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Chapter 2

Linear Equations & Systems

Solving linear equations and systems of linear equations in one or more variables.

Introduction

Solving linear equations and systems of linear equations in one or more variables.

A linear equation in one variable is the simplest nontrivial equation type in algebra. "Linear" means the variable appears only to the first power — no $x^{2}$ , no $\frac{1}{x}$ , no $2^{x}$ . It can always be written as $a x + b = 0$ with $a \neq = 0$ , and it has exactly one solution, namely $x = - \frac{b}{a}$ . (The condition $a \neq = 0$ matters: if $a = 0$ the $x$ disappears and it is no longer a linear equation in $x$ .)

Linear equations model countless real-world situations where change happens at a constant rate: distance at constant speed, cost with a fixed fee plus per-unit charge, temperature conversions, and many more.

2.1Standard Form and Equivalent Forms

Any linear equation $a x + b = 0$ can be rearranged into several useful forms.

Notation (Forms of a line).

Standard form: $a x + b y = c$ (for two variables, with $a$ and $b$ not both zero).
Slope-intercept: $y = m x + b$ .
Point-slope: $y - y_{1} = m (x - x_{1})$ .

All represent the same straight-line relationship — they are just different ways of writing the same set of points. You pick whichever form makes the task in front of you easiest: standard form for lining up systems, slope-intercept for graphing, point-slope when you know one point and the slope.

Intuition. The forms are like the same phone number written with or without dashes: the number never changes, only the layout.

Standard form is the "neat filing cabinet" version with all terms on one side.

Slope-intercept is the "ready to graph" version that tells you the tilt and where it crosses the y-axis.

Example 2.1. Rewrite $3 x - 6 = 0$ into slope-free solved form.

Solution.

Add $6$ to both sides: $3 x = 6$ .
Divide by $3$ : $x = 2$ .
Sanity check: $3 (2) - 6 = 6 - 6 = 0$ . ✓

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Example 2.2. Put the two-variable line $3 x - y = 6$ (standard form) into slope-intercept form.

Solution.

Isolate the $y$ term by subtracting $3 x$ : $- y = - 3 x + 6$ .
Multiply every term by $- 1$ so $y$ is positive: $y = 3 x - 6$ .
Now read off the slope $m = 3$ and the y-intercept $b = - 6$ , so the line crosses the vertical axis at $(0, - 6)$ .
Sanity check: at $x = 0$ , standard form gives $- y = 6$ , i.e. $y = - 6$ , matching the intercept. ✓

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Exercises for §2.1 →

2.2Solving Linear Equations

Proposition 2.3 (A linear equation has exactly one solution). If $a \neq = 0$ , the equation $a x + b = 0$ has exactly one solution, namely

x = - \frac{b}{a} .

The goal is to find it by isolating the variable using inverse operations while preserving equality. The one rule behind every step: whatever you do to one side of the equation you must do to the other, so the two sides stay equal. That rule is what keeps the solution unchanged.

Method 2.4 (Solving a linear equation).

Simplify both sides (distribute, combine like terms).
Move all terms containing the variable to one side.
Move constants to the other side.
Divide by the coefficient of the variable (this is valid because the coefficient of $x$ is nonzero in a genuine linear equation).

Intuition. Each step peels away one layer around the variable, exactly like unwrapping a gift without tearing the prize inside.

The balance-scale picture helps: whatever you do to one pan you must do to the other, or the scale tips and the statement stops being true.

Example 2.5. Solve $4 (x - 2) = 3 x + 5$ .

Solution.

Distribute the $4$ across the parentheses: $4 x - 8 = 3 x + 5$ .
Subtract $3 x$ from both sides to gather the variable on the left: $x - 8 = 5$ .
Add $8$ to both sides to isolate $x$ : $x = 13$ .
Sanity check: left side $4 (13 - 2) = 4 (11) = 44$ ; right side $3 (13) + 5 = 39 + 5 = 44$ . Both equal $44$ . ✓

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Example 2.6. Solve $3 (x + 2) = 4 x - 5$ , a case where the variable ends up on the right.

Solution.

Distribute: $3 x + 6 = 4 x - 5$ .
Subtract $3 x$ from both sides: $6 = x - 5$ .
Add $5$ to both sides: $x = 11$ .
Sanity check: left side $3 (11 + 2) = 3 (13) = 39$ ; right side $4 (11) - 5 = 44 - 5 = 39$ . ✓

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Exercises for §2.2 →

2.3Slope-Intercept Form

Definition 2.7 (Slope-intercept form). A non-vertical line solved for $y$ reads

y = m x + b,

where $m$ is the slope (rise over run: how much $y$ changes for a one-unit increase in $x$ ) and $b$ is the y-intercept (the $y$ -value where the line crosses the vertical axis, i.e. where $x = 0$ ).

This form is only available for non-vertical lines; a vertical line like $x = 4$ has undefined slope and cannot be written as $y = m x + b$ .

This form is ideal for graphing by hand: plot the intercept, then use the slope to find a second point.

Intuition. Slope $m$ is the "climb rate" of a staircase: for every step right, you rise $m$ (or fall, if $m$ is negative).

$b$ tells the "starting height" when $x = 0$ — where you begin before taking any steps.

Example 2.8. Graph $y = - 2 x + 4$ and find where it meets the axes.

Solution.

Read off the parts: slope $m = - 2$ (down $2$ for every $1$ to the right) and y-intercept $b = 4$ , so start by plotting $(0, 4)$ .
Use the slope to get a second point: from $(0, 4)$ go right $1$ and down $2$ to reach $(1, 2)$ . Going right once more lands on $(2, 0)$ .
Find the x-intercept by setting $y = 0$ : $0 = - 2 x + 4 \Rightarrow 2 x = 4 \Rightarrow x = 2$ , confirming the point $(2, 0)$ .
Sanity check: at $x = 1$ , $y = - 2 (1) + 4 = 2$ , matching the plotted point $(1, 2)$ . ✓

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Example 2.9. Write the equation of the line with slope $2$ and y-intercept $- 3$ .

Solution.

Substitute $m = 2$ and $b = - 3$ directly into $y = m x + b$ : $y = 2 x - 3$ .
Sanity check: this line should hit the y-axis at $- 3$ . At $x = 0$ , $y = 2 (0) - 3 = - 3$ . ✓

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Exercises for §2.3 →

2.4Applications of Linear Equations

Linear equations arise whenever a quantity changes at a constant rate — the "per unit" amount never changes as the quantity grows.

Cost $=$ fixed fee $+$ rate $\times$ quantity
Distance $=$ speed $\times$ time (when speed is constant)
Temperature conversion: $F = \frac{9}{5} C + 32$

Translate the English carefully into symbols, solve, and interpret the number in context. The fixed/one-time amount becomes the constant term; the "per unit" amount becomes the coefficient of the variable.

Intuition. Many everyday "if this, then how much?" questions are secretly linear equations in disguise.

Spot the two pieces: the fee you pay no matter what (the constant) and the amount that piles on per unit (the rate). Once you write the equation the algebra does the arithmetic for you.

Example 2.10. A taxi charges $$3$ to get in plus $$2.50$ per mile. How far can you ride for $$15.50$ ?

Solution.

Let $m$ be the miles. Fixed fee $3$ plus rate $2.5$ per mile equals the total: $3 + 2.5 m = 15.5$ .
Subtract the fixed fee: $2.5 m = 12.5$ .
Divide by the rate: $m = 5$ miles.
Sanity check: $3 + 2.5 (5) = 3 + 12.5 = 15.5$ . ✓

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Example 2.11. A phone plan costs $$10$ per month plus $$0.05$ per minute. Find the cost of $120$ minutes, then the equation that models total cost $C$ for $m$ minutes.

Solution.

Identify the pieces: fixed monthly cost $$10$ (constant) and rate $$0.05$ per minute (coefficient of $m$ ), so $C = 0.05 m + 10$ .
Substitute $m = 120$ : $C = 0.05 (120) + 10 = 6 + 10 = 16$ .
Sanity check: the units work — $0.05 \frac{dollars}{minute} \times 120 minutes = 6$ dollars, plus the $$10$ fixed fee gives $$16$ . ✓

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Exercises for §2.4 →

Common Mistakes

Distributing incorrectly across parentheses: 3(x − 2) = 3x − 6, never 3x − 2.
Moving a term across the equals sign without flipping its sign: x − 4 = 7 becomes x = 7 + 4, not x = 7 − 4.
Dividing only one side by the coefficient: Both sides must be divided.
Forgetting to check the solution: Especially after distributing or combining terms.
Assuming every linear equation must be written ax + b = 0 before solving: Any equivalent form can be solved directly.

A system of linear equations is a set of two or more equations that must be satisfied simultaneously by the same values of the variables. "Simultaneously" is the key word: a single pair of numbers has to make every equation true at once.

The solution is the intersection point (or points) of the lines (or planes) represented by the equations. In two variables the three possibilities are exactly one solution (lines cross once), no solution (parallel distinct lines), or infinitely many solutions (the two equations describe the same line).

2.5Definition and Geometric Meaning

Definition 2.12 (Solution of a system). Each equation in the system graphs as a straight line. A pair $(x, y)$ solves the system only if the point lies on every line in the system — being on just one line is not enough.

The number of solutions then depends on how the lines sit relative to each other.

Proposition 2.13 (How many solutions a two-variable system can have). Consistent systems have at least one solution; within these, if the lines have different slopes they cross exactly once (one solution), and if they are the same line they overlap everywhere (dependent, infinitely many solutions). Inconsistent systems have parallel-but-distinct lines (same slope, different intercept) that never meet, so there are no solutions.

Intuition. Think of two roads that cross at one town: that town is the solution.

If the roads are parallel they never meet (no solution). If they are the exact same road, every point on the road works (infinitely many).

Example 2.14. Verify that the system $x + y = 5$ and $x - y = 1$ has the unique solution $(3, 2)$ .

Solution.

The two lines have different slopes ( $y = - x + 5$ has slope $- 1$ ; $y = x - 1$ has slope $1$ ), so exactly one intersection exists.
Check the point in the first equation: $3 + 2 = 5$ . ✓
Check the point in the second equation: $3 - 2 = 1$ . ✓
Because $(3, 2)$ satisfies both equations, it is the solution — and since the slopes differ, it is the only one.

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Exercises for §2.5 →

2.6Solving by Substitution

Substitution turns a two-variable system into a single one-variable equation you already know how to solve. The trick is to express one variable in terms of the other, then replace it.

Method 2.15 (Solving a system by substitution).

Solve one equation for one variable (choose the easiest — ideally a variable with coefficient $1$ ).
Substitute that expression into the remaining equation(s).
Solve the resulting equation in one variable.
Back-substitute to find the other variable(s).

Intuition. Solve for one "mystery box" and drop the expression into the other equation like a replacement part.

You reduce the number of unknowns step by step: two equations with two unknowns become one equation with one unknown.

Example 2.16. Solve $y = 3 x - 1$ and $2 x + y = 9$ .

Solution.

The first equation is already solved for $y$ , so substitute $3 x - 1$ in place of $y$ in the second: $2 x + (3 x - 1) = 9$ .
Combine like terms: $5 x - 1 = 9$ , so $5 x = 10$ and $x = 2$ .
Back-substitute into $y = 3 x - 1$ : $y = 3 (2) - 1 = 5$ . Solution $(2, 5)$ .
Sanity check in the second equation: $2 (2) + 5 = 4 + 5 = 9$ . ✓

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Example 2.17. Solve $y = 2 x + 1$ and $x + y = 7$ , where you substitute into an equation that also contains $x$ .

Solution.

Replace $y$ in the second equation with $2 x + 1$ : $x + (2 x + 1) = 7$ .
Combine: $3 x + 1 = 7$ , so $3 x = 6$ and $x = 2$ .
Back-substitute: $y = 2 (2) + 1 = 5$ . Solution $(2, 5)$ .
Sanity check: $x + y = 2 + 5 = 7$ . ✓

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Exercises for §2.6 →

2.7Solving by Elimination

Elimination cancels one variable by combining the two equations, so you never have to isolate a variable first. It shines when the coefficients already line up or are easy to match.

Method 2.18 (Solving a system by elimination).

Multiply one or both equations by constants so that the coefficients of one variable become opposites (or equal). Remember to multiply every term, including the constant on the right.
Add the equations (or subtract, if you made them equal); that variable cancels.
Solve for the remaining variable.
Back-substitute.

Intuition. Line up the like terms and make the numbers in front of one variable cancel when you add, like removing one color of blocks from two piles.

What remains is a simpler equation.

Example 2.19. Solve $3 x + 2 y = 7$ and $5 x - 2 y = 1$ , where the $y$ -coefficients are already opposites.

Solution.

Add the two equations term by term so the $y$ terms cancel: $(3 x + 5 x) + (2 y - 2 y) = 7 + 1$ , giving $8 x = 8$ , so $x = 1$ .
Back-substitute into the first equation: $3 (1) + 2 y = 7 \Rightarrow 2 y = 4 \Rightarrow y = 2$ . Solution $(1, 2)$ .
Sanity check in the second equation: $5 (1) - 2 (2) = 5 - 4 = 1$ . ✓

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Example 2.20. Solve $2 x + 3 y = 7$ and $2 x - y = 1$ , where you must eliminate by subtracting because the $x$ -coefficients are equal.

Solution.

Subtract the second equation from the first so the $x$ terms cancel: $(2 x - 2 x) + (3 y - (- y)) = 7 - 1$ , giving $4 y = 6$ , so $y = \frac{6}{4} = \frac{3}{2}$ .
Back-substitute into $2 x - y = 1$ : $2 x - \frac{3}{2} = 1 \Rightarrow 2 x = \frac{5}{2} \Rightarrow x = \frac{5}{4}$ .
Sanity check in the first equation: $2 (\frac{5}{4}) + 3 (\frac{3}{2}) = \frac{5}{2} + \frac{9}{2} = \frac{14}{2} = 7$ . ✓

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Exercises for §2.7 →

2.8Applications

Systems appear whenever two conditions must be satisfied at once: "the sum is 15 and the difference is 3", "cost and revenue both depend on quantity and we want break-even", "two distances add to a total with different speeds." As a rule of thumb, two unknowns need two independent facts to pin them down.

Translate each English sentence into an equation, solve the system, and check that the numbers make sense together.

Intuition. Two facts about the same pair of numbers give two equations. Solving finds the unique pair that satisfies both facts.

One clue about two unknowns is not enough; a second clue narrows it to a single answer, like two witnesses locking down one suspect.

Example 2.21. The sum of two numbers is $15$ ; their difference is $3$ . Let the larger be $x$ and the smaller be $y$ .

Solution.

Translate the two facts into equations:
$x + y = 15$
$x - y = 3$
Eliminate $y$ by adding the equations: $2 x = 18$ , so $x = 9$ .
Back-substitute into $x + y = 15$ : $9 + y = 15$ , so $y = 6$ .
Sanity check: sum $9 + 6 = 15$ ✓ and difference $9 - 6 = 3$ ✓. The larger number is $9$ .

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Example 2.22. A theater sells adult tickets at $$8$ and child tickets at $$5$ . It sells $200$ tickets for a total of $$1300$ . How many of each?

Solution.

Let $a$ be adult tickets and $c$ child tickets. The count and the money give two equations:
$a + c = 200$
$8 a + 5 c = 1300$
Solve the first for $c$ : $c = 200 - a$ , then substitute into the second: $8 a + 5 (200 - a) = 1300$ .
Expand: $8 a + 1000 - 5 a = 1300$ , so $3 a = 300$ and $a = 100$ .
Back-substitute: $c = 200 - 100 = 100$ .
Sanity check: revenue $8 (100) + 5 (100) = 800 + 500 = 1300$ ✓, and $100 + 100 = 200$ tickets ✓.

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Exercises for §2.8 →

Common Mistakes

Solving only one equation and forgetting the second: The pair must work in every equation of the system.
Arithmetic sign errors during elimination: A minus sign mistake when subtracting equations ruins the cancellation.
Not multiplying both sides of an equation when scaling: You must multiply every term, including the constant.
Assuming every system has exactly one solution: Check for parallel or identical lines (same slope and intercept).
Mixing up the variables after back-substitution: Label clearly which number belongs to which variable.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

Practice this chapter