Linear Functions and Systems of Equations

A linear function is the simplest non-constant function you will meet, yet it is the backbone of almost every later topic: rates of change, tangent lines, approximations, and the simplest possible model relating two quantities. This chapter builds the linear function from its defining property, develops the standard ways of writing its equation, and then extends to systems of two or three linear equations — the algebra needed whenever several linear conditions must hold simultaneously.

You need confident algebraic manipulation and the coordinate plane from earlier study. Nothing here requires calculus, though the idea of "rate of change" previewed here becomes the derivative in Block D.

1. The Defining Property: Constant Rate of Change

A function $f$ is linear if it can be written $f(x) = mx+c$ for constants $m,c$. Its defining geometric property is that the gradient (slope) between any two points on its graph is always the same number:

This is what "linear" means: equal changes in $x$ always produce equal changes in $y$, regardless of where on the line you measure. $c=f(0)$ is the $y$-intercept, the value where the line crosses the $y$-axis.

Definition. The gradient $m$ of a line through $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\dfrac{y_2-y_1}{x_2-x_1}$ (with $x_1\neq x_2$). A vertical line has no defined gradient — it is not the graph of a function at all, since it fails the vertical line test.

2. Forms of a Linear Equation

Three equivalent forms of a linear equation are used, each convenient in different circumstances.

Slope-intercept form: $y=mx+c$ — immediately shows the gradient and $y$-intercept; best for graphing quickly.

Point-slope form: $y-y_1=m(x-x_1)$ — built directly from a known gradient $m$ and one known point $(x_1,y_1)$; best when you are given a point and a gradient, rather than two points.

General form: $ax+by+c=0$ — treats $x$ and $y$ symmetrically; useful for systems of equations (Sections 4–5) and for lines with undefined gradient ($b=0$, a vertical line).

Worked example 1. Find the equation, in slope-intercept form, of the line through $(2,5)$ and $(6,-3)$.

Using point-slope form with $(2,5)$: $y-5=-2(x-2) \implies y-5=-2x+4 \implies y=-2x+9$.

3. Parallel and Perpendicular Lines

Key distinction. Two non-vertical lines are parallel exactly when their gradients are equal: $m_1=m_2$. They are perpendicular exactly when the product of their gradients is $-1$: $m_1 m_2 = -1$, equivalently $m_2 = -\dfrac{1}{m_1}$.

The perpendicularity condition is not arbitrary: rotating a line by $90°$ turns a "rise of $m$ per unit run" into a "run of $m$ per unit rise in the opposite direction," which is exactly the negative reciprocal relationship.

Worked example 2. Find the equation of the line through $(4,1)$ perpendicular to $y=3x-2$.

The given line has gradient $3$, so the perpendicular gradient is $-\dfrac13$. Using point-slope form:

4. Solving a Linear System by Substitution and Elimination

A system of linear equations is a set of equations that must hold simultaneously; a solution is any point(s) satisfying all of them at once. For two equations in two unknowns, this is geometrically the intersection point of two lines.

Substitution solves one equation for one variable and substitutes that expression into the other equation.

Worked example 3. Solve $\begin{cases} 2x+y=11 \\ x-y=1\end{cases}$ by substitution.

From the second equation, $y=x-1$. Substituting into the first:

Elimination adds or subtracts multiples of the equations to cancel a variable directly.

Worked example 4. Solve $\begin{cases} 3x+2y=16 \\ 5x-2y=8 \end{cases}$ by elimination.

The $y$-coefficients are already opposite, so adding the equations eliminates $y$ directly:

Substituting back: $3(3)+2y=16 \implies 2y=7 \implies y=3.5$.

5. The Three Cases: Unique Solution, No Solution, Infinitely Many

A system of two linear equations in two unknowns falls into exactly one of three cases, determined by comparing the gradients (and intercepts) of the two lines.

Definition. Writing both equations as $y=m_1x+c_1$ and $y=m_2x+c_2$: if $m_1\neq m_2$, the lines meet at exactly one point (a unique solution). If $m_1=m_2$ and $c_1\neq c_2$, the lines are parallel and distinct, giving no solution (an inconsistent system). If $m_1=m_2$ and $c_1=c_2$, the two equations describe the same line, giving infinitely many solutions (a dependent system).

Worked example 5 (the edge case). Solve $\begin{cases} 4x-2y=6 \\ -6x+3y=-9\end{cases}$.

Rewriting: the first gives $y=2x-3$; the second gives $3y=6x-9 \implies y=2x-3$ as well — identical lines. Algebraically, multiplying the first equation by $-\tfrac32$ gives exactly the second equation, confirming the system is dependent with infinitely many solutions, all of the form $(t, 2t-3)$ for any real $t$. A student who only checks that elimination "doesn't immediately fail" might miss that the entire equation has vanished (e.g. subtracting $-1.5\times$ the first from the second gives $0=0$), which is the signature of a dependent system rather than an error.

6. Systems of Three Linear Equations in Three Unknowns

The same elimination idea extends to three equations in $x,y,z$: eliminate one variable between pairs of equations to reduce to a $2\times2$ system, solve that, then back-substitute.

Worked example 6. Solve:

To eliminate $z$, add equations (1) and (3): $(x+y+z)+(x+2y-z)=6+5 \implies 2x+3y=11$. Eliminate $z$ a second time by subtracting equation (2) from equation (1): $(x+y+z)-(2x-y+z)=6-3 \implies -x+2y=3$. These two results form a genuine $2\times2$ system in $x$ and $y$ alone, solvable by the same substitution or elimination methods from Section 4. Rearranging the second equation gives $x=2y-3$; substituting into the first, $2(2y-3)+3y=11 \implies 7y=17 \implies y=\tfrac{17}{7}$. Back-substituting, $x=2\left(\tfrac{17}{7}\right)-3=\tfrac{13}{7}$, and from equation (1), $z=6-x-y=6-\tfrac{13}{7}-\tfrac{17}{7}=\tfrac{12}{7}$. As this example shows, a perfectly correct elimination can still produce fractional values — always re-check the method by substituting all three results back into all three original equations, rather than expecting "nice" integers as a sign of correctness.

Common Exam Pitfalls

• Computing the gradient as $\dfrac{x_2-x_1}{y_2-y_1}$ (upside down). Gradient is rise over run: change in $y$ divided by change in $x$, never the reverse.
• Using $m_2=m_1$ for perpendicular lines instead of $m_2=-\dfrac{1}{m_1}$. Equal gradients give parallel lines; perpendicularity requires the negative reciprocal.
• Declaring "no solution" the moment elimination produces $0=0$. This signature means the system is dependent (infinitely many solutions), the opposite conclusion from "no solution" (which instead gives a false statement like $0=5$).
• Forgetting to substitute back and solve for the second variable after finding the first one in a system — a system isn't solved until every unknown has a numerical value.
• Sign errors when multiplying an entire equation by a negative number during elimination — multiply every single term, including the constant on the right-hand side.
• Assuming a vertical line has gradient $0$. A vertical line has an undefined gradient (division by zero in the slope formula); a horizontal line is the one with gradient $0$.

Summary

• A linear function $f(x)=mx+c$ has constant gradient $m=\dfrac{y_2-y_1}{x_2-x_1}$ between any two points, and $y$-intercept $c$.
• Three equivalent forms: slope-intercept $y=mx+c$, point-slope $y-y_1=m(x-x_1)$, general form $ax+by+c=0$.
• Parallel lines share the same gradient ($m_1=m_2$); perpendicular lines have gradients multiplying to $-1$ ($m_1m_2=-1$).
• A $2\times2$ linear system is solved by substitution or elimination, giving exactly one of: a unique solution ($m_1\neq m_2$), no solution ($m_1=m_2$, $c_1\neq c_2$), or infinitely many solutions ($m_1=m_2$, $c_1=c_2$).
• Eliminating a variable and reaching $0=0$ signals a dependent (infinitely many solutions) system; reaching a false numerical statement (e.g. $0=5$) signals an inconsistent (no solution) system.
• A $3\times3$ system is reduced to a $2\times2$ system by eliminating one variable between pairs of equations, then solved and back-substituted, with a final check against all three original equations.