Visual walkthrough — Simultaneous equations — substitution, elimination, cross-multiplication
This page builds the whole machinery of solving two simultaneous linear equations, from the very first squiggle, using pictures. By the end you will see why the same denominator shows up in the parent note no matter which method you pick — substitution, elimination, or cross-multiplication.
We will use nothing but arithmetic and one picture-idea: two straight lines meeting at a point. If you have met Linear Equations in Two Variables you are ready. If the word determinant scares you, don't worry — we will grow it in front of you and only name it at the end.
Step 1 — What a single equation even means
WHAT. Take one equation like . A solution is any pair of numbers — one for , one for — that makes the two sides equal.
WHY. Before we solve two equations we must be crystal clear what one asks. One linear equation does not pin down a single point; it allows infinitely many pairs. If then ; if then ; if then . Plot every such pair and they line up perfectly straight — that is why we call it a line.
PICTURE. Each dot below is one allowed pair. They fall on a straight orange line. The label on the line is the equation itself.
Here every symbol earns its place:
Step 2 — Two equations = two lines = one crossing
WHAT. Now add a second rule, . It has its own line of allowed pairs.
WHY. A pair that must obey both rules must sit on both lines at once. The only place that can happen is where the two lines cross. So "solve the system" literally means "find the crossing point."
PICTURE. Two lines, one crossing marked with a magenta dot. That dot's coordinates are the answer.
Cross-link: seeing this crossing directly on a graph is the Graphical Method for Simultaneous Equations; here we compute it exactly instead of eyeballing it.
Step 3 — Turn a picture-problem into a number-problem (isolate )
WHAT. From equation (1), rewrite all by itself:
WHY. The picture shows the crossing but not its exact coordinates. To get numbers we make depend on : pick any , and this formula hands you the matching on line 1. We have turned line 1 into a little machine "give me , I give you ."
PICTURE. Slide a vertical dashed line at position ; where it hits line 1 is the height this formula outputs.
Symbol by symbol, using the letters from the parent's general system:
Step 4 — Force the second line to agree (substitution)
WHAT. Feed that machine's output into equation (2), replacing its :
WHY. We just demanded: "of all points on line 1, which one also satisfies line 2?" Because we plugged line 1's into line 2, the only unknown left is — one equation, one unknown, solvable.
PICTURE. The dashed vertical line now slides until the point on line 1 lands exactly on line 2. That special slide position is the crossing.
Clear the fraction (multiply every term by ) and gather the 's:
\;\Longrightarrow\; \underbrace{(a_2 b_1 - a_1 b_2)}_{\text{coefficient of } x}\, x \;=\; \underbrace{c_2 b_1 - b_2 c_1}_{\text{plain number}}.$$ --- ## Step 5 — The formula for $x$ appears **WHAT.** Divide across: $$x = \frac{c_2 b_1 - b_2 c_1}{a_2 b_1 - a_1 b_2}.$$ **WHY.** This is the exact horizontal position of the crossing — no guessing off a graph. Notice the **bottom**: $a_2 b_1 - a_1 b_2$. Flip its sign and it is $a_1 b_2 - a_2 b_1$, the exact number the parent note calls the key denominator. It was *born here*, from cancelling the extra $y$. **PICTURE.** Term-by-term colouring of the fraction so you can see which coefficients built the top and which built the bottom. $$x = \frac{\overbrace{c_2 b_1}^{\text{line 2 target} \times b_1} - \overbrace{b_2 c_1}^{b_2 \times \text{line 1 target}}}{\underbrace{a_2 b_1 - a_1 b_2}_{\text{the “cross” of the $x,y$ weights}}}$$ > [!recall]- Why do the *targets* $c_1,c_2$ only sit on top, never on the bottom? > The bottom measures **whether the two lines point different ways** — that depends only on the weights $a,b$, not on where the lines are shifted to. The $c$'s only move the lines around; they decide *where* the crossing is, not *whether* one exists. ::: Because the denominator is built purely from the direction-weights $a_1,b_1,a_2,b_2$; the constants $c$ only appear in numerators. --- ## Step 6 — Get $y$ the same way (elimination view) **WHAT.** Instead of substituting back, kill $x$ symmetrically: multiply (1) by $a_2$, (2) by $a_1$, subtract: $$(a_2 b_1 - a_1 b_2)\, y = a_2 c_1 - a_1 c_2 \;\Longrightarrow\; y = \frac{a_2 c_1 - a_1 c_2}{a_2 b_1 - a_1 b_2}.$$ **WHY.** Substitution isolated one variable; **elimination** scales the two lines so the $x$-parts are identical, then subtracts them away — same crossing, mirror-image bookkeeping. Crucially the **same denominator** returns. That is not luck: both routes are asking the identical geometric question. **PICTURE.** Two scaled equations stacked; the $a_1 a_2 x$ terms line up and annihilate on subtraction, leaving a pure-$y$ equation. $$\underbrace{a_2 a_1 x}_{\text{equal}} + a_2 b_1 y = a_2 c_1 \quad\ominus\quad \underbrace{a_1 a_2 x}_{\text{equal}} + a_1 b_2 y = a_1 c_2 \quad\Rightarrow\quad (a_2 b_1 - a_1 b_2) y = a_2 c_1 - a_1 c_2$$ --- ## Step 7 — Same result, third packaging (cross-multiplication) **WHAT.** Line up the two answers so the shared denominator becomes one ratio-chain: $$\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}.$$ **WHY.** Cross-multiplication is **not a new idea** — it is Steps 5 and 6 written as a memorable diagonal pattern. Cover the column of the variable you want; the surviving two columns cross-multiply (top-left × bottom-right minus top-right × bottom-left). **PICTURE.** The coefficient grid with the diagonal arrows drawn for each of $x$, $y$, and the shared $1$. $$\underbrace{a_1 b_2 - a_2 b_1}_{\text{this number} =\, \text{the determinant}}$$ This shared bottom number is exactly what [[Determinants]], [[Cramer's Rule]] and [[Matrix Methods]] later formalise — you have just met it early, hand-built. > [!example] Worked check (Example 3 from the parent, done correctly) > System $\;2x+3y=8,\;\;5x-4y=3$. So $a_1{=}2,b_1{=}3,c_1{=}8,\;a_2{=}5,b_2{=}-4,c_2{=}3$. > Denominator: $a_1 b_2 - a_2 b_1 = 2(-4)-5(3) = -8-15 = -23$. > $$x=\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}=\frac{3(3)-(-4)(8)}{-23}=\frac{41}{-23}=\ ?$$ > Careful with the *sign convention*: cross-multiplication solves for the **standard form** $a x + b y + c = 0$, i.e. constants moved left as $c_1=-8,\ c_2=-3$. Re-running with those signs gives $x=\frac{41}{23},\ y=\frac{34}{23}$. Verify: $2(\tfrac{41}{23})+3(\tfrac{34}{23})=\tfrac{82+102}{23}=\tfrac{184}{23}=8$ ✓ and $5(\tfrac{41}{23})-4(\tfrac{34}{23})=\tfrac{205-136}{23}=\tfrac{69}{23}=3$ ✓. > [!mistake] The trap the parent note fell into > Cross-multiplication's tidy formula assumes you wrote the system as $ax+by+c=0$ (constant on the left, so $c$ carries a minus). Mixing that with the $ax+by=c$ layout flips a sign and you "lose" the answer. **Pick one layout and keep the signs of $c$ consistent.** --- ## Step 8 — The degenerate cases: when the crossing hides **WHAT.** What if the shared denominator $D = a_1 b_2 - a_2 b_1 = 0$? **WHY.** $D$ measures whether the two lines point in **different directions**. If $D=0$, they point the *same* way. Then either they are the *same line* (every point solves both — infinite solutions) or *parallel but shifted* (they never meet — no solution). Dividing by $D=0$ is the algebra screaming "there is no single crossing point." **PICTURE.** Three side-by-side panels: one clean crossing ($D\neq0$), one identical overlapping line (infinite), one parallel pair (none). > [!formula] The full verdict > $$D = a_1 b_2 - a_2 b_1$$ > $$\begin{aligned} > D \neq 0 &: \text{ one unique solution (lines cross once)}\\ > D = 0,\ \text{same ratios with } c &: \text{ infinitely many (same line)}\\ > D = 0,\ \text{ratios differ with } c &: \text{ none (parallel)} > \end{aligned}$$ Concretely with the $ax+by=c$ layout: $D=0$ together with $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$ means *identical* lines; if instead $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$ the lines are *parallel*. This is where [[Systems of Linear Inequalities]] and half-plane pictures take over from crossing-points. --- ## The one-picture summary Every method — substitution, elimination, cross-multiplication — is the **same journey**: two lines → one crossing → one shared denominator $D$ that decides everything. > [!recall]- Feynman retelling — say it back in plain words > I have two straight-line rules. A point that follows *both* has to sit exactly where the lines cross. To find that spot with numbers, I take one line and turn it into a machine "give me $x$, I hand you the $y$ that keeps this line happy." I feed that $y$ into the other line's rule — now only $x$ is unknown, so I solve it, and the $y$ pops out too. If I'd rather, I can scale both lines until their $x$-parts match and subtract them so $x$ disappears — I get the exact same answer, and the exact same number sitting on the bottom of every fraction. That bottom number, $a_1 b_2 - a_2 b_1$, is the truth-teller: if it isn't zero, there's one clean crossing; if it *is* zero, the lines point the same way, so they're either the very same line (endless answers) or parallel strangers (no answer). Cross-multiplication is just this whole story written as a neat diagonal pattern so I don't have to re-derive it each time. > [!recall]- Which single quantity decides one / none / infinite solutions? > Question ::: The determinant $D = a_1 b_2 - a_2 b_1$ — nonzero gives one solution; zero gives either none (parallel) or infinite (identical).