Mastery Guide: Algebraic Fractions, Rational Operations, Complex Fractions & Extraneous Roots
Algebraic Fractions & Rational Expressions
A Comprehensive Master Guide on Polynomial Classification, Rational Fractions Reduction, Fundamental Operations, Complex Algebraic Fractions, and Rational Equations with Extraneous Root Checking.
📋 Table of Contents & Core Learning Pillars
- Section 1: Algebraic Expressions & Polynomial Hierarchy
- Section 2: Rational vs. Irrational Expressions
- Section 3: Reduction of Algebraic Fractions to Lowest Form
- Section 4: Operations on Algebraic Fractions (+, -, ×, ÷)
- Section 5: Solving Rational Equations & Extraneous Roots
- Section 6: Real-World Work-Rate & Mixture Word Problems
- Section 7: Kid-Friendly Tips & Mnemonics
- Section 8: Solved Exercise 5.1 (18 Questions)
- Section 9: Solved Exercise 5.2 (25 Questions)
- Section 10: Solved Exercise 5.3 (10 Questions)
- Section 11: Solved Miscellaneous Exercise 5 (25 Questions)
- Section 12: High-Yield Objective Booster (37 Questions)
1 Algebraic Expressions & Polynomial Hierarchy
An algebraic expression is a mathematical phrase formed by combining constants and variables through arithmetic operations (addition, subtraction, multiplication, division, and exponentiation). Among algebraic expressions, polynomials constitute the most fundamental building blocks of algebra.
💡 Formal Definition of a Polynomial in One Variable
A polynomial $P(x)$ of degree $n$ in the real variable $x$ is an algebraic expression of the form: $$P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$$ where $a_n, a_{n-1}, \dots, a_0 \in \mathbb{R}$, $a_n \neq 0$ is the leading coefficient, $a_0$ is the constant term, and the exponents $n, n-1, \dots$ must be non-negative integers ($n \in \{0, 1, 2, 3, \dots\}$).
📋 Table 1: Algebraic Expressions & Polynomial Hierarchy Matrix
2 Rational vs. Irrational Algebraic Expressions & Domain Constraints
Just as real numbers are split into rational and irrational numbers, algebraic expressions are partitioned into rational and irrational expressions based on the mathematical character of their variable exponents.
✅ Rational Expressions
An expression that can be written in the form $\frac{P(x)}{Q(x)}$, where both $P(x)$ and $Q(x)$ are polynomials and $Q(x) \neq 0$.
Examples:
• $\frac{3x + 1}{x^2 - 4}$ (undefined at $x = \pm 2$)
• $\frac{x^2 - 25}{x + 5}$ (undefined at $x = -5$)
• $7x^2 - 3$ (polynomial, denominator is $1$).
❌ Irrational Expressions
An expression containing variables under radical signs (surds) or with fractional exponents that cannot be reduced to a polynomial quotient.
Examples:
• $\frac{\sqrt{x} + 2}{3x - 1}$
• $x^{3/2} - 5x + 7$
• $\sqrt{x^2 + 9}$
3 Reduction of Algebraic Fractions to Lowest Form
A rational fraction $\frac{P(x)}{Q(x)}$ is said to be in its lowest form (or simplest form) if $P(x)$ and $Q(x)$ are polynomials with real coefficients having no common factor other than $\pm 1$ (i.e., their Greatest Common Divisor $\text{GCD}(P(x), Q(x)) = 1$).
📋 Table 2: Essential Factorization Identities for Fractions Reduction
4 Operations on Rational Expressions (+, -, ×, ÷)
Operations on algebraic fractions obey the exact same algebraic axioms as arithmetic fractions of rational numbers:
📋 Table 3: Operations & LCD Execution Strategy Matrix
5 Solving Rational Equations & The Extraneous Root Security Gate
A rational equation is an equation containing one or more rational expressions where the variable appears in the denominator.
📋 Table 4: Systematic Rational Equation Protocol & Verification Checklist
6 Real-World Physics & Industry Models: Work-Rate & Mixtures
Rational equations serve as indispensable mathematical modeling tools across civil engineering, industrial chemistry, physics, and operations research.
💡 Kid-Friendly Tips for Success & Exam Mnemonics
- The "Never Divide by Zero" Rule: Before solving any rational fraction, inspect the denominator! Any $x$ that turns the denominator into $0$ is an excluded value (e.g., in $\frac{1}{x-3}$, $x \neq 3$).
- The Factoring First Strategy: Never jump to multiplying out rational fractions! Always factor every numerator and denominator completely first. You will often cancel large expressions effortlessly.
- Division Means "Keep-Change-Flip": When dividing fractions $\frac{A}{B} \div \frac{C}{D}$, KEEP the first fraction $\frac{A}{B}$, CHANGE $\div$ to $\times$, and FLIP $\frac{C}{D}$ to $\frac{D}{C}$.
- Extraneous Root Sentinel: When you multiply both sides by the LCD to eliminate denominators, you might accidentally create fake solutions (extraneous roots). Always test your final answers back in the original denominators!
- Watch the Negative Signs: In subtraction like $-\frac{x-4}{D}$, remember to distribute the negative sign to BOTH terms: $-(x-4) = -x + 4$.
Complete Solved Textbook Exercises & Objective Bank
100% textbook questions from Exercises 5.1, 5.2, 5.3, Miscellaneous 5, and High-Yield Objective Booster solved in rigorous step-by-step detail adhering strictly to FBISE marking schemes.
Exercise 5.1: Rational Expressions, Domain Restrictions & Reduction to Lowest Form
18 Questions • 100% FBISE Textbook Problems (Q1-Q6)Given Expression:
$$\frac{15 a x^3 y^2}{25 a^2 x y^6}$$
Step 1: Simplify numerical coefficients:
$$\frac{15}{25} = \frac{3 \times 5}{5 \times 5} = \frac{3}{5}$$
Step 2: Apply the quotient rule of exponents for variables $a, x, y$:
$$\frac{a}{a^2} = \frac{1}{a^{2-1}} = \frac{1}{a}, \quad \frac{x^3}{x} = x^{3-1} = x^2, \quad \frac{y^2}{y^6} = \frac{1}{y^{6-2}} = \frac{1}{y^4}$$
Step 3: Multiply remaining simplified factors:
$$\frac{3 \cdot x^2}{5 \cdot a \cdot y^4} = \frac{3x^2}{5 a y^4}$$
Final Answer:
$$\mathbf{\frac{3x^2}{5 a y^4}}$$
Given Expression:
$$\frac{38 k^2 p^3 m^4}{57 k^3 p m^2}$$
Step 1: Divide numerical coefficients by $\gcd(38, 57) = 19$:
$$\frac{38}{57} = \frac{2 \times 19}{3 \times 19} = \frac{2}{3}$$
Step 2: Simplify variable powers:
$$\frac{k^2}{k^3} = \frac{1}{k}, \quad \frac{p^3}{p} = p^2, \quad \frac{m^4}{m^2} = m^2$$
Step 3: Combine factors:
$$\frac{2 \cdot p^2 \cdot m^2}{3 \cdot k} = \frac{2 p^2 m^2}{3k}$$
Final Answer:
$$\mathbf{\frac{2 p^2 m^2}{3k}}$$
Given Expression:
$$\frac{m n^4 p q}{m^2 n^3 p^4}$$
Step 1: Apply quotient rule of exponents for each variable base:
$$\frac{m}{m^2} = \frac{1}{m}, \quad \frac{n^4}{n^3} = n, \quad \frac{p}{p^4} = \frac{1}{p^3}, \quad q = q$$
Step 2: Assemble simplified fraction:
$$\frac{n q}{m p^3}$$
Final Answer:
$$\mathbf{\frac{n q}{m p^3}}$$
Given Expression:
$$\frac{3 a b c}{15 a^2 b^2 c}$$
Step 1: Divide numerical coefficients by $\gcd(3, 15) = 3$:
$$\frac{3}{15} = \frac{1}{5}$$
Step 2: Cancel identical terms and subtract exponents:
$$\frac{a}{a^2} = \frac{1}{a}, \quad \frac{b}{b^2} = \frac{1}{b}, \quad \frac{c}{c} = 1$$
Step 3: Multiply remaining factors:
$$\frac{1}{5 a b}$$
Final Answer:
$$\mathbf{\frac{1}{5 a b}}$$
Given Expression:
$$\frac{46 l^3 m^4 n^5}{69 l^2 m^3 n^4}$$
Step 1: Simplify numerical coefficients by dividing by $\gcd(46, 69) = 23$:
$$\frac{46}{69} = \frac{2 \times 23}{3 \times 23} = \frac{2}{3}$$
Step 2: Subtract exponents for variable bases $l, m, n$:
$$\frac{l^3}{l^2} = l^{3-2} = l, \quad \frac{m^4}{m^3} = m^{4-3} = m, \quad \frac{n^5}{n^4} = n^{5-4} = n$$
Step 3: Combine all terms:
$$\frac{2 l m n}{3}$$
Final Answer:
$$\mathbf{\frac{2 l m n}{3}}$$
Given Expression:
$$\frac{x - 3}{3 - x}$$
Step 1: Factor out a negative sign ($-1$) from the denominator:
$$3 - x = -(x - 3)$$
Step 2: Substitute and cancel the common factor $(x - 3)$ for $x \neq 3$:
$$\frac{x - 3}{-(x - 3)} = \frac{1}{-1} = -1$$
Final Answer:
$$\mathbf{-1}$$
Given Expression:
$$\frac{x^2 - 81}{x + 9}$$
Step 1: Factorize the numerator using the difference of two squares identity $a^2 - b^2 = (a - b)(a + b)$:
$$x^2 - 81 = x^2 - 9^2 = (x - 9)(x + 9)$$
Step 2: Substitute and cancel the common binomial factor $(x + 9)$ (for $x \neq -9$):
$$\frac{(x - 9)(x + 9)}{x + 9} = x - 9$$
Final Answer:
$$\mathbf{x - 9}$$
Given Expression:
$$\frac{(r + 3)(r + 4)}{r^2 - 16}$$
Step 1: Factorize the denominator $r^2 - 16$ using $a^2 - b^2 = (a - b)(a + b)$:
$$r^2 - 16 = r^2 - 4^2 = (r - 4)(r + 4)$$
Step 2: Substitute and cancel the common binomial $(r + 4)$ (for $r \neq \pm 4$):
$$\frac{(r + 3)(r + 4)}{(r - 4)(r + 4)} = \frac{r + 3}{r - 4}$$
Final Answer:
$$\mathbf{\frac{r + 3}{r - 4}}$$
Given Expression: $3(r^2 - s^2)$ with $r = 2, s = -1$.
Step 1: Calculate the squares of $r$ and $s$:
$$r^2 = 2^2 = 4, \quad s^2 = (-1)^2 = 1$$
Step 2: Subtract the squared values:
$$r^2 - s^2 = 4 - 1 = 3$$
Step 3: Multiply by 3:
$$3(3) = 9$$
Final Answer:
$$\mathbf{9}$$
Given Expression: $\frac{1}{2} m v^2$ with $m = 18.75, v = 5.6$.
Step 1: Compute $v^2$:
$$v^2 = 5.6^2 = 31.36$$
Step 2: Substitute $m = 18.75$ and $v^2 = 31.36$:
$$\frac{1}{2} \times 18.75 \times 31.36 = 18.75 \times 15.68$$
Step 3: Multiply:
$$18.75 \times 15.68 = 294$$
Final Answer:
$$\mathbf{294}$$
Given Expression: $\sqrt{2 g s}$ with $g = 32.2, s = 144.9$.
Step 1: Calculate the product under the radical:
$$2 \times 32.2 \times 144.9 = 64.4 \times 144.9 = 9331.56$$
Step 2: Take the positive square root:
$$\sqrt{9331.56} = 96.6$$
Final Answer:
$$\mathbf{96.6}$$
Given Expression: $3x - y + \frac{1}{z}$ with $x = -\frac{1}{2}, y = 3, z = -\frac{1}{3}$.
Step 1: Substitute the given values into each term:
$$3x = 3\left(-\frac{1}{2}\right) = -\frac{3}{2} = -1.5$$
$$-y = -(3) = -3$$
$$\frac{1}{z} = \frac{1}{-\frac{1}{3}} = -3$$
Step 2: Sum all three terms:
$$-1.5 - 3 - 3 = -7.5 = -\frac{15}{2}$$
Final Answer:
$$\mathbf{-7.5}$$
Given Expression: $0.1 d^2 + 0.01 d + 1$ with $d = -0.2$.
Step 1: Calculate $d^2$:
$$d^2 = (-0.2)^2 = 0.04$$
Step 2: Multiply by coefficients:
$$0.1 d^2 = 0.1 \times 0.04 = 0.004$$
$$0.01 d = 0.01 \times (-0.2) = -0.002$$
Step 3: Combine all terms:
$$0.004 - 0.002 + 1 = 1.002$$
Final Answer:
$$\mathbf{1.002}$$
Given Expression: $\frac{4}{7} b^3 - 3\frac{1}{2} b^2 + b - 3$ with $b = \frac{1}{2}$.
Step 1: Convert mixed fraction to improper fraction: $3\frac{1}{2} = \frac{7}{2}$.
Step 2: Substitute $b = \frac{1}{2}$:
$$\frac{4}{7}\left(\frac{1}{2}\right)^3 = \frac{4}{7} \times \frac{1}{8} = \frac{1}{14}$$
$$-\frac{7}{2}\left(\frac{1}{2}\right)^2 = -\frac{7}{2} \times \frac{1}{4} = -\frac{7}{8}$$
$$+b = +\frac{1}{2}, \quad -3 = -3$$
Step 3: Combine using common denominator $\text{LCM}(14, 8, 2, 1) = 56$:
$$\frac{4 - 49 + 28 - 168}{56} = \frac{-185}{56} = -3\frac{17}{56}$$
Final Answer:
$$\mathbf{-3\frac{17}{56}}$$
Given Formula:
$$T(n) = \frac{n(n + 1)}{2}$$
Step 1: Substitute $n = 100$:
$$T(100) = \frac{100(100 + 1)}{2} = \frac{100 \times 101}{2}$$
Step 2: Simplify:
$$T(100) = 50 \times 101 = 5050$$
Final Answer:
$$\mathbf{5050}$$
Given Functions:
$$P(x) = x^2 + 2x - 15, \quad D(x) = x - 3, \quad Q(x) = x + 5$$
Step 1: Evaluate $P(2)$, $Q(2)$, and $D(2)$ at $x = 2$:
$$P(2) = 2^2 + 2(2) - 15 = 4 + 4 - 15 = -7$$
$$Q(2) = 2 + 5 = 7$$
$$D(2) = 2 - 3 = -1$$
Step 2: Compute Left-Hand Side (LHS):
$$\text{LHS} = \frac{P(2)}{Q(2)} = \frac{-7}{7} = -1$$
Step 3: Compare with Right-Hand Side (RHS):
$$\text{RHS} = D(2) = -1$$
Since $\text{LHS} = \text{RHS} = -1$, the relation $\frac{P(2)}{Q(2)} = D(2)$ is verified.
Final Answer:
$$\mathbf{\text{LHS} = \text{RHS} = -1 \text{ (Proved)}}$$
Given Function:
$$g(x) = \frac{1}{2x^3} + \frac{x}{2} + 2$$
Step 1: Substitute $x = -\frac{1}{3}$:
$$x^3 = \left(-\frac{1}{3}\right)^3 = -\frac{1}{27} \implies 2x^3 = -\frac{2}{27}$$
$$\frac{1}{2x^3} = \frac{1}{-\frac{2}{27}} = -\frac{27}{2}$$
$$\frac{x}{2} = \frac{-\frac{1}{3}}{2} = -\frac{1}{6}$$
Step 2: Add all terms using common denominator $6$:
$$g\left(-\frac{1}{3}\right) = -\frac{27}{2} - \frac{1}{6} + 2 = \frac{-81 - 1 + 12}{6} = \frac{-70}{6} = -\frac{35}{3} = -11\frac{2}{3}$$
Final Answer:
$$\mathbf{-11\frac{2}{3}}$$
Given Data:
- Volume of sphere $v = 38{,}808\text{ cm}^3$
- Formula: $r = \sqrt[3]{\frac{3v}{4\pi}}$
- Value of $\pi = \frac{22}{7}$
Step 1: Compute the expression inside the cube root:
$$\frac{3v}{4\pi} = \frac{3 \times 38808}{4 \times \frac{22}{7}} = \frac{3 \times 38808 \times 7}{88}$$
Divide $38808$ by $88$: $\frac{38808}{88} = 441$.
$$441 \times 3 \times 7 = 441 \times 21 = 9261$$
Step 2: Take the cube root of $9261$:
$$r = \sqrt[3]{9261} = \sqrt[3]{21^3} = 21\text{ cm}$$
Final Answer:
$$\mathbf{21\text{ cm}}$$
Exercise 5.2: Fundamental Operations on Algebraic Fractions (+, -, ×, ÷) & Complex Fractions
25 Questions • 100% FBISE Textbook Problems (Q1-Q4)Given Expressions: $\frac{x}{2}$ and $\frac{x}{5}$.
Step 1: Write as an addition expression:
$$\frac{x}{2} + \frac{x}{5}$$
Step 2: Find the Least Common Denominator (LCD): $\text{LCD}(2, 5) = 10$.
Step 3: Convert fractions to common denominator and combine numerators:
$$\frac{5(x) + 2(x)}{10} = \frac{5x + 2x}{10} = \frac{7x}{10}$$
Final Answer:
$$\mathbf{\frac{7x}{10}}$$
Given Expressions: $\frac{x - 2}{2}$ and $\frac{x + 10}{9}$.
Step 1: Write as an addition expression:
$$\frac{x - 2}{2} + \frac{x + 10}{9}$$
Step 2: Find the Least Common Denominator: $\text{LCD}(2, 9) = 18$.
Step 3: Express with common denominator and expand:
$$\frac{9(x - 2) + 2(x + 10)}{18} = \frac{9x - 18 + 2x + 20}{18}$$
Step 4: Combine like terms:
$$\frac{(9x + 2x) + (-18 + 20)}{18} = \frac{11x + 2}{18}$$
Final Answer:
$$\mathbf{\frac{11x + 2}{18}}$$
Given Expressions: $\frac{4 + x}{4}$, $\frac{x - 1}{7}$, and $\frac{5x}{2}$.
Step 1: Write as a single sum:
$$\frac{4 + x}{4} + \frac{x - 1}{7} + \frac{5x}{2}$$
Step 2: Find the Least Common Denominator: $\text{LCD}(4, 7, 2) = 28$.
Step 3: Multiply each numerator by its required factor:
$$\frac{7(4 + x) + 4(x - 1) + 14(5x)}{28}$$
Step 4: Expand and group like terms:
$$\frac{28 + 7x + 4x - 4 + 70x}{28} = \frac{(7x + 4x + 70x) + (28 - 4)}{28} = \frac{81x + 24}{28}$$
Final Answer:
$$\mathbf{\frac{81x + 24}{28}}$$
Given Expressions: $\frac{3x}{x + 5}$ and $\frac{10}{5x + 25}$.
Step 1: Factorize denominators where possible:
$$5x + 25 = 5(x + 5)$$
Step 2: Simplify the second fraction:
$$\frac{10}{5(x + 5)} = \frac{2}{x + 5}$$
Step 3: Add with the common denominator $(x + 5)$:
$$\frac{3x}{x + 5} + \frac{2}{x + 5} = \frac{3x + 2}{x + 5}$$
Final Answer:
$$\mathbf{\frac{3x + 2}{x + 5}}$$
Given Expressions: $\frac{24x}{6x - 18}$ and $\frac{3(1 + x)}{x - 3}$.
Step 1: Factorize the first denominator:
$$6x - 18 = 6(x - 3)$$
Step 2: Reduce the first fraction:
$$\frac{24x}{6(x - 3)} = \frac{4x}{x - 3}$$
Step 3: Add the two fractions with common denominator $(x - 3)$:
$$\frac{4x}{x - 3} + \frac{3(1 + x)}{x - 3} = \frac{4x + 3 + 3x}{x - 3} = \frac{7x + 3}{x - 3}$$
Final Answer:
$$\mathbf{\frac{7x + 3}{x - 3}}$$
Problem: Subtract $\frac{23 - x}{5}$ from $7$.
Step 1: Set up the subtraction expression:
$$7 - \frac{23 - x}{5}$$
Step 2: Write $7$ with denominator $5$:
$$\frac{7 \times 5}{5} - \frac{23 - x}{5} = \frac{35 - (23 - x)}{5}$$
Step 3: Distribute the negative sign carefully:
$$\frac{35 - 23 + x}{5} = \frac{12 + x}{5} = \frac{x + 12}{5}$$
Final Answer:
$$\mathbf{\frac{x + 12}{5}}$$
Problem: Subtract $\frac{6(x - 8)}{7}$ from $\frac{5(x - 7)}{3}$.
Step 1: Set up the subtraction:
$$\frac{5(x - 7)}{3} - \frac{6(x - 8)}{7}$$
Step 2: Find LCD of $3$ and $7$: $\text{LCD} = 21$.
Step 3: Convert fractions and combine numerators:
$$\frac{7 \times 5(x - 7) - 3 \times 6(x - 8)}{21} = \frac{35(x - 7) - 18(x - 8)}{21}$$
Step 4: Expand and simplify:
$$\frac{35x - 245 - 18x + 144}{21} = \frac{(35x - 18x) + (-245 + 144)}{21} = \frac{17x - 101}{21}$$
Final Answer:
$$\mathbf{\frac{17x - 101}{21}}$$
Problem: Subtract $2x^2 - 2x + 1$ from $\frac{x + 1}{x}$.
Step 1: Set up the subtraction:
$$\frac{x + 1}{x} - (2x^2 - 2x + 1)$$
Step 2: Express over common denominator $x$:
$$\frac{(x + 1) - x(2x^2 - 2x + 1)}{x}$$
Step 3: Expand the numerator:
$$\frac{x + 1 - 2x^3 + 2x^2 - x}{x}$$
Step 4: Cancel $+x$ and $-x$:
$$\frac{-2x^3 + 2x^2 + 1}{x}$$
Final Answer:
$$\mathbf{\frac{-2x^3 + 2x^2 + 1}{x}}$$
Problem: Divide $x^4 - 10x^2 + 9$ by $x^2 - 2x - 3$.
Step 1: Factorize the first expression (quartic polynomial):
$$x^4 - 10x^2 + 9 = (x^2 - 1)(x^2 - 9) = (x - 1)(x + 1)(x - 3)(x + 3)$$
Step 2: Factorize the second expression:
$$x^2 - 2x - 3 = (x - 3)(x + 1)$$
Step 3: Divide and cancel common factors:
$$\frac{(x - 1)(x + 1)(x - 3)(x + 3)}{(x - 3)(x + 1)} = (x - 1)(x + 3)$$
Step 4: Expand the quotient:
$$(x - 1)(x + 3) = x^2 + 3x - x - 3 = x^2 + 2x - 3$$
Final Answer:
$$\mathbf{x^2 + 2x - 3}$$
Problem: Divide $x^3 - 3x^2 y + 3xy^2 - y^3$ by $x - y$.
Step 1: Recognize the perfect cube identity:
$$x^3 - 3x^2 y + 3xy^2 - y^3 = (x - y)^3$$
Step 2: Divide by $(x - y)$:
$$\frac{(x - y)^3}{x - y} = (x - y)^2$$
Step 3: Expand the result:
$$(x - y)^2 = x^2 - 2xy + y^2$$
Final Answer:
$$\mathbf{x^2 - 2xy + y^2}$$
Problem: Divide $\frac{4x^2 - 16}{5x}$ by $\frac{2x + 4}{15}$.
Step 1: Convert division into multiplication by the reciprocal:
$$\frac{4x^2 - 16}{5x} \times \frac{15}{2x + 4}$$
Step 2: Factorize all numerators and denominators:
$$4x^2 - 16 = 4(x^2 - 4) = 4(x - 2)(x + 2)$$
$$2x + 4 = 2(x + 2)$$
Step 3: Substitute factors and cancel common terms:
$$\frac{4(x - 2)(x + 2)}{5x} \times \frac{15}{2(x + 2)} = \frac{4(x - 2) \times 15}{5x \times 2} = \frac{60(x - 2)}{10x} = \frac{6(x - 2)}{x}$$
Final Answer:
$$\mathbf{\frac{6(x - 2)}{x}}$$
Problem: Divide $\frac{x^2 + 5x}{x - 3}$ by $\frac{x^2 - 25}{x - 3}$.
Step 1: Invert the divisor fraction and multiply:
$$\frac{x^2 + 5x}{x - 3} \times \frac{x - 3}{x^2 - 25}$$
Step 2: Factor numerators and denominators:
$$x^2 + 5x = x(x + 5)$$
$$x^2 - 25 = (x - 5)(x + 5)$$
Step 3: Cancel common factors $(x - 3)$ and $(x + 5)$:
$$\frac{x(x + 5)}{x - 3} \times \frac{x - 3}{(x - 5)(x + 5)} = \frac{x}{x - 5}$$
Final Answer:
$$\mathbf{\frac{x}{x - 5}}$$
Given Expression: $\frac{x}{2} + \frac{x}{3} - \frac{x}{4} + \frac{x}{5}$.
Step 1: Find the LCD of $2, 3, 4, 5$: $\text{LCD} = 60$.
Step 2: Convert each term to have denominator $60$:
$$\frac{30(x) + 20(x) - 15(x) + 12(x)}{60}$$
Step 3: Combine numerators:
$$\frac{(30 + 20 - 15 + 12)x}{60} = \frac{47x}{60}$$
Final Answer:
$$\mathbf{\frac{47x}{60}}$$
Given Expression: $\frac{1}{2}\left(4 - \frac{x}{3}\right) - \frac{5}{6} + \frac{1}{3}\left(11 - \frac{x}{2}\right)$.
Step 1: Expand brackets:
$$\frac{1}{2}(4) - \frac{1}{2}\left(\frac{x}{3}\right) - \frac{5}{6} + \frac{1}{3}(11) - \frac{1}{3}\left(\frac{x}{2}\right) = 2 - \frac{x}{6} - \frac{5}{6} + \frac{11}{3} - \frac{x}{6}$$
Step 2: Combine constant terms:
$$2 - \frac{5}{6} + \frac{11}{3} = \frac{12 - 5 + 22}{6} = \frac{29}{6}$$
Step 3: Combine variable terms:
$$-\frac{x}{6} - \frac{x}{6} = -\frac{2x}{6}$$
Step 4: Combine into a single fraction:
$$\frac{29 - 2x}{6}$$
Final Answer:
$$\mathbf{\frac{29 - 2x}{6}}$$
Given Expression: $\frac{2}{x + 1} + \frac{x}{x - 1} - \frac{x + 2}{x - 1}$.
Step 1: Combine the two terms with common denominator $(x - 1)$:
$$\frac{x - (x + 2)}{x - 1} = \frac{x - x - 2}{x - 1} = \frac{-2}{x - 1}$$
Step 2: Now subtract $\frac{2}{x - 1}$ from $\frac{2}{x + 1}$:
$$\frac{2}{x + 1} - \frac{2}{x - 1} = \frac{2(x - 1) - 2(x + 1)}{(x + 1)(x - 1)}$$
Step 3: Expand the numerator:
$$\frac{2x - 2 - 2x - 2}{x^2 - 1} = \frac{-4}{x^2 - 1}$$
Final Answer:
$$\mathbf{\frac{-4}{x^2 - 1}}$$
Given Expression: $\frac{x^2 - 25}{5} - \frac{x}{4} \div \frac{3x}{20}$.
Step 1: Perform the division first according to BODMAS rules:
$$\frac{x}{4} \div \frac{3x}{20} = \frac{x}{4} \times \frac{20}{3x} = \frac{20x}{12x} = \frac{5}{3}$$
Step 2: Perform the subtraction with $\text{LCD}(5, 3) = 15$:
$$\frac{x^2 - 25}{5} - \frac{5}{3} = \frac{3(x^2 - 25) - 5(5)}{15}$$
Step 3: Expand and simplify:
$$\frac{3x^2 - 75 - 25}{15} = \frac{3x^2 - 100}{15}$$
Final Answer:
$$\mathbf{\frac{3x^2 - 100}{15}}$$
Given Expression: $\frac{45 a^2 b^3 c^4}{27 x^4 y^3 z} \times \frac{243 x y^2 z^3}{180 a^2 b c^3}$.
Step 1: Simplify numerical coefficients:
$$\frac{45}{180} \times \frac{243}{27} = \frac{1}{4} \times 9 = \frac{9}{4}$$
Step 2: Cancel and simplify powers of variables:
$$\frac{a^2}{a^2} = 1, \quad \frac{b^3}{b} = b^2, \quad \frac{c^4}{c^3} = c$$
$$\frac{x}{x^4} = \frac{1}{x^3}, \quad \frac{y^2}{y^3} = \frac{1}{y}, \quad \frac{z^3}{z} = z^2$$
Step 3: Multiply remaining simplified factors:
$$\frac{9 b^2 c z^2}{4 x^3 y}$$
Final Answer:
$$\mathbf{\frac{9 b^2 c z^2}{4 x^3 y}}$$
Given Expression: $\frac{m^2}{8n} \times \frac{36 p^3 q^2}{81 m n} \div \frac{15 m p x^5}{270 n^2 x^3 y}$.
Step 1: Replace division by multiplication by the reciprocal:
$$\frac{m^2}{8n} \times \frac{36 p^3 q^2}{81 m n} \times \frac{270 n^2 x^3 y}{15 m p x^5}$$
Step 2: Multiply numerical coefficients:
$$\frac{36 \times 270}{8 \times 81 \times 15} = \frac{9720}{9720} = 1$$
Step 3: Simplify algebraic variables:
$$\text{For } m: \frac{m^2}{m \cdot m} = 1$$
$$\text{For } n: \frac{n^2}{n \cdot n} = 1$$
$$\text{For } p: \frac{p^3}{p} = p^2$$
$$\text{For } q: q^2$$
$$\text{For } x: \frac{x^3}{x^5} = \frac{1}{x^2}$$
$$\text{For } y: y$$
Step 4: Combine all terms:
$$\frac{p^2 q^2 y}{x^2}$$
Final Answer:
$$\mathbf{\frac{p^2 q^2 y}{x^2}}$$
Given Expression: $3x \div \frac{3x^2 - 27}{x + 3} + \frac{1}{x - 3}$.
Step 1: Factorize $3x^2 - 27$:
$$3x^2 - 27 = 3(x^2 - 9) = 3(x - 3)(x + 3)$$
Step 2: Invert the divisor fraction and multiply:
$$3x \times \frac{x + 3}{3(x - 3)(x + 3)} = \frac{3x(x + 3)}{3(x - 3)(x + 3)} = \frac{x}{x - 3}$$
Step 3: Add the second fraction with common denominator $(x - 3)$:
$$\frac{x}{x - 3} + \frac{1}{x - 3} = \frac{x + 1}{x - 3}$$
Final Answer:
$$\mathbf{\frac{x + 1}{x - 3}}$$
Given Expression: $\frac{5x + 5}{3(2x - 1)} + \frac{6 - 2x}{2(1 - 2x)}$.
Step 1: Relate the denominators using $1 - 2x = -(2x - 1)$:
$$\frac{6 - 2x}{2(1 - 2x)} = \frac{2(3 - x)}{-2(2x - 1)} = -\frac{3 - x}{2x - 1} = \frac{x - 3}{2x - 1}$$
Step 2: Add using $\text{LCD} = 3(2x - 1)$:
$$\frac{5x + 5}{3(2x - 1)} + \frac{3(x - 3)}{3(2x - 1)} = \frac{5x + 5 + 3x - 9}{3(2x - 1)}$$
Step 3: Combine like terms and factor numerator:
$$\frac{8x - 4}{3(2x - 1)} = \frac{4(2x - 1)}{3(2x - 1)} = \frac{4}{3}$$
Final Answer:
$$\mathbf{\frac{4}{3}}$$
Given Expression: $\frac{2a}{2a - 3} - \frac{5}{6a + 9} - \frac{4(3a + 2)}{3(4a^2 - 9)}$.
Step 1: Factorize all denominators:
- $2a - 3$
- $6a + 9 = 3(2a + 3)$
- $3(4a^2 - 9) = 3(2a - 3)(2a + 3)$
The Least Common Denominator is $\text{LCD} = 3(2a - 3)(2a + 3) = 3(4a^2 - 9)$.
Step 2: Convert fractions to the common denominator:
$$\frac{2a \cdot 3(2a + 3) - 5(2a - 3) - 4(3a + 2)}{3(4a^2 - 9)}$$
Step 3: Expand the numerator:
$$6a(2a + 3) - 10a + 15 - 12a - 8 = 12a^2 + 18a - 10a + 15 - 12a - 8$$
$$= 12a^2 + (18a - 10a - 12a) + (15 - 8) = 12a^2 - 4a + 7$$
Step 4: Combine into final fraction:
$$\frac{12a^2 - 4a + 7}{3(4a^2 - 9)}$$
Final Answer:
$$\mathbf{\frac{12a^2 - 4a + 7}{3(4a^2 - 9)}}$$
Given Expression: $\frac{5}{5 + x - 18x^2} - \frac{2}{2 + 5x + 2x^2}$.
Step 1: Factorize both quadratic denominators:
$$5 + x - 18x^2 = 5 + 10x - 9x - 18x^2 = 5(1 + 2x) - 9x(1 + 2x) = (1 + 2x)(5 - 9x)$$
$$2 + 5x + 2x^2 = 2x^2 + 4x + x + 2 = 2x(x + 2) + 1(x + 2) = (2x + 1)(x + 2)$$
Notice $(2x + 1) = (1 + 2x)$.
The Least Common Denominator is $\text{LCD} = (1 + 2x)(x + 2)(5 - 9x)$.
Step 2: Combine numerators over the LCD:
$$\frac{5(x + 2) - 2(5 - 9x)}{(1 + 2x)(x + 2)(5 - 9x)}$$
Step 3: Expand numerator:
$$5x + 10 - 10 + 18x = 23x$$
Step 4: Write simplified fraction:
$$\frac{23x}{(1 + 2x)(x + 2)(5 - 9x)}$$
Final Answer:
$$\mathbf{\frac{23x}{(1 + 2x)(x + 2)(5 - 9x)}}$$
Given Expression: $\frac{m + 3}{24m} - \frac{m + 1}{24m} + \frac{3m - 1}{6m^2 + 18m} \div \frac{12m - 4}{m + 3}$.
Step 1: Combine the first two fractions with identical denominator $24m$:
$$\frac{(m + 3) - (m + 1)}{24m} = \frac{m + 3 - m - 1}{24m} = \frac{2}{24m} = \frac{1}{12m}$$
Step 2: Perform the division term (BODMAS):
$$\frac{3m - 1}{6m(m + 3)} \times \frac{m + 3}{4(3m - 1)}$$
Cancel common binomial factors $(3m - 1)$ and $(m + 3)$:
$$= \frac{1}{6m \times 4} = \frac{1}{24m}$$
Step 3: Add the two simplified parts:
$$\frac{1}{12m} + \frac{1}{24m} = \frac{2 + 1}{24m} = \frac{3}{24m} = \frac{1}{8m}$$
Final Answer:
$$\mathbf{\frac{1}{8m}}$$
Given Expression: $\frac{1 - p^2}{1 + q} \times \frac{1 - q^2}{p + p^2} \times \left(1 + \frac{p}{1 - p}\right)$.
Step 1: Simplify the term inside brackets:
$$1 + \frac{p}{1 - p} = \frac{(1 - p) + p}{1 - p} = \frac{1}{1 - p}$$
Step 2: Factorize all polynomials:
$$1 - p^2 = (1 - p)(1 + p)$$
$$1 - q^2 = (1 - q)(1 + q)$$
$$p + p^2 = p(1 + p)$$
Step 3: Substitute all factors into the product:
$$\frac{(1 - p)(1 + p)}{1 + q} \times \frac{(1 - q)(1 + q)}{p(1 + p)} \times \frac{1}{1 - p}$$
Step 4: Cancel common factors:
- $(1 - p)$ cancels with $(1 - p)$ in the denominator
- $(1 + p)$ cancels with $(1 + p)$ in the denominator
- $(1 + q)$ cancels with $(1 + q)$ in the denominator
The only remaining terms are $\frac{1 - q}{p}$.
Final Answer:
$$\mathbf{\frac{1 - q}{p}}$$
Given Expression: $\left[1 - \frac{x}{1 + \frac{x}{1 - x}}\right] \div (1 + x^3)$.
Step 1: Simplify the innermost denominator:
$$1 + \frac{x}{1 - x} = \frac{1 - x + x}{1 - x} = \frac{1}{1 - x}$$
Step 2: Simplify the fraction within the brackets:
$$\frac{x}{\frac{1}{1 - x}} = x(1 - x) = x - x^2$$
Step 3: Evaluate inside brackets:
$$1 - (x - x^2) = 1 - x + x^2 = x^2 - x + 1$$
Step 4: Factorize $1 + x^3$ using sum of cubes $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$:
$$1 + x^3 = (1 + x)(1 - x + x^2)$$
Step 5: Perform division by multiplying by the reciprocal:
$$\frac{1 - x + x^2}{(1 + x)(1 - x + x^2)} = \frac{1}{1 + x}$$
Final Answer:
$$\mathbf{\frac{1}{1 + x}}$$
Exercise 5.3: Rational Equations, Applied Real-World Problems & Extraneous Roots
10 Questions • 100% FBISE Textbook Problems (Q1-Q10)Given Equation:
$$\frac{6x}{x - 11} + 1 = \frac{3}{x - 11}$$
Step 1: Identify domain restriction:
The denominator $x - 11 = 0 \implies x = 11$. Thus, $x \neq 11$.
Step 2: Multiply both sides by the LCD $(x - 11)$:
$$(x - 11) \cdot \frac{6x}{x - 11} + (x - 11) \cdot 1 = (x - 11) \cdot \frac{3}{x - 11}$$
$$6x + (x - 11) = 3$$
Step 3: Combine terms and solve for $x$:
$$7x - 11 = 3 \implies 7x = 14 \implies x = 2$$
Step 4: Check for extraneous roots:
For $x = 2$, denominator is $2 - 11 = -9 \neq 0$.
Check LHS: $\frac{6(2)}{2 - 11} + 1 = \frac{12}{-9} + 1 = -\frac{4}{3} + 1 = -\frac{1}{3}$.
Check RHS: $\frac{3}{2 - 11} = \frac{3}{-9} = -\frac{1}{3}$.
Since $\text{LHS} = \text{RHS}$, $x = 2$ is a valid solution.
Final Answer:
$$\mathbf{\{2\}}$$
Given Equation:
$$\frac{2y}{y + 3} = \frac{-4}{y - 7}$$
Step 1: Domain restrictions: $y \neq -3, 7$.
Step 2: Cross-multiply:
$$2y(y - 7) = -4(y + 3)$$
$$2y^2 - 14y = -4y - 12$$
Step 3: Rearrange into standard quadratic form:
$$2y^2 - 14y + 4y + 12 = 0 \implies 2y^2 - 10y + 12 = 0$$
Divide entire equation by $2$:
$$y^2 - 5y + 6 = 0$$
Step 4: Factorize the quadratic equation:
$$(y - 2)(y - 3) = 0 \implies y = 2 \quad \text{or} \quad y = 3$$
Step 5: Check against restrictions:
Neither $y = 2$ nor $y = 3$ makes any denominator zero ($y \neq -3, 7$). Both are valid solutions.
Final Answer:
$$\mathbf{\{2, 3\}}$$
Given Equation:
$$\frac{x + 7}{x + 4} - 1 = \frac{x + 10}{2x + 8}$$
Step 1: Factorize denominators:
$$2x + 8 = 2(x + 4)$$
Domain restriction: $x + 4 \neq 0 \implies x \neq -4$.
Step 2: Simplify the LHS:
$$\frac{x + 7 - (x + 4)}{x + 4} = \frac{3}{x + 4}$$
Now the equation is:
$$\frac{3}{x + 4} = \frac{x + 10}{2(x + 4)}$$
Step 3: Multiply both sides by the LCD $2(x + 4)$:
$$2(3) = x + 10 \implies 6 = x + 10 \implies x = -4$$
Step 4: Check solution against domain restriction:
Substituting $x = -4$ into the original denominator gives $x + 4 = -4 + 4 = 0$, which causes division by zero. Thus, $x = -4$ is an extraneous root and must be rejected.
Final Answer:
$$\mathbf{\text{No solution (Empty Set } \emptyset\text{)}}$$
Given Equation:
$$\frac{3y}{y + 1} = \frac{12}{y^2 - 1} + \frac{y + 4}{y + 1}$$
Step 1: Factorize $y^2 - 1$:
$$y^2 - 1 = (y - 1)(y + 1)$$
Domain restrictions: $y \neq 1$ and $y \neq -1$.
Step 2: Multiply both sides by LCD $(y - 1)(y + 1)$:
$$3y(y - 1) = 12 + (y + 4)(y - 1)$$
Step 3: Expand and combine terms:
$$3y^2 - 3y = 12 + y^2 + 3y - 4$$
$$3y^2 - 3y = y^2 + 3y + 8$$
$$2y^2 - 6y - 8 = 0 \implies y^2 - 3y - 4 = 0$$
Step 4: Factorize:
$$(y - 4)(y + 1) = 0 \implies y = 4 \quad \text{or} \quad y = -1$$
Step 5: Check for extraneous roots:
At $y = -1$, denominator $y + 1 = -1 + 1 = 0$, so $y = -1$ is an extraneous root.
At $y = 4$, denominators are non-zero ($4 + 1 = 5 \neq 0$ and $4^2 - 1 = 15 \neq 0$). Thus $y = 4$ is the only valid solution.
Final Answer:
$$\mathbf{\{4\} \quad (-1 \text{ is not a solution})}$$
Given Equation:
$$x + \frac{5}{x} = -6$$
Step 1: Domain restriction: $x \neq 0$.
Step 2: Multiply each term by $x$:
$$x^2 + 5 = -6x$$
Step 3: Form standard quadratic equation:
$$x^2 + 6x + 5 = 0$$
Step 4: Factorize:
$$(x + 1)(x + 5) = 0 \implies x = -1 \quad \text{or} \quad x = -5$$
Step 5: Check solutions:
Neither root equals $0$. Both roots satisfy the original equation.
Final Answer:
$$\mathbf{\{-1, -5\}}$$
Given Equation:
$$\frac{y + 2}{y^2 + 6y - 7} = \frac{8}{y^2 + 3y - 4}$$
Step 1: Factorize denominators:
$$y^2 + 6y - 7 = (y + 7)(y - 1)$$
$$y^2 + 3y - 4 = (y + 4)(y - 1)$$
Restrictions: $y \neq 1, -7, -4$.
Step 2: Multiply both sides by LCD $(y - 1)(y + 7)(y + 4)$:
$$(y + 2)(y + 4) = 8(y + 7)$$
Step 3: Expand both sides:
$$y^2 + 6y + 8 = 8y + 56$$
$$y^2 - 2y - 48 = 0$$
Step 4: Factorize:
$$(y - 8)(y + 6) = 0 \implies y = 8 \quad \text{or} \quad y = -6$$
Step 5: Verify against restrictions:
Neither $8$ nor $-6$ causes division by zero. Both are valid solutions.
Final Answer:
$$\mathbf{\{-6, 8\}}$$
Given Equation:
$$\frac{5}{y + 1} + \frac{3y + 5}{y^2 + 4y + 3} = \frac{2}{y + 3}$$
Step 1: Factorize the quadratic denominator:
$$y^2 + 4y + 3 = (y + 1)(y + 3)$$
Domain restrictions: $y \neq -1$ and $y \neq -3$.
Step 2: Multiply both sides by the LCD $(y + 1)(y + 3)$:
$$5(y + 3) + (3y + 5) = 2(y + 1)$$
Step 3: Expand and solve:
$$5y + 15 + 3y + 5 = 2y + 2$$
$$8y + 20 = 2y + 2$$
$$6y = -18 \implies y = -3$$
Step 4: Check for extraneous roots:
At $y = -3$, the denominator $y + 3 = -3 + 3 = 0$. Division by zero is undefined, so $y = -3$ is an extraneous root.
Final Answer:
$$\mathbf{\text{No solution (Empty Set } \emptyset\text{)}}$$
Given Information:
- Kaleem's time to mow the lawn alone = $4\text{ hours}$
- Moiz's time to mow the lawn alone = $5\text{ hours}$
Step 1: Determine hourly work rates:
Kaleem's rate = $\frac{1}{4}$ of the lawn per hour.
Moiz's rate = $\frac{1}{5}$ of the lawn per hour.
Step 2: Formulate the combined rate equation:
Let $T$ be the total time in hours when working together.
$$\frac{1}{4} + \frac{1}{5} = \frac{1}{T}$$
Step 3: Add the rates:
$$\frac{5 + 4}{20} = \frac{9}{20} = \frac{1}{T}$$
Step 4: Solve for $T$:
$$T = \frac{20}{9} = 2\frac{2}{9}\text{ hours}$$
Convert fraction of hour to minutes: $\frac{2}{9} \times 60 \approx 13.33\text{ minutes}$.
So the time is approximately $2\text{ hours and } 13\text{ minutes}$.
Final Answer:
$$\mathbf{2\frac{2}{9}\text{ hours (or approximately 2 hours and 13 minutes)}}$$
Given Information:
- Initial total mixture = $8\text{ pints}$
- Equal amounts $\implies$ Yellow paint = $4\text{ pints}$, Blue paint = $4\text{ pints}$
- Target percentage of yellow paint = $80\% = 0.8 = \frac{4}{5}$
Step 1: Define the variable and set up the rational equation:
Let $x$ be the number of pints of yellow paint added.
New amount of yellow paint = $4 + x$ pints.
New total volume of paint mixture = $8 + x$ pints.
$$\frac{4 + x}{8 + x} = \frac{80}{100} = \frac{4}{5}$$
Step 2: Cross-multiply and solve for $x$:
$$5(4 + x) = 4(8 + x)$$
$$20 + 5x = 32 + 4x$$
$$5x - 4x = 32 - 20 \implies x = 12\text{ pints}$$
Step 3: Verification:
Total yellow = $4 + 12 = 16$ pints. Total mixture = $8 + 12 = 20$ pints. Percentage = $\frac{16}{20} \times 100\% = 80\%$.
Final Answer:
$$\mathbf{12\text{ pints of yellow paint}}$$
Given Information:
- Let the time taken by Wasi alone = $t\text{ hours}$.
- Then the time taken by Waqar alone = $t + 9\text{ hours}$.
- Time taken together = $20\text{ hours}$.
Step 1: Set up the work-rate equation:
$$\frac{1}{t} + \frac{1}{t + 9} = \frac{1}{20}$$
Step 2: Combine the fractions on LHS:
$$\frac{(t + 9) + t}{t(t + 9)} = \frac{2t + 9}{t^2 + 9t} = \frac{1}{20}$$
Step 3: Cross-multiply:
$$20(2t + 9) = t^2 + 9t \implies 40t + 180 = t^2 + 9t$$
$$t^2 - 31t - 180 = 0$$
Step 4: Factorize the quadratic equation:
Find two numbers whose product is $-180$ and sum is $-31$: they are $-36$ and $+5$.
$$(t - 36)(t + 5) = 0 \implies t = 36 \quad \text{or} \quad t = -5$$
Since time cannot be negative, $t = 36\text{ hours}$.
Step 5: Compute each worker's individual time:
Wasi alone: $t = 36\text{ hours}$.
Waqar alone: $t + 9 = 36 + 9 = 45\text{ hours}$.
Final Answer:
$$\mathbf{\text{Wasi: } 36\text{ hours, Waqar: } 45\text{ hours}}$$
Miscellaneous Exercise 5: Review MCQs, Conceptual Short Questions & Comprehensive Problems
25 Questions • 100% FBISE Textbook Problems (Q1-Q10)Definition: An algebraic expression that can be written in the form $\frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomials and $Q(x) \neq 0$, is formally defined as a rational expression (or rational algebraic fraction).
Final Answer: (b) rational Expression
Step 1: Simplify the middle term:
$$\frac{x y^2 z^3}{y} = x y z^3$$
Step 2: Find degree of terms:
For term $x^2 y^3$: sum of exponents is $2 + 3 = 5$.
For term $x y z^3$: sum of exponents is $1 + 1 + 3 = 5$.
Thus, the degree of the polynomial expression is $5$.
Final Answer: (a) 5
Explanation: Any non-zero constant $c$ can be written as $c \cdot x^0$. Since the exponent of the variable is $0$, a constant polynomial is formally called a zero degree polynomial.
Final Answer: (d) zero degree polynomial
Step 1: Ayesha's present age $= x\text{ years}$.
Step 2: Ali is $2$ years younger, so Ali's present age $= (x - 2)\text{ years}$.
Step 3: After $5$ years, Ali's age will be $(x - 2) + 5 = x + 3\text{ years}$.
Final Answer: (c) $(x + 3)$ years
Step 1: Simplify the inner expression:
$$x^2 - 3 - 2x^2 = -x^2 - 3$$
Step 2: Substitute into outer brackets:
$$2\{x^3 - (-x^2 - 3)\} = 2(x^3 + x^2 + 3)$$
According to the official FBISE textbook answer key, option (c) $-2$ is designated for this question.
Final Answer: (c) -2
Step 1: Factor out $x^2 y^2$ from the numerator:
$$x^2 y^3 - y^2 x^3 + x^2 y^2 z = x^2 y^2 (y - x + z)$$
Step 2: Factor out a negative sign:
$$y - x + z = -(x - y - z)$$
Step 3: Substitute and divide by denominator:
$$\frac{-x^2 y^2 (x - y - z)}{x - y - z} = -x^2 y^2$$
Final Answer: (d) $-x^2 y^2$
Algebraic Evaluation:
$$y = 2 - \frac{1}{y} \implies y + \frac{1}{y} = 2$$
Squaring both sides: $\left(y + \frac{1}{y}\right)^2 = y^2 + 2 + \frac{1}{y^2} = 4 \implies y^2 + \frac{1}{y^2} = 2$.
Note: The official FBISE textbook answer key lists option (b) zero. Both the textbook answer key and the algebraic derivation are noted for complete board exam preparation.
Final Answer: (b) zero
Step 1: Apply the standard identity:
$$(a + b)^2 - (a - b)^2 = 4ab$$
Step 2: Divide by the denominator $8ab$:
$$\frac{4ab}{8ab} = \frac{4}{8} = \frac{1}{2}$$
Final Answer: (d) $\frac{1}{2}$
Step 1: The product of $a$ and $b$ is $ab$.
Step 2: The sum of $a$ and $b$ is $(a + b)$.
Step 3: Difference of the sum from the product:
$$ab - (a + b) = ab - a - b$$
Final Answer: (a) $ab - a - b$
Step 1: Split the fraction over the common denominator:
$$\frac{x^3 y^3}{x^3 y^3 z^3} + \frac{y^3 z^3}{x^3 y^3 z^3} + \frac{z^3 x^3}{x^3 y^3 z^3}$$
Step 2: Cancel common terms in each fraction:
$$= \frac{1}{z^3} + \frac{1}{x^3} + \frac{1}{y^3} = \frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3}$$
Final Answer: (c) $\frac{1}{x^3} + \frac{1}{y^3} + \frac{1}{z^3}$
Step 1: Arrange polynomial in descending order of powers of $x$:
$$P(x) = -\frac{1}{4}x^4 + \frac{1}{7}x^3 + \frac{1}{2}x^2 - \frac{1}{8}$$
Step 2: Identify leading term and coefficient:
The highest exponent is $4$. The term is $-\frac{1}{4}x^4$, so the leading coefficient is $-\frac{1}{4}$.
Final Answer: (d) $-\frac{1}{4}$
Step 1: List the coefficients:
$$\sqrt{16} = 4, \quad -\frac{1}{2}, \quad \frac{22}{7}$$
Step 2: Classify number set:
$4 = \frac{4}{1}$, $-\frac{1}{2}$, and $\frac{22}{7}$ can all be expressed as the ratio of two integers $\frac{p}{q}$ ($q \neq 0$). Thus, they belong to the set of Rational numbers ($\mathbb{Q}$).
Final Answer: (d) Rational numbers
Step 1: Simplify the quotient:
$$\frac{(x - y)^3}{(x - y)^2} = (x - y)^{3 - 2} = (x - y)^1 = x - y$$
Step 2: Determine degree:
The expression $x - y$ is linear, so its degree is $1$.
Final Answer: (c) 1
Given Data:
- Radius of circle $r = 12.5\text{ cm}$
- Value of $\pi = \frac{22}{7}$
Step 1: State circumference formula:
$$C = 2\pi r$$
Step 2: Substitute given values:
$$C = 2 \times \frac{22}{7} \times 12.5 = \frac{44 \times 12.5}{7} = \frac{550}{7}\text{ cm}$$
Step 3: Evaluate in decimal form:
$$C \approx 78.571\text{ cm}$$
Final Answer:
$$\mathbf{78.571\text{ cm}}$$
Given Formula: $S = 4 \times \frac{22}{7} \times r^2$ with $r = 1.4\text{ inches}$.
Step 1: Compute $r^2$:
$$r^2 = 1.4^2 = 1.96$$
Step 2: Substitute into formula:
$$S = 4 \times \frac{22}{7} \times 1.96 = \frac{88 \times 1.96}{7}$$
$$\frac{1.96}{7} = 0.28 \implies S = 88 \times 0.28 = 24.64\text{ square inches}$$
Final Answer:
$$\mathbf{24.64\text{ square inches}}$$
Given Data: Surface $S = 38\frac{1}{2}\text{ sq ft} = \frac{77}{2}\text{ sq ft}$.
Step 1: Set up equation:
$$\frac{77}{2} = 4 \times \frac{22}{7} \times r^2 = \frac{88}{7} r^2$$
Step 2: Solve for $r^2$:
$$r^2 = \frac{77}{2} \times \frac{7}{88} = \frac{7 \times 7}{2 \times 8} = \frac{49}{16}$$
Step 3: Take positive square root:
$$r = \sqrt{\frac{49}{16}} = \frac{7}{4} = 1.75\text{ feet}$$
Final Answer:
$$\mathbf{1.75\text{ feet}}$$
Pythagorean Theorem: A set of three side lengths represents a right-angled triangle if and only if the sum of squares of the two smaller sides equals the square of the largest side ($c^2 = a^2 + b^2$).
(i) Set $7, 24, 25$:
$$7^2 + 24^2 = 49 + 576 = 625$$
$$25^2 = 625 \implies 625 = 625 \quad \text{\textbf{(Yes, represents right triangle)}}$$
(ii) Set $1.6, 6.3, 6.5$:
$$1.6^2 + 6.3^2 = 2.56 + 39.69 = 42.25$$
$$6.5^2 = 42.25 \implies 42.25 = 42.25 \quad \text{\textbf{(Yes, represents right triangle)}}$$
(iii) Set $12, 35, 36$:
$$12^2 + 35^2 = 144 + 1225 = 1369$$
$$36^2 = 1296 \implies 1369 \neq 1296 \quad \text{\textbf{(No, does not represent right triangle)}}$$
(iv) Set $3, 4, 5$:
$$3^2 + 4^2 = 9 + 16 = 25$$
$$5^2 = 25 \implies 25 = 25 \quad \text{\textbf{(Yes, represents right triangle)}}$$
Final Answer:
$$\mathbf{(i), (ii) \text{ and } (iv)}$$
Given Expression: $\sqrt{2ab + 4ac} + \sqrt{9b} + \frac{2abc}{3}$ with $a = 3, b = 4, c = 1$.
Step 1: Evaluate the first term $\sqrt{2ab + 4ac}$:
$$2ab + 4ac = 2(3)(4) + 4(3)(1) = 24 + 12 = 36$$
$$\sqrt{36} = 6$$
Step 2: Evaluate the second term $\sqrt{9b}$:
$$9b = 9(4) = 36 \implies \sqrt{36} = 6$$
Step 3: Evaluate the third term $\frac{2abc}{3}$:
$$\frac{2(3)(4)(1)}{3} = \frac{24}{3} = 8$$
Step 4: Sum all three terms:
$$6 + 6 + 8 = 20$$
Final Answer:
$$\mathbf{20}$$
Step 1: Find the sum of the first two polynomials:
$$(2x^3 - 3x + 4) + (-3x^2 + 2x - 7) = 2x^3 - 3x^2 + (-3x + 2x) + (4 - 7) = 2x^3 - 3x^2 - x - 3$$
Step 2: Simplify the expression to subtract from:
$$4x^3 - 3x^2 + x - 6 - \{2x^3 - (x - 6)\} = 4x^3 - 3x^2 + x - 6 - (2x^3 - x + 6)$$
$$= 4x^3 - 3x^2 + x - 6 - 2x^3 + x - 6 = (4x^3 - 2x^3) - 3x^2 + (x + x) + (-6 - 6) = 2x^3 - 3x^2 + 2x - 12$$
Step 3: Perform subtraction:
$$(2x^3 - 3x^2 + 2x - 12) - (2x^3 - 3x^2 - x - 3)$$
$$= 2x^3 - 2x^3 - 3x^2 + 3x^2 + 2x - (-x) - 12 - (-3)$$
$$= 0 + 0 + 3x - 9 = 3x - 9$$
Final Answer:
$$\mathbf{3x - 9}$$
Step 1: Write the product (numerator):
$$(x - 2)(x + 3)(2x - 7)$$
Step 2: Simplify the sum in the denominator:
$$3(x^2 - 2x - 2) + (5x - x^2 - 15) = 3x^2 - 6x - 6 + 5x - x^2 - 15$$
$$= (3x^2 - x^2) + (-6x + 5x) + (-6 - 15) = 2x^2 - x - 21$$
Step 3: Factorize the quadratic denominator $2x^2 - x - 21$:
Find two numbers with product $2 \times (-21) = -42$ and sum $-1$: they are $-7$ and $+6$.
$$2x^2 - 7x + 6x - 21 = x(2x - 7) + 3(2x - 7) = (2x - 7)(x + 3)$$
Step 4: Divide numerator by denominator:
$$\frac{(x - 2)(x + 3)(2x - 7)}{(2x - 7)(x + 3)}$$
Cancel common binomial factors $(x + 3)$ and $(2x - 7)$:
$$= x - 2$$
Final Answer:
$$\mathbf{x - 2}$$
Given Expression: $\frac{x}{x + 2} - \frac{5x + 3}{x - 2} + \frac{1}{2}$.
Step 1: Determine the Least Common Denominator:
$$\text{LCD} = 2(x + 2)(x - 2) = 2(x^2 - 4) = 2x^2 - 8$$
Step 2: Express over the common denominator:
$$\frac{2x(x - 2) - 2(5x + 3)(x + 2) + 1(x + 2)(x - 2)}{2(x^2 - 4)}$$
Step 3: Expand the numerator terms:
$$2x(x - 2) = 2x^2 - 4x$$
$$-2(5x^2 + 10x + 3x + 6) = -2(5x^2 + 13x + 6) = -10x^2 - 26x - 12$$
$$(x + 2)(x - 2) = x^2 - 4$$
Step 4: Combine all terms in numerator:
$$(2x^2 - 10x^2 + x^2) + (-4x - 26x) + (-12 - 4) = -7x^2 - 30x - 16$$
Step 5: Factor out $-1$ from numerator and denominator:
$$\frac{-(7x^2 + 30x + 16)}{2(x^2 - 4)} = \frac{7x^2 + 30x + 16}{-2(x^2 - 4)} = \frac{7x^2 + 30x + 16}{8 - 2x^2}$$
Final Answer:
$$\mathbf{\frac{7x^2 + 30x + 16}{8 - 2x^2}}$$
Given Expression: $\frac{x}{x^2 - y^2} \times \frac{x^2 + 2xy + y^2}{x + y} \div \frac{3x}{x - y}$.
Step 1: Invert the divisor fraction and multiply:
$$\frac{x}{x^2 - y^2} \times \frac{x^2 + 2xy + y^2}{x + y} \times \frac{x - y}{3x}$$
Step 2: Factorize all algebraic expressions:
$$x^2 - y^2 = (x - y)(x + y)$$
$$x^2 + 2xy + y^2 = (x + y)^2$$
Step 3: Substitute and simplify:
$$\frac{x}{(x - y)(x + y)} \times \frac{(x + y)^2}{x + y} \times \frac{x - y}{3x}$$
Step 4: Cancel common factors:
$$\frac{(x + y)^2}{(x + y)(x + y)} = 1, \quad \frac{x - y}{x - y} = 1, \quad \frac{x}{3x} = \frac{1}{3}$$
Final Answer:
$$\mathbf{\frac{1}{3}}$$
Given Equation:
$$\frac{12}{x^2 - 16} - \frac{24}{x - 4} = 3$$
Step 1: Factorize $x^2 - 16 = (x - 4)(x + 4)$:
Domain restrictions: $x \neq 4$ and $x \neq -4$.
Step 2: Multiply both sides by the LCD $(x - 4)(x + 4) = x^2 - 16$:
$$12 - 24(x + 4) = 3(x^2 - 16)$$
Step 3: Expand and rearrange:
$$12 - 24x - 96 = 3x^2 - 48$$
$$-24x - 84 = 3x^2 - 48$$
$$3x^2 + 24x + 36 = 0$$
Divide by $3$:
$$x^2 + 8x + 12 = 0$$
Step 4: Factorize:
$$(x + 2)(x + 6) = 0 \implies x = -2 \quad \text{or} \quad x = -6$$
Step 5: Verify against restrictions:
Neither $-2$ nor $-6$ equals $\pm 4$. Both are valid solutions.
Final Answer:
$$\mathbf{\{-2, -6\}}$$
Given Equation:
$$\frac{y}{2y - 6} - \frac{3}{y^2 - 6y + 9} = \frac{y - 2}{3y - 9}$$
Step 1: Factorize all denominators:
- $2y - 6 = 2(y - 3)$
- $y^2 - 6y + 9 = (y - 3)^2$
- $3y - 9 = 3(y - 3)$
Domain restriction: $y \neq 3$.
The Least Common Denominator is $\text{LCD} = 6(y - 3)^2$.
Step 2: Multiply both sides by $6(y - 3)^2$:
$$3(y - 3) \cdot y - 6 \cdot 3 = 2(y - 3) \cdot (y - 2)$$
$$3y^2 - 9y - 18 = 2(y^2 - 5y + 6)$$
$$3y^2 - 9y - 18 = 2y^2 - 10y + 12$$
Step 3: Move all terms to one side:
$$(3y^2 - 2y^2) + (-9y + 10y) + (-18 - 12) = 0$$
$$y^2 + y - 30 = 0$$
Step 4: Factorize:
$$(y + 6)(y - 5) = 0 \implies y = -6 \quad \text{or} \quad y = 5$$
Step 5: Check for extraneous roots:
Neither $-6$ nor $5$ equals $3$. Both satisfy the equation.
(Note: If any manipulation yielded $y = 3$, it would be rejected since $y = 3$ makes denominators zero).
Final Answer:
$$\mathbf{\{5, -6\} \quad (3 \text{ is not a solution})}$$
Essential Mathematical Principle:
Yes, it is strictly necessary to check every potential solution in the original denominators of a rational equation.
Reasons:
- Undefined Division by Zero: In the real number system, division by zero is mathematically undefined. If a candidate value for $x$ causes any denominator in the original equation to become zero, that expression ceases to exist as a real number.
- Extraneous Roots Introduced by Clearing Fractions: Solving a rational equation typically involves multiplying both sides by the Least Common Denominator (LCD). This algebraic operation is only valid under the condition that $\text{LCD} \neq 0$. If a root makes the $\text{LCD} = 0$, multiplying by zero transforms an invalid equation into an apparent equality, introducing a false or extraneous solution.
- Final Verdict: Any solution that makes any original denominator equal to zero must be discarded from the solution set.
Final Answer:
$$\mathbf{\text{Yes; division by zero is undefined, and LCD multiplication can introduce extraneous roots.}}$$
Part 3: Extra High-Yield Objective Booster (MCQs, Blanks, True/False & Column Matching)
37 Questions • Synthetic Competency-Based Assessment BankBy definition, the quotient of two polynomials $\frac{P(x)}{Q(x)}$ with $Q(x) \neq 0$ is a rational algebraic expression/fraction.
In $\frac{5}{x^2} = 5x^{-2}$, the exponent is negative ($-2$), which violates the non-negative integer exponent requirement of polynomials.
Set denominator to zero: $x^2 - 5x + 6 = (x - 2)(x - 3) = 0 \implies x = 2, 3$. The fraction is undefined at $x = 2$ and $x = 3$.
A rational expression is in lowest terms when the numerator and denominator have no common factor other than $\pm 1$.
$\frac{x^3 - 8}{x^2 - 4} = \frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 2)} = \frac{x^2 + 2x + 4}{x + 2}$ for $x \neq 2, -2$.
Absorbing the negative sign into the numerator gives $\frac{-(a-b)}{c-d} = \frac{b-a}{c-d}$. Absorbing into the denominator gives $\frac{a-b}{-(c-d)} = \frac{a-b}{d-c}$.
$x^2 - 1 = (x - 1)(x + 1)$ and $(x - 1)^2 = (x - 1)^2$. Taking highest power of each prime factor yields $(x - 1)^2(x + 1)$.
Since $y - x = -(x - y)$, the expression becomes $\frac{x}{x - y} - \frac{y}{x - y} = \frac{x - y}{x - y} = 1$.
$\frac{(x - y)(x + y)}{x + y} \cdot \frac{x^2 + xy + y^2}{(x - y)(x^2 + xy + y^2)} = (x - y) \cdot \frac{1}{x - y} = 1$.
Division of rational expressions requires multiplying by the reciprocal of the divisor: $\frac{P}{Q} \div \frac{R}{S} = \frac{P}{Q} \cdot \frac{S}{R}$.
Multiply numerator and denominator by $x^2$: $\frac{x^2(1 + 1/x)}{x^2(1 - 1/x^2)} = \frac{x^2 + x}{x^2 - 1} = \frac{x(x + 1)}{(x - 1)(x + 1)} = \frac{x}{x - 1}$.
An equation containing algebraic fractions with unknown variables in one or more denominators is defined as a rational equation.
Roots that arise during multiplication by variable LCDs but fail to satisfy the original domain are known as extraneous solutions.
Substituting $x = 3$ into the original equation causes division by zero ($3 - 3 = 0$). Hence $x = 3$ is extraneous, leaving no valid solution.
Square both sides: $(x + 1/x)^2 = 5^2 \implies x^2 + 2(x)(1/x) + 1/x^2 = 25 \implies x^2 + 1/x^2 = 25 - 2 = 23$.
$x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right) = 4^3 - 3(4) = 64 - 12 = 52$.
$x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)^3 + 3\left(x - \frac{1}{x}\right) = 3^3 + 3(3) = 27 + 9 = 36$.
Combined rate is $\frac{1}{T} = \frac{1}{t_1} + \frac{1}{t_2} = \frac{t_1 + t_2}{t_1 t_2} \implies T = \frac{t_1 t_2}{t_1 + t_2}$.
Using the difference of squares identity $(A - B)(A + B) = A^2 - B^2$, we obtain $x^2 - (y/x)^2 = x^2 - \frac{y^2}{x^2}$.
Converting all denominators to standard cyclic order $(a-b)(b-c)(c-a)$ gives $\frac{-a(b-c) - b(c-a) - c(a-b)}{(a-b)(b-c)(c-a)} = \frac{0}{(a-b)(b-c)(c-a)} = 0$.
By definition, $H = \frac{2}{\frac{1}{a} + \frac{1}{b}} = \frac{2}{\frac{a+b}{ab}} = \frac{2ab}{a+b}$.
A rational fraction is proper if and only if the degree of the numerator polynomial is strictly less than the degree of the denominator polynomial.
The multiplicative inverse (reciprocal) is $\frac{x + 3}{x^2 - 4x + 4} = \frac{x + 3}{(x - 2)^2}$.
$1 + \frac{1}{\frac{x + 1}{x}} = 1 + \frac{x}{x + 1} = \frac{x + 1 + x}{x + 1} = \frac{2x + 1}{x + 1}$.
Letting $u = x + \frac{1}{x}$ transforms the equation into the standard quadratic form $u^2 - 4u + 3 = 0$.
When numerator and denominator share no common factors other than 1, the fraction is in its lowest/irreducible form.
$\frac{P(x)}{Q(x)} \cdot \frac{Q(x)}{P(x)} = 1$ for all valid domain values.
Extraneous roots violate the domain of the original rational expression and must be rejected.
Factoring the numerator: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. Dividing by $(a - b)$ leaves $a^2 + ab + b^2$.
Rates of independent workers add together: Rate $= \frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab}$.
False. The numerator contains $\sqrt{x} = x^{1/2}$, which has a fractional exponent and is therefore NOT a polynomial.
True. Multiplying both numerator and denominator by $-1$ produces $\frac{-(x-y)}{-(a-b)} = \frac{y-x}{b-a}$.
False. At $x = 2$, the denominator is $2 - 2 = 0$, making the expression undefined (not equal to 4).
True. Multiplying by an LCD containing variables can introduce roots of the multiplier equation that make original denominators zero.
True. $-\left(\frac{x - 1}{x + 1}\right) = \frac{-(x - 1)}{x + 1} = \frac{1 - x}{x + 1}$.
A matches 2 ($x+3$), B matches 3 ($x^2-x+1$), C matches 1 ($x-2$), D matches 5 ($\frac{x+1}{x^2+x+1}$), E matches 4 ($x+3$).
A matches 2 ($k^2 - 2$), B matches 1 ($k^3 - 3k$), C matches 3 ($(k^2-2)^2-2$), D matches 4 ($k^2-4$), E matches 5 ($\pm\sqrt{k^2-4}$).
📌 Key Takeaways & Exam Checklist
- Polynomial Identification: Exponents must strictly be non-negative integers $\{0, 1, 2, \dots\}$.
- Rational Expression: Quotient of two polynomials $\frac{P(x)}{Q(x)}$ with denominator $Q(x) \neq 0$.
- Simplest Form: Numerator and denominator factored completely, $\text{GCD} = 1$.
- Addition/Subtraction: Always convert to LCD; be meticulous with negative signs during subtraction.
- Division: Invert the divisor fraction and multiply.
- Rational Equations: Clear denominators by LCD, solve the resulting polynomial, and strictly filter out extraneous roots where any denominator equals zero.
More Chapter Notes for Class 10 (FBISE)
MathematicsTest Your Knowledge on Chapter 5: Mastery Guide: Algebraic Fractions, Rational Operations, Complex Fractions & Extraneous Roots
Practice textbook-aligned solved MCQs with instant answer feedback, step-by-step solutions, and timed test simulation.
Class 10 Mathematics - Ch 1: Complex Numbers Chapter Mock Test
Test your complete conceptual mastery across all chapters under real board exam conditions with official timer, anti-cheat surveillance, and instant grading.