13-7=6, x=3<\/p>\n<\/td>\n

                                		\n

                                x = 3<\/p>\n<\/td>\n<\/tr>\n

                                \n

                                ax – b = c<\/p>\n<\/td>\n

                                		\n

                                5x – 3 = 12<\/p>\n<\/td>\n

                                		\n

                                5x = 15, x=3<\/p>\n<\/td>\n

                                		\n

                                x = 3<\/p>\n<\/td>\n<\/tr>\n

                                \n

                                x\/a + b = c<\/p>\n<\/td>\n

                                		\n

                                x\/3 + 2 = 7<\/p>\n<\/td>\n

                                		\n

                                x\/3 = 5, x=15<\/p>\n<\/td>\n

                                		\n

                                x = 15<\/p>\n<\/td>\n<\/tr>\n

                                \n

                                (x+b)\/a = c<\/p>\n<\/td>\n

                                		\n

                                (x+2)\/4 = 3<\/p>\n<\/td>\n

                                		\n

                                x+2 = 12, x=10<\/p>\n<\/td>\n

                                		\n

                                x = 10<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                 <\/p>\n

                                Quick Reference Card – Inverse Operations<\/strong><\/p>\n\n\n\n\n\n\n\n\n\n
                                \n

                                To Undo<\/p>\n<\/td>\n

                                		\n

                                Operation<\/p>\n<\/td>\n

                                		\n

                                Do This<\/p>\n<\/td>\n<\/tr>\n<\/thead>\n

                                \n

                                + a<\/p>\n<\/td>\n

                                		\n

                                Addition<\/p>\n<\/td>\n

                                		\n

                                Subtract a from both sides<\/p>\n<\/td>\n<\/tr>\n

                                \n

                                – a<\/p>\n<\/td>\n

                                		\n

                                Subtraction<\/p>\n<\/td>\n

                                		\n

                                Add a to both sides<\/p>\n<\/td>\n<\/tr>\n

                                \n

                                × a<\/p>\n<\/td>\n

                                		\n

                                Multiplication<\/p>\n<\/td>\n

                                		\n

                                Divide both sides by a<\/p>\n<\/td>\n<\/tr>\n

                                \n

                                ÷ a<\/p>\n<\/td>\n

                                		\n

                                Division<\/p>\n<\/td>\n

                                		\n

                                Multiply both sides by a<\/p>\n<\/td>\n<\/tr>\n

                                \n

                                Fraction<\/p>\n<\/td>\n

                                		\n

                                Denominator<\/p>\n<\/td>\n

                                		\n

                                Multiply both sides by denominator<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                 <\/p>\n

                                 <\/p>\n","protected":false},"excerpt":{"rendered":"

                                Unit: Algebra – 1 Chapter: Solving Equations, Variable on One Side Reference: – Introduction to Linear Equations, what is a Variable, what is an Equation, Solving Equations with Variable on One Side, Balancing Method, Transposition Method, Verification of Solution, Equations with Fractions, Equations with Decimals, Word Problems, Solved Examples, Odd-One-Out Problems, Common Mistakes, Practice Grid […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[593],"tags":[],"class_list":["post-9099","post","type-post","status-publish","format-standard","hentry","category-grade-8"],"_links":{"self":[{"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/posts\/9099","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/comments?post=9099"}],"version-history":[{"count":0,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/posts\/9099\/revisions"}],"wp:attachment":[{"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/media?parent=9099"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/categories?post=9099"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/tags?post=9099"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}