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Algebra - What is an Equation?

Grade 6CBSE

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

🔑Concepts

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An equation is a mathematical condition on a variable which states that two expressions are equal. It is characterized by the presence of an equality sign == between the Left Hand Side (LHS) and the Right Hand Side (RHS).

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Think of an equation as a balance scale. For the scale to be perfectly balanced (horizontal), the total weight on the LHS must be exactly equal to the total weight on the RHS. If we add or subtract something from one side, we must do the same to the other to keep it balanced.

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A variable is an unknown quantity represented by a letter like xx, yy, zz, ll, mm, or nn. In an equation, the variable represents a specific value that we need to find to make the statement true.

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An equation is different from an algebraic expression. For example, 2x+52x + 5 is an expression (it describes a value), whereas 2x+5=152x + 5 = 15 is an equation (it states a fact about the variable xx).

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The solution of an equation is the value of the variable for which the equation is satisfied (meaning LHS=RHSLHS = RHS). Only one specific value (or set of values) makes an equation true; for all other values, the LHS will not equal the RHS.

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Trial and Error Method: This is a fundamental way to find the solution by substituting different values for the variable and calculating the LHS. We continue this process until we find a value where the LHS matches the RHS provided in the equation.

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In Grade 6, we primarily deal with linear equations in one variable, where the highest power of the variable is 1. Visually, these represent simple relationships that can be plotted as points on a number line.

📐Formulae

LHS=RHSLHS = RHS

ax+b=cax + b = c (General form of a linear equation in one variable)

Value of LHS=Value of RHSValue\ of\ LHS = Value\ of\ RHS

x+a=b  ⟹  x=b−ax + a = b \implies x = b - a

x−a=b  ⟹  x=b+ax - a = b \implies x = b + a

💡Examples

Problem 1:

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

Solution:

Step 1: Try x=4x = 4. LHS=4+4=8LHS = 4 + 4 = 8. Since 8≠108 \neq 10, x=4x = 4 is not the solution. Step 2: Try x=5x = 5. LHS=5+4=9LHS = 5 + 4 = 9. Since 9≠109 \neq 10, x=5x = 5 is not the solution. Step 3: Try x=6x = 6. LHS=6+4=10LHS = 6 + 4 = 10. Since LHS=RHSLHS = RHS, the equation is satisfied.

Explanation:

We substituted different integers for xx and checked if the sum with 44 resulted in 1010. The process stopped when the equality was met at x=6x = 6.

Problem 2:

Check if n=5n = 5 is the solution to the equation 3n−2=133n - 2 = 13.

Solution:

Step 1: Identify the LHS, which is 3n−23n - 2. Step 2: Substitute n=5n = 5 into the LHS: 3×5−2=15−2=133 \times 5 - 2 = 15 - 2 = 13. Step 3: Compare with RHS. LHS=13LHS = 13 and RHS=13RHS = 13.

Explanation:

Since the value of the LHS after substitution is exactly equal to the RHS, n=5n = 5 is indeed the correct solution for the given equation.