---
title: "Equations of Lines and Linear Systems"
book: "High School Mathematics"
subject: math
language: en
chapter: 7
exercises: 10
source: https://one-course.com/books/math/2/en/chapter/7-equations-of-lines-and-linear-systems
---

# Chapter 7 — Equations of Lines and Linear Systems

A straight line in the coordinate plane is described by an [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation), and two lines together form a system of [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) whose solutions are their [intersection](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) points. This chapter connects three languages — geometric (lines), algebraic ([equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation)) and functional ([affine functions](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) from [Chapter 4](https://one-course.com/books/math/2/en/chapter/4-reference-functions#ch-g10-reffunc)) — and teaches the two standard ways of solving a [linear system](#def-g10-lines-system).

## 7.1 The equation of a line

**Theorem 7.1 (Reduced equation).**

In a [coordinate system](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system):

1. every non-vertical line has an [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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;
2. every vertical line has an [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $x = c$ ;
3. a point lies on the line exactly when its [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) satisfy the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) .

**Proof.** *Admitted at this level.* ∎

**Proposition 7.2 (Slope through two points).**

The [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) 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).](https://one-course.com/images/onecourse/chapters/math-2/g10-lines/fig-9dbde51774b7.svg)

*Three kinds of lines: slanted ($m \neq 0$), horizontal ($m = 0$), and vertical (no [reduced equation](#thm-g10-lines-reduced) — the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) is $x = c$).*

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

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

1. compute the [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) $m = \dfrac{y_B - y_A}{x_B - x_A}$ ;
2. write $y = mx + p$ and substitute the [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) of $A$ (or $B$ ) to find $p$ ;
3. check the final [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) with the *other* point.

If $x_A = x_B$, the line is vertical: its [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) is $x = x_A$.

**Example 7.4.**

Line through $A(1, 5)$ and $B(3, 1)$. [Slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine): $m = \frac{1 - 5}{3 - 1} = \frac{-4}{2} = -2$. Then $y = -2x + p$; the point $A$ gives $5 = -2 \times 1 + p$, so $p = 7$. [Equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine). Two distinct parallel lines have no common point; two lines with different [slopes](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) 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](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) point. If $m = m'$, the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) reads $0 = p' - p$: no solution when $p \neq p'$ (distinct parallels), every $x$ when $p = p'$ (same line). Parallelism (having no [intersection](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion), 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](https://one-course.com/books/math/2/en/chapter/6-vectors#def-g10-vectors-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](https://one-course.com/books/math/2/en/chapter/6-vectors#def-g10-vectors-vector) are [collinear](https://one-course.com/books/math/2/en/chapter/6-vectors#def-g10-vectors-collinear) ([Theorem 6.13](https://one-course.com/books/math/2/en/chapter/6-vectors#thm-g10-vectors-det)).

## 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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation)

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

to be satisfied *simultaneously*. A *solution* is a pair $(x, y)$ satisfying both. Each [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) describes a line, so solving the system means finding the [intersection](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) of two lines.

**Method 7.8 (Substitution).**

1. Solve one [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) for one unknown — pick the easiest, e.g. express $y$ in terms of $x$ ;
2. substitute this expression into the *other* [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) , which now contains only $x$ ;
3. solve for $x$ , then compute $y$ from step 1;
4. check the pair in both original [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) .

**Example 7.9.**

Solve $\begin{cases} x + 2y = 8 \\ 3x - y = 3 \end{cases}$ by substitution. The second [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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).**

1. Multiply each [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) by a well-chosen number so that one unknown gets opposite coefficients in the two [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) ;
2. add the [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) : that unknown disappears;
3. solve the remaining one-unknown [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) , then substitute back to find the other unknown;
4. check the pair in both original [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) .

**Example 7.11.**

Solve $\begin{cases} 2x + 3y = 7 \\ 5x - 2y = 8 \end{cases}$ by elimination. To eliminate $y$, multiply the first [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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.](https://one-course.com/images/onecourse/chapters/math-2/g10-lines/fig-f7a419123493.svg)

*A [linear system](#def-g10-lines-system) seen geometrically: each [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) is a line, and the solution $(2,1)$ of [Example 7.11](#ex-g10-lines-elimination) is their [intersection](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) point.*

**Remark 7.12 (How many solutions?).**

Like two lines, a [linear system](#def-g10-lines-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](https://one-course.com/books/math/2/en/chapter/6-vectors#thm-g10-vectors-det) 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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) and the $y$-intercept, and sketch it:

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

**Solution of Exercise 7.1.**

$y = 2x - 3$: [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) $2$, intercept $-3$. $y = -\frac13 x + 1$: [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) $-\frac13$, intercept $1$. $y = 4$: [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) $0$, intercept $4$ (horizontal line). $x = -2$: vertical line, no [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) and no [reduced equation](#thm-g10-lines-reduced).

**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 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](#thm-g10-lines-reduced) 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 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 $p = 2$: $y = -3x + 2$.

(c) $x_A = x_B = 4$: vertical line $x = 4$.

**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 of Exercise 7.4.**

$y = \frac{6x+2}{2} = 3x + 1$. The lines with [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) $3$ 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 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 of Exercise 7.6.**

First system: adding the [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) eliminates $y$: $2x = 14$, $x = 7$; then $7 + y = 10$ gives $y = 3$. Solution: $(7, 3)$.

Second system: multiply the first [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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 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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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 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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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 $A(1, 4)$.

1. Give the [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) of the line $d'$ parallel to $d$ passing through $A$ .
2. Compute the [intersection](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) point of $d'$ with the $x$ -axis.

**Solution of Exercise 7.9.**

*1.* Parallel means same [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) $2$: $y = 2x + p$ 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](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) point is $(-1, 0)$.

**Exercise 7.10 ★★★.**

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

1. Find the [equations](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) of the medians of the triangle $ABC$ issued from $A$ and from $B$ (a median joins a vertex to the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of the opposite side).
2. Compute their [intersection](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) point $G$ , and check that $G$ is also on the third median. (You should find that the [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) of $G$ are the averages of those of $A$ , $B$ , $C$ .)

**Solution of Exercise 7.10.**

*1.* [Midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of $[BC]$: $A' = \left(\frac{6+2}{2},
\frac{2+4}{2}\right) = (4, 3)$. The median from $A(0,0)$ has [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) $\frac{3 - 0}{4 - 0} = \frac34$: [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $y = \frac34 x$.

[Midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of $[AC]$: $B' = (1, 2)$. The median from $B(6, 2)$ has [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) $\frac{2 - 2}{1 - 6} = 0$: it is the horizontal line $y = 2$.

*2.* [Intersection](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion): $\frac34 x = 2$ gives $x = \frac83$, so $G\left(\frac83, 2\right)$. The third median joins $C(2,4)$ to the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of $[AB]$, $C' = (3, 1)$; its [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) is $\frac{1 - 4}{3 - 2} = -3$, [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) 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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation), two unknowns: the pair of lines they draw can cross once, never, or be one and the same line — and *every* [linear system](#def-g10-lines-system) inherits one of these three destinies. This problem classifies them, meets the number that decides (an old acquaintance from the [vectors](https://one-course.com/books/math/2/en/chapter/6-vectors#def-g10-vectors-vector) 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.**

1. Find the [reduced equation](#thm-g10-lines-reduced) of the line through $(1, 2)$ and $(3, 8)$ ( [Method 7.3](#met-g10-lines-through) ).
2. Among $y = 3x - 1$ , $y = 3x + 4$ and $y = -x + 7$ : which two lines are parallel ( [Proposition 7.5](#prop-g10-lines-parallel) ), and where do the non-parallel pairs meet? (Compute one [intersection](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-interunion) .)
3. The line through $(2, 1)$ and $(2, 5)$ : why does it have no [reduced equation](#thm-g10-lines-reduced) $y = mx + p$ , and what is its [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) ?
4. Is the point $(4, 11)$ on the line $y = 3x - 1$ ? And $(-1, -5)$ ?
5. The line through $(0, 3)$ with [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) $-2$ : give its [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) , its two axis intercepts, and the area of the triangle it cuts from the first quadrant.

**Part II — Three destinies.**

6. Solve by substitution ( [Method 7.8](#met-g10-lines-substitution) ): $\begin{cases} y = 2x - 3 \\ 3x + y = 7 \end{cases}$
7. Solve by elimination ( [Method 7.10](#met-g10-lines-elimination) ): $\begin{cases} 2x + 3y = 8 \\ 5x - 3y = -1 \end{cases}$
8. 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.
9. 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](https://one-course.com/books/math/2/en/chapter/6-vectors#pb-g10-vectors-1) . Explain the coincidence (which two [vectors](https://one-course.com/books/math/2/en/chapter/6-vectors#def-g10-vectors-vector) are [collinear](https://one-course.com/books/math/2/en/chapter/6-vectors#def-g10-vectors-collinear) exactly when the lines are parallel or identical?), and evaluate $ad - bc$ for the three systems of questions 7 and 8.
10. 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.**

11. 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?
12. 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?
13. 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.
14. 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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) and *subtracting* them each tell you directly?
15. 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.**

16. 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.
17. 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.
18. A workshop’s profit is $P = 3x + 2y$ on the region of question 17. Evaluate $P$ at the four corners. Admitting that the [maximum](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) 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.
19. 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.
20. Finale, the dictionary: one [equation](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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 of Problem 7.1.**

**1.** [Slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) $\frac{8 - 2}{3 - 1} = 3$; through $(1, 2)$: $y = 3x - 1$.

**2.** $y = 3x - 1$ and $y = 3x + 4$: same [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine), 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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-equation) $x = 2$. A reduced form $y = mx + p$ answers “one $y$ per $x$”, which a vertical line refuses: its [slope](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) 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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) $\to$ one crossing; equal [slopes](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine), different intercepts $\to$ parallel, none; equal everything $\to$ same line, infinitely many.

**9.** The lines’ direction [vectors](https://one-course.com/books/math/2/en/chapter/6-vectors#def-g10-vectors-vector) are $(-b, a)$ and $(-d, c)$ (or: [slopes](https://one-course.com/books/math/2/en/chapter/4-reference-functions#def-g10-reffunc-affine) $-\frac ab$ and $-\frac cd$), [collinear](https://one-course.com/books/math/2/en/chapter/6-vectors#def-g10-vectors-collinear) exactly when $ad - bc = 0$ — the determinant of [Problem 6.1](https://one-course.com/books/math/2/en/chapter/6-vectors#pb-g10-vectors-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](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-extrema) 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](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-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.