First-degree equations
Learning Objectives
- Solve a first-degree equation \(ax + b = 0\) and equations that reduce to it.
- Use equivalent transformations (adding to both sides, multiplying by a non-zero number) to isolate the unknown.
- Carry out the discussion of \(ax + b = 0\) by cases on \(a\) and \(b\).
- Recognise when an equation has exactly one solution, none, or every real number as a solution.
From the zero of a function to an equation
In Lesson 3 we found the zero of a linear function: the input where \(ax + b = 0\), namely \(x = -\tfrac{b}{a}\). Reading that statement from the other side, we were solving an equation. We now study first-degree equations in their own right: how to solve ones that do not arrive already in the form \(ax + b = 0\), and what happens to the coefficient \(a\) when it is allowed to be a parameter that might vanish.
First-degree equation
A first-degree equation in the unknown \(x\) is an equation that can be written in the form
\[ax + b = 0,\]
with \(a, b \in \mathbb{R}\). A solution is a real number that, put in place of \(x\), makes the equality true; the solution set is the set of all solutions. To solve the equation is to determine that set.
When \(a \neq 0\) the equation is genuinely of the first degree — the case tied to the linear function. We will see that allowing \(a = 0\) changes the story, which is exactly why the discussion below is needed.
Equivalent equations
Two equations are equivalent if they have the same solution set. We solve an equation by replacing it with simpler equivalent ones, using two transformations that never change the solution set:
add or subtract the same quantity on both sides;
multiply or divide both sides by the same non-zero number.
The word "non-zero" is essential: multiplying both sides by \(0\) turns every equation into \(0 = 0\) and destroys all information.
Equations that reduce to first degree
Many equations become \(ax + b = 0\) after tidying up.
Collecting the unknown. \(3x + 5 = 2x - 1\). Subtract \(2x\) from both sides: \(x + 5 = -1\). Subtract \(5\): \(x = -6\). Solution set \(\{-6\}\).
Expanding brackets. \(4(x - 1) = 2x + 6\). Expand: \(4x - 4 = 2x + 6\). Then \(2x = 10\), so \(x = 5\).
Clearing fractions. \(\tfrac{x}{3} + 2 = \tfrac{x}{2}\). Multiply both sides by \(6\) (non-zero!): \(2x + 12 = 3x\), hence \(x = 12\). Multiplying by the common denominator is the "multiply by a non-zero number" rule at work.
Discussion of \(ax + b = 0\)
Treat \(a\) and \(b\) as given real numbers. The solution set of \(ax + b = 0\) depends on them as follows:
| Case | The equation becomes | Solutions |
|---|---|---|
| \(a \neq 0\) | \(x = -\tfrac{b}{a}\) | exactly one: \(\left\{-\tfrac{b}{a}\right\}\) |
| \(a = 0,\; b \neq 0\) | \(b = 0\) (false) | none: \(\varnothing\) |
| \(a = 0,\; b = 0\) | \(0 = 0\) (true) | every real: \(\mathbb{R}\) |
If \(a \neq 0\), we may divide by \(a\): the equation \(ax = -b\) is equivalent to \(x = -\tfrac{b}{a}\). This single value is the one and only solution.
If \(a = 0\), the term \(ax\) vanishes and the equation reads \(b = 0\), which no longer contains \(x\). Either \(b \neq 0\), and the statement \(b = 0\) is false — no value of \(x\) can rescue it, so there is no solution; or \(b = 0\), and the statement \(0 = 0\) is true whatever \(x\) is, so every real number is a solution.
Discussing equations with a parameter
A parameter in the coefficient. Solve \(mx - 6 = 0\), where \(m\) is a real parameter. Here the coefficient of \(x\) is \(a = m\).
If \(m \neq 0\): \(x = \tfrac{6}{m}\) — one solution.
If \(m = 0\): the equation is \(-6 = 0\), which is false — no solution.
The all-reals case. Solve \((m - 3)x = m - 3\). Bring everything to one side: \((m - 3)x - (m - 3) = 0\), so the coefficient of \(x\) is \(a = m - 3\) and the constant is \(b = -(m - 3)\).
If \(m \neq 3\): divide by \(m - 3\) to get \(x = 1\) — one solution.
If \(m = 3\): the equation becomes \(0 = 0\), true for every \(x\) — every real number is a solution.
Connection to Computer Science
A program that solves \(ax + b = 0\) computes \(x = -b/a\) — but only after checking that \(a \neq 0\), because dividing by zero is undefined. The two situations it must handle separately, \(a = 0\) with \(b \neq 0\) (report "no solution") and \(a = 0\) with \(b = 0\) (report "every number works"), are precisely the degenerate cases of the discussion above. The mathematics and the code guard against the same thing.
"Solving" can mean none, one, or all
It is tempting to think that solving an equation always means pinning the unknown down to a single number. The discussion shows otherwise: a first-degree equation can have exactly one solution, no solution, or every real number as a solution.
The two extreme outcomes are two faces of the same fact. The equation \(0 = 0\) (from \(a = b = 0\)) is an identity — true whatever \(x\) is, so the solution set is all of \(\mathbb{R}\). The equation \(b = 0\) with \(b \neq 0\) (say \(5 = 0\)) is a contradiction — true for no \(x\), so the solution set is empty.
Notice that both unusual cases require \(a = 0\). A genuine first-degree equation, with \(a \neq 0\), always has exactly one solution. The surprises live precisely where the equation stops being first-degree.
Exercises
Exercise 1
Solve each equation for \(x\).
Exercise 2
For each equation, decide how many solutions it has.
Exercise 3
Consider \((2m - 4)x + 6 = 0\), where \(m\) is a real parameter.
Exercise 4
Decide whether each statement about first-degree equations is true or false.
Exercise 5
Decide whether each equation has exactly one solution.
Summary
A first-degree equation can be written \(ax + b = 0\). We solve it by equivalent transformations — adding the same quantity to both sides, or multiplying both sides by a non-zero number — until \(x\) is isolated.
Discussion of \(ax + b = 0\): if \(a \neq 0\), one solution \(x = -\tfrac{b}{a}\); if \(a = 0\) and \(b \neq 0\), no solution; if \(a = 0\) and \(b = 0\), every real number.
The empty and all-reals cases are a contradiction \((b = 0\text{ false})\) and an identity \((0 = 0)\) respectively, and both require \(a = 0\). A genuine first-degree equation always has exactly one solution.