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Simple Equations - Solving an equation

Grade 7CBSE

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

🔑Concepts

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An equation is a mathematical statement showing that two expressions are equal, much like a weighing scale in perfect equilibrium where the Left Hand Side (LHS) and Right Hand Side (RHS) are balanced.

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A variable is a symbol, usually a letter like x,y,x, y, or zz, that represents an unknown number, whereas a constant is a fixed value that does not change.

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The fundamental rule of solving equations is to keep the equation balanced by performing the exact same mathematical operation on both sides (addition, subtraction, multiplication, or division).

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Transposition is the process of moving a term from one side of the equal sign to the other. When a term is transposed, its sign changes: addition becomes subtraction and subtraction becomes addition.

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In multiplication and division transposition, a number multiplying the variable on one side becomes a divisor on the other side, and a number dividing the variable becomes a multiplier on the other side.

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The 'solution' or 'root' of an equation is the specific value of the variable that makes the LHS equal to the RHS. Visually, this is the point where the scales of the equation sit perfectly level.

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To solve an equation with multiple operations, generally follow the reverse order of operations to isolate the variable, often dealing with addition/subtraction before multiplication/division.

📐Formulae

LHS=RHSLHS = RHS

If x+a=bx + a = b, then x=b−ax = b - a

If x−a=bx - a = b, then x=b+ax = b + a

If ax=bax = b, then x=bax = \frac{b}{a} (where a≠0a \neq 0)

If \frac{x}{a} = b, then x=b×ax = b \times a (where a≠0a \neq 0)

💡Examples

Problem 1:

Solve for nn: 3n+7=253n + 7 = 25

Solution:

Step 1: Transpose +7+7 to the RHS by subtracting it from both sides: 3n=25−73n = 25 - 7 3n=183n = 18 Step 2: Transpose the multiplier 33 to the RHS by dividing: n=183n = \frac{18}{3} n=6n = 6 Verification: 3(6)+7=18+7=253(6) + 7 = 18 + 7 = 25. LHS = RHS.

Explanation:

We first isolate the term containing the variable by moving the constant, then we isolate the variable itself by dividing by its coefficient.

Problem 2:

Solve for xx: 2(x−3)=102(x - 3) = 10

Solution:

Step 1: Divide both sides by 22 to remove the bracket: x−3=102x - 3 = \frac{10}{2} x−3=5x - 3 = 5 Step 2: Transpose −3-3 to the RHS: x=5+3x = 5 + 3 x=8x = 8 Verification: 2(8−3)=2(5)=102(8 - 3) = 2(5) = 10. LHS = RHS.

Explanation:

In this problem, it is easier to divide the entire LHS by the outer multiplier first before handling the subtraction inside the parentheses.