Rational expressions — simplification, operations

2.1.21 · Maths › Algebra — Introduction & Intermediate

Domain restrictions: Yeh expression wahan undefined hoti hai jahan denominator zero ke barabar ho. Hume hamesha yeh restrictions identify karni aur batani chahiye.

First principles se step-by-step:

1. Numerator aur denominator ko completely factor karo

   • Kyun? Kyunki $\frac{ab}{ac}=\frac{b}{c}$ (common factor $a$ cancel karo)
   • Yeh tabhi kaam karta hai jab factors multiply ho rahe hon, add nahin

2. Cancel karne SE PEHLE restrictions identify karo

   • Kyun? Canceling factors ko remove karta hai, lekin original denominator ke zeros phir bhi matter karte hain
   • Example: $\frac{x-2}{(x-2)(x+3)}$ $x=2$ AUR $x=-3$ pe undefined hai, cancel karne ke baad bhi

3. Common factors cancel karo

   • Rule: $\frac{A\cdot C}{B\cdot C}=\frac{A}{B}$ for $C\ne 0$
   • Jo terms add/subtract ho rahe hain unhe KABHI cancel mat karo

4. Simplified form domain ke saath batao

General form: $\frac{P(x)}{Q(x)}=\frac{P(x)}{Q(x)}\text{where }Q(x)\ne 0$

Factoring ke baad: $\frac{A(x)\cdot C(x)}{B(x)\cdot C(x)}=\frac{A(x)}{B(x)}\text{where }B(x)\ne 0,C(x)\ne 0$

Solution har step pe WHY ke saath:

Step 1: Numerator factor karo $x^2-4=(x-2)(x+2)$ Difference of squares kyun? Kyunki $a^2-b^2=(a-b)(a+b)$

Step 2: Denominator factor karo $x^2+5x+6=(x+2)(x+3)$ Kyun? Do aisi numbers dhundho jo 6 se multiply hon aur 5 mein add hon: woh hain 2 aur 3

Step 3: Restrictions identify karo $x+2\ne 0\Rightarrow x\ne -2$ $x+3\ne 0\Rightarrow x\ne -3$ Cancel karne se pehle kyun? Yeh values ORIGINAL denominator ko zero banati thin

Step 4: Factored forms ke saath rewrite karo $\frac{(x-2)(x+2)}{(x+2)(x+3)}$

Step 5: Common factor $(x+2)$ cancel karo $=\frac{x-2}{x+3}$ Cancel kyun kar sakte hain? Kyunki $\frac{a\cdot b}{a\cdot c}=\frac{b}{c}$ jab $a\ne 0$

Final answer: $\frac{x-2}{x+3}$, jahan $x\ne -2,-3$

Multiplication rule ki derivation: $\frac{a}{b}\cdot \frac{c}{d}=\frac{a\cdot c}{b\cdot d}$ Kyun? Numerators ko aapas mein multiply karo, denominators ko aapas mein multiply karo (numeric fractions ki tarah)

Solution:

Step 1: Multiply karne SE PEHLE sab kuch factor karo $\frac{(x-1)(x+1)}{x(x+3)}\cdot \frac{x}{x-1}$ Pehle factor kyun karte hain? Taaki multiply karne se pehle fractions ke across cancel kar sakein

Step 2: Saari restrictions identify karo

• $x(x+3)$ se: $x\ne 0,-3$
• $x-1$ se: $x\ne 1$

Step 3: Single fraction ke roop mein likho $\frac{(x-1)(x+1)\cdot x}{x(x+3)\cdot (x-1)}$

Step 4: Common factors cancel karo

• Numerator aur denominator se $x$ cancel karo
• Numerator aur denominator se $(x-1)$ cancel karo

$=\frac{x+1}{x+3}$

Final answer: $\frac{x+1}{x+3}$, jahan $x\ne 0,1,-3$

Division rule ki derivation: $\frac{a}{b}÷\frac{c}{d}=\frac{a}{b}\cdot \frac{d}{c}$ Kyun? Fraction se divide karna = uske reciprocal se multiply karna Proof: $\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}\cdot \frac{d}{c}$ (upar aur neeche $\frac{d}{c}$ se multiply karo)

Solution:

Step 1: Division ko reciprocal se multiplication mein rewrite karo $\frac{x^2+2x}{x+3}\cdot \frac{x^2-9}{x}$

Step 2: Sab kuch factor karo $\frac{x(x+2)}{x+3}\cdot \frac{(x-3)(x+3)}{x}$ $x^2-9$ kyun? Difference of squares: $x^2-9=(x-3)(x+3)$

Step 3: Restrictions

• Denominators se: $x\ne -3,0$
• Original divisor ke denominator se: $x\ne 3$

Step 4: Combine karo aur cancel karo $\frac{x(x+2)(x-3)(x+3)}{x(x+3)}$ $x$ aur $(x+3)$ cancel karo: $=(x+2)(x-3)$ $=x^2-x-6$

Final answer: $x^2-x-6$, jahan $x\ne -3,0,3$

Like denominators ke liye rule: $\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}$ Kyun? Bilkul jaise $\frac{2}{7}+\frac{3}{7}=\frac{5}{7}$ — numerators combine karo, common denominator rakho

Solution: $\frac{3x+(2x+1)}{x-5}=\frac{5x+1}{x-5}$ jahan $x\ne 5$

Addition rule ki derivation: Alag-alag denominators wale fractions add karne ke liye, hume ek common denominator chahiye.

Kyun? Tum $\frac{1}{2}+\frac{1}{3}$ directly add nahin kar sakte — tumhe $\frac{3}{6}+\frac{2}{6}=\frac{5}{6}$ chahiye

LCD (Least Common Denominator) dhundhna:

• Har denominator ko factor karo
• LCD = sabhi factors ki highest powers ka product
• Yahan: $\text{LCD}=(x+1)(x-2)$

Solution:

Step 1: Har fraction ko LCD ke saath likho $\frac{2}{x+1}\cdot \frac{x-2}{x-2}+\frac{3}{x-2}\cdot \frac{x+1}{x+1}$ $\frac{x-2}{x-2}$ se multiply kyun karte hain? Yeh 1 ke barabar hai, isliye value nahin badal rahi

Step 2: Numerators expand karo $\frac{2(x-2)}{(x+1)(x-2)}+\frac{3(x+1)}{(x+1)(x-2)}$ $=\frac{2x-4}{(x+1)(x-2)}+\frac{3x+3}{(x+1)(x-2)}$

Step 3: Numerators combine karo $=\frac{2x-4+3x+3}{(x+1)(x-2)}=\frac{5x-1}{(x+1)(x-2)}$

Final answer: $\frac{5x-1}{(x+1)(x-2)}$, jahan $x\ne -1,2$

Solution:

Step 1: Denominators factor karo $x^2-4=(x-2)(x+2)$ $\frac{x}{(x-2)(x+2)}-\frac{2}{x+2}$

Step 2: LCD dhundho $\text{LCD}=(x-2)(x+2)$

Step 3: LCD ke saath rewrite karo $\frac{x}{(x-2)(x+2)}-\frac{2(x-2)}{(x+2)(x-2)}$ Doosre fraction ko $\frac{x-2}{x-2}$ se multiply kyun karte hain? Common denominator lene ke liye

Step 4: Combine karo $=\frac{x-2(x-2)}{(x-2)(x+2)}=\frac{x-2x+4}{(x-2)(x+2)}$ $=\frac{-x+4}{(x-2)(x+2)}$

Final answer: $\frac{4-x}{(x-2)(x+2)}$, jahan $x\ne 2,-2$

Kyun sahi lagta hai: Dono parts mein $x$ "ek jaisa" dikhta hai.

Mistake ko seriously lena: Tum $\frac{xa}{xb}=\frac{a}{b}$ se pattern match kar rahe ho, jo multiplication ke liye SACH MEIN correct hai. Confusion yeh hai ki addition aur multiplication ko mix up kar rahe ho.

Fix:

• Tum sirf factors cancel kar sakte ho (jo cheezein multiply ho rahi hain)
• Terms cancel NAHIN kar sakte (jo cheezein add ho rahi hain)
• $\frac{x\cdot 3}{x\cdot 5}=\frac{3}{5}$ ✓ (factors)
• $\frac{x+3}{x+5}\ne \frac{3}{5}$ ✗ (terms)

Numbers se test karo: $x=1$ lo: $\frac{1+3}{1+5}=\frac{4}{6}=\frac{2}{3}\ne \frac{3}{5}$

Kyun sahi lagta hai: $(x-3)$ cancel karne ke baad woh "chala gaya" toh uska zikr kyun karein?

Mistake ko seriously lena: Tum algebraically soch rahe ho — ek baar simplify kar diya toh restricted form "naya" function hai.

Fix:

• Original expression $x=3$ pe undefined thi (denominator 0 ban jaata tha)
• Simplified form $\frac{1}{x+2}$ bhale hi $x=3$ pe defined hogi, lekin hum original restriction maintain karte hain
• Soch lo: hum expression simplify kar rahe hain, uska domain nahin badal rahe

Sahi answer: $\frac{1}{x+2}$ jahan $x\ne 3,-2$

Visual test: Dono functions graph karo — original mein $x=3$ pe ek "hole" hai, jo remain karta hai.

Kyun sahi lagta hai: "Fractions mein addition, denominator mein addition" — pattern matching galat jagah gayi.

Mistake ko seriously lena: Tum "operations combine hote hain" rule ko overgeneralize kar rahe ho. Shayad soch rahe ho ki $\frac{a}{b}+\frac{a}{c}$ mein $(b+c)$ kisi tarah involve hoga.

Fix:

• LCD product $(x-2)(x+3)$ hai, sum nahin
• Product kyun? Har fraction ko apne original denominator ka multiple hona chahiye
• $\frac{1}{x-2}=\frac{x+3}{(x-2)(x+3)}$ — LCD tak pahunchne ke liye hum multiply karte hain

Sahi: $\frac{1}{x-2}+\frac{1}{x+3}=\frac{x+3}{(x-2)(x+3)}+\frac{x-2}{(x-2)(x+3)}=\frac{2x+1}{(x-2)(x+3)}$

Simplifying bilkul $\frac{6}{8}$ ko $\frac{3}{4}$ mein reduce karne jaisi hai — common 2 cancel karo. Lekin yahan hum common polynomial factors cancel karte hain jaise $(x+2)$. Trick yeh hai: tum sirf un cheezein ko cancel kar sakte ho jo aapas mein multiply hoon, add nahin.

Adding in expressions bilkul $\frac{1}{2}+\frac{1}{3}$ add karne jaisa hai — pehle common denominator chahiye! Polynomials ke saath, tum denominators ka "least common multiple" dhundhte ho (unhe factor karo, phir unique factors ko aapas mein multiply karo).

Bada rule: Kabhi zero se divide mat karo! Isliye agar tumhare denominator mein $(x-5)$ hai, toh $x$ 5 nahin ho sakta (woh bottom ko zero bana dega, aur $\frac{\text{kuch bhi}}{0}$ undefined hai/maths tod deta hai). Hum inhe "restrictions" ke roop mein likhte hain.

Rational expressions ko "fraction algebra" samjho — numeric fractions ke har rule (common denominator dhundho, factor karo aur cancel karo, seedha across multiply karo) bilkul waise hi kaam karte hain, lekin polynomials ke saath!

Addition/subtraction ke liye: "LCD ko SABHI factors chahiye, har ek HIGHEST power tak raised"

Canceling ke liye: "Sirf FACTORS (×) cancel karo, TERMS (+) kabhi nahin"

Memory hook: "Can't FRaED without the RED" — Restrictions aur Execution must be Done right.

Summary

Rational expressions woh fractions hain jinme polynomial numerator aur denominator hote hain. Core operations:

1. Simplification: Completely factor karo, restrictions identify karo, common factors cancel karo
2. Multiplication: Factor karo, seedha across multiply karo, cancel karo
3. Division: Reciprocal se multiply karo, phir multiplication rules follow karo
4. Addition/Subtraction: LCD dhundho, har fraction rewrite karo, numerators combine karo

Critical rule: Har final answer ke saath domain restrictions (jo values KISI BHI denominator ko zero banayein) batao.

Connections

• Polynomial factoring — simplification ke liye zaroori prerequisite
• Domain and range — restrictions domain define karti hain
• Complex fractions — nested rational expressions
• Rational equations — jab rational expressions equal set ki jayein toh solve karna
• Polynomial long division — improper rational expressions ke liye
• Limits and continuity — canceled factors se bane "holes" ka calculus perspective
• Partial fraction decomposition — complex rational expressions ko todna

#flashcards/maths

Rational expression kya hota hai? :: Do polynomials ka ratio $\frac{P(x)}{Q(x)}$ jahan $Q(x)\ne 0$

$\frac{x+5}{x+3}$ mein cancel kyun nahin kar sakte? :: Kyunki $x$ ek TERM hai (add ho raha hai), FACTOR nahin (multiply nahin ho raha). Sirf factors cancel kar sakte hain.

Concept Map