{"id":9103,"date":"2026-06-01T21:33:48","date_gmt":"2026-06-01T21:33:48","guid":{"rendered":"https:\/\/kapdec.com\/help\/?p=9103"},"modified":"2026-06-01T21:33:48","modified_gmt":"2026-06-01T21:33:48","slug":"solving-linear-equations-algebraic-method","status":"publish","type":"post","link":"https:\/\/kapdec.com\/help\/solving-linear-equations-algebraic-method\/","title":{"rendered":"Solving Linear Equations, Algebraic Method"},"content":{"rendered":"

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

Chapter: <\/strong>Solving Linear Equations – Algebraic Methods<\/strong><\/h3>\n

Reference: – Introduction to Solving Linear Equations in Two Variables, Systems of Linear Equations, Substitution Method, Elimination Method, Cross Multiplication Method, Comparison Method, Consistent vs Inconsistent Systems, Dependent Systems, Solved Examples, Odd-One-Out Problems, Common Mistakes<\/em><\/p>\n

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

    \n
  • What is a System of Linear Equations in Two Variables<\/em><\/li>\n
  • Substitution Method<\/em><\/li>\n
  • Elimination Method<\/em><\/li>\n
  • Cross Multiplication Method<\/em><\/li>\n
  • Identifying Consistent, Inconsistent, and Dependent Systems<\/em><\/li>\n<\/ul>\n

    Introduction to Solving Linear Equations in Two Variables<\/strong><\/p>\n

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

    When we have two linear equations in two variables (usually x and y), we call it a system of linear equations. Solving the system means finding the ordered pair (x, y) that satisfies both equations simultaneously.<\/p>\n

    In real life, we often need two equations to solve for two unknowns. For example, if you buy apples and oranges, knowing the total cost and the total number of items gives two equations.<\/p>\n

    When we solve a system of linear equations, we essentially ask:<\/p>\n

    "What is the common point (x, y) that lies on both lines?"<\/p>\n

    Geometrically, the solution is the intersection point of the two lines.<\/p>\n

    Importance of Solving Systems<\/u><\/strong><\/p>\n

      \n
    • Solves real-world problems involving two unknowns (prices, speeds, mixtures, etc.)<\/li>\n
    • Foundation for higher algebra, linear programming, and optimization<\/li>\n
    • Used extensively in economics, engineering, physics, and computer science<\/li>\n<\/ul>\n

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

      System:<\/strong>
      \nx + y = 5
      \nx – y = 1<\/p>\n

      Solution:<\/strong> Adding the two equations gives 2x = 6 → x = 3, then y = 2. So the solution is (3, 2), which satisfies both equations.<\/p>\n

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

      1. System of Linear Equations – Types of Solutions<\/strong><\/p>\n

      When we solve two linear equations in two variables, three possibilities exist:<\/p>\n

      Case 1 – Unique Solution (Consistent System):<\/strong>
      \nThe two lines intersect at exactly one point. There is one unique pair (x, y) that satisfies both equations.
      \nExample: x + y = 5 and x – y = 1 → solution (3, 2)<\/p>\n

      Case 2 – No Solution (Inconsistent System):<\/strong>
      \nThe two lines are parallel and never intersect. There is no pair (x, y) that satisfies both equations.
      \nExample: x + y = 5 and x + y = 7 (same left side, different right side)<\/p>\n

      Case 3 – Infinite Solutions (Dependent System):<\/strong>
      \nThe two lines coincide (are the same line). Every point on the line is a solution.
      \nExample: x + y = 5 and 2x + 2y = 10 (second equation is just twice the first)<\/p>\n

      General Form for Checking:<\/strong> For equations a\u2081x + b\u2081y + c\u2081 = 0 and a\u2082x + b\u2082y + c\u2082 = 0<\/p>\n

        \n
      • Unique solution if a\u2081\/a\u2082 ≠ b\u2081\/b\u2082<\/li>\n
      • No solution if a\u2081\/a\u2082 = b\u2081\/b\u2082 ≠ c\u2081\/c\u2082<\/li>\n
      • Infinite solutions if a\u2081\/a\u2082 = b\u2081\/b\u2082 = c\u2081\/c\u2082<\/li>\n<\/ul>\n

         <\/p>\n

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

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

        In the substitution method, we solve one equation for one variable in terms of the other, and then substitute that expression into the second equation.<\/p>\n

        Steps of Substitution Method:<\/strong><\/p>\n

          \n
        • Step 1: From one equation, express one variable (say y) in terms of the other variable (x).<\/li>\n
        • Step 2: Substitute this expression into the other equation.<\/li>\n
        • Step 3: Solve the resulting single-variable equation for x.<\/li>\n
        • Step 4: Substitute x back to find y.<\/li>\n
        • Step 5: Verify the solution in both original equations.<\/li>\n<\/ul>\n

          Example 1:<\/strong> Solve by substitution: x + y = 10 and 2x – y = 5<\/p>\n

          Solution:<\/strong>
          \nFrom the first equation: y = 10 – x
          \nSubstitute into second equation: 2x – (10 – x) = 5
          \nSimplify: 2x – 10 + x = 5 → 3x – 10 = 5 → 3x = 15 → x = 5
          \nNow find y: y = 10 – 5 = 5
          \nVerification: First: 5+5=10 \u2713; Second: 2(5)-5=10-5=5 \u2713
          \nAnswer:<\/strong> (5, 5)<\/p>\n

          Example 2:<\/strong> Solve by substitution: 3x + 2y = 12 and y = x + 1<\/p>\n

          Solution:<\/strong>
          \nSince y = x + 1 is already solved for y, substitute into first equation:
          \n3x + 2(x + 1) = 12 → 3x + 2x + 2 = 12 → 5x + 2 = 12 → 5x = 10 → x = 2
          \nNow find y: y = x + 1 = 2 + 1 = 3
          \nVerification: First: 3(2)+2(3)=6+6=12 \u2713; Second: 3=2+1 \u2713
          \nAnswer:<\/strong> (2, 3)<\/p>\n

          Example 3:<\/strong> Solve by substitution: 2x + y = 7 and x – y = 2<\/p>\n

          Solution:<\/strong>
          \nFrom the second equation: x = y + 2
          \nSubstitute into first: 2(y + 2) + y = 7 → 2y + 4 + y = 7 → 3y + 4 = 7 → 3y = 3 → y = 1
          \nThen x = 1 + 2 = 3
          \nAnswer:<\/strong> (3, 1)<\/p>\n

           <\/p>\n

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

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

          In the elimination method, we add or subtract the two equations to eliminate one variable. Sometimes we need to multiply one or both equations by constants to make the coefficients of one variable opposites.<\/p>\n

          Steps of Elimination Method:<\/strong><\/p>\n

            \n
          • Step 1: Write both equations in standard form (ax + by = c).<\/li>\n
          • Step 2: Make the coefficients of one variable (say x) equal in magnitude (opposite signs for addition).<\/li>\n
          • Step 3: Add or subtract the equations to eliminate that variable.<\/li>\n
          • Step 4: Solve for the remaining variable.<\/li>\n
          • Step 5: Substitute back to find the other variable.<\/li>\n
          • Step 6: Verify the solution.<\/li>\n<\/ul>\n

            Example 1:<\/strong> Solve by elimination: x + y = 10 and x – y = 4<\/p>\n

            Solution:<\/strong>
            \nAdd the two equations: (x + y) + (x – y) = 10 + 4 → 2x = 14 → x = 7
            \nSubstitute x = 7 into first equation: 7 + y = 10 → y = 3
            \nVerification: Second: 7 – 3 = 4 \u2713
            \nAnswer:<\/strong> (7, 3)<\/p>\n

            Example 2:<\/strong> Solve by elimination: 2x + 3y = 13 and 4x – 3y = 11<\/p>\n

            Solution:<\/strong>
            \nNotice y coefficients are +3 and -3 (already opposites). Add the equations:
            \n(2x + 3y) + (4x – 3y) = 13 + 11 → 6x = 24 → x = 4
            \nSubstitute x = 4 into first equation: 2(4) + 3y = 13 → 8 + 3y = 13 → 3y = 5 → y = 5\/3
            \nAnswer:<\/strong> (4, 5\/3)<\/p>\n

            Example 3:<\/strong> Solve by elimination: 3x + 2y = 8 and 2x + 3y = 7<\/p>\n

            Solution:<\/strong>
            \nTo eliminate x, multiply first equation by 2 and second equation by 3:
            \nFirst ×2: 6x + 4y = 16
            \nSecond ×3: 6x + 9y = 21
            \nSubtract second from first: (6x + 4y) – (6x + 9y) = 16 – 21 → -5y = -5 → y = 1
            \nSubstitute y = 1 into first original: 3x + 2(1) = 8 → 3x + 2 = 8 → 3x = 6 → x = 2
            \nAnswer:<\/strong> (2, 1)<\/p>\n

            Example 4:<\/strong> Solve by elimination: 5x – 3y = 11 and 2x + 4y = 6<\/p>\n

            Solution:<\/strong>
            \nTo eliminate x, multiply first equation by 2 and second equation by 5:
            \nFirst ×2: 10x – 6y = 22
            \nSecond ×5: 10x + 20y = 30
            \nSubtract: (10x – 6y) – (10x + 20y) = 22 – 30 → -26y = -8 → y = 8\/26 = 4\/13
            \nSubstitute y = 4\/13 into second original: 2x + 4(4\/13) = 6 → 2x + 16\/13 = 6 → Multiply by 13: 26x + 16 = 78 → 26x = 62 → x = 62\/26 = 31\/13
            \nAnswer:<\/strong> (31\/13, 4\/13)<\/p>\n

             <\/p>\n

            Method 3: Cross Multiplication Method<\/strong><\/p>\n

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

            The cross-multiplication method is a formula-based approach for solving systems of the form a\u2081x + b\u2081y + c\u2081 = 0 and a\u2082x + b\u2082y + c\u2082 = 0.<\/p>\n

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

            x \/ (b\u2081c\u2082 – b\u2082c\u2081) = y \/ (c\u2081a\u2082 – c\u2082a\u2081) = 1 \/ (a\u2081b\u2082 – a\u2082b\u2081)<\/p>\n

            From this, we get:<\/p>\n

            x = (b\u2081c\u2082 – b\u2082c\u2081) \/ (a\u2081b\u2082 – a\u2082b\u2081)<\/p>\n

            y = (c\u2081a\u2082 – c\u2082a\u2081) \/ (a\u2081b\u2082 – a\u2082b\u2081)<\/p>\n

            Note:<\/strong> This method is efficient when the coefficients are large or when rote memorization saves time.<\/p>\n

            Example:<\/strong> Solve using cross multiplication: 2x + 3y – 8 = 0 and 3x + 2y – 7 = 0<\/p>\n

            Solution:<\/strong>
            \nHere a\u2081=2, b\u2081=3, c\u2081=-8; a\u2082=3, b\u2082=2, c\u2082=-7<\/p>\n

            First compute the denominator (a\u2081b\u2082 – a\u2082b\u2081) = (2×2 – 3×3) = 4 – 9 = -5<\/p>\n

            Now find x: x = (b\u2081c\u2082 – b\u2082c\u2081) \/ (a\u2081b\u2082 – a\u2082b\u2081) = (3×(-7) – 2×(-8)) \/ (-5) = (-21 + 16) \/ (-5) = (-5)\/(-5) = 1<\/p>\n

            Now find y: y = (c\u2081a\u2082 – c\u2082a\u2081) \/ (a\u2081b\u2082 – a\u2082b\u2081) = ((-8)×3 – (-7)×2) \/ (-5) = (-24 + 14) \/ (-5) = (-10)\/(-5) = 2<\/p>\n

            Answer:<\/strong> (1, 2)<\/p>\n

             <\/p>\n

            Comparison of Methods<\/strong><\/p>\n

            Substitution Method:<\/strong> Best when one equation is already solved for a variable (like y = 2x + 3) or can be easily solved. Works well for all systems but can get messy with fractions.<\/p>\n

            Elimination Method:<\/strong> Best when coefficients line up nicely or can be easily matched. Very systematic and avoids fractions until the end. Often the fastest method for most problems.<\/p>\n

            Cross Multiplication Method:<\/strong> Best when you need a quick formula-based solution, especially for large coefficients. Less intuitive but very efficient once memorized.<\/p>\n

             <\/p>\n

            Special Cases – Consistent, Inconsistent, Dependent<\/strong><\/p>\n

            Case 1 – Unique Solution (Consistent System):<\/strong>
            \nThe lines intersect at one point. The algebraic condition is a\u2081\/a\u2082 ≠ b\u2081\/b\u2082.<\/p>\n

            Example:<\/strong> 2x + y = 5 and x – y = 1
            \nHere 2\/1 ≠ 1\/(-1) → 2 ≠ -1 → Unique solution (2, 1)<\/p>\n

            Case 2 – No Solution (Inconsistent System):<\/strong>
            \nThe lines are parallel and distinct. Condition: a\u2081\/a\u2082 = b\u2081\/b\u2082 ≠ c\u2081\/c\u2082.<\/p>\n

            Example:<\/strong> x + y = 5 and x + y = 8
            \nHere 1\/1 = 1\/1 ≠ 5\/8 → No solution<\/p>\n

            Case 3 – Infinite Solutions (Dependent System):<\/strong>
            \nThe lines are coincident (same line). Condition: a\u2081\/a\u2082 = b\u2081\/b\u2082 = c\u2081\/c\u2082.<\/p>\n

            Example:<\/strong> x + y = 5 and 2x + 2y = 10
            \nHere 1\/2 = 1\/2 = 5\/10 → Infinite solutions<\/p>\n

             <\/p>\n

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

            Example 1 – Substitution Method:<\/strong> Solve 2x + 3y = 8 and x – y = -1<\/p>\n

            Solution:<\/strong> From second equation: x = y – 1
            \nSubstitute into first: 2(y – 1) + 3y = 8 → 2y – 2 + 3y = 8 → 5y – 2 = 8 → 5y = 10 → y = 2
            \nThen x = 2 – 1 = 1
            \nAnswer:<\/strong> (1, 2)<\/p>\n

             <\/p>\n

            Example 2 – Elimination Method:<\/strong> Solve 4x – 5y = 3 and 3x + 2y = 8<\/p>\n

            Solution:<\/strong> Multiply first by 2 and second by 5 to eliminate y:
            \nFirst ×2: 8x – 10y = 6
            \nSecond ×5: 15x + 10y = 40
            \nAdd: 23x = 46 → x = 2
            \nSubstitute into second original: 3(2) + 2y = 8 → 6 + 2y = 8 → 2y = 2 → y = 1
            \nAnswer:<\/strong> (2, 1)<\/p>\n

             <\/p>\n

            Example 3 – No Solution Case:<\/strong> Solve x + 2y = 5 and 2x + 4y = 12<\/p>\n

            Solution:<\/strong> Divide second equation by 2: x + 2y = 6
            \nNow we have x + 2y = 5 and x + 2y = 6. These are contradictory.
            \nAnswer:<\/strong> No solution (Inconsistent system)<\/p>\n

             <\/p>\n

            Example 4 – Infinite Solutions Case:<\/strong> Solve 3x – y = 4 and 6x – 2y = 8<\/p>\n

            Solution:<\/strong> Divide second equation by 2: 3x – y = 4, which is exactly the same as the first equation.
            \nAnswer:<\/strong> Infinite solutions (Dependent system)<\/p>\n

             <\/p>\n

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

            Mistake 1 – Forgetting to multiply the entire equation<\/strong>
            \nWhen using elimination, if you multiply an equation by a number, multiply EVERY term, not just one.
            \nCorrect approach: 2(x + y = 5) → 2x + 2y = 10 (both terms multiplied)<\/p>\n

            Mistake 2 – Making sign errors when subtracting equations<\/strong>
            \nWhen subtracting, be careful with negative signs.
            \nCorrect approach: (2x + 3y) – (2x – y) = 2x + 3y – 2x + y = 4y (not 2y)<\/p>\n

            Mistake 3 – Substituting back into the same equation<\/strong>
            \nAfter finding x, substitute into the OTHER equation to verify or find y.
            \nCorrect approach: Use an equation you haven't used yet for verification.<\/p>\n

            Mistake 4 – Mixing up the condition for no solution vs infinite solutions<\/strong>
            \nRemember: a\u2081\/a\u2082 = b\u2081\/b\u2082 ≠ c\u2081\/c\u2082 → no solution; a\u2081\/a\u2082 = b\u2081\/b\u2082 = c\u2081\/c\u2082 → infinite solutions.
            \nCorrect approach: Check all three ratios carefully.<\/p>\n

            Mistake 5 – Giving only one value when there are infinite solutions<\/strong>
            \nFor dependent systems, the answer is "infinite solutions" or "all points on the line," not a single pair.
            \nCorrect approach: Recognize the condition and state appropriately.<\/p>\n

             <\/p>\n

            Quick Reference Summary<\/strong><\/p>\n

            Substitution Method:<\/strong>
            \nSolve for one variable → substitute into other equation → solve → back-substitute<\/p>\n

            Elimination Method:<\/strong>
            \nMake coefficients opposites → add\/subtract to eliminate → solve → back-substitute<\/p>\n

            Cross Multiplication Method:<\/strong>
            \nUse formula x\/(b\u2081c\u2082 – b\u2082c\u2081) = y\/(c\u2081a\u2082 – c\u2082a\u2081) = 1\/(a\u2081b\u2082 – a\u2082b\u2081)<\/p>\n

            Types of Systems:<\/strong>
            \nUnique solution → a\u2081\/a\u2082 ≠ b\u2081\/b\u2082
            \nNo solution → a\u2081\/a\u2082 = b\u2081\/b\u2082 ≠ c\u2081\/c\u2082
            \nInfinite solutions → a\u2081\/a\u2082 = b\u2081\/b\u2082 = c\u2081\/c\u2082<\/p>\n

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

            Unit: Algebra – 1 Chapter: Solving Linear Equations – Algebraic Methods Reference: – Introduction to Solving Linear Equations in Two Variables, Systems of Linear Equations, Substitution Method, Elimination Method, Cross Multiplication Method, Comparison Method, Consistent vs Inconsistent Systems, Dependent Systems, Solved Examples, Odd-One-Out Problems, Common Mistakes After studying this chapter, you should be able to […]<\/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-9103","post","type-post","status-publish","format-standard","hentry","category-grade-8"],"_links":{"self":[{"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/posts\/9103","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=9103"}],"version-history":[{"count":0,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/posts\/9103\/revisions"}],"wp:attachment":[{"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/media?parent=9103"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/categories?post=9103"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kapdec.com\/help\/wp-json\/wp\/v2\/tags?post=9103"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}