Simultaneous equations — substitution, elimination, cross-multiplication

Overview

Simultaneous equations are systems where we need to find values that satisfy multiple equations at the same time. The most common case: two equations with two unknowns. We explore three methods to solve them, each revealing different structural insights about how equations interact.

Why multiple methods matter: Different methods suit different equation structures. Substitution works when one variable is already isolated; elimination shines when coefficients align nicely; cross-multiplication gives a symmetric formula-based approach.

Core Intuition

Three possible scenarios:

• One unique solution: Lines intersect at exactly one point (most common)
• No solution: Lines are parallel (inconsistent system)
• Infinite solutions: Lines are identical (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}$$ **Step 1**: Isolate $y$ from equation (1): $$b_1 y = c_1 - a_1 x$$ $$y = \frac{c_1 - a_1 x}{b_1} \quad \text{...(3)}$$ **Why this step?** We create a "definition" of $y$ that depends only on $x$. Now wherever we see $y$, we can replace it with this expression. **Step 2**: Substitute (3) into equation (2): $$a_2 x + b_2 \left( \frac{c_1 - a_1 x}{b_1} \right) = c_2$$ **Why this step?** Equation (2) now contains only $x$. We've eliminated $y$ by replacing it with its expression in terms of $x$. **Step 3**: Solve for $x$: $$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}$$ **Step 4**: Substitute $x$ back into (3) to find $y$. > [!example] Worked Example 1: Substitution > Solve: > $$\begin{cases} > 2x + y = 7 \quad \text{...(1)} \\ > 3x - 2y = 4 \quad \text{...(2)} > \end{cases}$$ **Solution**: From (1): $y = 7 - 2x$ ... *Why? We isolate the simpler variable with coefficient 1* Substitute into (2): $3x - 2(7 - 2x) = 4$ ... *Why? We're replacing every $y$ with its equivalent in terms of $x$* Expand: $3x - 14 + 4x = 4$ ... *Why? Distribute the $-2$ through the parentheses* Combine: $7x = 18$ ... *Why? Collect like terms and add14 to both sides* Therefore: $x = \frac{18}{7}$ Substitute back: $y = 7 - 2 \cdot \frac{18}{7} = 7 - \frac{36}{7} = \frac{49 - 36}{7} = \frac{13}{7}$ ... *Why? Use the original expression for $y$ with our found value of $x$* **Verification**: - Equation (1): $2 \cdot \frac{18}{7} + \frac{13}{7} = \frac{36 + 13}{7} = \frac{49}{7} = 7$ ✓ - Equation (2): $3 \cdot \frac{18}{7} - 2 \cdot \frac{13}{7} = \frac{54 - 26}{7} = \frac{28}{7} = 4$ ✓ ## Method 2: Elimination > [!definition] Elimination Method > **Multiply equations by suitable numbers** so that coefficients of one variable become equal (or opposites), then **add or subtract** the equations to eliminate that variable. ### Derivation from First Principles Given the same system: $$\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}$$ **Strategy**: Make the coefficient of $x$ (or $y$) the same in both equations. **Step 1**: Multiply equation (1) by $a_2$ and equation (2) by $a_1$: $$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)}$$ **Why this step?** Both equations now have $x$ with coefficient $a_1 a_2$. When we subtract, $x$ will vanish. **Step 2**: Subtract (4) from (3): $$(a_2 b_1 - a_1 b_2)y = a_2 c_1 - a_1 c_2$$ **Why this step?** The x terms cancel completely: a_1 a_2 x - a_1 a_2 x = 0. We're left with only y. **Step 3**: Solve for y:$$y = \frac{a_2 c_1 - a_1 c_2}{a_2 b_1 - a_1 b_2}$$ > **Step 4**: Substitute y into either original equation to find x. > > [!formula] Key Insight > The denominator (a_2 b_1 - a_1 b_2) appears in both substitution and elimination. This is the **determinant** of the coefficient matrix. When it equals zero, the lines are parallel or identical. > > [!example] Worked Example 2: Elimination > Solve: > $$\begin{cases} > 3x + 4y = 10 \quad \text{...(1)} \\ > 5x - 2y = 4 \quad \text{...(2)} > \end{cases}$$ **Solution**: **Strategy**: Eliminate y by making its coefficients opposites. Multiply (1) by : 3x + 4y = 10 ... *(keep as is)* Multiply (2) by 2: 10x - 4y = 8 ... *Why? Now y has coefficients +4 and -4* Add the equations:$$3x + 4y + 10x - 4y = 10 + 8$$ $$13x = 18$$ ... *Why add? The $+4y$ and $-4y$ cancel out completely* Therefore: $x = \frac{18}{13}$ Substitute into (1): $$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}$$ **Verification**: Substitute both values into (2): $$5 \cdot \frac{18}{13} - 2 \cdot \frac{19}{13} = \frac{90 - 38}{13} = \frac{52}{13} = 4$$ ✓ ## Method 3: Cross-Multiplication > [!definition] Cross-Multiplication Method > A **formula-based** approach that uses the pattern of coefficients to write solutions directly. Works best when equations are in standard form. ### Derivation from First Principles Start with: $$\begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases}$$ Write the coefficients and constants in a special arrangement: $$\begin{array}{cc} x & y & 1 \\ \hline a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}$$ The pattern (derived from elimination): $$\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}$$ **Why this pattern?** Each numerator is a cross-product of coefficients: - For $x$: cross-multiply $b$ and $c$ columns - For $y$: cross-multiply $c$ and $a$ columns - Denominator: cross-multiply $a$ and $b$ columns > [!formula] Cross-Multiplication Formula > $$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}$$ **Where does this come from?** It's the result of elimination, packaged as a memorizable pattern. The denominator is always the same determinant $(a_1 b_2 - a_2 b_1)$. > [!mnemonic] Cross-Multiplication Memory Aid > Draw a diagram with two rows. Cover the column you're solving for, then cross-multiply the remaining two columns: top-left × bottom-right minus top-right × bottom-left. The denominator is always $a$ and $b$ cross-multiplied. > [!example] Worked Example 3: Cross-Multiplication > Solve: > $$\begin{cases} > 2x + 3y = 8 \\ > 5x - 4y = 3 > \end{cases}$$ **Solution**: Identify: $a_1 = 2, b_1 = 3, c_1 = 8$ and $a_2 = 5, b_2 = -4, c_2 = 3$ **For denominator** (always the same): $$a_1 b_2 - a_2 b_1 = 2(-4) - 5(3) = -8 - 15 = -23$$ **For $x$** (cross-multiply $b$ and $c$): $$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}$$ ... *Why? We're using the pattern: $b_1 c_2 - b_2 c_1$ in the numerator* **For $y$** (cross-multiply $c$ and $a$): $$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}$$ ... *Why? We're using the pattern: $c_1 a_2 - c_2 a_1$ in the numerator* **Verification** in first equation: $$2 \left( -\frac{41}{23} \right) + 3 \left( -\frac{34}{23} \right) = \frac{-82 - 102}{23} = \frac{-184}{23} = -8$$ Wait! We should get $+8$. Let me recalculate... Actually, $\frac{-184}{23} = -8$, but the RHS is $+8$. There's an error. **Recalculation**: Let me redo this carefully. $$x = \frac{3\cdot 3 - (-4) \cdot 8}{2 \cdot (-4) - 5 \cdot 3} = \frac{9 + 32}{-8 - 15} = \frac{41}{-23}$$ Check in first equation: $2 \cdot \frac{41}{-23} + 3y = 8$ $$\frac{-82}{23} + 3y = 8$$ $$3y = 8 + \frac{82}{23} = \frac{184 + 82}{23} = \frac{266}{23}$$ $$y = \frac{266}{69}$$ This doesn't match. Let me use elimination to verify: Multiply first by 4: $8x + 12y = 32$ Multiply second by 3: $15x - 12y = 9$ Add: $23x = 41$, so $x = \frac{41}{23}$ Ah! The sign error: $x = +\frac{41}{23}$ (positive). **Corrected**: $$x = \frac{41}{23}, \quad y = \frac{266 - 164}{69} = \frac{34}{23}$$ From equation 1: $2 \cdot \frac{41}{23} + 3y = 8 \Rightarrow \frac{82}{23} + 3y = \frac{184}{23} \Rightarrow y = \frac{102}{69} = \frac{34}{23}$ ✓ ## Common Mistakes > [!mistake] Mistake 1: Sign Errors in Elimination > **Wrong approach**: When subtracting equations, forgetting to distribute the negative sign to all terms. Example: Subtracting $3x + 4y = 10$ from $5x - 2y = 4$: **Wrong**: $2x - 6y = -6$ ❌ **Right**: $(5x - 2y) - (3x + 4y) = 5x - 2y - 3x - 4y =2x - 6y$ and $4- 10 = -6$ ✓ **Why the wrong idea feels right**: When we see subtraction, we instinctively subtract the first terms only, forgetting that subtraction means "add the opposite of each term." **The fix**: Rewrite subtraction as adding the negative: $(5x - 2y) + (-3x - 4y)$. This forces you to change every sign in the second equation. > [!mistake] Mistake 2: Cross-Multiplication Pattern Confusion > **Wrong approach**: Mixing up which cross-products go in the numerator for $x$ vs $y$. **Why the wrong idea feels right**: All three fractions look similar, and it's easy to rotate the pattern incorrectly. **The fix**: Always write the column headers ($x$, $y$, $1$) above your coefficient table. Cover the column you're solving for, then cross-multiply what remains. The denominator is always $a \times b$ cross-products. > [!mistake] Mistake 3: Forgetting to Verify > **Wrong approach**: Solving for both variables and stopping without checking them in the original equations. **Why the wrong idea feels right**: After all that algebra, you're tired and assume you must be correct. **The fix**: Verification takes30 seconds and catches90% of arithmetic errors. Substitute both values into **both** original equations. If even one equation doesn't work, you have an error. > [!mistake] Mistake 4: Dividing by Zero (Parallel Lines) > **Wrong approach**: When $a_1 b_2 - a_2 b_1 = 0$, trying to compute $x$ and $y$ anyway. **Why the wrong idea feels right**: You follow the formula mechanically without checking conditions. **The fix**: Always compute the determinant $(a_1 b_2 - a_2 b_1)$ first. If it's zero, stop—the lines are parallel (no solution) or identical (infinite solutions). Check if the equations are multiples of each other to distinguish the two cases. ## When to Use Which Method | Method | Best When | Advantage | |--------|-----------|--------| | **Substitution** | One variable has coefficient $\pm 1$ | Minimal arithmetic | | **Elimination** | Coefficients are small integers with common multiples | Avoids fractions longer | | **Cross-multiplication** | You want a formula or both coefficients are mesy | Symmetric, systematic | **General strategy**: Scan the coefficients first. If one variable is already isolated or has coefficient 1, use substitution. If you spot an easy common multiple, use elimination. For formal exams where speed matters, practice cross-multiplication for muscle memory. > [!recall]- Feynman Technique: Explain to a12-Year-Old > Imagine you and your friend are each thinking of a secret number. But you give each other clues: > - You say: "Twice my number plus your number equals 7" > - Your friend says: "Three times my number minus twice yours equals 4" You need to find both secret numbers that make both clues true at the same time—that's why they're called "simultaneous." **Substitution** is like solving one clue for your number (like "my number must be (7 minus your number) divided by 2"), then using that formula in the other clue to find your friend's number. **Elimination** is like making the clues match up so one number disappears. If you double one clue and triple another, sometimes when you add them together, one variable cancels out magically—like it vanishes! **Cross-multiplication** is a shortcut formula that looks fancy but is just a faster way to do the same thing. It's like having a calculator pattern you memorize once you understand the long way. The coolest part? You can think of these as two lines on a graph, and the answer is where they cross! ## Connections - [[Linear Equations in Two Variables]] — the foundation; simultaneous equations are systems of these - [[Graphical Method for Simultaneous Equations]] — visualizing solutions as intersection points - [[Determinants]] — the $(a_1 b_2 - a_2 b_1)$ term is a $2 \times 2$ determinant - [[Cramer's Rule]] — cross-multiplication is Cramer's rule for $2 \times 2$ systems - [[Matrix Methods]] — simultaneous equations generalize to matrix equations $AX = B$ - [[Systems of Linear Inequalities]] — graphical regions where inequalities overlap --- ## Flashcards #flashcards/maths What are simultaneous equations? :: A system of multiple equations that must be satisfied by the same set of variable values at the same time. Most commonly: two linear equations in two unknowns. What does the solution to two linear simultaneous equations represent graphically? ::: The coordinates of the point where the two lines intersect on a graph. Describe the substitution method in 3 steps :: 1) Isolate one variable in one equation, 2) Substitute that expression into the other equation, 3) Solve for the remaining variable, then back-substitute. When is the substitution method most efficient? ::: When one variable already has a coefficient of ±1, making isolation simple without introducing fractions immediately. Describe the elimination method in 3 steps ::: 1) Multiply equations by suitable numbers to make one variable's coefficients equal or opposite, 2) Add or subtract equations to eliminate that variable, 3) Solve for the remaining variable, then substitute back. What is the key advantage of elimination over substitution? ::: It can avoid fractions longer when coefficients have nice common multiples, making arithmetic cleaner. What is the determinant in a $2 \times 2$ system $a_1 x + b_1 y = c_1$ and $a_2 x + b_2 y = c_2$? ::: $a_1 b_2 - a_2 b_1$ (the cross-product of the $x$ and $y$ coefficients). What does it mean if the determinant $(a_1 b_2 - a_2 b_1) = 0$? ::: The lines are either parallel (no solution) or identical (infinite solutions). The system has no unique solution. State the cross-multiplication formula for $x$ :: $x = \frac{b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1}$ State the cross-multiplication formula for $y$ ::: $y = \frac{c_1 a_2 - c_2 a_1}{a_1 b_2 - a_2 b_1}$ What is the mnemonic for cross-multiplication numerators? ::: Cover the column you're solving for (x, y, or 1), then cross-multiply the remaining two columns: top-left × bottom-right minus top-right × bottom-left. Why must you always verify your solution? ::: To catch arithmetic errors; substitute both values into both original equations to confirm they work. What common mistake happens when subtracting equations in elimination? ::: Forgetting to distribute the negative sign to all terms of the equation being subtracted. If $3x + 2y = 12$ and $6x + 4y = 20$, what type of system is this? ::: Inconsistent (no solution) because the left sides are proportional ($6x + 4y = 2(3x + 2y) = 24\neq 20$), so the lines are parallel. If $3x + 2y = 12$ and $6x + 4y = 24$, what type of system is this? ::: Dependent (infinite solutions) because the second equation is exactly twice the first—they represent the same line. ## 🖼️ 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 ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Simultaneous equations matlab do yazyada equations hain jo **ek sath** satisfy hone chahiye—same values of variables se dono equations true ho jani chahiye. Sabse common case: do linear equations, do unknowns (x aur y). > > Teen tarike hain solve karne ke: **Substitution** mein ek equation sek variable ko isolate karo (jaise $y = 7 - 2x$), phir dosri equation mein replace kar do—ab sirf ek variable bacha. **Elimination** mein equations ko multiply karke coefficients match karo, phir add ya subtract karke ek variable ko cancel kar do (elimination ho gaya!). **Cross-multiplication** ek formula-based shortcut hai jisme pattern yad rakhte ho—coefficients ko cross-multiply karo aur answer directly nikal lo. > > Graphically socho toh: do linear equations matlab do straight lines. Solution woh point hai jahan dono lines intersect karti hain—wahi $(x, y)$ coordinates dono equations ko satisfy karte hain. Agar lines parallel hain toh no solution (koi intersection nahi). Agar dono lines same hain toh infinite solutions (har point pe intersection!). Cross-multiplication ka denominator $(a_1 b_2 - a_2 b_1)$ agar zero ho toh pata chal jata hai ki parallel ya same line hai. > > Real life mein bhi useful hai: agar do conditionsiye ho (jaise "total cost aur total items"), simultaneous equations se dono unknowns (price of each item) nikal sakte ho. Practice karo teen methods, phir apne ap samajh aa jayega ki kaunsa method kab fast hai! ![[audio/2.1.10-Simultaneous-equations-—-substitution,-elimination,-cross-multiplication.mp3]]