Linear Equations in 2-Space and 3-Space

You’ve been graphing equations like $2x + y = 7$ for years: the solutions form a straight line. In this unit you’ll see what happens when you add a third variable. The same kind of equation now describes a flat plane in 3-space, and two planes usually meet in a line. This page builds the pictures you’ll need for the rest of the unit on lines and planes.

Key ideas

One equation in two variables: a line

An equation of the form $Ax + By = C$ (with $A$ and $B$ not both zero) is a linear equation in two variables. Every solution is an ordered pair $(x, y)$, and when you plot all of them in 2-space (the $xy$-plane), you get a straight line.

A point is on the line exactly when its coordinates make the equation true. For example, $(1, 5)$ is on $2x + y = 7$ because $2(1) + 5 = 7$.

Two equations in two variables: three possibilities

A system of two linear equations asks for the points that are on both lines. Two lines in a plane can meet in three ways:

How the lines sit	Number of solutions	What you notice in the equations
They cross	exactly one point	$x$ and $y$ coefficients are not in the same ratio
Parallel and distinct	none	$x$ and $y$ coefficients in the same ratio, constant term not
Coincident (the same line)	infinitely many	the whole equation is a multiple of the other

For example, in $3x - 6y = 9$ and $x - 2y = 5$, the left sides are in the ratio $3 : 1$, but $9$ is not $3 \times 5$. Same slope, different lines: parallel, no solution.

Two lines in 2-space meet in one point, in no points, or in every point.

One equation in three variables: a plane

In 3-space, each point has three coordinates $(x, y, z)$ (see vectors in 3-space). An equation of the form

$$Ax + By + Cz = D \qquad (A, B, C \text{ not all zero})$$

is a linear equation in three variables, and its solution points form a plane: a flat surface that goes on forever in every direction.

The same equation can mean different things depending on the space you’re in. That’s the big idea of this page:

Equation	Solutions in 2-space	Solutions in 3-space
$x = 0$	the $y$-axis (a line)	the $yz$-plane
$y = 0$	the $x$-axis (a line)	the $xz$-plane
$y = x$	a line through the origin	a vertical plane containing the $z$-axis
$z = 5$	(no $z$ in 2-space)	a horizontal plane 5 units above the $xy$-plane
$x + z = 2$	(no $z$ in 2-space)	a plane parallel to the $y$-axis

Why is $x = 0$ a whole plane in 3-space? The equation only restricts $x$. The other coordinates, $y$ and $z$, can be anything, so every point $(0, y, z)$ is a solution, and those points fill the $yz$-plane.

The same reasoning gives a handy rule: if a variable is missing from the equation, the plane is parallel to that variable’s axis. In $x + z = 2$, $y$ is missing, so you can slide any solution point in the $y$-direction and it stays a solution.

The plane $x + z = 2$ contains the line $x + z = 2$ in the $xz$-plane, slid along the $y$-axis.

Sketching a plane with intercepts

To sketch a plane like $2x + 3y + 6z = 12$, find where it crosses each axis by setting the other two variables to $0$:

• $x$-intercept: $2x = 12$, so $(6, 0, 0)$
• $y$-intercept: $3y = 12$, so $(0, 4, 0)$
• $z$-intercept: $6z = 12$, so $(0, 0, 2)$

Join the three intercepts to get a triangle. That triangle is the part of the plane in the first octant, and it shows how the plane is tilted.

Two equations in three variables

A system of two linear equations in $x$, $y$ and $z$ asks for the points on both planes. Two planes can:

• intersect in a line (the usual case),
• be parallel and distinct: no solution, or
• be coincident (the same plane): infinitely many solutions, a whole plane of them.

You can spot the last two the same way as in 2-space: compare the coefficients of $x$, $y$ and $z$. If they are in the same ratio, the planes are parallel; if the constant terms are in that ratio too, the planes are the same.

Notice what doesn’t happen: two equations in three unknowns never pin down a single point. You’ll always have at least one “free” variable left over. To find the line of intersection, let one variable be a parameter $t$ (any real number) and solve for the other two in terms of $t$. You’ll learn to write these lines as vector equations in lines in 3-space.

Worked examples

Example 1: Two lines that cross

Solve the system and describe it geometrically.

$$2x + y = 7 \qquad x - 3y = 7$$

Solution. Multiply the first equation by $3$ and add the second to eliminate $y$:

$$\begin{aligned} 6x + 3y &= 21 \\ x - 3y &= 7 \\ 7x &= 28 && \text{add} \\ x &= 4 \end{aligned}$$

Then $2(4) + y = 7$ gives $y = -1$.

The two lines intersect at the single point $(4, -1)$.

Check: $2(4) + (-1) = 7$ ✓ and $4 - 3(-1) = 7$ ✓.

Example 2: Parallel or coincident?

Without solving, decide how many solutions each system has.

• (a) $3x - 6y = 9$ and $x - 2y = 5$
• (b) $4x + 10y = 6$ and $2x + 5y = 3$

Solution.

(a) The $x$ and $y$ coefficients are in the ratio $3 : 1$ ($3 = 3 \times 1$ and $-6 = 3 \times (-2)$). If the lines were the same, the constants would be in that ratio too, but $9 \ne 3 \times 5 = 15$. The lines are parallel and distinct, so there is no solution.

(b) The first equation is exactly $2$ times the second: $2(2x + 5y) = 2(3)$. The equations describe the same line, so there are infinitely many solutions: every point on $2x + 5y = 3$, such as $(4, -1)$ and $(-1, 1)$.

Example 3: Same equation, two spaces

Describe the solutions of each equation in 2-space and in 3-space.

• (a) $y = 2$
• (b) $x + y = 4$

Solution.

(a) In 2-space, $y = 2$ is a horizontal line 2 units above the $x$-axis. In 3-space, only $y$ is fixed; $x$ and $z$ are free. The solutions $(x, 2, z)$ form a plane parallel to the $xz$-plane, 2 units from it in the positive $y$-direction.

(b) In 2-space, $x + y = 4$ is the line through $(4, 0)$ and $(0, 4)$. In 3-space, $z$ is missing, so the plane is parallel to the $z$-axis. It’s the “wall” that stands on the line $x + y = 4$ in the $xy$-plane, passing through $(4, 0, 0)$ and $(0, 4, 0)$ and going straight up and down.

Example 4: The line where two planes meet

Find the line of intersection of the planes

$$x + 2y - z = 3 \qquad 2x + y + z = 6$$

Solution. The coefficients $1, 2, -1$ and $2, 1, 1$ are not in the same ratio, so the planes aren’t parallel. They meet in a line.

Add the equations to eliminate $z$:

$$3x + 3y = 9 \quad\Rightarrow\quad x = 3 - y$$

Two equations, three unknowns: one variable stays free. Let $y = t$. Then $x = 3 - t$, and from the first equation

$$z = x + 2y - 3 = (3 - t) + 2t - 3 = t$$

The line of intersection is

$$x = 3 - t, \qquad y = t, \qquad z = t, \qquad t \in \mathbb{R}$$

Each value of $t$ gives a point: $t = 0$ gives $(3, 0, 0)$ and $t = 1$ gives $(2, 1, 1)$.

Check with $t = 1$: $2 + 2(1) - 1 = 3$ ✓ and $2(2) + 1 + 1 = 6$ ✓.

Common mistakes

Calling x = 3 a line no matter what. In 2-space it’s a vertical line, but in 3-space it’s a plane (every point $(3, y, z)$). Always ask which space you’re working in before describing a graph.

Checking only the left sides for coincident lines or planes. Matching coefficient ratios only tells you the lines (or planes) are parallel. They’re the same only if the constant terms are in that ratio too.

Expecting two planes to meet in a point. Two equations in three unknowns can’t pin down a single point. If they intersect at all, they share a whole line (or a whole plane).

Stopping when you get to 0 = 0. When elimination wipes out an equation completely ($0 = 0$), it doesn’t mean “no solution”. It means the equations were really the same, and there are infinitely many solutions.

Forgetting to check your line. After finding a line of intersection, substitute one point (say $t = 1$) into both original equations. A sign error usually shows up right away.

Practice

1. (Warm-up) Which of these points are on the line $3x - 2y = 12$: $(2, -3)$ or $(4, 1)$?

Solution

For $(2, -3)$: $3(2) - 2(-3) = 6 + 6 = 12$ ✓. It’s on the line.

For $(4, 1)$: $3(4) - 2(1) = 12 - 2 = 10 \ne 12$. It’s not on the line.

2. (Warm-up) Describe the solutions of $x = -2$ in 2-space and in 3-space.

Solution

In 2-space: a vertical line through $(-2, 0)$, parallel to the $y$-axis.

In 3-space: $y$ and $z$ are free, so the solutions $(-2, y, z)$ form a plane parallel to the $yz$-plane, crossing the $x$-axis at $(-2, 0, 0)$.

3. (Warm-up) Is the point $(1, -2, 3)$ on the plane $2x + y - z = -3$?

Solution

$$2(1) + (-2) - 3 = 2 - 2 - 3 = -3$$

Yes, the point is on the plane.

4. (Core) Solve the system $x + 4y = 2$ and $3x - 2y = 20$, and describe the result geometrically.

Solution

Multiply the first equation by $3$ and subtract the second:

$$\begin{aligned} 3x + 12y &= 6 \\ 3x - 2y &= 20 \\ 14y &= -14 && \text{subtract} \\ y &= -1 \end{aligned}$$

Then $x + 4(-1) = 2$, so $x = 6$.

The lines intersect at the single point $(6, -1)$. Check: $3(6) - 2(-1) = 20$ ✓.

5. (Core) Consider the system $2x + ky = 5$ and $6x - 9y = 12$.

• (a) Find the value of $k$ that makes the lines parallel.
• (b) For that value of $k$, how many solutions does the system have? Is there any $k$ that gives infinitely many solutions?

Solution

(a) The $x$-coefficients are in the ratio $2 : 6 = 1 : 3$. For parallel lines the $y$-coefficients need the same ratio: $\dfrac{k}{-9} = \dfrac{1}{3}$, so $k = -3$.

(b) With $k = -3$, the left side of the second equation is $3$ times the first, but $12 \ne 3 \times 5 = 15$. The lines are parallel and distinct, so there is no solution. The constants are never in the ratio $1 : 3$, so no value of $k$ makes the lines coincident: the system never has infinitely many solutions. (For every $k \ne -3$ it has exactly one.)

6. (Core) Find the intercepts of the plane $3x + 4y + 6z = 12$ and describe how you would sketch it.

Solution

Set two variables to $0$ each time:

• $x$-intercept: $3x = 12$, so $(4, 0, 0)$
• $y$-intercept: $4y = 12$, so $(0, 3, 0)$
• $z$-intercept: $6z = 12$, so $(0, 0, 2)$

Plot the three points on the axes and join them to form a triangle. The triangle is the part of the plane in the first octant.

7. (Core) Describe the solutions of each equation in 3-space.

• (a) $z = -1$
• (b) $2y + z = 4$

Solution

(a) $x$ and $y$ are free, so this is a horizontal plane (parallel to the $xy$-plane) 1 unit below it, crossing the $z$-axis at $(0, 0, -1)$.

(b) $x$ is missing, so the plane is parallel to the $x$-axis. In the $yz$-plane it contains the line through $(0, 2, 0)$ and $(0, 0, 4)$; the plane is that line slid along the $x$-direction.

8. (Challenge) Find the line of intersection of the planes $x + y + z = 2$ and $x - y + 3z = 8$. Give two points on it.

Solution

Add the equations to eliminate $y$:

$$2x + 4z = 10 \quad\Rightarrow\quad x = 5 - 2z$$

Let $z = t$. Then $x = 5 - 2t$, and from the first equation

$$y = 2 - x - z = 2 - (5 - 2t) - t = -3 + t$$

The line is $x = 5 - 2t$, $y = -3 + t$, $z = t$, for $t \in \mathbb{R}$.

$t = 0$ gives $(5, -3, 0)$ and $t = 1$ gives $(3, -2, 1)$.

Check $(3, -2, 1)$: $3 - 2 + 1 = 2$ ✓ and $3 + 2 + 3 = 8$ ✓.

9. (Challenge) Show that the planes $2x - y + 4z = 3$ and $-4x + 2y - 8z = 5$ have no points in common. Then write the equation of a plane that is coincident with the first plane but looks different.

Solution

Multiply the first equation by $-2$:

$$-4x + 2y - 8z = -6$$

The left side now matches the second plane exactly, but the right sides are $-6$ and $5$. No point can make $-4x + 2y - 8z$ equal to both, so the planes are parallel and distinct, with no common points.

Any non-zero multiple of the whole first equation is the same plane, for example $6x - 3y + 12z = 9$ (multiply by $3$).