Linear equations in two variables — graphical and algebraic solutions: Exercises

Common mistake Edge case you MUST hold in mind — vertical lines (

$b = 0$ ) The ratio test compares $\frac{a _{1}}{a _{2}}, \frac{b _{1}}{b _{2}}, \frac{c _{1}}{c _{2}}$ , and the slope story says slope $= - a_{i} / b_{i}$ . But if $b_{i} = 0$ the equation is $a_{i} x = c_{i}$ , i.e. $x = \frac{c _{i}}{a _{i}}$ — a vertical line with NO finite slope, and $\frac{b _{1}}{b _{2}} = \frac{0}{0}$ is undefined. Fix / how to read it safely: the ratio test still works if you treat $\frac{0}{0}$ as "match" ONLY when both $b$ 's are zero. Two vertical lines $x = p$ and $x = q$ are parallel (no solution) when $p \neq = q$ and the same line (infinite) when $p = q$ ; they never cross. If exactly one line is vertical and the other isn't, slopes can't match, so they cross once (unique). Whenever you see a $y$ -coefficient of $0$ , drop the slope formula and reason directly from $x =$ constant. We flag this again in Problem 5.4.

Level 1 — Recognition

(Can you read off the type of a system without solving it?)

Problem 1.1

Is $3 x - 5 y + 7 = 0$ a linear equation in two variables? State $a, b, c$ .

Recall Solution 1.1

WHAT we check: every variable has degree $1$ — no $x^{2}$ , no $x y$ , no $x$ . Here we only see $x$ and $y$ to the first power. ✓ Comparing with $a x + b y + c = 0$ : $a = 3$ , $b = - 5$ , $c = 7$ . And $(a, b) = (3, - 5) \neq = (0, 0)$ . ✓ Answer: Yes. It is linear; $a = 3, b = - 5, c = 7$ .

Problem 1.2

Without solving, classify the system as unique / infinite / no solution: $2 x + 3 y = 6, 4 x + 6 y = 18.$

Recall Solution 1.2

WHY ratios: the parent note showed slope of $a_{i} x + b_{i} y = c_{i}$ is $- a_{i} / b_{i}$ . Comparing $\frac{a _{1}}{a _{2}}, \frac{b _{1}}{b _{2}}, \frac{c _{1}}{c _{2}}$ tells us slope-and-shift at a glance. (Here every $b_{i} \neq = 0$ , so slopes are finite and the test applies cleanly.) $\frac{a _{1}}{a _{2}} = \frac{2}{4} = \frac{1}{2}, \frac{b _{1}}{b _{2}} = \frac{3}{6} = \frac{1}{2}, \frac{c _{1}}{c _{2}} = \frac{6}{18} = \frac{1}{3} .$ First two equal (same slope) but third differs (different shift): $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ . Answer: No solution — parallel lines.

Problem 1.3

A single equation $x + y = 5$ . How many solutions does it have, and name three of them.

Recall Solution 1.3

WHAT: one linear equation in two unknowns pins down a whole line, not a point. HOW: pick any $x$ , solve for $y = 5 - x$ . $x = 0 \Rightarrow (0, 5)$ ; $x = 2 \Rightarrow (2, 3)$ ; $x = 5 \Rightarrow (5, 0)$ . Answer: infinitely many — every point on the line $y = 5 - x$ . Three examples: $(0, 5), (2, 3), (5, 0)$ .

Level 2 — Application

(Turn the crank: substitution and elimination.)

The picture below is your mental model for the whole level: substitution slides one line's rule into the other; elimination stacks the two lines so a matched term cancels. Both are just two ways of hunting the single crossing point.

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

Problem 2.1

Solve by substitution: $y = 2 x - 1, 3 x + y = 9.$

Recall Solution 2.1

WHY substitution here: the first equation already gives $y$ alone, so we can drop it straight into the second and collapse two unknowns to one. Geometrically (left panel of the figure): we walk along the blue line, where $y$ is forced to be $2 x - 1$ , until we also land on the yellow line. Replace $y$ in eqn 2: $3 x + (2 x - 1) = 9 \Rightarrow 5 x - 1 = 9 \Rightarrow 5 x = 10 \Rightarrow x = 2.$ Back-substitute: $y = 2 (2) - 1 = 3.$ Check both: $y = 2 x - 1$ : $3 = 4 - 1$ ✓. $3 x + y$ : $6 + 3 = 9$ ✓. Answer: $(2, 3)$ .

Problem 2.2

Solve by elimination: $3 x + 2 y = 16, 5 x - 2 y = 8.$

Recall Solution 2.2

WHY elimination here: the $y$ -coefficients are already $+ 2$ and $- 2$ — opposite. Adding the equations kills $y$ instantly (legal because adding equal things to both sides keeps truth). Geometrically (right panel of the figure): stacking the two rules cancels the $y$ -direction, leaving one equation in $x$ alone — a single vertical "where is the crossing" line. Add: $(3 x + 5 x) + (2 y - 2 y) = 16 + 8 \Rightarrow 8 x = 24 \Rightarrow x = 3.$ Back-substitute into eqn 1: $3 (3) + 2 y = 16 \Rightarrow 9 + 2 y = 16 \Rightarrow y = \frac{7}{2} .$ Check: eqn 2: $5 (3) - 2 (3.5) = 15 - 7 = 8$ ✓. Answer: $(3, \frac{7}{2})$ .

Problem 2.3

Solve: $2 x + 3 y = 13, 4 x - y = 5.$ (One coefficient is $1$ — choose the smart method.)

Recall Solution 2.3

WHY: in eqn 2 the $y$ -coefficient is $- 1$ , so isolating $y$ is clean: $y = 4 x - 5$ . Substitute into eqn 1: $2 x + 3 (4 x - 5) = 13 \Rightarrow 2 x + 12 x - 15 = 13 \Rightarrow 14 x = 28 \Rightarrow x = 2.$ Then $y = 4 (2) - 5 = 3.$ Check both: $2 (2) + 3 (3) = 4 + 9 = 13$ ✓; $4 (2) - 3 = 5$ ✓. Answer: $(2, 3)$ .

Level 3 — Analysis

(Now the numbers hide something — read the structure.)

Problem 3.1

For what value of $k$ does the system have no solution? $x + 2 y = 3, 2 x + k y = 7.$

Recall Solution 3.1

WHAT no-solution means: parallel — same slope, different shift: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ . (Both $b$ 's here are nonzero for the $k$ we find, so the finite-slope test is valid.) Slope match: $\frac{a _{1}}{a _{2}} = \frac{1}{2}$ must equal $\frac{b _{1}}{b _{2}} = \frac{2}{k}$ , so $\frac{1}{2} = \frac{2}{k} \Rightarrow k = 4.$ Now confirm the shift differs: $\frac{c _{1}}{c _{2}} = \frac{3}{7}$ , and $\frac{1}{2} \neq = \frac{3}{7}$ . ✓ Answer: $k = 4$ (and only $k = 4$ ).

Problem 3.2

For what value of $k$ does the system have infinitely many solutions? $3 x + 4 y = 12, 6 x + 8 y = k .$

Recall Solution 3.2

WHAT infinite means: ALL three ratios equal — one equation is a multiple of the other. $\frac{a _{1}}{a _{2}} = \frac{3}{6} = \frac{1}{2}$ , $\frac{b _{1}}{b _{2}} = \frac{4}{8} = \frac{1}{2}$ (already matched). For coincidence we need $\frac{c _{1}}{c _{2}} = \frac{12}{k} = \frac{1}{2} \Rightarrow k = 24.$ Answer: $k = 24$ . Then eqn 2 is exactly $2 \times$ eqn 1.

Problem 3.3

Use the Cramer-style formula from the parent note to solve $5 x + 3 y = 11, 2 x + 4 y = 10,$ and confirm the denominator $D \neq = 0$ .

Recall Solution 3.3

WHY Cramer: it packages elimination into one shot; the denominator $D = a_{1} b_{2} - a_{2} b_{1}$ doubles as the "unique-solution detector." Picture the determinant (the figure below): write the coefficients of $x$ and $y$ as a little $2 \times 2$ box $D = a_{1} a_{2} b_{1} b_{2} = 5234 = ↘ main diagonal 5 \cdot 4 - ↗ anti-diagonal 2 \cdot 3 = 20 - 6 = 14.$ The value $D$ is the area of the parallelogram spanned by the two coefficient rows (as arrows); when that area is $0$ the rows point the same way — parallel lines, no unique answer.

Here $a_{1} = 5, b_{1} = 3, c_{1} = 11, a_{2} = 2, b_{2} = 4, c_{2} = 10$ and $D = 14 \neq = 0 \Rightarrow$ unique solution. $x = \frac{c _{1} b _{2} - c _{2} b _{1}}{D} = \frac{11 \cdot 4 - 10 \cdot 3}{14} = \frac{44 - 30}{14} = \frac{14}{14} = 1.$ $y = \frac{a _{1} c _{2} - a _{2} c _{1}}{D} = \frac{5 \cdot 10 - 2 \cdot 11}{14} = \frac{50 - 22}{14} = \frac{28}{14} = 2.$ Check: $5 (1) + 3 (2) = 11$ ✓; $2 (1) + 4 (2) = 10$ ✓. Answer: $(1, 2)$ , $D = 14$ . (See Simultaneous equations by matrices & determinants for where $D$ becomes a determinant.)

Level 4 — Synthesis

(Build the equations yourself from a story, then solve.)

Problem 4.1 (boat & stream)

A boat covers $24$ km downstream in $2$ h and returns the same $24$ km upstream in $3$ h. Find the boat's speed $b$ and the stream's speed $s$ in still water.

Recall Solution 4.1

WHY add/subtract $s$ : downstream the current helps, so effective speed is $b + s$ ; upstream it opposes, so $b - s$ . Downstream speed $= \frac{24}{2} = 12 \Rightarrow b + s = 12$ . Upstream speed $= \frac{24}{3} = 8 \Rightarrow b - s = 8$ . Eliminate $s$ by adding: $2 b = 20 \Rightarrow b = 10$ . Then $s = 12 - b = 2$ . Check: $b + s = 12$ ✓, $b - s = 8$ ✓. Answer: boat $10$ km/h, stream $2$ km/h. More of this type in Word problems — age, boat-stream, mixture.

Problem 4.2 (ages)

A father is presently $4$ times as old as his son. In $10$ years he will be only $2 \frac{1}{2}$ times as old. Find their present ages.

Recall Solution 4.2

WHAT variables: let father $= F$ , son $= S$ (years, now). " $4$ times now": $F = 4 S$ . "In 10 years, $\frac{5}{2}$ times": $F + 10 = \frac{5}{2} (S + 10)$ . Substitute $F = 4 S$ into the second: $4 S + 10 = \frac{5}{2} (S + 10)$ . Multiply through by $2$ : $8 S + 20 = 5 (S + 10) = 5 S + 50$ . $3 S = 30 \Rightarrow S = 10$ , so $F = 4 (10) = 40$ . Check: now $40 = 4 \times 10$ ✓; in 10 yr $50$ vs $2.5 \times 20 = 50$ ✓. Answer: son $10$ , father $40$ .

Problem 4.3 (mixture / two items)

$5$ pens and $3$ pencils cost ₹ $34$ ; $2$ pens and $6$ pencils cost ₹ $28$ . Find the cost of one pen $p$ and one pencil $q$ .

Recall Solution 4.3

Equations: $5 p + 3 q = 34$ and $2 p + 6 q = 28$ . WHY elimination on $q$ : $3$ and $6$ share factor $6$ . Multiply eqn 1 by $2$ : $10 p + 6 q = 68$ . Subtract eqn 2: $(10 p - 2 p) + (6 q - 6 q) = 68 - 28 \Rightarrow 8 p = 40 \Rightarrow p = 5.$ Back-substitute in eqn 1: $5 (5) + 3 q = 34 \Rightarrow 25 + 3 q = 34 \Rightarrow q = 3.$ Check: $2 (5) + 6 (3) = 10 + 18 = 28$ ✓. Answer: pen ₹ $5$ , pencil ₹ $3$ .

Level 5 — Mastery

(Graphical reading, parameters, a vertical-line edge case, and a system that isn't quite linear until you look.)

Problem 5.1 (graphical)

The two lines $x + y = 4$ and $x - y = 2$ are drawn in the figure below. Read the intersection from the graph, then confirm algebraically.

How to read the figure: the horizontal axis is $x$ , the vertical axis is $y$ , and the thin white cross through $(0, 0)$ marks the axes. The blue line is $x + y = 4$ (it falls as $x$ grows, since $y = 4 - x$ ). The yellow line is $x - y = 2$ (it rises, since $y = x - 2$ ). The single green dot is the one point lying on both lines — the solution. The green label states its coordinates.

Recall Solution 5.1

Reading the graph: the blue line $x + y = 4$ and the yellow line $x - y = 2$ cross at the marked green point $(3, 1)$ . Confirm algebraically (elimination): add the two equations: $(x + y) + (x - y) = 4 + 2 \Rightarrow 2 x = 6 \Rightarrow x = 3$ . Then $y = 4 - 3 = 1$ . Check: $x - y = 3 - 1 = 2$ ✓. Answer: $(3, 1)$ — graph and algebra agree, as they must (see Graphing lines and intercepts).

Problem 5.2 (parameter with three outcomes)

For the system $x + y = 5, 2 x + 2 y = m$ , describe how the solution set changes as $m$ varies. Give the special value of $m$ and the two regimes.

Recall Solution 5.2

WHY one parameter splits into cases: $\frac{a _{1}}{a _{2}} = \frac{1}{2} = \frac{b _{1}}{b _{2}}$ always (slopes always match here). Only $\frac{c _{1}}{c _{2}} = \frac{5}{m}$ moves.

If $\frac{5}{m} = \frac{1}{2}$ , i.e. $m = 10$ : all three ratios equal ⇒ same line, infinitely many solutions (eqn 2 is exactly $2 \times$ eqn 1).
If $m \neq = 10$ : slopes match but constants don't ⇒ parallel, no solution. Because slopes are locked equal, there is no $m$ giving a unique solution. Answer: $m = 10$ → infinite solutions; every other $m$ → no solution; unique solution impossible.

Problem 5.3 (reciprocal substitution — mastery)

Solve $\frac{2}{x} + \frac{3}{y} = 13, \frac{5}{x} - \frac{4}{y} = - 2, x \neq = 0, y \neq = 0.$

Recall Solution 5.3

WHY a swap: these aren't linear in $x, y$ (the unknowns sit in denominators). But let $u = \frac{1}{x}$ , $v = \frac{1}{y}$ — then everything becomes linear in $u, v$ , and all our machinery works. System: $2 u + 3 v = 13$ and $5 u - 4 v = - 2$ . Eliminate $v$ : multiply first by $4$ , second by $3$ : $8 u + 12 v = 52$ , $15 u - 12 v = - 6$ . Add: $23 u = 46 \Rightarrow u = 2$ . Then $2 (2) + 3 v = 13 \Rightarrow 3 v = 9 \Rightarrow v = 3$ . Undo the swap: $u = \frac{1}{x} = 2 \Rightarrow x = \frac{1}{2}$ ; $v = \frac{1}{y} = 3 \Rightarrow y = \frac{1}{3}$ . Both nonzero ✓ (so the original was well-defined). Check: $\frac{2}{1/2} + \frac{3}{1/3} = 4 + 9 = 13$ ✓; $\frac{5}{1/2} - \frac{4}{1/3} = 10 - 12 = - 2$ ✓. Answer: $x = \frac{1}{2}, y = \frac{1}{3}$ .

Problem 5.4 (vertical-line edge case)

Classify and, where possible, solve each system. Note carefully where the standard slope/ratio test needs adapting. $(a) x = 3, 2 x + y = 10 (b) x = 3, x = 5$

Recall Solution 5.4

WHY the caution box applies: both first equations are $x = const$ , i.e. $b = 0$ — vertical lines with no finite slope. Do NOT plug into $- a / b$ ; reason directly. (a) Line $x = 3$ is vertical; line $2 x + y = 10$ has finite slope $- 2$ . A vertical line and a non-vertical line always cross exactly once. Put $x = 3$ into the second: $2 (3) + y = 10 \Rightarrow y = 4$ . Unique solution $(3, 4)$ . (b) Both $x = 3$ and $x = 5$ are vertical, so both have "the same (undefined) slope" — they are parallel. Since the constants differ ( $3 \neq = 5$ ) they never meet. No solution. (Had they been $x = 3$ and $x = 3$ , they'd be the same line — infinitely many solutions.) Takeaway: with a $0$ $y$ -coefficient the ratio $\frac{b _{1}}{b _{2}} = \frac{0}{0}$ is meaningless — switch to "vertical vs vertical vs slanted" reasoning.

Connections

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