How are all solutions related to one solution?

In the previous lesson, every solution of

x+y+z=6,\qquad y-z=2

had the form $(4-2t,\;2+t,\;t)$ . For example, $\mathbf p=(4,2,0)$ and $\mathbf q=(2,3,1)$ both solve it.

Subtract them before continuing. Does $\mathbf q-\mathbf p$ solve the same equations? If not, what do the two left-hand sides produce?

Keep a record of your investigation

Use start $\mathbf p=(4,2,0)$ and $t=0$ as the reference. Write your predicted candidate and both equation totals before changing $t$ . Keep the equations, starting point, and change direction fixed while varying only $t$ . For the later starting-point comparison, keep $t$ fixed instead. Record which setting changed in each trial, and explain the evidence before opening the comparison.

Separate a starting solution from a change

The difference is $(-2,1,1)$ . The explorer begins at $\mathbf p$ and adds scaled copies of this difference. Predict the candidate at $t=2$ , then test it.

Keep $t=2$ fixed. Before selecting Use another starting solution, predict the candidate if the start changes from $(4,2,0)$ to $(2,3,1)$ . Then make the change and check both equations.

One solution plus an allowed change

x + y + z = 6; y − z = 2

Parameter t

Coordinates of the starting point, change, and candidate
Vector	x	y
Start	4	2
Change	0	0
Candidate	4	2

Candidate = (4, 2, 0).

x + y + z = 6; y − z = 2. Both equations hold.

Change alone: (0, 0, 0) gives 0 in both left-hand sides.

The controls sample t from −3 to 3. The equations allow every real t. Changing the starting solution changes the parameter labels, not the whole solution family.

Compare the changes and the candidates

Starting at $(4,2,0)$ and using $t=2$ gives $(0,4,2)$ , which solves both equations. The change $(-4,2,2)$ alone produces zero in both left-hand sides.

Starting instead at $(2,3,1)$ gives another solution for every $t$ . For a fixed $t$ the candidate changes, but letting $t$ vary over all real numbers covers the same entire family.

Why subtraction removes the right-hand sides

For the two known solutions, the first left-hand side gives 6 in both cases. Subtracting those results gives $6-6=0$ . The second gives $2-2=0$ .

Directly checking the difference $\mathbf d=(-2,1,1)$ :

-2+1+1=0,\qquad 1-1=0.

The corresponding homogeneous system has the same coefficients but zero right-hand sides:

x+y+z=0,\qquad y-z=0.

The zero vector always solves a homogeneous system. There may also be nonzero solutions, as $\mathbf d$ demonstrates here.

Select Zero right-hand sides. What starting vector now works? How does this family differ from the original one?

Compare the families

Zero is now a valid starting vector, and every $t(-2,1,1)$ solves the homogeneous system. This family passes through the origin. The original family is shifted by $(4,2,0)$ and does not contain the origin, because zero cannot produce right-hand sides 6 and 2.

Why a zero-output change preserves a solution

Take a known solution and add a homogeneous solution. In each equation, the left-hand side adds their contributions: the required right-hand side plus zero. The result still satisfies the original system.

For this example,

\begin{bmatrix}x\\y\\z\end{bmatrix} =\begin{bmatrix}4\\2\\0\end{bmatrix} +t\begin{bmatrix}-2\\1\\1\end{bmatrix},\qquad t\in\mathbb R.

This generates valid solutions, but why does it generate all of them? Because any solution minus the known starting solution solves the homogeneous system. Solving that homogeneous system gives $z=t$ , $y=t$ , and $x=-2t$ . There are no other allowed changes.

A known starting solution is called a particular solution. It need not be unique. If the original system is inconsistent, there is no particular solution to start from, so this construction cannot be applied.

Adding two original solutions is different from adding a homogeneous change. For example, $\mathbf p+\mathbf q=(6,5,1)$ produces right-hand sides 12 and 4, so it does not solve the original system.

Discover the family for a new constraint

For $x+y=5$ , use $(5,0)$ as a starting solution. Find all changes that preserve the equation. Then use $(2,3)$ as a different start and explain why the full family is unchanged.

Compare your construction

Every change $(s,-s)$ contributes zero. The solutions are $(5,0)+s(1,-1)$ . Using $(2,3)$ gives $(2,3)+t(1,-1)$ ; these describe the same set because $s=t-3$ . The parameter value depends on the chosen starting point, while the set of solutions does not.

Design your own comparison

Choose your own nonzero parameter value to generate a new starting solution on paper. Keep both original equations and the direction $(-2,1,1)$ fixed. Predict whether using your new start changes the whole solution family or only the parameter labels.

Choose one target solution from the original family and find the parameter that reaches it from each start. Then explain how to translate the parameter for any target in the family.

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

If your new start is $\mathbf p+s\mathbf d$ with $\mathbf d=(-2,1,1)$ , then $\mathbf p+t\mathbf d=(\mathbf p+s\mathbf d)+(t-s)\mathbf d$ . The same target receives parameter $t-s$ from the new start. One matched point is evidence; this identity establishes that the entire families agree.

Question 1 of 30 checked · 0 correct

For x+y=7, a known solution is (2, 5). Which change preserves the equation?(1, 1)(3, −3)(0, 7)