Linear equations in two variables — graphical and algebraic solutions: Worked examples

Before we start, one reminder in plain words. A system of two linear equations is just two straight lines drawn on the same graph paper. "Solving" it means finding the single point $(x, y)$ where both lines pass through — the crossing point. Everything below is a variation on: do the lines cross once, never, or everywhere?

The scenario matrix

Every problem this topic throws at you lands in exactly one of these cells. The last column names the worked example that nails it.

#	Case class	What makes it special	Answer shape	Example
C1	Clean unique solution, all positive	Two lines cross once, tidy numbers	one point	Ex 1
C2	Unique solution with negative values	The crossing sits in quadrant II/III/IV	one point (some coords $< 0$ )	Ex 2
C3	A zero coefficient (vertical / horizontal line)	One variable is missing → line is $x = k$ or $y = k$	one point	Ex 3
C4	No solution (parallel lines)	Same slope, different intercept	none	Ex 4
C5	Infinite solutions (same line)	One equation is a multiple of the other	a whole line	Ex 5
C6	Fractional / messy unique solution	Numbers don't cancel nicely — use the formula	one point (fractions)	Ex 6
C7	Real-world word problem	You must build the two equations first	one point (with units)	Ex 7
C8	Exam twist — solve for a parameter	Find the $k$ that makes a case happen	value(s) of $k$	Ex 8

The three shapes, once, as a picture

Before the examples, hold the three outcomes in your eye. Every case above is one of these three pictures.

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

Cross once (teal + orange lines meet at a plum dot): a unique $(x, y)$ .
Parallel (two lines, same steepness, gap between them): they never meet → no solution.
Same line (one line drawn twice): they meet everywhere → infinite solutions.

Keep this figure in mind: every worked example below is really asking "which of these three am I?"

Ex 1 — Cell C1: clean unique solution

Ex 2 — Cell C2: the crossing is negative

Textbook answers are not always cheerful positive numbers. The crossing point can sit anywhere — including where $x$ or $y$ is negative. If you assume answers are positive, this case will fool you.

The figure shows both lines meeting at the plum dot $(- 2, 3)$ — safely to the left of the $y$ -axis. Nothing wrong with that; the algebra simply reported where the roads actually cross.

Ex 3 — Cell C3: a zero coefficient (a vertical or horizontal line)

The teal vertical line $x = 3$ meets the orange slanted line $x + 2 y = 7$ at the plum dot $(3, 2)$ . A vertical line has no slope number (it's "infinitely steep"), which is exactly why the ratio test that uses slopes is easiest to sidestep here: just read $x$ off first.

Ex 4 — Cell C4: no solution (parallel lines)

Worked example Solve

$3 x + 6 y = 9, x + 2 y = 5.$ Forecast: Divide the first equation by 3. Compare what you get to the second equation. Do you smell a contradiction coming?

Run the ratio test from the parent note. Match $a_{1} x + b_{1} y = c_{1}$ to $3 x + 6 y = 9$ and $a_{2} x + b_{2} y = c_{2}$ to $x + 2 y = 5$ : $\frac{a _{1}}{a _{2}} = \frac{3}{1} = 3, \frac{b _{1}}{b _{2}} = \frac{6}{2} = 3, \frac{c _{1}}{c _{2}} = \frac{9}{5} .$ Why the ratios? Equal $a$ - and $b$ -ratios mean equal slope (parallel). The constant ratio decides same line vs parallel gap.
Compare: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = 3$ but $\frac{c _{1}}{c _{2}} = \frac{9}{5} \neq = 3$ . So $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ → parallel, no solution.
Sanity-check by force. Divide equation 1 by 3: $x + 2 y = 3$ . But equation 2 says $x + 2 y = 5$ . The same left side cannot equal both 3 and 5 — a contradiction. Why this confirms it: $x + 2 y$ is a single number for any given point; asking it to be 3 and 5 at once is impossible → no point exists.

Answer: No solution. Verify (via the determinant): $D = a_{1} b_{2} - a_{2} b_{1} = 3 \cdot 2 - 1 \cdot 6 = 0$ . A zero denominator blocks a unique answer — consistent with "parallel."

Ex 5 — Cell C5: infinite solutions (same line)

Worked example Solve

$x - 3 y = 2, 4 x - 12 y = 8.$ Forecast: Multiply the first equation by 4 in your head. What does it become? Is that new equation... familiar?

Ratio test: $\frac{a _{1}}{a _{2}} = \frac{1}{4}$ , $\frac{b _{1}}{b _{2}} = \frac{- 3}{- 12} = \frac{1}{4}$ , $\frac{c _{1}}{c _{2}} = \frac{2}{8} = \frac{1}{4}$ . Why: all three ratios equal $\frac{1}{4}$ → each coefficient of equation 1 is exactly $\frac{1}{4}$ of equation 2. One equation is a scaled copy of the other.
Confirm directly: multiply equation 1 by 4: $4 x - 12 y = 8$ — that is equation 2. Why this matters: the two "different" equations describe the same line. Every point on that line satisfies both.
Describe the answer set. There is no single point; the solutions are all pairs on $x - 3 y = 2$ , i.e. $x = 2 + 3 t, y = t$ for any real $t$ . Why parameterise? "Infinitely many" still needs a precise description — pick $y = t$ free, then $x$ follows.

Verify: Take $t = 1$ : $(x, y) = (5, 1)$ . Check both: $5 - 3 (1) = 2$ ✓ and $4 (5) - 12 (1) = 20 - 12 = 8$ ✓. Take $t = 0$ : $(2, 0)$ → $2 - 0 = 2$ ✓, $8 - 0 = 8$ ✓. Both random points work → truly infinite solutions.

Ex 6 — Cell C6: messy fractions, use the formula

When adding/subtracting won't cancel anything nicely, don't fight it — reach for the Cramer formula derived in the parent note. It never fails as long as $D \neq = 0$ .

Worked example Solve

$3 x + 4 y = 5, 2 x - 5 y = 7.$ Forecast: No coefficient matches or is an easy multiple of another. Do you expect a whole-number answer, or fractions?

Compute the denominator $D = a_{1} b_{2} - a_{2} b_{1} = (3) (- 5) - (2) (4) = - 15 - 8 = - 23$ . Why first? If $D = 0$ we'd stop (parallel/same). Here $D = - 23 \neq = 0$ → a unique crossing exists.
Compute $x$ : $x = \frac{c _{1} b _{2} - c _{2} b _{1}}{D} = \frac{( 5 ) ( - 5 ) - ( 7 ) ( 4 )}{- 23} = \frac{- 25 - 28}{- 23} = \frac{- 53}{- 23} = \frac{53}{23}$ . Why the formula and not elimination? The numbers share no common factor, so scaling to match coefficients would create ugly products anyway — the formula does that bookkeeping for us.
Compute $y$ : $y = \frac{a _{1} c _{2} - a _{2} c _{1}}{D} = \frac{( 3 ) ( 7 ) - ( 2 ) ( 5 )}{- 23} = \frac{21 - 10}{- 23} = \frac{11}{- 23} = - \frac{11}{23}$ .

Answer: $(\frac{53}{23}, - \frac{11}{23})$ . Verify: Equation 1: $3 \cdot \frac{53}{23} + 4 \cdot (- \frac{11}{23}) = \frac{159 - 44}{23} = \frac{115}{23} = 5$ ✓. Equation 2: $2 \cdot \frac{53}{23} - 5 \cdot (- \frac{11}{23}) = \frac{106 + 55}{23} = \frac{161}{23} = 7$ ✓. Fractions are a perfectly good answer.

Ex 7 — Cell C7: a real-world word problem

The hard part of a word problem is building the two equations. Once they exist, it's just Ex 1 again. See Word problems — age, boat-stream, mixture for the full family.

Worked example Two pencils and three erasers cost ₹18. Four pencils and one eraser cost ₹16. Find the price of one pencil (

$p$ ) and one eraser ( $e$ ). Forecast: Before solving, guess: is a pencil cheaper or dearer than an eraser? (Look at the second deal — four pencils for only ₹16.)

Translate into equations. "Two pencils and three erasers cost ₹18": $2 p + 3 e = 18$ . "Four pencils and one eraser cost ₹16": $4 p + e = 16$ . Why: each item's total cost = (price) × (quantity); the "and" means add the parts.
Isolate $e$ from the second equation (its coefficient is 1 → substitution is cleanest): $e = 16 - 4 p$ . Why substitution here? A coefficient of $1$ means no fractions when we isolate — the parent note's rule "coefficient is 1 → substitute."
Substitute into equation 1: $2 p + 3 (16 - 4 p) = 18 \Rightarrow 2 p + 48 - 12 p = 18 \Rightarrow - 10 p = - 30 \Rightarrow p = 3$ .
Back-substitute: $e = 16 - 4 (3) = 16 - 12 = 4$ . Why finish here: once $p$ is known, $e$ drops straight out.

Answer: pencil = ₹3, eraser = ₹4 (so a pencil is cheaper — forecast check!). Verify with units: $2 (₹3) + 3 (₹4) = ₹6 + ₹12 = ₹18$ ✓ and $4 (₹3) + ₹4 = ₹12 + ₹4 = ₹16$ ✓. Both money totals match.

Ex 8 — Cell C8: the exam twist (solve for a parameter)

The sneakiest exam question doesn't ask you to solve a system — it asks for which value of $k$ the system behaves a certain way. You must run the case tests backwards.

Worked example For what value of

$k$ does $2 x + 3 y = 7, (k) x + 9 y = 21$ have infinitely many solutions? Forecast: Infinite solutions = same line. That needed all three ratios equal. Which of the three ratios is already forced, and which one you get to tune with $k$ ?

Write the target condition. Infinite solutions ⇔ $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ , i.e. $\frac{2}{k} = \frac{3}{9} = \frac{7}{21}$ . Why all three: two equal ratios only guarantee parallel; the third equality is what makes them the same line.
Check the fixed ratios first: $\frac{3}{9} = \frac{1}{3}$ and $\frac{7}{21} = \frac{1}{3}$ . Good — these two already agree, so an infinite-solution case is possible. Why check: if $\frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ , no $k$ could ever give infinite solutions.
Solve the remaining ratio for $k$ : $\frac{2}{k} = \frac{1}{3} \Rightarrow k = 6$ . Why: cross-multiplying, $2 \cdot 3 = 1 \cdot k$ , so $k = 6$ .

Answer: $k = 6$ . Verify: With $k = 6$ the second equation is $6 x + 9 y = 21$ . Divide by 3: $2 x + 3 y = 7$ — identical to equation 1. Same line → infinitely many solutions ✓. (As a bonus: any $k \neq = 6$ with these numbers gives no solution, since then $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}}$ fails while $\frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ holds — parallel.)

Recap: which cell, which move

Recall Which case is

$D = a_{1} b_{2} - a_{2} b_{1} = 0$ ? Parallel or same line — never a unique solution. Then use the constant ratio $\frac{c _{1}}{c _{2}}$ to decide no solution vs infinite.

Recall A crossing point turned out to be

$(- 2, 3)$ . Is that valid? Yes — solutions can have negative coordinates. The crossing simply lives in a different quadrant (Ex 2). Always verify in both equations regardless of sign.

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
Word problems — age, boat-stream, mixture
Graphing lines and intercepts
Linear inequalities in two variables