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Chapter 4 — Simple Equations

Class 7 · Mathematics

Overview

This chapter introduces simple linear equations in one variable and builds the first formal bridge from arithmetic to algebra. Students learn what an equation is, how an unknown (variable) is represented, and why equality must be maintained on both sides. The chapter emphasizes the balancing (inverse-operation) method and transposition to isolate the unknown, and it teaches systematic steps: simplify both sides, collect like terms, perform inverse operations on both sides, and verify the solution. Importance: mastering simple equations develops logical thinking, procedural fluency, and problem‑solving skills needed for higher algebra, geometry, and real‑life quantitative reasoning. Key themes include variables and constants, coefficients, terms and expressions, equality and operations on both sides, solving one‑step and multi‑step linear equations (including those with parentheses, negatives and fractions), framing equations from word problems, and checking solutions. By the end of the chapter a student will be able to form an equation from a given statement, solve various types of linear equations using appropriate inverse operations or transposition, verify answers by…

Learning Objectives

  • Define the terms variable, constant, coefficient and equation in the context of simple equations
  • Explain the balance method and the principle of transposition for solving simple equations
  • Identify the unknown in a given statement and represent it by a suitable variable
  • Translate word problems into linear equations in one variable
  • Formulate simple equations from algebraic statements and everyday situations (ages, money, consecutive numbers)
  • Solve one-step linear equations using addition, subtraction, multiplication and division
  • Solve two-step linear equations of the form ax + b = c and check the solution by substitution
  • Solve linear equations with variables on both sides by collecting like terms and transposing

Topics in this chapter

10 topics · tap a topic title to jump straight to it.

🟰1

Introduction to Simple Equations

📐 MATHEMATICAL FORMULA / THEOREM

Introduction to Simple Equations

Key Point: If ax + b = c then ax = c - b and x = (c - b)/a (a ≠ 0).

What is a simple equation? A simple equation (or linear equation in one variable) is a mathematical sentence that shows two expressions are equal and contains a variable (unknown) usually denoted by letters like x. Example: x + 5 = 12.

Parts of an equation: the variable (unknown, e.g. x), the coefficient (number multiplied by variable), and the constant (a fixed number).

Goal: Find the value of the variable that makes the equality true. That value is called the solution or root of the equation.

Basic idea (balance method): Think of an equation as a balanced scale. Whatever you do to one side, you must do to the other side to keep it balanced. Use inverse operations to isolate the variable: subtract/add to remove constants, divide/multiply to remove coefficients.

Steps to solve ax + b = c: (1) Use addition/subtraction to move constants to the right: ax = c - b. (2) Divide by a to get x = (c - b)/a. (3) Check by substituting the value back into the original equation.

📌 Examples
  • 1) Solve x + 5 = 12. Subtract 5 from both sides: x = 12 - 5 = 7. Check: 7 + 5 = 12 ✓.
  • 2) Solve 3x = 15. Divide both sides by 3: x = 15/3 = 5. Check: 3×5 = 15 ✓.
  • 3) Solve 2x + 7 = 19. Subtract 7: 2x = 12. Divide by 2: x = 6. Check: 2×6 + 7 = 19 ✓.
  • 4) Word problem: Sarah has some stickers. After giving away 8, she has 14. How many did she have? Let x be original number. x - 8 = 14 ⇒ x = 14 + 8 = 22 stickers.
  • 5) Negative and division: Solve -4x = 20. Divide: x = 20 / (-4) = -5. Check: -4×(-5) = 20 ✓.
  • 6) Using transposition: Solve x - 9 = 3. Transpose -9 to other side as +9: x = 3 + 9 = 12.
🧮 Formulas
  1. \[If ax + b = c then ax = c - b and x = (c - b)/a (a ≠ 0).\]
  2. \[Addition/Subtraction property: If A = B then A ± k = B ± k (use to remove constants).\]
  3. \[Multiplication/Division property: If A = B then A×k = B×k and A/k = B/k for k ≠ 0 (use to remove coefficients).\]
  4. \[Transposition rule (shortcut): Move a term from one side to the other by changing its sign (e.g.\]
    \[x - 5 = 7 ⇒ x = 7 + 5).\]
  5. \[Check solution by substitution: replace variable with found value and verify both sides are equal.\]
🟰2

Basic Principles of Solving Equations

📐 MATHEMATICAL FORMULA / THEOREM

Basic Principles of Solving Equations

Key Point: Properties of equality: If a = b, then a + c = b + c; a - c = b - c; a × c = b × c; a ÷ c = b ÷ c (c ≠ 0).

What is an equation? An equation is a statement that two expressions are equal, e.g. 3x + 5 = 20. The unknown (usually x) is a number that makes the two sides equal.

Basic principle (Balance method): An equation is like a balanced scale. Whatever you do to one side, you must do to the other side to keep it balanced. This ensures equality remains true.

Inverse operations: To isolate the unknown, use inverse (opposite) operations: subtraction undoes addition, division undoes multiplication, and vice versa. Example: to remove +5, subtract 5 from both sides; to remove ×3, divide both sides by 3.

Step-by-step method:

  1. Simplify each side (expand brackets, combine like terms).
  2. Collect terms containing the unknown on one side and constants on the other.
  3. Use inverse operations to isolate the unknown (undo addition/subtraction first, then multiplication/division).
  4. Make sure not to divide by zero. If the coefficient of the unknown becomes zero, check for special cases (no solution or infinitely many solutions).
  5. Check the solution by substituting it back into the original equation.

Special cases:

  • If you get a true statement without the variable (e.g., 0 = 0) then the equation has infinitely many solutions.
  • If you get a false statement without the variable (e.g., 0 = 5) then the equation has no solution.

Why it works: The rules follow from properties of equality: adding, subtracting, multiplying (by non-zero), or dividing (by non-zero) the same quantity to both sides preserves equality. These properties let us transform an equation step by step to a simpler equivalent equation whose solution is the same as the original.

Common linear forms: ax + b = c or ax + b = cx + d. Solve by using the balance principle to isolate x.

📌 Examples
  • Simple: Solve x + 7 = 12. Subtract 7 from both sides: x = 12 - 7 = 5. Check: 5 + 7 = 12 (true).
  • With coefficient: Solve 3x - 4 = 11. Add 4 to both sides: 3x = 15. Divide both sides by 3: x = 5. Check: 3(5)-4 = 11.
  • Real-life (money): A notebook and a pen cost ₹65 together. If the notebook costs ₹50, find the price of the pen. Equation: 50 + p = 65 → p = 65 - 50 = 15. So the pen costs ₹15.
  • Age puzzle: Twice my age plus 5 equals 25. Let x = my age. 2x + 5 = 25 → 2x = 20 → x = 10 years. Check: 2×10+5 = 25.
🧮 Formulas
  1. \[Properties of equality: If a = b\]
    \[then a + c = b + c\]
    \[a - c = b - c\]
    \[a × c = b × c\]
    \[a ÷ c = b ÷ c (c ≠ 0).\]
  2. \[Standard linear equation: ax + b = c → x = (c - b) / a (a ≠ 0).\]
  3. \[Two-sided linear form: ax + b = cx + d → (a - c)x = d - b → x = (d - b) / (a - c) (a ≠ c).\]
  4. \[Check rule: Substitute found x back into the original equation to verify the equality.\]
3

Solving One-step and Two-step Equations (Addition/Subtraction)

📐 MATHEMATICAL FORMULA / THEOREM

Solving One-step and Two-step Equations (Addition/Subtraction)

Key Point: If x + a = b, then x = b - a.

What is an equation? An equation is a statement that two expressions are equal, e.g. x + 5 = 12. The goal is to find the value of the unknown (usually x) that makes the statement true.

Key idea — use inverse operations: To isolate the unknown, perform the inverse (opposite) operation to both sides of the equation. For addition, the inverse is subtraction; for subtraction, the inverse is addition. Always do the same operation to both sides so the equality stays true.

One-step equations (addition or subtraction)
These require one inverse operation to solve.

  • If the equation is x + a = b, subtract a from both sides: x = b - a.
  • If the equation is x - a = b, add a to both sides: x = b + a.

Two-step equations (involving addition/subtraction)
Two-step equations need two stages of work. There are two common situations:

  1. Equations where you first simplify by combining like terms (or removing parentheses), then perform one inverse operation. Example: x + 6 - 3 = 14 → simplify to x + 3 = 14 → subtract 3 → x = 11.
  2. Equations where the unknown also has a coefficient (e.g. 3x + 5 = 20). You usually first undo addition/subtraction, then undo multiplication/division. Example: 3x + 5 = 20 → subtract 5 → 3x = 15 → divide by 3 → x = 5.

General method (steps)

  1. Simplify each side (remove parentheses, combine like terms).
  2. Use inverse operations to isolate the variable (undo addition/subtraction first when both occur).
  3. If needed, use inverse of multiplication/division to get x alone.
  4. Check the solution by substituting back into the original equation.

Checking answers: Substitute the found value into the original equation and verify both sides are equal.

📌 Examples
  • 1) One-step (addition): Solve x + 7 = 12. Subtract 7 from both sides: x = 12 - 7 = 5. Check: 5 + 7 = 12 ✓.
  • 2) One-step (subtraction): Solve x - 9 = 4. Add 9 to both sides: x = 4 + 9 = 13. Check: 13 - 9 = 4 ✓.
  • 3) Two-step (addition/subtraction only): Solve x + 6 - 3 = 14. First simplify: x + 3 = 14. Subtract 3: x = 11. Check: 11 + 6 - 3 = 17 - 3 = 14 ✓.
  • 4) Two-step (addition then division): Solve 3x + 5 = 20. Subtract 5: 3x = 15. Divide by 3: x = 5. Check: 3(5) + 5 = 15 + 5 = 20 ✓.
  • 5) Variable on both sides (requires moving terms): Solve x + 4 = 2 + 7. Simplify right side: x + 4 = 9. Subtract 4: x = 5. Check: 5 + 4 = 9 ✓.
  • 6) Real-life numeric example: A shopkeeper had x rupees. After selling some goods he had x - 120 = 380. Add 120 to both sides: x = 500. So he started with 500 rupees.
🧮 Formulas
  1. \[If x + a = b\]
    \[then x = b - a.\]
  2. \[If x - a = b\]
    \[then x = b + a.\]
  3. \[To keep equality: do the same operation to both sides.\]
  4. \[For two-step with coefficient: to solve ax + b = c → first subtract b (ax = c - b)\]
    \[then divide by a (x = (c - b)/a).\]
  5. \[Always check: substitute the found x back into the original equation.\]
✖️4

Solving Equations Using Multiplication and Division

📐 MATHEMATICAL FORMULA / THEOREM

Solving Equations Using Multiplication and Division

Key Point: If a ≠ 0 and ax = b, then x = b/a.

An equation is a statement that two expressions are equal. In many simple equations the unknown (say x) is multiplied or divided by a number. To solve such equations we use inverse operations so that x is isolated on one side.

Key idea: If x is multiplied by a nonzero number a (ax = b), divide both sides by a to get x = b/a. If x is divided by a (x/a = b), multiply both sides by a to get x = a·b. Always perform the same operation on both sides — think of a balanced scale: whatever you do to one pan, do to the other.

Steps to solve:

  • Identify whether the unknown is multiplied or divided by a number.
  • Apply the inverse operation to both sides (divide if multiplied, multiply if divided).
  • Simplify and find x.
  • Check the solution by substituting back into the original equation.

Caution: You cannot divide by zero. In equations of the form ax = b, a must be nonzero to get a unique solution.

📌 Examples
  • 1) 4x = 20. Divide both sides by 4: x = 20 ÷ 4 = 5. Check: 4×5 = 20 ✓.
  • 2) x/5 = 7. Multiply both sides by 5: x = 7×5 = 35. Check: 35/5 = 7 ✓.
  • 3) (1/3)x = 4. Multiply both sides by 3: x = 4×3 = 12. Check: (1/3)×12 = 4 ✓.
  • 4) -3x = 9. Divide both sides by -3: x = 9 ÷ (-3) = -3. Check: -3×(-3) = 9 ✓.
  • 5) 0.5x = 6. Divide both sides by 0.5 (or multiply by 2): x = 6 ÷ 0.5 = 12. Check: 0.5×12 = 6 ✓.
🧮 Formulas
  1. \[If a ≠ 0 and ax = b\]
    \[then x = b/a.\]
  2. \[If x/a = b (with a ≠ 0)\]
    \[then x = a·b.\]
  3. \[If (m/n)x = b (m ≠ 0)\]
    \[then x = (n·b)/m.\]
  4. \[Always check the solution by substituting x back into the original equation.\]
🟰5

Equations with Variables on Both Sides

📐 MATHEMATICAL FORMULA / THEOREM

Equations with Variables on Both Sides

Key Point: General linear form: ax + b = cx + d (a, b, c, d are constants).

An equation with variables on both sides is an equation in which the unknown (variable) appears in expressions on both the left-hand side (LHS) and the right-hand side (RHS). Example: 3x + 5 = x + 11.

Goal: find the value(s) of the variable that make the two sides equal.

General method (step-by-step):

  1. Expand both sides (remove brackets) if needed.
  2. Bring all terms containing the variable to one side (use addition/subtraction on both sides).
  3. Bring constant terms to the other side.
  4. Simplify and solve for the variable by dividing or multiplying as required.
  5. Check the solution by substituting back into the original equation.

Transposition idea (useful shortcut): to move a term from one side to the other, change its sign and perform the same operation on both sides (this is just applying addition/subtraction to both sides).

Special cases:

  • If after simplifying you get a true statement like 0 = 0, the equation is an identity and has infinitely many solutions.
  • If you get a false statement like 0 = 9, the equation has no solution (it is inconsistent).

Example of the general form and how to solve it: For ax + b = cx + d (where a, b, c, d are numbers): bring variable terms together: (a - c)x = d - b. If a ≠ c then x = (d - b) / (a - c). If a = c then either infinite solutions (if d = b) or no solution (if d ≠ b).

Always verify the result by substituting the value of the variable into the original equation.

📌 Examples
  • 1) Solve: 3x + 5 = x + 11. Steps: 3x - x = 11 - 5 ⇒ 2x = 6 ⇒ x = 3. Check: 3(3)+5 = 14 and 3+11 = 14 ✓.
  • 2) Solve: 2(x + 3) = 4x - 2. Steps: 2x + 6 = 4x - 2 ⇒ 6 + 2 = 4x - 2x ⇒ 8 = 2x ⇒ x = 4. Check: LHS = 2(7) = 14, RHS = 16 - 2 = 14 ✓.
  • 3) Solve: 5x - 7 = 5x + 2. Steps: 5x - 5x = 2 + 7 ⇒ 0 = 9. This is impossible ⇒ No solution (inconsistent).
  • 4) Solve: 4x + 6 = 4(x + 1.5). Steps: RHS = 4x + 6 ⇒ equation becomes 4x + 6 = 4x + 6 ⇒ 0 = 0. This is always true ⇒ infinitely many solutions (identity).
  • 5) Real-life example: Two taxis. Taxi A charges Rs 40 plus Rs 10 per km; Taxi B charges Rs 25 plus Rs 12 per km. After how many km are fares equal? 40 + 10x = 25 + 12x ⇒ 15 = 2x ⇒ x = 7.5 km. At 7.5 km both cost Rs 115.
🧮 Formulas
  1. \[General linear form: ax + b = cx + d (a\]
    \[b\]
    \[c\]
    \[d are constants).\]
  2. \[Bring variables together: (a - c)x = d - b.\]
  3. \[Solution (if a ≠ c): x = (d - b) / (a - c).\]
  4. \[If a = c and d = b → infinitely many solutions (identity).\]
  5. \[If a = c and d ≠ b → no solution (contradiction).\]
  6. \[Valid operations: add/subtract same number from both sides\]
    \[multiply/divide both sides by same non-zero number.\]
🟰6

Equations Involving Brackets and Simplification

📐 MATHEMATICAL FORMULA / THEOREM

Equations Involving Brackets and Simplification

Key Point: Distributive property: a(b + c) = ab + ac and a(b - c) = ab - ac

What it means: An equation involving brackets contains expressions in parentheses (brackets) such as a(b + c) or (x + 3). To solve such equations you must first simplify by removing brackets (using distributive property), then collect like terms, and finally use inverse operations to isolate the variable.

Step-by-step method:

  1. Use the distributive property to remove brackets: a(b + c) = ab + ac and a(b - c) = ab - ac.
  2. Simplify each side: combine like terms (add/subtract coefficients of same variable and constant terms).
  3. Bring variable terms to one side and constants to the other (use addition/subtraction).
  4. Use multiplication/division to make the variable coefficient 1 (inverse operations).
  5. Check the solution by substituting back into the original equation.

Important idea: The balancing principle — whatever you do to one side of the equation, do the same to the other side.

Worked outline (general): For an equation like p(qx + r) + s = t(ux + v) + w, first expand both sides to get linear expressions of the form Ax + B = Cx + D, then follow the steps above to find x.

📌 Examples
  • Example 1: Solve 3(x + 4) = 21 Step 1: Expand: 3x + 12 = 21 Step 2: Subtract 12 from both sides: 3x = 9 Step 3: Divide by 3: x = 3 Check: 3(3 + 4) = 3*7 = 21 ✓
  • Example 2: Solve 2(x - 5) + 3 = 3x + 1 Step 1: Expand: 2x - 10 + 3 = 3x + 1 → 2x - 7 = 3x + 1 Step 2: Subtract 2x from both sides: -7 = x + 1 Step 3: Subtract 1: x = -8 Check: 2(-8 - 5)+3 = 2(-13)+3 = -26+3 = -23 and 3(-8)+1 = -24+1 = -23 ✓
  • Example 3: Solve 4(x + 2) - 3(2x - 1) = 5 Step 1: Expand: 4x + 8 - 6x + 3 = 5 → -2x + 11 = 5 Step 2: Subtract 11: -2x = -6 Step 3: Divide by -2: x = 3 Check: 4(3+2)-3(6-1)=4*5-3*5=20-15=5 ✓
🧮 Formulas
  1. \[Distributive property: a(b + c) = ab + ac and a(b - c) = ab - ac\]
  2. \[Combine like terms: (mx + nx) = (m + n)x and (a + b) are constants added together\]
  3. \[Inverse operations: use addition/subtraction to move terms and multiplication/division to isolate the variable\]
  4. \[Balancing principle: perform the same operation on both sides of the equation\]
🟰7

Formulating Equations from Word Problems

📐 MATHEMATICAL FORMULA / THEOREM

Formulating Equations from Word Problems

Key Point: General linear form: ax + b = c. Solve by: x = (c − b)/a (provided a ≠ 0).

What it means
Formulating equations from word problems means reading a situation, choosing a variable for the unknown, converting words into a mathematical equation (usually a linear equation in one variable), solving it, and checking the answer.

Step-by-step method

  1. Read carefully: Understand exactly what is asked and identify known quantities and the unknown.
  2. Choose a variable: Let x (or any letter) represent the unknown quantity.
  3. Translate words to symbols: Convert phrases into mathematical expressions using a list of common keyword translations (see below).
  4. Form the equation: Combine expressions to make an equation (usually of the form ax + b = c).
  5. Solve the equation: Use inverse operations (add/subtract, multiply/divide) to find the value of the variable.
  6. Check: Substitute the value back into the original words to ensure it makes sense.

Common keyword translations

  • "sum", "increased by", "more than" → +
  • "difference", "decreased by", "less than" → − (note: "b less than a" means a − b)
  • "product", "times", "twice" → ×
  • "quotient", "divided by" → ÷
  • "is", "are", "gives", "equals" → =
  • "consecutive integers" → n, n+1, n+2,...

Tips for tricky phrasing

  • For "7 more than a number is 12" translate as x + 7 = 12 (not 7 + x = 12 — same mathematically but keep order to check meaning).
  • For "7 less than a number is 12" translate as x - 7 = 12. For "7 less than 12 is a number" translate as 12 - 7 = x.
  • Always identify which quantity is the subject when two quantities are compared (who is being compared to whom).

Why this helps
Formulating equations turns a real-world story into a precise mathematical statement. Once the equation is correct, algebra gives the answer reliably.

📌 Examples
  • Example 1 — Simple increase: Problem: A number increased by 7 is 15. Form equation and solve. Let x = the number. Then x + 7 = 15. Solve: x = 15 − 7 = 8. Check: 8 increased by 7 is 15.
  • Example 2 — Age problem: Problem: Five years ago, Ravi was 3 times as old as his son. If Ravi is 41 now, find his son's present age. Let s = son's present age. Five years ago son's age = s − 5, Ravi's age five years ago = 41 − 5 = 36. Equation: 36 = 3(s − 5). Solve: 36 = 3s − 15 ⇒ 3s = 51 ⇒ s = 17. Check: Five years ago son was 12 and Ravi 36; 36 is 3 × 12.
  • Example 3 — Money/sharing: Problem: Meena had rupees x. She spent 120 and had one-fourth of her money left. Find x. Let x = original money. After spending 120 she has (x − 120) and this equals (1/4)x. Equation: x − 120 = (1/4)x. Solve: x − (1/4)x = 120 ⇒ (3/4)x = 120 ⇒ x = 160. Check: 160 − 120 = 40 which is one-fourth of 160.
  • Example 4 — Consecutive integers: Problem: The sum of three consecutive integers is 48. Find them. Let n = first integer. Then n + (n+1) + (n+2) = 48 ⇒ 3n + 3 = 48 ⇒ 3n = 45 ⇒ n = 15. The integers: 15, 16, 17.
🧮 Formulas
  1. \[General linear form: ax + b = c\]
    \[Solve by: x = (c − b)/a (provided a ≠ 0).\]
  2. \[Consecutive integers: n\]
    \[n+1\]
    \[n+2, ..\]
    \[Use n for the first integer.\]
  3. \[Consecutive even (or odd) integers: n\]
    \[n+2\]
    \[n+4, ...\]
  4. \[Fractional parts: If after operation the remaining part is p/q of original\]
    \[e.g.\]
    \[remainder = (p/q)·x\]
    \[then form x − (taken) = (p/q)·x and solve.\]
  5. \[Age relation template: If present ages are A and B and a years ago relation is given\]
    \[use (A − a) and (B − a) in equation.\]
  6. \[Percent template: &quot\]
    \[After spending r% of x\]
    \[left is (100−r)% of x → (100−r)/100 · x = given amount&quot\]
    \[.\]
🧴8

Verification and Checking Solutions

📐 MATHEMATICAL FORMULA / THEOREM

Verification and Checking Solutions

Key Point: Substitution principle: If x = a is a proposed solution, then replace x by a in the original equation and evaluate both sides.

What it means: Verification (or checking) a solution means substituting the value you found for the unknown into the original equation to see whether both sides become equal. It confirms the solution is correct and helps catch errors.

Why it is important: Solving steps may have arithmetic slips or sign errors. Checking ensures the answer truly satisfies the equation. For some equations (with fractions or when squaring), checking prevents accepting extraneous answers.

Steps to verify a solution:

  1. Take the value found for the variable (say x = a).
  2. Substitute x = a into the original equation everywhere x appears.
  3. Simplify both sides separately to compute the left-hand side (LHS) and the right-hand side (RHS).
  4. Compare LHS and RHS. If they are equal, x = a is a correct solution; otherwise it is not.

Tips: Show the substitution clearly (write the equation with the number plugged in). Always simplify carefully (handle signs and fractions). If LHS ≠ RHS, rework the solving steps or try solving again.

📌 Examples
  • Example 1 (simple): Solve x + 5 = 12. Solution: x = 7. Check: Substitute x = 7 → LHS = 7 + 5 = 12, RHS = 12. Since LHS = RHS, x = 7 is correct.
  • Example 2 (with coefficient): Solve 3x - 4 = 11. Solution: 3x = 15 → x = 5. Check: Substitute x = 5 → LHS = 3(5) - 4 = 15 - 4 = 11, RHS = 11. Correct.
  • Example 3 (fraction): Solve x/4 + 2 = 6. Solution: x/4 = 4 → x = 16. Check: Substitute x = 16 → LHS = 16/4 + 2 = 4 + 2 = 6, RHS = 6. Correct.
  • Example 4 (negative coefficients): Solve -2x + 3 = 7. Solution: -2x = 4 → x = -2. Check: Substitute x = -2 → LHS = -2(-2) + 3 = 4 + 3 = 7, RHS = 7. Correct.
  • Example 5 (wrong claim): For (x + 3)/2 = 5 someone claims x = 4. Check: Substitute x = 4 → LHS = (4 + 3)/2 = 7/2 = 3.5, RHS = 5. LHS ≠ RHS so x = 4 is incorrect. Real solution: x + 3 = 10 → x = 7; check gives (7+3)/2 = 5.
🧮 Formulas
  1. \[Substitution principle: If x = a is a proposed solution\]
    \[then replace x by a in the original equation and evaluate both sides.\]
  2. \[Property of equality: Whatever operation (addition\]
    \[subtraction\]
    \[multiplication\]
    \[division by nonzero) you perform on one side of an equation\]
    \[perform it on the other side to keep equality while solving.\]
  3. \[General linear solution: For ax + b = c (a ≠ 0)\]
    \[x = (c - b) / a\]
    \[After computing x\]
    \[verify by substitution: a((c - b)/a) + b = c.\]
  4. \[Checking rule: x = a is correct ⇔ LHS(x=a) = RHS(x=a).\]
🔢9

Problem-Solving Strategies and Tips

📐 MATHEMATICAL FORMULA / THEOREM

Problem-Solving Strategies and Tips

Key Point: General linear form: ax + b = c → x = (c - b) / a , provided a ≠ 0

Simple equations involve finding the value of a single unknown that makes an equality true. Effective problem solving combines careful translation of words into symbols, algebraic manipulation, and checking. The following clear strategy and tips help solve most one-variable linear equations encountered in Class 7.

  • Step 1: Read and understand - Read the problem twice. Identify what is known (numbers, relationships) and what you must find (the unknown).
  • Step 2: Translate words to an equation - Introduce a variable (commonly x). Convert phrases: "sum of", "difference", "product", "quotient", "is", "equals" into +, -, ×, ÷, =. Use brackets for clarity when needed.
  • Step 3: Simplify both sides - Expand brackets, collect like terms and combine numbers on each side so each side is as simple as possible.
  • Step 4: Isolate the variable - Use inverse operations to get the variable alone. Use addition/subtraction first to move constants, then multiplication/division to remove coefficients. Always do the same operation to both sides to keep the equation balanced.
  • Step 5: Clear fractions if present - Multiply both sides by the LCM of denominators before simplifying to avoid working with fractions from the start.
  • Step 6: Solve and check - Compute the value of the variable and substitute it back into the original equation to verify correctness and units.
  • Helpful models and techniques
    • Balance model - Think of the equation as a balance scale: what you do to one side do to the other.
    • Bar model - Draw bars representing parts of the quantity for word problems (useful for comparison or parts-of-a-whole problems).
    • Number line - Use for small integer solutions and to show relative position of solution.
  • Common tips and cautions
    • Keep signs carefully. When moving a term across the equals sign, change its sign.
    • Always perform identical operations on both sides.
    • Prefer simplifying each side completely before isolating the variable.
    • Check answers by substitution; this catches sign and arithmetic mistakes.
    • For fractional coefficients, clear denominators early to reduce errors.

Following a consistent step-by-step approach reduces mistakes, builds confidence, and makes solving word problems systematic.

📌 Examples
  • Example 1 — Simple addition: 'A number increased by 7 is 15.' Equation: x + 7 = 15. Solve: x = 15 - 7 = 8. Check: 8 + 7 = 15.
  • Example 2 — Using transposition: 'Three times a number minus 5 equals 16.' Equation: 3x - 5 = 16. Add 5 to both sides: 3x = 21. Divide by 3: x = 7. Check: 3(7)-5 = 21-5 =16.
  • Example 3 — Clearing fractions: 'One half of a number plus 3 is 11.' Equation: (1/2)x + 3 = 11. Subtract 3: (1/2)x = 8. Multiply both sides by 2: x = 16. Check: (1/2)*16 + 3 = 8 + 3 =11.
  • Example 4 — Word problem with bar model: 'Rita has 4 more pencils than Kiran. Together they have 20 pencils. How many does Kiran have?' Let Kiran = x, Rita = x + 4. Equation: x + (x + 4) = 20 → 2x + 4 = 20 → 2x = 16 → x = 8. Kiran has 8, Rita has 12. Check: 8 + 12 = 20.
🧮 Formulas
  1. \[General linear form: ax + b = c → x = (c - b) / a\]
    \[provided a ≠ 0\]
  2. \[Addition/Subtraction property: If A = B then A ± k = B ± k (perform same operation on both sides)\]
  3. \[Multiplication/Division property: If A = B then A × k = B × k and A ÷ k = B ÷ k (k ≠ 0)\]
  4. \[Transposition rule (informal): Moving a term across = changes its sign (e.g.\]
    \[x - 5 = 7 → x = 7 + 5)\]
  5. \[Clearing denominators: Multiply both sides by LCM of denominators to remove fractions before simplifying\]
  6. \[Check rule: Substitute found value into original equation to verify both sides are equal\]
🔢10

Practice Exercises and Example Types

📐 MATHEMATICAL FORMULA / THEOREM

Practice Exercises and Example Types

Key Point: For ax + b = c ⇒ ax = c − b ⇒ x = (c − b)/a (provided a ≠ 0).

What this topic covers
Practice Exercises and Example Types in Simple Equations (Class 7) help students apply methods to solve linear equations in one variable. Exercises range from one-step and two-step equations to equations with variables on both sides, and word problems that model real-life situations.

General approach / steps to solve

  • Read the equation carefully and identify the unknown (variable).
  • Simplify each side separately (collect like terms, remove parentheses if any).
  • Use inverse operations to isolate the variable: add/subtract to remove constants, multiply/divide to remove coefficients.
  • If variable appears on both sides, bring variable terms to one side (use transposition: moving a term across the equals sign changes its sign).
  • Perform arithmetic to find the variable. Always verify by substituting the value back into the original equation.

Types of practice examples

  • One-step equations: e.g. x + a = b or ax = b. Solve by single inverse operation.
  • Two-step equations: e.g. ax + b = c. First remove b by subtraction/addition, then divide by a.
  • Variables on both sides: e.g. ax + b = cx + d. Bring like terms together and solve.
  • Identity and contradiction: Some equations are true for all x (identity) or for no x (contradiction). Example: 3(x+1)=3x+3 (identity), 2(x+1)=2x+5 (contradiction).
  • Word problems: Translate a real situation into an equation (age, money, length/perimeter, shares) and solve.

Tips and common mistakes

  • Always perform the same operation on both sides.
  • When transposing, change the sign: moving +5 to other side becomes −5.
  • Watch for distributive property before simplifying: 2(x+3) = 2x + 6.
  • Check the solution by substitution into the original equation (not into a simplified form only).
📌 Examples
  • 1) One-step: Solve x + 7 = 12. Solution: x = 12 − 7 = 5. Check: 5 + 7 = 12.
  • 2) Two-step: Solve 3x + 4 = 19. Solution: 3x = 19 − 4 = 15 ⇒ x = 15/3 = 5. Check: 3(5)+4 = 19.
  • 3) Variable both sides: Solve 5x − 2 = 2x + 7. Solution: 5x − 2x = 7 + 2 ⇒ 3x = 9 ⇒ x = 3. Check: 5(3)−2 = 13 and 2(3)+7 = 13.
  • 4) Identity/Contradiction: 2(x + 3) = 2x + 6 ⇒ 2x + 6 = 2x + 6 (identity) so every real x satisfies it. Example of contradiction: 2(x + 1) = 2x + 5 ⇒ 2x + 2 = 2x + 5 ⇒ 2 = 5 (false), so no solution.
  • 5) Word problem (money): Rahul has some rupees. If he spends 45 rupees he has 70 left. Find original amount. Equation: x − 45 = 70 ⇒ x = 70 + 45 = 115 rupees.
  • 6) Word problem (age): Five years ago, A was three times as old as B. Now sum of their ages is 44. Let current ages be a and b. Given: (a−5) = 3(b−5) and a + b = 44. Solve: a = 44 − b. Substitute: (44−b−5)=3(b−5) ⇒ 39−b = 3b−15 ⇒ 39+15 = 4b ⇒ 54 = 4b ⇒ b = 13.5 years, a = 30.5 years. (Shows using two equations and substitution.)
🧮 Formulas
  1. \[For ax + b = c ⇒ ax = c − b ⇒ x = (c − b)/a (provided a ≠ 0).\]
  2. \[Inverse operations: to remove +k use −k on both sides\]
    \[to remove −k use +k\]
    \[to remove ×k use ÷k\]
    \[to remove ÷k use ×k.\]
  3. \[Transposition rule (informal): moving a term across '=' changes its sign\]
    \[Example: x + 5 = 12 ⇒ x = 12 − 5.\]
  4. \[Distributive property: a(b + c) = ab + ac\]
    \[use before combining like terms.\]
  5. \[Checking: substitute the found x back into the original equation to verify equality.\]

Key Concepts

Variable
A symbol (usually a letter) that represents an unknown number.
Constant
A fixed number that does not change its value in an expression or equation.
Coefficient
The numerical factor multiplied by a variable in a term.
Term
A single number, a variable, or the product of numbers and variables separated by + or − signs.
Algebraic Expression
A combination of terms (numbers, variables, products) without an equality sign.
Equation
A mathematical statement that two expressions are equal, containing an = sign.
Simple Equation
An equation of the first degree in one variable that can be solved by basic operations.
Linear Equation
An equation in which the highest power of the variable is 1 (degree one).
Unknown
The specific value the variable stands for that makes an equation true.
Solution (Root)
A value of the variable that satisfies the equation (makes both sides equal).
Forming Equations
Writing an equation from a word problem by translating phrases into mathematical expressions.
Balance Method
Solving by performing equal operations on both sides of the equation to keep it balanced.
Transposition
Moving a term from one side of an equation to the other by changing its sign.
Addition/Subtraction Principle
If two sides of an equation are equal, adding or subtracting the same number to both sides preserves equality.
Multiplication/Division Principle
If two sides of an equation are equal, multiplying or dividing both sides by the same nonzero number preserves equality.
Like Terms
Terms that have the same variable factors (same variables raised to the same powers).
Degree (of an Equation)
The highest power of the variable in an equation; for simple (linear) equations the degree is 1.
Identity
An equation that is true for all values of the variable involved.
Contradiction
An equation that has no solution; it is never true for any value of the variable.
Verification
Checking by substituting the found value into the original equation to confirm both sides are equal.

Practice Questions

  1. Which of the following is a simple (linear) equation in one variable? / निम्नलिखित में से कौन-सा एक चर में रैखिक (सरल) समीकरण है? (a) x² + 5 = 14 (b) 3x + 7 = 19 (c) xy = 10 (d) x + y = 5
    Show answer

    (b) 3x + 7 = 19 — It has only one variable x raised to the first power, making it a linear equation in one variable. / क्योंकि इसमें केवल एक चर x है जिसकी घात 1 है, इसलिए यह एक चर में रैखिक समीकरण है।

  2. Solve: 2x + 5 = 17. What is the value of x? / हल करें: 2x + 5 = 17। x का मान क्या है? (a) 6 (b) 7 (c) 11 (d) 4
    Show answer

    (a) 6 — Subtract 5 from both sides: 2x = 12; divide by 2: x = 6. Check: 2(6)+5 = 17 ✓ / दोनों पक्षों से 5 घटाएं: 2x = 12; 2 से भाग दें: x = 6।

  3. If five times a number is decreased by 3, the result is 22. Which equation represents this? / यदि किसी संख्या का पाँच गुना 3 से घटाया जाए तो परिणाम 22 है। इसे दर्शाने वाला समीकरण कौन-सा है? (a) 5 – 3x = 22 (b) 5x + 3 = 22 (c) 5x – 3 = 22 (d) 5(x – 3) = 22
    Show answer

    (c) 5x – 3 = 22 — 'Five times a number' is 5x, 'decreased by 3' means –3, and the result is 22. / 'संख्या का पाँच गुना' = 5x, '3 से घटाया' = –3, परिणाम = 22।

  4. Fill in the blank: In the equation 4x – 9 = 11, the value of x is ______. / रिक्त स्थान भरें: समीकरण 4x – 9 = 11 में x का मान ______ है।
    Show answer

    x = 5 — Add 9 to both sides: 4x = 20; divide by 4: x = 5. Check: 4(5)–9 = 11 ✓ / दोनों पक्षों में 9 जोड़ें: 4x = 20; 4 से भाग दें: x = 5।

  5. Fill in the blank: Transposition means moving a term to the other side of the equation with ______ sign. / रिक्त स्थान भरें: पक्षान्तरण का अर्थ है किसी पद को समीकरण के दूसरी तरफ ______ चिह्न के साथ स्थानांतरित करना।
    Show answer

    Opposite / विपरीत — When a term moves across the '=' sign, its sign changes (e.g., +5 becomes –5). / जब कोई पद '=' चिह्न को पार करता है, तो उसका चिह्न बदल जाता है।

  6. True or False: If x = 4 satisfies an equation, substituting x = 4 in the equation makes both sides equal. / सत्य या असत्य: यदि x = 4 किसी समीकरण को संतुष्ट करता है, तो x = 4 रखने पर समीकरण के दोनों पक्ष समान हो जाते हैं।
    Show answer

    True / सत्य — This is the definition of a solution (root). Verification means substituting the found value and checking that LHS = RHS. / यह हल (मूल) की परिभाषा है। सत्यापन में प्राप्त मान रखकर LHS = RHS जाँचते हैं।

  7. Riya had some money. After spending ₹45 she had ₹80 left. If x is the original amount, form an equation and find x. / रिया के पास कुछ पैसे थे। ₹45 खर्च करने के बाद उसके पास ₹80 बचे। यदि x मूल राशि है, तो समीकरण बनाकर x ज्ञात करें।
    Show answer

    Equation: x – 45 = 80 → x = 80 + 45 = ₹125 / समीकरण: x – 45 = 80 → x = 80 + 45 = ₹125। The original amount was ₹125. / मूल राशि ₹125 थी।

  8. Solve for x: 3(x + 4) = 21. Show the steps. / x के लिए हल करें: 3(x + 4) = 21। चरण दिखाएं।
    Show answer

    Step 1 – Expand brackets: 3x + 12 = 21. Step 2 – Subtract 12: 3x = 9. Step 3 – Divide by 3: x = 3. Check: 3(3+4)=3×7=21 ✓ / चरण 1 – कोष्ठक हटाएं: 3x + 12 = 21। चरण 2 – 12 घटाएं: 3x = 9। चरण 3 – 3 से भाग: x = 3।

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