Class 6 Mathematics - Ch 6: Mastery Guide: Linear Equations in One Variable, Algebraic Sentences, Fractional & Decimal Equations, and Applied Word Problems (FBISE)
Instructional Guide: Unit 6 Linear Equations
- Distinguish between open sentences, true sentences, and false sentences.
- Define a linear equation in one variable ($ax + b = c, a \ne 0$) and identify its highest exponent as 1.
- Construct linear equations from everyday English sentences and real-world scenarios.
- Solve simple linear equations using the balance scale principle (adding/subtracting/multiplying/dividing same numbers on both sides).
- Solve linear equations involving integers, fractions, brackets, and decimals.
- Solve real-life word problems involving ages, consecutive integers, geometry perimeters, and money.
- Algebraic Expressions: Combining like terms, evaluating expressions by numerical substitution (from Unit 5).
- Integers & Fractions: Addition and subtraction of signed integers, finding LCM to clear fractional denominators.
- Distributive Property: Multiplying outside constants into brackets: $a(bx + c) = abx + ac$.
- One-Sided Operation Error: Performing an operation on only one side of the equation, destroying the balance: e.g., $x + 5 = 12 \implies x = 12 + 5$ (WRONG!).
- Sign Reversal on Transposition: Forgetting that moving a term across the equal sign changes its operation ($+ \leftrightarrow -$, $\times \leftrightarrow \div$).
- Consecutive Numbers Setup: Confusing consecutive integers ($x, x+1, x+2$) with consecutive even/odd integers ($x, x+2, x+4$).
- Bracket Distribution Error: Failing to multiply all terms inside: e.g., $5(1.2x - 4) \ne 6x - 4$ (Must be $6x - 20$).
Kid-Friendly Rhymes & Memory Tricks
"What you do to the Left pan, do to the Right,
Keep the beam balanced, level and tight!"
"Cross over the equals bridge, take a flip and turn,
Plus becomes minus, a quick lesson to learn!"
"Without a value for $x$, I cannot say True or False,
I am an Open Sentence, waiting for your call!"
"Multiply by the LCM across every term,
Fractions disappear, make your answer firm!"
Why Linear Equations Matter in Everyday Life
"A father is 3 times as old as his son; their combined age is 40." $\implies x + 3x = 40 \implies 4x = 40$ $\implies x = 10\text{ (Son)}, 30\text{ (Father)}$.
Buying a notebook and geometry box for Rs. 130 where the notebook costs Rs. 10 more than twice the box: $g + (2g+10) = 130 \implies g = 40\text{ Rs}$.
Finding length and width of a fenced garden when perimeter is 24m and length is twice the width:
$2(2w + w) = 24 $
$\implies 6w = 24 $
$\implies w = 4\text{m}, l = 8\text{m}$.
Dividing a 27m electrical cable into two pieces such that one is 9m longer: $x + (x+9) = 27 \implies 2x = 18$ $ \implies x = 9\text{m}, 18\text{m}$.
4. Comprehensive Conceptual Theory
4.1 Types of Sentences in Mathematics
In mathematics, statements are classified based on whether their truth value is definite or variable:
| Sentence Type | Definition | Examples |
|---|---|---|
| True Sentence (Closed) | A mathematical statement that is universally correct. | $2 + 3 = 5$, $13 + 12 = 25$, $2x + 3 = 3 + 2x$ |
| False Sentence (Closed) | A mathematical statement that is incorrect. | $5 - 4 = 9$, $4 + 15 = 20$, $3x + 6x - x = 7x$ |
| Open Sentence | A statement containing one or more unknown variables whose truth cannot be decided until values are substituted. | $x + 3 = 9$, $5x - 7 = 3$, $x > 9$ |
4.2 Linear Equations in One Variable
An equation is an open sentence joined by the equality sign ($=$). A linear equation in one variable is an algebraic equation in which the highest exponent/power of the variable is 1.
Standard Form of a Linear Equation in One Variable:
4.3 Properties of Equality & Balance Scale Principle
An equation remains true and balanced when identical arithmetic operations are applied to both sides:
If $A = B$, then $A + c = B + c$.
e.g., $x - 8 = 20 $
$\implies x - 8 + 8 = 20 + 8 $
$\implies x = 28$.
If $A = B$, then $A - c = B - c$.
e.g., $x + 9 = 15 $
$\implies x + 9 - 9 = 15 - 9 $
$\implies x = 6$.
If $A = B$, then $A \times c = B \times c$.
e.g., $\frac{1}{3}x = 4 $
$\implies 3 \times \frac{1}{3}x = 3 \times 4 $
$\implies x = 12$.
If $A = B$, then $\frac{A}{c} = \frac{B}{c}$ ($c \ne 0$).
e.g., $5x = 10 \implies \frac{5x}{5} = \frac{10}{5} \implies x = 2$.
4.4 Step-by-Step Method for Solving Linear Equations
- Clear Fractions & Decimals: Multiply all terms by the Least Common Denominator (LCD) or powers of 10.
- Remove Brackets / Parentheses: Apply the distributive law: $a(bx + c) = abx + ac$.
- Combine Like Terms: Simplify like terms separately on the Left-Hand Side (LHS) and Right-Hand Side (RHS).
- Collect Variable Terms on One Side: Add/subtract variable terms so that the variable appears only on one side.
- Isolate the Variable: Multiply or divide by the variable's coefficient to get $x = \text{number}$.
- Check / Verify: Substitute the solution back into the original equation to confirm $\text{LHS} = \text{RHS}$.
5. Exercise 6.1 Step-by-Step Solutions
Question 1: Write algebraic sentences for the following word sentences:
$$\mathbf{x + 14 = 17}$$
$$\mathbf{x + 24 > 50}$$
$$\mathbf{10x = 150}$$
$$\mathbf{x + 10 < 45}$$
$$\mathbf{x + 1.83 = 11.08}$$
$$\mathbf{19 - 2x \le 10}$$
$$\mathbf{\frac{x}{6} = 4}$$
$$\mathbf{2x + 12 = 16}$$
$$\mathbf{12 > 10 - \frac{x}{2}}$$
$$\mathbf{19 + \frac{x}{10} < 29}$$
Question 2: Write the word sentences for the following algebraic sentences:
- (i) $10x = 60$: 10 times of a number is equal to 60.
- (ii) $20 - 2x = 0$: The difference of 20 and twice of a number is equal to 0.
- (iii) $x + 23 = 40$: A number added to 23 is equal to 40.
- (iv) $4x < 180$: 4 times of a number is less than 180.
- (v) $2x - 5 \le 25$: 5 subtracted from twice of a number is less than or equal to 25.
- (vi) $\frac{p}{6} \ge 30$: A number $p$ divided by 6 is greater than or equal to 30.
Question 3: Classify each statement as a True, False, or Open sentence:
Question 4: Determine whether the statement is True or False for the given variable value:
$7(21) = 147 \ne 28 \implies \mathbf{False}$
$-(6) + 9 = 3 \ne 15 \implies \mathbf{False}$
$5 - 3 = 2 \ne 10 \implies \mathbf{False}$
$23 - 12 = 11 = 11 \implies \mathbf{True}$
$19 - 2(7) = 19 - 14 = 5 \ne 6 \implies \mathbf{False}$
$\frac{123}{3} + 4.9 = 41 + 4.9 = 45.9 > 8.5 \implies \mathbf{True}$
$10 - \frac{6}{2} = 10 - 3 = 7 \ngtr 8 \implies \mathbf{False}$
6. Exercise 6.2 Step-by-Step Solutions
Question 1: Write an equation for each of the following statements:
- (i) Four times a number is 20: Let number be $x \implies \mathbf{4x = 20}$
- (ii) Six subtracted from a number is equal to 6: Let number be $y \implies \mathbf{y - 6 = 6}$
- (iii) Two times a number taken away from 10 is 4: Let number be $x \implies \mathbf{10 - 2x = 4}$
- (iv) Three times a number increased by 5 is 17: Let number be $y \implies \mathbf{3y + 5 = 17}$
- (v) Four times a number subtracted from 28 is 8: Let number be $x \implies \mathbf{28 - 4x = 8}$
Question 2: Complete the table with word descriptions:
| Equation | Word Description |
|---|---|
| $6x + 3 = 9$ | 3 added to 6 times a number gives 9 |
| $2x - 6 = 12$ | Six subtracted from two times a number is 12 |
| $x + 4 = 7$ | Four added to a number is 7 |
| $3x = 5$ | Three times a number is 5 |
| $\frac{x}{9} = 4$ | A number divided by 9 is 4 |
Question 3: Solve the following basic equations:
$x = 13 - 6 \implies \mathbf{x = 7}$
$x = -4 - 11 \implies \mathbf{x = -15}$
$x = 14 + 4 \implies \mathbf{x = 18}$
$y = -5 + 6 \implies \mathbf{y = 1}$
$z = 1.8 - 0.2 \implies \mathbf{z = 1.6}$
$a = -3.2 + 6 \implies \mathbf{a = 2.8}$
Question 4: Solve the following 2-step equations:
$x = \frac{-26}{2} \implies \mathbf{x = -13}$
$x = \frac{64}{-8} \implies \mathbf{x = -8}$
$2x = 7 + 3 = 10 \implies \mathbf{x = 5}$
$16x = 44 - 4 = 40 \implies x = \frac{40}{16} = \mathbf{2.5}$
$-5x = 3 - 18 = -15 \implies \mathbf{x = 3}$
$-3y = 8.6 + 6.4 = 15 \implies \mathbf{y = -5}$
Question 5: Solve equations with variables on both sides:
$y + y = 8 - 4 $ $\implies 2y = 4 $ $\implies \mathbf{y = 2}$
$x + 9x = 4 + 6 $ $\implies 10x = 10 $ $\implies \mathbf{x = 1}$
$2y - 7y = -27 + 7 $ $\implies -5y = -20 $ $\implies \mathbf{y = 4}$
$2x - 9x = -4 - 3 $ $\implies -7x = -7 $ $\implies \mathbf{x = 1}$
Question 6: Solve equations containing brackets:
$2y + 6 = 8 \implies 2y = 2 \implies \mathbf{y = 1}$
$5x - 35 = -15 \implies 5x = 20 \implies \mathbf{x = 4}$
$-14y + 28 = -4y \implies -10y = -28 \implies \mathbf{y = 2.8}$
$6 - 12z = -18 \implies -12z = -24 \implies \mathbf{z = 2}$
7. Exercise 6.3 Step-by-Step Solutions
Question 1: Solve fractional linear equations:
Multiply entire equation by $\text{LCM}(2, 3) = 6$:
$$6\left(\frac{3x}{2}\right) + 6\left(\frac{x}{3}\right) = 6(11) \implies 9x + 2x = 66 \implies 11x = 66 \implies \mathbf{x = 6}$$
Cross-multiply:
$$3(x+1) = 2(x-1) \implies 3x + 3 = 2x - 2 \implies 3x - 2x = -2 - 3 \implies \mathbf{x = -5}$$
Rearrange variable and constant terms:
$$\frac{1}{5} + \frac{1}{3} = 2x - x \implies x = \frac{3 + 5}{15} = \mathbf{\frac{8}{15}}$$
Multiply by $\text{LCM}(2, 3) = 6$:
$$3x - 18 = 24 - 4x \implies 3x + 4x = 24 + 18 \implies 7x = 42 \implies \mathbf{x = 6}$$
Equate fractions and cross-multiply:
$$\frac{2y-1}{5} = \frac{y+3}{7} \implies 7(2y-1) = 5(y+3) \implies 14y - 7 = 5y + 15 \implies 9y = 22 \implies \mathbf{y = \frac{22}{9}}$$
Cross-multiply:
$$5(x-3) = 2(x+4) \implies 5x - 15 = 2x + 8 \implies 3x = 23 \implies \mathbf{x = \frac{23}{3}}$$
Question 2: Solve decimal linear equations:
$0.5x = 10 \implies x = \frac{10}{0.5} = \mathbf{20}$
$1.5x - 0.8x = 2.1 \implies 0.7x = 2.1 \implies \mathbf{x = 3}$
$3.5x - 1.5x = 8 \implies 2x = 8 \implies \mathbf{x = 4}$
$7x + 6x - 20 = 32 \implies 13x = 52 \implies \mathbf{x = 4}$
$0.8x = 8 \implies x = \frac{8}{0.8} = \mathbf{10}$
$0.9x - 0.8x = -9 - 12 \implies 0.1x = -21 \implies \mathbf{x = -210}$
8. Exercise 6.4 Step-by-Step Solutions (Word Problems)
Questions 1 to 5: Applied Number & Geometry Problems
Q1: If 3 times a number is added to 18, it becomes 36. What is the number?
Let number $= x \implies 3x + 18 = 36 \implies 3x = 18 \implies \mathbf{x = 6}$.
Q2: The length of a rectangle is twice its width. The perimeter of the rectangle is 24m. Find length and width.
Let width $= w$, length $= 2w$.
$$\text{Perimeter} = 2(l + w) = 2(2w + w) = 6w = 24 \implies \mathbf{w = 4\text{ m}}, \mathbf{l = 8\text{ m}}.$$
Q3: Rafay is 2 years older than Saleh. The sum of their ages is 28. Find both their ages.
Let Saleh's age $= s$, Rafay's age $= s + 2$.
$$s + (s + 2) = 28 \implies 2s = 26 \implies \mathbf{s = 13\text{ years (Saleh)}}, \mathbf{\text{Rafay} = 15\text{ years}}.$$
Q4: The sum of two consecutive integers is 25. Find the integers.
Let integers be $x$ and $x+1$.
$$x + (x + 1) = 25 \implies 2x = 24 \implies x = 12 \implies \text{Integers are } \mathbf{12 \text{ and } 13}.$$
Q5: Think of a number, multiply it by 3 and then subtract 5. The result is 22. What is the number?
Let number $= x \implies 3x - 5 = 22 \implies 3x = 27 \implies \mathbf{x = 9}$.
Questions 6 to 10: Advanced Applied Problems
Q6: The sum of 4 consecutive odd numbers is 56. Find the greatest of the 4 numbers.
Let numbers be $x, x+2, x+4, x+6$.
$$x + (x+2) + (x+4) + (x+6) = 56 \implies 4x + 12 = 56 \implies 4x = 44 \implies x = 11.$$
$$\text{Greatest number} = x + 6 = 11 + 6 = \mathbf{17}.$$
Q7: The sum of half of a number and 49 is $2\frac{1}{4}$ of the number. Find the number.
Let number be $x$:
$$\frac{x}{2} + 49 = \frac{9}{4}x \implies 49 = \frac{9}{4}x - \frac{2}{4}x \implies 49 = \frac{7}{4}x \implies x = \frac{49 \times 4}{7} = \mathbf{28}.$$
Q8: Mr. Farooq is 3 times as old as his son. Their combined age is 40 years. What are their ages?
Let son's age $= s$, Mr. Farooq $= 3s$.
$$s + 3s = 40 \implies 4s = 40 \implies \mathbf{s = 10\text{ years (Son)}}, \mathbf{\text{Father} = 30\text{ years}}.$$
Q9: Munazza spent Rs. 95. She buys a Mathematics book and a diary. If the cost of the diary is Rs. 5 more than the Mathematics book, find their prices.
Let Math book $= b$, Diary $= b + 5$.
$$b + (b + 5) = 95 \implies 2b = 90 \implies \mathbf{b = \text{Rs. } 45\text{ (Maths book)}}, \mathbf{\text{Diary} = \text{Rs. } 50}.$$
Q10: Umer buys a rope 27 meters long. He cuts it into two pieces in such a way that one piece is 9 meters longer than the other. Find the length of the two pieces.
Let shorter piece $= x$, longer piece $= x + 9$.
$$x + (x + 9) = 27 \implies 2x = 18 \implies \mathbf{x = 9\text{ m}}, \mathbf{\text{Longer piece} = 18\text{ m}}.$$
9. Review Exercise 6 Solutions & Mastery Guide
Question 1: Multiple Choice Questions (MCQs) with Full Mathematical Proofs
$$\mathbf{5 + 2x = 13}$$ Correct Option: (b) $5 + 2x = 13$
$13 + 12 = 25$ is mathematically true.
Correct Option: (d) $13 + 12 = 25$
$4 + 15 = 19 \ne 20$.
Correct Option: (d) $4 + 15 = 20$
$x + 3 = 9$ contains an unknown variable.
Correct Option: (d) $x + 3 = 9$
$9x = 9$ has variable $x$ with exponent 1.
Correct Option: (d) $9x = 9$
An open sentence containing an equality symbol.
Correct Option: (b) an open sentence
$9x = 54 \implies x = \frac{54}{9} = 6$.
Correct Option: (b) $6$
$3x = 12 \implies x = 4$.
Correct Option: (d) $4$
$7x + y = 9$ contains TWO distinct variables ($x$ and $y$).
Correct Option: (a) $7x + y = 9$
By definition, degree / exponent is 1.
Correct Option: (a) $1$
$$\mathbf{x - 7 = 5}$$ Correct Option: (d) $x - 7 = 5$
$$\mathbf{\frac{y}{4} = 5}$$ Correct Option: (b) $\frac{y}{4} = 5$
$\frac{x}{6} - 4 = 0 \implies \frac{x}{6} = 4 \implies x = 24$.
Correct Option: (b) $24$
In $4x - 6 = 2 \implies 4(2) - 6 = 8 - 6 = 2$.
Correct Option: (c) $4x - 6 = 2$
Question 2: Solve the following equations:
$3x = 6 $ $\implies \mathbf{x = 2}$
$5y - 3 = 12 $ $\implies 5y = 15 $ $\implies \mathbf{y = 3}$
$3(z+3) = 2(3z-5) $ $\implies 3z+9 = 6z-10 $ $\implies 3z = 19 $ $\implies \mathbf{z = \frac{19}{3} = 6\frac{1}{3}}$
$8a + 6 - 4.5a = 30 - a $ $\implies 3.5a + 6 = 30 - a $ $\implies 4.5a = 24 \implies \mathbf{a = \frac{16}{3} = 5\frac{1}{3}}$
$0.1b = -21 \implies \mathbf{b = -210}$
$3(2x-3) = 4(x-3) $ $\implies 6x-9 = 4x-12 $ $\implies 2x = -3 $ $\implies \mathbf{x = -\frac{3}{2}}$
Question 3: Divide Rs. 63 between two students so that one gets 6 times as much as the other.
• 2nd student's share $= 6x$
$$x + 6x = 63 \implies 7x = 63 \implies x = \text{Rs. } 9$$
Shares: 1st student gets $\mathbf{\text{Rs. } 9}$, 2nd student gets $6(9) = \mathbf{\text{Rs. } 54}$.
Question 4: Aslam is five times as old as his son. Difference of their ages is 40 years. Find the age of both.
• Aslam's age $= 5s$
$$5s - s = 40 \implies 4s = 40 \implies s = 10\text{ years}$$
Ages: Son is $\mathbf{10\text{ years}}$, Aslam is $5(10) = \mathbf{50\text{ years}}$.
Question 5: Ayesha bought a note book and a geometry box for Rs. 130. If the note book costs Rs. 10 more than 2 times the cost of geometry box, what is the price of geometry box?
• Price of note book $= 2g + 10$
$$g + (2g + 10) = 130 \implies 3g + 10 = 130 \implies 3g = 120 \implies g = 40$$
Answer: Price of geometry box is $\mathbf{\text{Rs. } 40}$ (and note book is $\text{Rs. } 90$).
Unit 6 Mastery Self-Assessment Checklist
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