High School Mathematics · Grades 10–12

7Equations of Lines and Linear Systems

A straight line in the coordinate plane is described by an equation, and two lines together form a system of equations whose solutions are their intersection points. This chapter connects three languages — geometric (lines), algebraic (equations) and functional (affine functions from Chapter 4) — and teaches the two standard ways of solving a linear system.

7.1 The equation of a line

Theorem 7.1 (Reduced equation)

In a coordinate system:

every non-vertical line has an equation of the form

$y = mx + p ,$

with $m$ (the slope) and $p$ (the $y$ -intercept) uniquely determined — this is the reduced equation of the line;
every vertical line has an equation $x = c$ ;
a point lies on the line exactly when its coordinates satisfy the equation.

Proof. Admitted at this level. ∎

Proposition 7.2 (Slope through two points)

The slope of the (non-vertical) line through $A(x_A, y_A)$ and $B(x_B, y_B)$ , with $x_A \neq x_B$ , is

m = \frac{y_B - y_A}{x_B - x_A} \qquad \text{(``change in $y$ over change in $x$'').}

Proof. Both points satisfy $y = mx + p$ , so $y_B - y_A = (mx_B + p) - (mx_A + p) = m(x_B - x_A)$ ; divide by $x_B - x_A \neq 0$ . ∎

Three kinds of lines: slanted (m ≠ 0), horizontal (m = 0), and vertical (no reduced equation — the equation is x = c). — Three kinds of lines: slanted ( $m \neq 0$ ), horizontal ( $m = 0$ ), and vertical (no reduced equation — the equation is $x = c$ ).

Method 7.3 (Finding the equation of a line through two points)

Given $A$ and $B$ with $x_A \neq x_B$ :

compute the slope $m = \dfrac{y_B - y_A}{x_B - x_A}$ ;
write $y = mx + p$ and substitute the coordinates of $A$ (or $B$ ) to find $p$ ;
check the final equation with the other point.

If $x_A = x_B$ , the line is vertical: its equation is $x = x_A$ .

Example 7.4

Line through $A(1, 5)$ and $B(3, 1)$ . Slope: $m = \frac{1 - 5}{3 - 1} = \frac{-4}{2} = -2$ . Then $y = -2x + p$ ; the point $A$ gives $5 = -2 \times 1 + p$ , so $p = 7$ . Equation: $y = -2x + 7$ . Check with $B$ : $-2 \times 3 + 7 = 1$ .

Proposition 7.5 (Parallel lines)

Two non-vertical lines are parallel if and only if they have the same slope. Two distinct parallel lines have no common point; two lines with different slopes have exactly one.

Proof. Let $y = mx + p$ and $y = m'x + p'$ be the two lines. A common point solves $mx + p = m'x + p'$ , i.e. $(m - m')x = p' - p$ . If $m \neq m'$ , this has exactly one solution $x = \frac{p' - p}{m - m'}$ : one intersection point. If $m = m'$ , the equation reads $0 = p' - p$ : no solution when $p \neq p'$ (distinct parallels), every $x$ when $p = p'$ (same line). Parallelism (having no intersection, or being equal) is thus equivalent to $m = m'$ . ∎

Remark 7.6 (Direction vector)

The line $y = mx + p$ goes up $m$ units for each unit to the right, so the vector $\vec u\,(1, m)$ is a direction vector of the line: the line is parallel to $\vec u$ . Two lines are parallel exactly when their direction vectors are collinear (Theorem 6.13).

7.2 Linear systems

Definition 7.7 (Linear system)

A system of two linear equations in the unknowns $x$ and $y$ is a pair of equations

\begin{cases} a x + b y = c \\ a' x + b' y = c' \end{cases}

to be satisfied simultaneously. A solution is a pair $(x, y)$ satisfying both. Each equation describes a line, so solving the system means finding the intersection of two lines.

Method 7.8 (Substitution)

Solve one equation for one unknown — pick the easiest, e.g. express $y$ in terms of $x$ ;
substitute this expression into the other equation, which now contains only $x$ ;
solve for $x$ , then compute $y$ from step 1;
check the pair in both original equations.

Example 7.9

Solve $\begin{cases} x + 2y = 8 \\ 3x - y = 3 \end{cases}$ by substitution. The second equation gives $y = 3x - 3$ . Substitute into the first:

x + 2(3x - 3) = 8 \ \Longleftrightarrow\ x + 6x - 6 = 8 \ \Longleftrightarrow\ 7x = 14 \ \Longleftrightarrow\ x = 2 ,

then $y = 3 \times 2 - 3 = 3$ . The solution is the pair $(2, 3)$ . Check: $2 + 2 \times 3 = 8$ and $3 \times 2 - 3 = 3$ .

Method 7.10 (Elimination)

Multiply each equation by a well-chosen number so that one unknown gets opposite coefficients in the two equations;
add the equations: that unknown disappears;
solve the remaining one-unknown equation, then substitute back to find the other unknown;
check the pair in both original equations.

Example 7.11

Solve $\begin{cases} 2x + 3y = 7 \\ 5x - 2y = 8 \end{cases}$ by elimination. To eliminate $y$ , multiply the first equation by $2$ and the second by $3$ :

\begin{cases} 4x + 6y = 14 \\ 15x - 6y = 24 \end{cases}

Adding them: $19x = 38$ , so $x = 2$ . Substituting into $2x + 3y = 7$ : $4 + 3y = 7$ , so $y = 1$ . Solution: $(2, 1)$ . Check in the second equation: $5 \times 2 - 2 \times 1 = 8$ .

A linear system seen geometrically: each equation is a line, and the solution (2,1) of is their intersection point. — A linear system seen geometrically: each equation is a line, and the solution $(2,1)$ of Example 7.11 is their intersection point.

Remark 7.12 (How many solutions?)

Like two lines, a linear system has exactly one solution in general (non-parallel lines), and exceptionally none (distinct parallel lines) or infinitely many (twice the same line). The criterion of Theorem 6.13 decides: the system has a unique solution exactly when $ab' - a'b \neq 0$ .

Example 7.13 (Modeling with a system)

Three notebooks and two pens cost $8$ ; one notebook and four pens cost $6$ . Let $x$ be the price of a notebook and $y$ that of a pen:

\begin{cases} 3x + 2y = 8 \\ x + 4y = 6 . \end{cases}

From the second equation, $x = 6 - 4y$ ; substituting, $3(6 - 4y) + 2y = 8$ , so $18 - 10y = 8$ , giving $y = 1$ and then $x = 2$ . A notebook costs $2$ and a pen costs $1$ .

7.3 Exercises

Exercise 7.1 ★

For each line, read off the slope and the $y$ -intercept, and sketch it:

y = 2x - 3, \qquad y = -\tfrac13 x + 1, \qquad y = 4, \qquad x = -2 .

Solution

Solution of Exercise 7.1.

$y = 2x - 3$ : slope $2$ , intercept $-3$ . $y = -\frac13 x + 1$ : slope $-\frac13$ , intercept $1$ . $y = 4$ : slope $0$ , intercept $4$ (horizontal line). $x = -2$ : vertical line, no slope and no reduced equation.

Exercise 7.2 ★

Does the point $A(2, 7)$ belong to the line $y = 3x + 1$ ? And $B(-1, -1)$ ? And the point $C(4, 13)$ ?

Solution

Solution of Exercise 7.2.

$3 \times 2 + 1 = 7$ : yes, $A$ is on the line. $3 \times (-1) + 1 = -2 \neq -1$ : $B$ is not. $3 \times 4 + 1 = 13$ : yes, $C$ is on the line.

Exercise 7.3 ★

Find the reduced equation of the line through:

\text{(a) } A(0, 2) \text{ and } B(3, 8); \qquad \text{(b) } A(-1, 5) \text{ and } B(2, -4); \qquad \text{(c) } A(4, 1) \text{ and } B(4, 6).

Solution

Solution of Exercise 7.3.

(a) $m = \frac{8 - 2}{3 - 0} = 2$ and $p = y_A = 2$ (the point $A$ is on the $y$ -axis): $y = 2x + 2$ .

(b) $m = \frac{-4 - 5}{2 - (-1)} = \frac{-9}{3} = -3$ ; then $5 = -3 \times (-1) + p$ gives $p = 2$ : $y = -3x + 2$ .

Exercise 7.4 ★

Among the lines $y = 3x - 1$ , $y = -3x + 2$ , $y = 3x + 5$ , $y = \frac{6x + 2}{2}$ , which are parallel to each other? Which are actually equal?

Solution

Solution of Exercise 7.4.

$y = \frac{6x+2}{2} = 3x + 1$ . The lines with slope $3$ are $y = 3x - 1$ , $y = 3x + 5$ and $y = 3x + 1$ : those three are parallel to each other; no two of them are equal (different intercepts), and none is parallel to $y = -3x + 2$ .

Exercise 7.5 ★

Solve by substitution:

\begin{cases} y = 2x - 1 \\ 3x + y = 9 \end{cases} \qquad\text{then}\qquad \begin{cases} x - y = 4 \\ 2x + 3y = 3 . \end{cases}

Solution

Solution of Exercise 7.5.

First system: substitute $y = 2x - 1$ into $3x + y = 9$ : $3x + 2x - 1 = 9$ , so $5x = 10$ , $x = 2$ , then $y = 3$ . Solution: $(2, 3)$ .

Second system: from $x - y = 4$ , $x = y + 4$ . Then $2(y + 4) + 3y = 3$ , so $5y = -5$ , $y = -1$ , then $x = 3$ . Solution: $(3, -1)$ .

Exercise 7.6 ★

Solve by elimination:

\begin{cases} x + y = 10 \\ x - y = 4 \end{cases} \qquad\text{then}\qquad \begin{cases} 3x + 2y = 12 \\ 2x + 5y = 8 . \end{cases}

Solution

Solution of Exercise 7.6.

First system: adding the equations eliminates $y$ : $2x = 14$ , $x = 7$ ; then $7 + y = 10$ gives $y = 3$ . Solution: $(7, 3)$ .

Second system: multiply the first equation by $5$ and the second by $-2$ :

\begin{cases} 15x + 10y = 60 \\ -4x - 10y = -16 \end{cases}

Adding: $11x = 44$ , so $x = 4$ ; then $3 \times 4 + 2y = 12$ gives $y = 0$ . Solution: $(4, 0)$ .

Exercise 7.7 ★★

Determine, without solving them, how many solutions each system has:

\begin{cases} 2x - y = 3 \\ -4x + 2y = -6 \end{cases} \qquad \begin{cases} 2x - y = 3 \\ -4x + 2y = 1 \end{cases} \qquad \begin{cases} 2x - y = 3 \\ x + y = 0 . \end{cases}

Solution

Solution of Exercise 7.7.

Compute $ab' - a'b$ in each case.

First: $2 \times 2 - (-4) \times (-1) = 4 - 4 = 0$ , and the second equation is $-2$ times the first: twice the same line, infinitely many solutions.

Second: $ab' - a'b = 0$ again, but the right-hand sides are not in the ratio $-2$ ( $1 \neq -6$ ): two distinct parallel lines, no solution.

Third: $2 \times 1 - 1 \times (-1) = 3 \neq 0$ : exactly one solution.

Exercise 7.8 ★★

Two hundred tickets were sold for a school play, some at $5$ (children) and some at $9$ (adults), for a total of $1352$ . How many tickets of each kind were sold?

Solution

Solution of Exercise 7.8.

Let $x$ be the number of child tickets and $y$ the number of adult tickets:

\begin{cases} x + y = 200 \\ 5x + 9y = 1352 . \end{cases}

From the first equation $x = 200 - y$ ; substituting, $5(200 - y) + 9y = 1352$ , so $1000 + 4y = 1352$ , $y = 88$ , then $x = 112$ . So $112$ child tickets and $88$ adult tickets. Check: $5 \times 112 + 9 \times 88 = 560 + 792 = 1352$ .

Exercise 7.9 ★★

Let $d$ be the line $y = 2x - 3$ and $A(1, 4)$ .

Give the equation of the line $d'$ parallel to $d$ passing through $A$ .
Compute the intersection point of $d'$ with the $x$ -axis.

Solution

Solution of Exercise 7.9.

1. Parallel means same slope $2$ : $y = 2x + p$ with $4 = 2 \times 1 + p$ , so $p = 2$ : $d'\colon y = 2x + 2$ .

2. On the $x$ -axis, $y = 0$ : $2x + 2 = 0$ gives $x = -1$ . The intersection point is $(-1, 0)$ .

Exercise 7.10 ★★★

Let $A(0, 0)$ , $B(6, 2)$ and $C(2, 4)$ .

Find the equations of the medians of the triangle $ABC$ issued from $A$ and from $B$ (a median joins a vertex to the midpoint of the opposite side).
Compute their intersection point $G$ , and check that $G$ is also on the third median. (You should find that the coordinates of $G$ are the averages of those of $A$ , $B$ , $C$ .)

Solution

Solution of Exercise 7.10.

1. Midpoint of $[BC]$ : $A' = \left(\frac{6+2}{2}, \frac{2+4}{2}\right) = (4, 3)$ . The median from $A(0,0)$ has slope $\frac{3 - 0}{4 - 0} = \frac34$ : equation $y = \frac34 x$ .

Midpoint of $[AC]$ : $B' = (1, 2)$ . The median from $B(6, 2)$ has slope $\frac{2 - 2}{1 - 6} = 0$ : it is the horizontal line $y = 2$ .

2. Intersection: $\frac34 x = 2$ gives $x = \frac83$ , so $G\left(\frac83, 2\right)$ . The third median joins $C(2,4)$ to the midpoint of $[AB]$ , $C' = (3, 1)$ ; its slope is $\frac{1 - 4}{3 - 2} = -3$ , equation $y = -3x + 10$ . At $x = \frac83$ : $y = -8 + 10 = 2$ : $G$ is on it. And indeed $G = \left(\frac{0 + 6 + 2}{3}, \frac{0 + 2 + 4}{3}\right)$ : the centroid averages the coordinates of the vertices.

7.4 Problem: One solution, none, or infinitely many

Problem 7.1

Weekend problem — two lines have three possible destinies, a determinant decides between them, and the world’s oldest textbook already knew

Two linear equations, two unknowns: the pair of lines they draw can cross once, never, or be one and the same line — and every linear system inherits one of these three destinies. This problem classifies them, meets the number that decides (an old acquaintance from the vectors chapter), solves the two-thousand-year-old pheasants-and-rabbits of the Chinese Nine Chapters, and ends with corners of regions — the first step of the optimization used by every airline and factory today.

Part I — The line, fluently.

Find the reduced equation of the line through $(1, 2)$ and $(3, 8)$ (Method 7.3).
Among $y = 3x - 1$ , $y = 3x + 4$ and $y = -x + 7$ : which two lines are parallel (Proposition 7.5), and where do the non-parallel pairs meet? (Compute one intersection.)
The line through $(2, 1)$ and $(2, 5)$ : why does it have no reduced equation $y = mx + p$ , and what is its equation?
Is the point $(4, 11)$ on the line $y = 3x - 1$ ? And $(-1, -5)$ ?
The line through $(0, 3)$ with slope $-2$ : give its equation, its two axis intercepts, and the area of the triangle it cuts from the first quadrant.

Part II — Three destinies.

Solve by substitution (Method 7.8): $\begin{cases} y = 2x - 3 \\ 3x + y = 7 \end{cases}$
Solve by elimination (Method 7.10): $\begin{cases} 2x + 3y = 8 \\ 5x - 3y = -1 \end{cases}$
Classify — and interpret with lines — the two systems

$\begin{cases} 2x - y = 3 \\ 4x - 2y = 6 \end{cases} \qquad\text{and}\qquad \begin{cases} 2x - y = 3 \\ 4x - 2y = 1 . \end{cases}$

State the complete trichotomy: crossing, parallel, identical — one solution, none, infinitely many.
For the general system $ax + by = e$ , $cx + dy = f$ , the deciding number is $ad - bc$ — the same determinant as the collinearity test of Problem 6.1. Explain the coincidence (which two vectors are collinear exactly when the lines are parallel or identical?), and evaluate $ad - bc$ for the three systems of questions 7 and 8.
For which value of $m$ does $\begin{cases} x + my = 3 \\ 2x + 6y = 7 \end{cases}$ have no solution? Solve the system for all other values of $m$ … or at least explain how you would.

Part III — The Nine Chapters.

From the Nine Chapters on the Mathematical Art (China, about 2000 years ago): a cage holds pheasants and rabbits — $35$ heads and $94$ legs in all. How many of each?
A juice stand blends a $60\,\%$ -fruit syrup with a $20\,\%$ -fruit drink to obtain $30$ L at $35\,\%$ fruit. How many liters of each?
A two-digit number has digit sum $12$ ; swapping its digits decreases it by $18$ . Find it, using the digit algebra of the digit-algebra weekend problem of the Middle School volume to set up the system.
At the bakery, $3$ coffees and $2$ croissants cost $9.10$ euros; $2$ coffees and $3$ croissants cost $8.90$ euros. Solve elegantly: what do adding the two equations and subtracting them each tell you directly?
A market stall claims: $2$ apples $+ 1$ pear for $5$ euros, and $4$ apples $+ 2$ pears for $9$ euros. Solve — or rather, explain what the system’s destiny reveals about the stall’s arithmetic.

Part IV — Corners rule.

The line $y = 3x - 1$ splits the plane in two. Which side is the origin on? Describe the set $y > 3x - 1$ and how one tests a point against it.
Draw the region defined by the four constraints

$x \geq 0, \qquad y \geq 0, \qquad x + y \leq 4, \qquad y \leq 2x + 1,$

and compute its four corner points.
A workshop’s profit is $P = 3x + 2y$ on the region of question 17. Evaluate $P$ at the four corners. Admitting that the maximum of such a linear quantity on such a region is always reached at a corner (picture the lines $3x + 2y =$ constant sweeping across), give the optimal production plan.
The two taxis of the two-thermometers weekend problem of the Middle School volume as a system: write and solve $\begin{cases} y = 1.5x + 3 \\ y = 2x \end{cases}$ and interpret the solution.
Finale, the dictionary: one equation $\leftrightarrow$ one line; one system $\leftrightarrow$ two lines with three destinies, decided by $ad - bc$ ; many inequalities $\leftrightarrow$ a region whose corners carry the optimum. State it in three sentences — you have just toured the entrance hall of linear algebra and linear programming, both built in full in the university volumes.

Solution

Solution of Problem 7.1.

1. Slope $\frac{8 - 2}{3 - 1} = 3$ ; through $(1, 2)$ : $y = 3x - 1$ .

2. $y = 3x - 1$ and $y = 3x + 4$ : same slope, parallel (and distinct). Crossing $y = 3x - 1$ with $y = -x + 7$ : $3x - 1 = -x + 7$ gives $x = 2$ , $y = 5$ : point $(2, 5)$ .

3. Both points share $x = 2$ : the line is vertical, equation $x = 2$ . A reduced form $y = mx + p$ answers “one $y$ per $x$ ”, which a vertical line refuses: its slope would be infinite.

4. $3 \times 4 - 1 = 11$ : yes, $(4, 11)$ is on the line. $3 \times (-1) - 1 = -4 \neq -5$ : no.

5. $y = -2x + 3$ ; crosses the axes at $(0, 3)$ and $\left(\frac32, 0\right)$ . Triangle area: $\frac12 \times \frac32 \times 3 = \frac94 = 2.25$ .

6. Substituting: $3x + 2x - 3 = 7$ , $x = 2$ , $y = 1$ : solution $(2, 1)$ .

7. Adding: $7x = 7$ , $x = 1$ ; then $3y = 6$ , $y = 2$ : solution $(1, 2)$ .

8. First system: the second equation is twice the first: one line counted twice — infinitely many solutions (all points of $2x - y = 3$ ). Second system: same left sides proportional, right sides not ( $2 \times 3 = 6 \neq 1$ ): two parallel lines — no solution. Trichotomy: distinct slopes $\to$ one crossing; equal slopes, different intercepts $\to$ parallel, none; equal everything $\to$ same line, infinitely many.

9. The lines’ direction vectors are $(-b, a)$ and $(-d, c)$ (or: slopes $-\frac ab$ and $-\frac cd$ ), collinear exactly when $ad - bc = 0$ — the determinant of Problem 6.1 in a new job. Values: question 7: $2 \times (-3) - 3 \times 5 = -21 \neq 0$ : unique solution. Question 8: $2 \times (-2) - (-1) \times 4 = 0$ for both: parallel-or-identical, and the constants separate the two cases.

10. $ad - bc = 6 - 2m$ : zero for $m = 3$ . There $x + 3y = 3$ against $2x + 6y = 7$ : doubling the first gives $2x + 6y = 6 \neq 7$ : parallel, no solution. For $m \neq 3$ , elimination gives the unique solution $y = \frac{1}{6 - 2m}$ , then $x = 3 - \frac{m}{6 - 2m}$ .

11. $p + r = 35$ , $2p + 4r = 94$ : halving, $p + 2r = 47$ , so $r = 12$ and $p = 23$ : twenty-three pheasants, twelve rabbits — the Nine Chapters’ own answer.

12. $x + y = 30$ and $0.6x + 0.2y = 0.35 \times 30 = 10.5$ : subtracting $0.2 \times$ the first, $0.4x = 4.5$ : $x = 11.25$ L of syrup, $y = 18.75$ L of drink.

13. Digits $a$ , $b$ : $a + b = 12$ and $(10a + b) - (10b + a) = 9(a - b) = 18$ , so $a - b = 2$ : $a = 7$ , $b = 5$ : the number is $75$ .

14. Adding: $5c + 5r = 18$ , so $c + r = 3.60$ — one coffee plus one croissant. Subtracting: $c - r = 0.20$ . Hence $c = 1.90$ and $r = 1.70$ euros: the symmetric shortcut solved it without any substitution.

15. Doubling the first claim gives $4$ apples $+ 2$ pears $= 10$ euros, but the stall charges $9$ : the system is inconsistent — parallel lines, empty solution set. Verdict: the two price claims cannot both be right; either a discount is hiding, or the arithmetic is.

16. At the origin: $0 > 3 \times 0 - 1 = -1$ : true, so $(0,0)$ lies in the region $y > 3x - 1$ (above the line). Test any point by plugging it in: above/below according to the inequality’s truth.

17. Corners: $(0, 0)$ ; $(4, 0)$ (axes and $x + y = 4$ ); $(0, 1)$ (axis and $y = 2x + 1$ ); and $x + y = 4$ with $y = 2x + 1$ : $x = 1$ , $y = 3$ : $(1, 3)$ .

18. $P(0,0) = 0$ ; $P(4, 0) = 12$ ; $P(0, 1) = 2$ ; $P(1, 3) = 9$ . Maximum at the corner $(4, 0)$ : the plan $x = 4$ , $y = 0$ earns $12$ — corners rule, as the sweeping parallel lines $3x + 2y = c$ make visible.

19. $1.5x + 3 = 2x$ gives $x = 6$ , $y = 12$ : at six kilometers both taxis charge twelve euros — the break-even crossing of the two-thermometers weekend problem of the Middle School volume, now a system’s unique solution.

20. A linear equation’s solutions draw a line; a system superposes two lines, and $ad - bc$ tells at a glance whether they cross once (nonzero) or fall into the parallel-or-identical cases (zero). Pile up inequalities instead and the solutions form a polygonal region whose corners carry any linear optimum. Linear algebra generalizes the determinant to any number of unknowns; linear programming industrializes the corners.