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Algebra - Solution of an Equation by Trial and Error

Grade 6CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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An equation is a mathematical statement that asserts the equality of two expressions, separated by an equal sign (==). Visually, think of an equation as a balanced weighing scale where the weight on the Left Hand Side (LHS) is exactly equal to the weight on the Right Hand Side (RHS).

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A variable is a symbol, usually a letter like x,y,orzx, y, or z, that represents an unknown number. In the trial and error method, we treat the variable as a placeholder that we fill with different numbers to see which one 'fits' the balance of the equation.

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The solution of an equation is the specific value of the variable that makes the equation true (i.e., makes LHS=RHSLHS = RHS). If we imagine the equation as a locked door, the solution is the specific key that turns the lock.

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The Trial and Error Method involves substituting various numerical values for the variable one by one and calculating the result for the LHS. We continue this process until we find a value where the calculated LHS matches the given RHS.

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To stay organized, we use a systematic table with columns for the 'Assumed Value of Variable', the 'Calculated LHS', the 'Fixed RHS', and a 'Conclusion' (Is LHS=RHSLHS = RHS?). This visual grid helps track which numbers have been tried and how close we are to the answer.

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During the trial process, if the calculated LHS is much smaller than the RHS, we should try a significantly larger number for the variable. If the LHS is slightly larger than the RHS, we should try a slightly smaller number. This directional guessing makes the process faster.

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The equality sign (==) acts as a fulcrum. If the value substituted makes the LHS greater than the RHS (LHS>RHSLHS > RHS), the scale tips to the left. If it makes the LHS smaller (LHS<RHSLHS < RHS), it tips to the right. We only stop when the scale is perfectly horizontal.

📐Formulae

General form of a linear equation: ax+b=cax + b = c

Condition for solution: LHS=RHSLHS = RHS

Verification: If x=kx = k is a solution, then a(k)+b=ca(k) + b = c

💡Examples

Problem 1:

Solve the equation x+4=10x + 4 = 10 using the trial and error method.

Solution:

We will try different values for xx:

  1. Let x=2x = 2: LHS=2+4=6LHS = 2 + 4 = 6. Since 6≠106 \neq 10, x=2x=2 is not the solution.
  2. Let x=4x = 4: LHS=4+4=8LHS = 4 + 4 = 8. Since 8≠108 \neq 10, x=4x=4 is not the solution.
  3. Let x=5x = 5: LHS=5+4=9LHS = 5 + 4 = 9. Since 9≠109 \neq 10, x=5x=5 is not the solution.
  4. Let x=6x = 6: LHS=6+4=10LHS = 6 + 4 = 10. Since 10=1010 = 10, the condition LHS=RHSLHS = RHS is satisfied. Therefore, the solution is x=6x = 6.

Explanation:

We systematically substituted increasing values for xx and observed that the LHS sum was getting closer to 10. Once we reached x=6x = 6, the equation balanced perfectly.

Problem 2:

Find the value of mm in 3m=153m = 15 by trial and error.

Solution:

We will substitute values for mm and check if the product with 3 equals 15:

  1. Try m=1m = 1: LHS=3×1=3≠15LHS = 3 \times 1 = 3 \neq 15.
  2. Try m=3m = 3: LHS=3×3=9≠15LHS = 3 \times 3 = 9 \neq 15.
  3. Try m=4m = 4: LHS=3×4=12≠15LHS = 3 \times 4 = 12 \neq 15.
  4. Try m=5m = 5: LHS=3×5=15=15LHS = 3 \times 5 = 15 = 15. Since LHS=RHSLHS = RHS when m=5m = 5, the solution is m=5m = 5.

Explanation:

In this multiplicative equation, we tested integers for mm. By observing that 3×43 \times 4 was 12 (too small) and 3×53 \times 5 was 15 (correct), we identified the solution.