Linear equations in two variables — graphical and algebraic solutions: Visual walkthrough

We start from the very first idea: what a single equation even looks like.

Step 1 — One equation is a fence of dots that lines up

WHAT. Take one equation, say $2 x + 3 y = 12$ . A "solution" is any pair of numbers $(x, y)$ that makes the two sides equal.

WHY. Before we hunt for the crossing point of two lines, we must see clearly that one equation already carries a whole shape's worth of answers — not a single answer.

PICTURE. Below, each blue dot is a pair that works: $(0, 4)$ , $(3, 2)$ , $(6, 0)$ . Notice they don't scatter randomly — they fall on a perfectly straight track.

Figure — Linear equations in two variables — graphical and algebraic solutions

$x$ ::: the horizontal coordinate we are free to choose.
$y$ ::: the vertical coordinate forced on us once $x$ is chosen, so the equation stays true.
Why straight? ::: Solving for $y$ gives $y = - \frac{2}{3} x + 4$ — a fixed slope $- \frac{2}{3}$ , so equal steps sideways cause equal steps up/down (see Slope-intercept form y = mx + c).

WHAT. Now draw two equations on the same grid: $a_{1} x + b_{1} y = c_{1} and a_{2} x + b_{2} y = c_{2} .$

WHY. One line = infinitely many answers, too vague to be an "answer". A second line usually crosses the first at exactly one dot. That shared dot is the unique pair satisfying both — the thing we actually want.

PICTURE. The lavender line and the coral line cross at one mint dot. That mint dot lives on both tracks at once.

$a_{1}, b_{1}, c_{1}$ ::: the three numbers that fix the lavender line's tilt and position.
$a_{2}, b_{2}, c_{2}$ ::: the three numbers that fix the coral line.
the mint dot ::: the pair $(x, y)$ that makes both equations true simultaneously.

Recall Why "on both lines" is the whole game

A point on line 1 satisfies equation 1. A point on line 2 satisfies equation 2. The crossing point is the only place that manages both — so its coordinates are the solution of the system.

Step 3 — Kill $y$ by making its two coefficients twins

WHAT. We want the $x$ -coordinate of the crossing. To find it, we make $y$ disappear. Multiply equation 1 by $b_{2}$ and equation 2 by $b_{1}$ : $b_{2} \cdot (a_{1} x + b_{1} y) = b_{2} c_{1} \Rightarrow a_{1} b_{2} x + b_{1} b_{2} y = c_{1} b_{2}$ $b_{1} \cdot (a_{2} x + b_{2} y) = b_{1} c_{2} \Rightarrow a_{2} b_{1} x + b_{1} b_{2} y = c_{2} b_{1}$

WHY this tool — scaling then subtracting? Multiplying a whole equation by a number is legal (both sides grow equally, the line is unchanged — same set of solutions). We choose the scalings so the $y$ -terms become identical ( $b_{1} b_{2} y$ in both). Identical things subtract to zero — that is exactly how we erase a variable. We pick $y$ to erase first because then only $x$ survives.

PICTURE. Watch the $y$ -columns: both are now the same block $b_{1} b_{2} y$ (mint), ready to cancel. The $x$ -columns and constants (coral/lavender) stay distinct.

$b_{2} \cdot (\dots)$ ::: scaling equation 1 so its $y$ -coefficient becomes $b_{1} b_{2}$ .
$b_{1} \cdot (\dots)$ ::: scaling equation 2 so its $y$ -coefficient also becomes $b_{1} b_{2}$ .
Why these exact multipliers? ::: they force the two $y$ -blocks to match, which is the only way subtraction wipes $y$ out.

Step 4 — Subtract, and out pops the $x$ -formula

WHAT. Subtract the second scaled line from the first. The twin $y$ -blocks cancel: $leftover x -coefficient (a_{1} b_{2} - a_{2} b_{1}) x = leftover constant c_{1} b_{2} - c_{2} b_{1} .$ Divide by that leftover coefficient: $x = \frac{c _{1} b _{2} - c _{2} b _{1}}{a _{1} b _{2} - a _{2} b _{1}}$

WHY. After the $y$ -blocks vanish, we are left with a single equation in a single unknown $x$ — and a one-unknown linear equation is trivially solved by dividing. That division delivers the exact $x$ of the crossing point.

PICTURE. The mint $y$ -blocks annihilate (shown fading to zero), leaving a clean $(number) \cdot x = (number)$ bar that we split to isolate $x$ .

$a_{1} b_{2} - a_{2} b_{1}$ ::: the coefficient of $x$ that survived — call it $D$ , the denominator.
$c_{1} b_{2} - c_{2} b_{1}$ ::: the constant that survived — the numerator for $x$ .
the cancelled $b_{1} b_{2} y$ ::: gone, because we engineered it to be identical in both lines.

Step 5 — By perfect symmetry, get $y$ too

WHAT. Repeat Steps 3–4 but this time kill $x$ : scale equation 1 by $a_{2}$ , equation 2 by $a_{1}$ , subtract. The $x$ -blocks become twins and cancel, leaving $y = \frac{a _{1} c _{2} - a _{2} c _{1}}{a _{1} b _{2} - a _{2} b _{1}}$

WHY. The two variables play mirror-image roles. Whatever trick isolates $x$ isolates $y$ if we just swap which coefficient we match. Notice the same denominator $D$ appears — because the lines' tilts (not the constants) decide both coordinates.

PICTURE. The bookkeeping is the mirror of Step 3–4: the coral $x$ -columns become identical and cancel, the mint $y$ survives.

$a_{1} c_{2} - a_{2} c_{1}$ ::: the numerator for $y$ — note the constants $c$ now sit where the $c b$ pair sat before.
same $D$ underneath ::: the crossing point shares one denominator for both coordinates.

Step 6 — The degenerate case: what if $D = 0$ ?

WHAT. The formula divides by $D$ . Division by zero is forbidden. So we must ask: when is $D = a_{1} b_{2} - a_{2} b_{1} = 0$ , and what does that look like?

WHY. A formula is only trustworthy once you know exactly where it breaks. $D = 0$ is precisely the boundary between "one crossing" and "no crossing / infinite crossings".

PICTURE. $D = 0$ means $a_{1} b_{2} = a_{2} b_{1}$ , i.e. $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}}$ — the two lines have equal slope. Equal slope means they are either the same line (mint, overlapping — infinite solutions) or parallel (coral, never touching — no solution). Neither has a single crossing, so the formula correctly refuses to give one.

$D = 0$ ::: the algebra's alarm bell for "no unique crossing".
same line ::: all three ratios match, $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ — infinitely many shared points.
parallel ::: the tilts match but constants don't, $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ — zero shared points.

Step 7 — A concrete run, watched end to end

WHAT. Solve $2 x + 3 y = 13$ and $3 x - y = 3$ with the boxed formula.

WHY. To see the letters turn into real numbers and land on the crossing dot.

PICTURE. Read off $a_{1} = 2, b_{1} = 3, c_{1} = 13, a_{2} = 3, b_{2} = - 1, c_{2} = 3$ .

$D = a_{1} b_{2} - a_{2} b_{1} = (2) (- 1) - (3) (3) = - 2 - 9 = - 11.$ $x = \frac{c _{1} b _{2} - c _{2} b _{1}}{D} = \frac{( 13 ) ( - 1 ) - ( 3 ) ( 3 )}{- 11} = \frac{- 13 - 9}{- 11} = \frac{- 22}{- 11} = 2.$ $y = \frac{a _{1} c _{2} - a _{2} c _{1}}{D} = \frac{( 2 ) ( 3 ) - ( 3 ) ( 13 )}{- 11} = \frac{6 - 39}{- 11} = \frac{- 33}{- 11} = 3.$

Answer $(2, 3)$ . Both checks pass: $2 (2) + 3 (3) = 13$ ✓ and $3 (2) - 3 = 3$ ✓ — and the mint dot in the figure sits exactly where the two lines cross.

The one-picture summary

Two lines → engineer twin coefficients → subtract to erase one variable → divide → both coordinates fall out with a shared denominator $D$ ; and $D = 0$ is the "no unique crossing" siren.

Recall Feynman retelling — the whole walkthrough in plain words

I have two straight roads. Each road is all the pairs that make its own equation true, so alone a road is too vague — infinitely many answers. Where the two roads cross is the one spot that obeys both rules at once; that's my answer. To compute it without drawing, I play a trick: I stretch each equation by just the right number so the $y$ -parts become identical, then I subtract one from the other — the identical $y$ -parts vanish and I'm left with a plain "(number) times $x$ = number", which I solve by dividing. Doing the mirror trick (stretching to match the $x$ -parts) gives $y$ . Both answers share the same bottom number $D = a_{1} b_{2} - a_{2} b_{1}$ , which is made only from the road tilts. If $D$ turns out to be zero, the tilts were equal — the roads are the same road (touch everywhere) or parallel train tracks (never touch) — so there is no single crossing, and the formula rightly refuses to divide by zero.

Connections

2.1.09 Linear equations in two variables — graphical and algebraic solutions (Hinglish)
Slope-intercept form y = mx + c
Simultaneous equations by matrices & determinants
Consistency and rank of linear systems
Graphing lines and intercepts
Word problems — age, boat-stream, mixture
Linear inequalities in two variables

Visual walkthrough — Linear equations in two variables — graphical and algebraic solutions