What does a free variable let us choose?

In the previous lesson, a column without a pivot left a variable free. Here we will use that choice to describe every solution.

You have seen how elimination can reveal a unique answer. What happens when the constraints leave a choice?

Consider:

\begin{aligned} x+y+z&=6,\\ y-z&=2. \end{aligned}

Choose $z=0$ . The second equation then gives $y=2$ , and the first gives $x=4$ .

Now choose $z=2$ . Calculate $y$ first, then $x$ . Do both equations still hold? Before using the explorer, predict whether $z=-1$ will work too.

Keep a record of your investigation

Use $t=0$ , giving $(4,2,0)$ , as your reference. Before each trial, write the predicted triple and the two equation totals. Keep both equations fixed and change only the parameter $t$ ; $x$ and $y$ must follow from that choice. Record the triples and totals, then explain why several coordinates can change even though you made only one independent choice.

Make a choice and check both equations

Try t = 0, t = 2, and t = -1. The letter $t$ records your choice for $z$ .

Watch how $x$ and $y$ respond. When $z$ increases by 1, how much does each of the other coordinates change?

One choice moves three coordinates.

x + y + z = 6 and y − z = 2

Choose z = t 0

x = 4 − 2t4

y = 2 + t2

z = t0

Candidate: (4, 2, 0)

First equation: (4) + (2) + (0) = 6 ✓

Second equation: (2) − (0) = 2 ✓

Both equations hold. Each different value of t gives another solution.

The slider samples −2 ≤ t ≤ 3 in steps of 0.25. With just the first two equations, every real value of t is allowed.

Compare your predictions

The three choices give $(4,2,0)$ , $(0,4,2)$ , and $(6,1,-1)$ . All satisfy both equations. Increasing $z$ by 1 increases $y$ by 1 and decreases $x$ by 2.

Replace the list with a rule

We cannot list infinitely many triples. Instead, let $z=t$ and solve in terms of that choice:

y-z=2\quad\Longrightarrow\quad y=2+t.

Then:

x=6-y-z=6-(2+t)-t=4-2t.

All solutions can be written as

(x,y,z)=(4-2t,\ 2+t,\ t),\qquad t\in\mathbb R.

Here $z$ is a free variable: we can choose it, and the equations determine the remaining variables. The letter $t$ is a parameter used to describe that choice.

Only one coordinate is free. Choosing $x$ , $y$ , and $z$ independently would usually break the equations.

Does the rule really describe every solution?

First, substitute the formulas:

(4-2t)+(2+t)+t=6,\qquad (2+t)-t=2.

Both identities hold for every real $t$ , not just the values the slider offers.

Conversely, take any solution and call its $z$ coordinate $t$ . The second equation forces $y=2+t$ ; the first then forces $x=4-2t$ . So no solution is missing from our description.

The slider’s range and step size are interface limits. Values such as $t=100$ or $t=\sqrt{2}$ are also allowed in the original system.

Can another equation remove the choice?

Predict what happens if we require $x=0$ . Select Add x = 0, then move the slider. Which value satisfies all three equations?

Since $x=4-2t$ , the extra requirement gives

4-2t=0\quad\Longrightarrow\quad t=2.

Only $(0,4,2)$ remains. Points at other slider positions still satisfy the first two equations, but fail the new one.

Now imagine adding $2y-2z=4$ instead. It is just twice an existing equation, so it removes no freedom. Counting equations alone does not tell us whether an answer is uniquely determined.

In a model-fitting problem, this distinction matters: several parameter settings can satisfy the same constraints when those constraints leave freedom.

Design your own comparison

Choose a parameter value not used in the worked examples. Predict its triple, then test it in the explorer or by substitution. Keep the original two equations fixed.

On paper, design one extra equation that selects your chosen triple and a different extra equation that leaves every original solution valid. Predict the effect of each separately, then substitute the whole parameter family to justify your predictions.

Write a short conclusion: what you predicted, what you observed, and how the evidence changed or supported your explanation.

Check the reasoning in your investigation

For your chosen value $t_0$ , adding $z=t_0$ selects exactly that triple. A multiple of $y-z=2$ adds no restriction. Other equations are valid answers if substitution shows the first selects just one parameter value and the second holds for every real $t$ . Checking only your chosen triple is insufficient for the second claim.

Check your understanding

Question 1 of 30 checked · 0 correct

For x = 4 − 2t, y = 2 + t, z = t, which solution corresponds to t = −2?(8, 0, −2)(0, 0, −2)(4, 2, −2)

Keep learning

Next: How are all solutions related to one solution?. We will separate one starting solution from the changes that preserve every equation.