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3.1 Introduction to Linear Systems

Two equations of two variables

\[ a_1 x + b_1 y = c_1 \]\[ a_2 x + b_2 y = c_2 \]

To "solve" the system → find \((x, y)\) that satisfies BOTH equations.

Geometric view: each is a line

\((x, y)\) working in BOTH eqs is equivalent to finding point(s) that is/are on BOTH lines.

intersect once → only one \((x, y)\) on both lines

one solution

Graph showing two lines intersecting at a single point on a Cartesian coordinate system.
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Graph showing two parallel lines that never intersect on a Cartesian coordinate system.

parallel, never intersect

no points on BOTH lines

no solution

Graph showing two lines overlapping perfectly, appearing as a single line on a Cartesian coordinate system.

lines on top of each other

every point on one line is a point on the other line

infinitely-many solutions

Example

\[ x + 3y = 9 \]- ∞
\[ 2x + y = 8 \]- ∞

One method — method of elimination

Add multiple of one equation to the other such that one variable vanishes.

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If we choose to eliminate \(y\), we can multiply equation ② by \(-3\) and add to equation ①.

\(-3② + ① \rightarrow\)

\[\begin{aligned}-6x - 3y &= -24 \\x + 3y &= 9 \end{aligned}\]

add

\[-5x = -15\]

\(x = 3\)

Finding the value of \(y\)

Go to any equation to find \(y\).

From equation ①:

\[\begin{aligned}3y &= 9 - x \\y &= 3 - \frac{1}{3}x \\&= 3 - \frac{1}{3}(3) \end{aligned}\]

\(y = 2\)

These two lines intersect once at \((3, 2)\).

A coordinate graph showing two intersecting lines, one red and one green, meeting at a single point.
Intersection of two lines
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Example

\[\begin{aligned}x + 2y &= 4 \quad - ① \\2x + 4y &= 9 \quad - ② \end{aligned}\]

To eliminate \(x\), calculate \(-2① + ②\):

\[\begin{aligned}-2x - 4y &= -8 \\2x + 4y &= 9 \end{aligned}\]

\((-2①)\)

\((②)\)

add

\[0 = 1\]

Clearly false

No \((x, y)\) works in BOTH equations.

\(\rightarrow\) lines are parallel (no intersection)

A coordinate graph showing two parallel lines, one red and one green, that never intersect.
Parallel lines with no intersection
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Example: Systems of Linear Equations

\[ \begin{aligned} x + 2y &= 4 \quad \text{--- (1)} \\ 2x + 4y &= 8 \quad \text{--- (2)} \end{aligned} \]

Performing the operation \(-2(1) + (2) \rightarrow 0 = 0\).

Clearly true regardless of \((x, y)\).

  • ANY \((x, y)\) on one line also works on the other
  • Two identical lines
  • Infinitely-many solutions
A 2D coordinate graph showing two overlapping identical lines passing through the first and second quadrants.

Expressing the Solutions

To express the solutions, pick one variable at will, use either equation to solve the other.

Let \(y = t\) (whatever we want)

Then (1): \(x = 4 - 2y = 4 - 2t\)

Solutions: \((4 - 2t, t)\)

Or, alternatively:

Let \(x = t\)

Then (1): \(y = 2 - \frac{1}{2}t\)

Solutions: \((t, 2 - \frac{1}{2}t)\)

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3D: 3 Equations with 3 Variables

\[ \left. \begin{aligned} a_1 x + b_1 y + c_1 z &= d_1 \\ a_2 x + b_2 y + c_2 z &= d_2 \\ a_3 x + b_3 y + c_3 z &= d_3 \end{aligned} \right\} \text{ 3 planes} \]

Possibilities:

One solution

Intersection of planes (1), (2), (3). (Intersection of ceiling and two walls) or floor.

A 3D cube diagram where three planes (labeled 1, 2, 3) intersect at a single corner point.

No solution

Ceiling, floor, and wall. Two planes intersect but the 3rd one only intersects one.

A 3D diagram showing three planes where two are parallel (ceiling and floor) and one intersects them.
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Systems of Linear Equations in Three Variables

Geometric Interpretations

When all three planes are the same, there are infinitely-many solutions.

A single parallelogram representing three coincident planes.

When three planes intersect at one line, there are infinitely-many solutions.

Three planes intersecting along a single common line.

all parallel
no solution

Solution method: elimination (same idea)

Example

\[\begin{aligned}x + 5y + z &= 2 & \text{--- (1)} \\ 2x + y - 2z &= 1 & \text{--- (2)} \\ x + 7y + 2z &= 3 & \text{--- (3)}\end{aligned}\]

Work with two at a time to eliminate a variable.

For example, let's eliminate \(x\) from (1), (2).

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Step 1: Eliminate \(x\)

\[\begin{aligned}-2(1) + (2) &\rightarrow -9y - 4z = -3 & \text{--- (4)} \\ -1(1) + (3) &\rightarrow 2y + z = 1 & \text{--- (5)}\end{aligned}\]
two eqs.
two variables
(2D)

Step 2: Solve the 2D System

\[ +4(5) + (4) \implies -y = 1 \rightarrow \boxed{y = -1} \]
\[ (5): z = 1 - 2y = 1 - 2(-1) \rightarrow \boxed{z = 3} \]

Step 3: Back-substitute to find \(x\)

\[\begin{aligned}(1): x &= 2 - 5y - z \\ &= 2 - 5(-1) - (3) \\ &\boxed{x = 4}\end{aligned}\]

one solution

three planes meet at one point