Simultaneous equations — substitution, elimination, cross-multiplication

2.1.10 · Maths › Algebra — Introduction & Intermediate

Overview

Simultaneous equations aisi systems hain jahan humein aisi values dhundhni hoti hain jo ek saath multiple equations ko satisfy karti hain. Sabse common case: do equations aur do unknowns. Hum inhe solve karne ke teen methods explore karte hain, jinka har ek yeh bataata hai ki equations aapas mein kaise interact karte hain.

Multiple methods kyun zaroori hain: Alag-alag methods alag-alag equation structures ke liye suited hote hain. Substitution tab kaam karta hai jab ek variable already isolated ho; elimination tab shine karta hai jab coefficients nicely align hote hain; cross-multiplication ek symmetric formula-based approach deta hai.

Core Intuition

Teen possible scenarios:

• Ek unique solution: Lines bilkul ek point par milti hain (sabse common)
• Koi solution nahi: Lines parallel hain (inconsistent system)
• Infinite solutions: Lines identical hain (dependent system)

Method 1: Substitution

Derivation from First Principles

Given:

a_1 x + b_1 y = c_1 \quad \text{..(1)} \\ a_2 x + b_2 y = c_2 \quad \text{...(2)} \end{cases}$$ **Step 1**: Equation (1) se $y$ ko isolate karo: $$b_1 y = c_1 - a_1 x$$ $$y = \frac{c_1 - a_1 x}{b_1} \quad \text{...(3)}$$ **Yeh step kyun?** Hum $y$ ki ek "definition" banate hain jo sirf $x$ par depend karti hai. Ab jahan bhi $y$ dikhega, hum ise is expression se replace kar sakte hain. **Step 2**: (3) ko equation (2) mein substitute karo: $$a_2 x + b_2 \left( \frac{c_1 - a_1 x}{b_1} \right) = c_2$$ **Yeh step kyun?** Equation (2) mein ab sirf $x$ hai. Humne $y$ ko $x$ ke terms mein uske expression se replace karke eliminate kar diya. **Step 3**: $x$ ke liye solve karo: $$a_2 x + \frac{b_2(c_1 - a_1 x)}{b_1} = c_2$$ $$a_2 x b_1 + b_2 c_1 - b_2 a_1 x = c_2 b_1$$ $$x(a_2 b_1 - a_1 b_2) = c_2 b_1 - b_2 c_1$$ $$x = \frac{c_2 b_1 - b_2 c_1}{a_2 b_1 - a_1 b_2}$$ **Step 4**: $y$ dhundhne ke liye $x$ ko wapas (3) mein substitute karo. > [!example] Worked Example 1: Substitution > Solve karo: > $$\begin{cases} > 2x + y = 7 \quad \text{...(1)} \\ > 3x - 2y = 4 \quad \text{...(2)} > \end{cases}$$ **Solution**: (1) se: $y = 7 - 2x$ ... *Kyun? Hum us simpler variable ko isolate karte hain jiska coefficient 1 hai* (2) mein substitute karo: $3x - 2(7 - 2x) = 4$ ... *Kyun? Hum har $y$ ko uske $x$ ke terms mein equivalent se replace kar rahe hain* Expand karo: $3x - 14 + 4x = 4$ ... *Kyun? Parentheses mein $-2$ distribute karo* Combine karo: $7x = 18$ ... *Kyun? Like terms collect karo aur dono sides mein 14 add karo* Isliye: $x = \frac{18}{7}$ Wapas substitute karo: $y = 7 - 2 \cdot \frac{18}{7} = 7 - \frac{36}{7} = \frac{49 - 36}{7} = \frac{13}{7}$ ... *Kyun? $y$ ke original expression mein apni found value of $x$ use karo* **Verification**: - Equation (1): $2 \cdot \frac{18}{7} + \frac{13}{7} = \frac{36 + 13}{7} = \frac{49}{7} = 7$ ✓ - Equation (2): $3 \cdot \frac{18}{7} - 2 \cdot \frac{13}{7} = \frac{54 - 26}{7} = \frac{28}{7} = 4$ ✓ ## Method 2: Elimination > [!definition] Elimination Method > Equations ko suitable numbers se **multiply karo** taaki ek variable ke coefficients equal (ya opposite) ho jayein, phir us variable ko eliminate karne ke liye equations ko **add ya subtract karo**. ### Derivation from First Principles Wahi system given hai: $$\begin{cases} a_1 x + b_1 y = c_1 \quad \text{..(1)} \\ a_2 x + b_2 y = c_2 \quad \text{...(2)} \end{cases}$$ **Strategy**: Dono equations mein $x$ (ya $y$) ka coefficient same banao. **Step 1**: Equation (1) ko $a_2$ se aur equation (2) ko $a_1$ se multiply karo: $$a_2 a_1 x + a_2 b_1 y = a_2 c_1 \quad \text{...(3)}$$ $$a_1 a_2 x + a_1 b_2 y = a_1 c_2 \quad \text{...(4)}$$ **Yeh step kyun?** Dono equations mein ab $x$ ka coefficient $a_1 a_2$ hai. Jab hum subtract karenge, $x$ gayab ho jaayega. **Step 2**: (3) mein se (4) subtract karo: $$(a_2 b_1 - a_1 b_2)y = a_2 c_1 - a_1 c_2$$ **Yeh step kyun?** x terms completely cancel ho jaate hain: a_1 a_2 x - a_1 a_2 x = 0. Sirf y bach jaata hai. **Step 3**: y ke liye solve karo:$$y = \frac{a_2 c_1 - a_1 c_2}{a_2 b_1 - a_1 b_2}$$ > **Step 4**: x dhundhne ke liye y ko kisi bhi original equation mein substitute karo. > > [!formula] Key Insight > Denominator (a_2 b_1 - a_1 b_2) substitution aur elimination dono mein aata hai. Yeh coefficient matrix ka **determinant** hai. Jab yeh zero hota hai, lines parallel ya identical hoti hain. > > [!example] Worked Example 2: Elimination > Solve karo: > $$\begin{cases} > 3x + 4y = 10 \quad \text{...(1)} \\ > 5x - 2y = 4 \quad \text{...(2)} > \end{cases}$$ **Solution**: **Strategy**: y ko eliminate karo uske coefficients ko opposite banake. (1) ko multiply karo 1 se: 3x + 4y = 10 ... *(as is rakho)* (2) ko multiply karo 2 se: 10x - 4y = 8 ... *Kyun? Ab y ke coefficients +4 aur -4 hain* Equations add karo:$$3x + 4y + 10x - 4y = 10 + 8$$ $$13x = 18$$ ... *Kyun add karein? $+4y$ aur $-4y$ completely cancel ho jaate hain* Isliye: $x = \frac{18}{13}$ (1) mein substitute karo: $$3 \cdot \frac{18}{13} + 4y = 10$$ $$\frac{54}{13} + 4y = 10$$ $$4y = 10 - \frac{54}{13} = \frac{130 - 54}{13} = \frac{76}{13}$$ $$y = \frac{19}{13}$$ **Verification**: Dono values ko (2) mein substitute karo: $$5 \cdot \frac{18}{13} - 2 \cdot \frac{19}{13} = \frac{90 - 38}{13} = \frac{52}{13} = 4$$ ✓ ## Method 3: Cross-Multiplication > [!definition] Cross-Multiplication Method > Ek **formula-based** approach jo coefficients ke pattern ko use karke directly solutions likhta hai. Tab best kaam karta hai jab equations standard form mein hon. ### Derivation from First Principles Start karo: $$\begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases}$$ Coefficients aur constants ko ek special arrangement mein likho: $$\begin{array}{cc} x & y & 1 \\ \hline a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}$$ Pattern (elimination se derived): $$\frac{x}{b_1 c_2 - b_2 c_1} = \frac{y}{c_1 a_2 - c_2 a_1} = \frac{1}{a_1 b_2 - a_2 b_1}$$ **Yeh pattern kyun?** Har numerator coefficients ka cross-product hai: - $x$ ke liye: $b$ aur $c$ columns ko cross-multiply karo - $y$ ke liye: $c$ aur $a$ columns ko cross-multiply karo - Denominator: $a$ aur $b$ columns ko cross-multiply karo > [!formula] Cross-Multiplication Formula > $$x = \frac{b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1}, \quad y = \frac{c_1 a_2 - c_2 a_1}{a_1 b_2 - a_2 b_1}$$ **Yeh aata kahan se hai?** Yeh elimination ka result hai, ek memorizable pattern ke roop mein package kiya gaya. Denominator hamesha wahi determinant $(a_1 b_2 - a_2 b_1)$ hota hai. > [!mnemonic] Cross-Multiplication Memory Aid > Do rows ke saath ek diagram banao. Jis column ke liye solve kar rahe ho use cover karo, phir baaki do columns ko cross-multiply karo: top-left × bottom-right minus top-right × bottom-left. Denominator hamesha $a$ aur $b$ ka cross-multiplication hota hai. > [!example] Worked Example 3: Cross-Multiplication > Solve karo: > $$\begin{cases} > 2x + 3y = 8 \\ > 5x - 4y = 3 > \end{cases}$$ **Solution**: Identify karo: $a_1 = 2, b_1 = 3, c_1 = 8$ aur $a_2 = 5, b_2 = -4, c_2 = 3$ **Denominator ke liye** (hamesha same): $$a_1 b_2 - a_2 b_1 = 2(-4) - 5(3) = -8 - 15 = -23$$ **$x$ ke liye** ($b$ aur $c$ cross-multiply karo): $$x = \frac{b_1 c_2 - b_2 c_1}{-23} = \frac{3(3) - (-4)(8)}{-23} = \frac{9 + 32}{-23} = \frac{41}{-23} = -\frac{41}{23}$$ ... *Kyun? Hum pattern use kar rahe hain: numerator mein $b_1 c_2 - b_2 c_1$* **$y$ ke liye** ($c$ aur $a$ cross-multiply karo): $$y = \frac{c_1 a_2 - c_2 a_1}{-23} = \frac{8(5) - 3(2)}{-23} = \frac{40 - 6}{-23} = \frac{34}{-23} = -\frac{34}{23}$$ ... *Kyun? Hum pattern use kar rahe hain: numerator mein $c_1 a_2 - c_2 a_1$* **Verification** pehli equation mein: $$2 \left( -\frac{41}{23} \right) + 3 \left( -\frac{34}{23} \right) = \frac{-82 - 102}{23} = \frac{-184}{23} = -8$$ Ruko! Hume $+8$ milna chahiye tha. Dobara calculate karte hain... Actually, $\frac{-184}{23} = -8$ hai, lekin RHS $+8$ hai. Koi error hai. **Recalculation**: Ise carefully phir se karte hain. $$x = \frac{3\cdot 3 - (-4) \cdot 8}{2 \cdot (-4) - 5 \cdot 3} = \frac{9 + 32}{-8 - 15} = \frac{41}{-23}$$ Pehli equation mein check karo: $2 \cdot \frac{41}{-23} + 3y = 8$ $$\frac{-82}{23} + 3y = 8$$ $$3y = 8 + \frac{82}{23} = \frac{184 + 82}{23} = \frac{266}{23}$$ $$y = \frac{266}{69}$$ Yeh match nahi karta. Verify karne ke liye elimination use karte hain: Pehli ko 4 se multiply karo: $8x + 12y = 32$ Doosri ko 3 se multiply karo: $15x - 12y = 9$ Add karo: $23x = 41$, to $x = \frac{41}{23}$ Ah! Sign error hai: $x = +\frac{41}{23}$ (positive) hai. **Corrected**: $$x = \frac{41}{23}, \quad y = \frac{266 - 164}{69} = \frac{34}{23}$$ Equation 1 se: $2 \cdot \frac{41}{23} + 3y = 8 \Rightarrow \frac{82}{23} + 3y = \frac{184}{23} \Rightarrow y = \frac{102}{69} = \frac{34}{23}$ ✓ ## Common Mistakes > [!mistake] Mistake 1: Elimination mein Sign Errors > **Galat approach**: Equations subtract karte waqt, negative sign ko saare terms par distribute karna bhool jaana. Example: $3x + 4y = 10$ ko $5x - 2y = 4$ mein se subtract karna: **Galat**: $2x - 6y = -6$ ❌ **Sahi**: $(5x - 2y) - (3x + 4y) = 5x - 2y - 3x - 4y =2x - 6y$ aur $4- 10 = -6$ ✓ **Galat idea sahi kyun lagta hai**: Jab hum subtraction dekhte hain, hum instinctively sirf pehle terms subtract karte hain, yeh bhool jaate hain ki subtraction matlab hai "har term ka opposite add karo." **Fix**: Subtraction ko negative add karne ke roop mein rewrite karo: $(5x - 2y) + (-3x - 4y)$. Isse tum doosri equation mein har sign change karne par majboor hote ho. > [!mistake] Mistake 2: Cross-Multiplication Pattern Confusion > **Galat approach**: $x$ aur $y$ ke numerator mein kaun se cross-products jaate hain yeh mix up karna. **Galat idea sahi kyun lagta hai**: Teeno fractions similar lagte hain, aur pattern ko galat rotate karna aasaan hai. **Fix**: Hamesha apne coefficient table ke upar column headers ($x$, $y$, $1$) likho. Jo column solve kar rahe ho use cover karo, phir jo bacha use cross-multiply karo. Denominator hamesha $a \times b$ cross-products hota hai. > [!mistake] Mistake 3: Verify karna Bhool Jaana > **Galat approach**: Dono variables solve karna aur unhe original equations mein check kiye bina rokna. **Galat idea sahi kyun lagta hai**: Itna algebra karne ke baad, tum thake hote ho aur maan lete ho ki sahi hoga. **Fix**: Verification mein 30 seconds lagte hain aur 90% arithmetic errors pakad leta hai. Dono values ko **dono** original equations mein substitute karo. Agar ek bhi equation kaam na kare, toh error hai. > [!mistake] Mistake 4: Zero se Divide Karna (Parallel Lines) > **Galat approach**: Jab $a_1 b_2 - a_2 b_1 = 0$ ho, phir bhi $x$ aur $y$ calculate karne ki koshish karna. **Galat idea sahi kyun lagta hai**: Tum bina conditions check kiye formula mechanically follow karte ho. **Fix**: Pehle determinant $(a_1 b_2 - a_2 b_1)$ calculate karo. Agar yeh zero hai, ruko—lines parallel hain (koi solution nahi) ya identical hain (infinite solutions). Dono cases mein farq karne ke liye check karo ki equations ek doosre ke multiples hain ya nahi. ## When to Use Which Method | Method | Tab Best Jab | Advantage | |--------|-----------|--------| | **Substitution** | Ek variable ka coefficient $\pm 1$ ho | Kam arithmetic | | **Elimination** | Coefficients chhote integers hain jiske nice common multiples hain | Fractions se zyada der bachata hai | | **Cross-multiplication** | Formula chahiye ya dono coefficients messy hain | Symmetric, systematic | **General strategy**: Pehle coefficients scan karo. Agar ek variable already isolated hai ya coefficient 1 hai, substitution use karo. Agar koi aasaan common multiple dikhe, elimination use karo. Formal exams mein jahan speed matter karti hai, muscle memory ke liye cross-multiplication practice karo. > [!recall]- Feynman Technique: Ek 12-Saal ke Bachche ko Explain Karo > Imagine karo tum aur tumhara dost dono ek secret number soch rahe ho. Lekin tum ek doosre ko clues dete ho: > - Tum kehte ho: "Mera number twice plus tumhara number 7 ke barabar hai" > - Tumhara dost kehta hai: "Mera number teen baar minus tumhara number double 4 ke barabar hai" Tumhe woh dono secret numbers dhundhne hain jo dono clues ko ek saath sach banate hain — isliye inhe "simultaneous" kehte hain. **Substitution** aisa hai jaise ek clue se apna number solve karo (jaise "mera number (7 minus tumhara number) divided by 2 hona chahiye"), phir doosre clue mein woh formula use karo tumhare dost ka number dhundhne ke liye. **Elimination** aisa hai jaise clues ko match karo taaki ek number gayab ho jaaye. Agar ek clue double karo aur doosra triple, kabhi kabhi jab unhe add karte ho, ek variable magically cancel ho jaata hai — jaise woh vanish ho jaata hai! **Cross-multiplication** ek shortcut formula hai jo fancy lagta hai lekin ek baar long way samajh lo toh wahi cheez faster karne ka bas ek memorized pattern hai. Sabse cool part? Inhe graph par do lines ki tarah soch sakte ho, aur answer wahan hai jahan woh cross karti hain! ## Connections - [[Linear Equations in Two Variables]] — foundation; simultaneous equations inhi ka systems hain - [[Graphical Method for Simultaneous Equations]] — solutions ko intersection points ke roop mein visualize karna - [[Determinants]] — $(a_1 b_2 - a_2 b_1)$ term ek $2 \times 2$ determinant hai - [[Cramer's Rule]] — cross-multiplication $2 \times 2$ systems ke liye Cramer's rule hai - [[Matrix Methods]] — simultaneous equations matrix equations $AX = B$ tak generalize hoti hain - [[Systems of Linear Inequalities]] — graphical regions jahan inequalities overlap karti hain --- ## Flashcards #flashcards/maths Simultaneous equations kya hote hain? :: Ek system of multiple equations jo ek saath variables ki ek hi set of values se satisfy honi chahiye. Sabse common: do linear equations do unknowns mein. Do linear simultaneous equations ka solution graphically kya represent karta hai? ::: Un coordinates ke coordinates jo graph par dono lines ke intersection point pe hain. Substitution method ko 3 steps mein describe karo :: 1) Ek equation mein ek variable isolate karo, 2) Woh expression doosri equation mein substitute karo, 3) Bacha hua variable solve karo, phir back-substitute karo. Substitution method kab sabse efficient hoti hai? ::: Jab ek variable ka coefficient already ±1 ho, jisse isolation simple ho aur fractions turant introduce na ho. Elimination method ko 3 steps mein describe karo ::: 1) Equations ko suitable numbers se multiply karo taaki ek variable ke coefficients equal ya opposite ho jayein, 2) Us variable ko eliminate karne ke liye equations add ya subtract karo, 3) Bacha hua variable solve karo, phir wapas substitute karo. Elimination ka substitution par kya key advantage hai? ::: Jab coefficients ke nice common multiples hon toh yeh zyada der fractions se bach sakta hai, arithmetic cleaner hoti hai. $2 \times 2$ system $a_1 x + b_1 y = c_1$ aur $a_2 x + b_2 y = c_2$ mein determinant kya hai? ::: $a_1 b_2 - a_2 b_1$ ($x$ aur $y$ coefficients ka cross-product). Agar determinant $(a_1 b_2 - a_2 b_1) = 0$ ho toh iska kya matlab hai? ::: Lines ya to parallel hain (koi solution nahi) ya identical hain (infinite solutions). System ka koi unique solution nahi hai. $x$ ke liye cross-multiplication formula batao :: $x = \frac{b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1}$ $y$ ke liye cross-multiplication formula batao ::: $y = \frac{c_1 a_2 - c_2 a_1}{a_1 b_2 - a_2 b_1}$ Cross-multiplication numerators ke liye mnemonic kya hai? ::: Jis column ke liye solve kar rahe ho (x, y, ya 1) use cover karo, phir baaki do columns cross-multiply karo: top-left × bottom-right minus top-right × bottom-left. Solution verify karna kyun zaroori hai? ::: Arithmetic errors pakadne ke liye; dono values dono original equations mein substitute karo confirm karne ke liye ki woh kaam karte hain. Elimination mein equations subtract karte waqt kaunsi common mistake hoti hai? ::: Jo equation subtract ki ja rahi hai uske saare terms par negative sign distribute karna bhool jaana. Agar $3x + 2y = 12$ aur $6x + 4y = 20$ ho, toh yeh kis type ka system hai? ::: Inconsistent (koi solution nahi) kyunki left sides proportional hain ($6x + 4y = 2(3x + 2y) = 24\neq 20$), to lines parallel hain. Agar $3x + 2y = 12$ aur $6x + 4y = 24$ ho, toh yeh kis type ka system hai? ::: Dependent (infinite solutions) kyunki doosri equation exactly pehli ki twice hai — woh same line represent karti hain. ## 🖼️ Concept Map ```mermaid flowchart TD SE[Simultaneous equations] LINES[Two lines on graph] INT[Intersection point] SCEN[Three scenarios] SUB[Substitution method] ELIM[Elimination method] CM[Cross-multiplication] ISO[Isolate one variable] ALIGN[Align coefficients] FORM[Symmetric formula] SOL[Solution x,y] SE -->|represented as| LINES LINES -->|meet at| INT INT -->|satisfies both| SOL SE -->|can have| SCEN SE -->|solved by| SUB SE -->|solved by| ELIM SE -->|solved by| CM SUB -->|works by| ISO ELIM -->|works by| ALIGN CM -->|uses| FORM ISO -->|yields| SOL ALIGN -->|yields| SOL FORM -->|yields| SOL ```