Systems of Equations and Inequalities: Star problems

Let $n=x+y$.

If all $n$ packages were small they would contain $7n$ units; changing a small package to a large one adds $4$ units.

Thus $7n+4y=200$.

Since $y\ge0$, we have $n\le200/7<29$, while the shipment condition gives $n\ge20$.

The equation also says $7n$ differs from $200$ by a multiple of $4$.

Because $7n$ and $-n$ have the same remainder on division by $4$, $n$ must be a multiple of $4$.

Within $20\le n\le28$, only $n=20,24,28$ remain.

For these totals, $y=(200-7n)/4$ gives $15,8,1$, respectively; subtracting from $n$ gives $x=5,16,27$.

All six counts are nonnegative integers, their package totals are at least $20$, and their unit totals are $200$, so every candidate works.

The total-number substitution reduced two integer unknowns to three possible totals; the finite bound and divisibility condition prove that no other shipment is possible.

Doubling the second equation and subtracting the first gives

Since both $x$ and $z$ are nonnegative, both terms must be zero.

Thus $x=z=0$, and the total equation gives $y=3/2$.

This triple also satisfies the first equation, so it is the unique permitted solution.

Without the sign conditions, let $z=t$.

The derived equation gives $x=-3t$, and the total then gives $y=3/2+2t$.

Direct substitution verifies both original equations for every real $t$, yielding the entire line of unrestricted solutions.

Along that line, $z\ge0$ requires $t\ge0$, while $x\ge0$ requires $t\le0$.

Their intersection is only $t=0$, where $y$ is also nonnegative.

The student’s dimension intuition describes the solution set of the equations alone.

It does not account for additional inequalities, which can meet an entire solution line at just one boundary point.

A free variable in elimination can therefore become fixed after the domain restrictions are imposed.

For (a), suppose some two lines are nonparallel.

They meet at a unique point $P$.

Each remaining line belongs to a triple containing those two, so by the hypothesis it must pass through their only possible common point $P$.

Thus all four share $P$.

If no two are nonparallel, all four lines are parallel, allowing coincidence.

The triple hypothesis implies that every pair is consistent: include any third equation to obtain a solution.

Two parallel consistent lines must coincide.

Therefore all four lines are the same line, and again they have common solutions.

For (b), the four lines $x=0$, $y=0$, $x+y=1$, and $x+2y=3$ have pairwise distinct directions, so every pair meets.

The first three already have no common point, because $x=y=0$ contradicts $x+y=1$.

Hence the four together are inconsistent.

For (c), take the four equations in the answer.

The first three meet at $(0,0,0)$.

A triple containing $x+y+z=1$ and two coordinate equations is solved by setting the remaining coordinate to $1$.

But all four would require both $x+y+z=0$ and $x+y+z=1$.

The number of unknowns changes how much partial consistency can guarantee.

Subtracting the equations gives

Substitute $y=2+e_1-e_2-x$ into the first equation to obtain

These formulas determine a unique real solution for every allowed error pair.

The triangle inequality now gives

and

Take $e_1=-1/100$ and $e_2=1/100$.

Then $x-1=199/100$, $y-1=-201/100$, and

Thus one permitted error pair attains all three absolute bounds, proving sharpness.

The two equations have almost the same relative coefficients of $x$ and $y$.

A large increase in one component can be nearly canceled by a large decrease in the other.

Elimination consequently multiplies the small measurement errors by coefficients near $100$.

Their sum behaves differently: it is obtained just by subtracting the two readings, so its error is at most the sum of the original error bounds.

If $D=\pm1$, Cramer’s formulas give integer $x,y$ for every integer right-hand side, and the nonzero determinant guarantees real uniqueness.

Thus the condition is sufficient.

Conversely, the assumed real uniqueness forces $D\ne0$.

For right-hand sides $(1,0)$ and $(0,1)$, the solution formulas show that $d/D,-c/D,-b/D,a/D$ are all integers.

Hence write $a=D\alpha$, $b=D\beta$, $c=D\gamma$, and $d=D\delta$ with integer $\alpha,\beta,\gamma,\delta$.

Substituting into the determinant gives

so

Both factors are integers, forcing $D=\pm1$.

This proves necessity; testing a few right-hand sides was enough only because they controlled every coefficient in the inverse formulas.

For the construction,

All four entries are at least $N$.

Cramer’s rule gives the displayed integer solution formulas.

Substitution, or reversing the same elimination, verifies both equations for arbitrary integers $m,n$.

Large coefficients therefore do not prevent universal integrality; the determinant, not their individual size, governs it.

Subtracting equation 1 from equation 2 gives

Subtracting equation 2 from equation 3 gives

For $t\ne0,1$, divide these by $t-1$ and $t(t-1)$, respectively, to obtain the two equations in Hint 2.

Subtracting those reduced equations gives

For $t\notin\{-1,0,1\}$, this forces $z=t^2+t+1$, then $y=-t(t^2+t+1)$, and finally $x=t^3$.

All steps are reversible under these restrictions, so this is the unique solution.

The needed identity $t^3-1=(t-1)(t^2+t+1)$ follows directly by expansion.

Now inspect the original system.

At $t=0$, it reduces to $x=0$ and $y+z=1$.

At $t=1$, all equations reduce to $x+y+z=1$.

At $t=-1$, the first and third coincide and the second is $x-y+z=-1$, giving $y=1$ and $x+z=0$.

These are precisely the listed families.

Expanding the determinant after subtracting the first row from the other rows gives $t(t-1)^3(t+1)$.

Its nonzero values certify uniqueness, and its zeros locate the cases needing separate inspection.

A zero determinant alone does not distinguish inconsistency from infinitely many solutions; here all three exceptional systems are consistent.

Necessity follows because factory $i$ can use only the other warehouses, whose combined demand is $T-c_i$.

Thus

For sufficiency, let the shipment from factory 1 to warehouse 2 be $t$.

Its row total forces the shipment to warehouse 3 to be $r_1-t$.

Successively using column 3, row 2, column 1, and row 3 forces exactly the matrix in the answer.

Its remaining column total also holds because total supply equals total demand.

Hence every plan has this form, and every such matrix with nonnegative entries is a plan.

Nonnegativity gives $L\le t\le U$ with the stated maximum and minimum.

It remains to prove $L\le U$.

Compare each of the three lower bounds with each of the three upper bounds.

Six comparisons follow just from nonnegative supplies, demands, and their equal totals.

The other three are

These are exactly $r_3+c_3\le T$, $r_2+c_2\le T$, and

Thus the assumed conditions make every lower bound at most every upper bound, proving sufficiency constructively.

For the numerical data, substitution gives the displayed matrix and interval

All six entries are integers exactly when $t$ is an integer, because one entry is $t$ itself.

Therefore $t=0,1,2,3,4,5,6$ gives exactly seven whole-unit plans.

The boundary lines meet at

Write $x=5/3+2s-t$ and

Then

Thus every $s,t\ge0$ gives a feasible point, and conversely every feasible point has this representation by taking $s=(2x+y-4)/3$ and

At such a point,

If both variable coefficients are nonnegative, its least value is the value at $s=t=0$, namely $(5a+2b)/3$.

Therefore the three listed conditions are sufficient.

They are also necessary.

If $c>(5a+2b)/3$, the feasible point $P$ already violates the inequality.

If $2a-b<0$, keep $t=0$ and let $s$ grow: the displayed expression decreases without bound, so choose any nonnegative $s$ large enough to make it less than $c$.

If $2b-a<0$, use $s=0$ and sufficiently large $t$ instead.

These rays provide explicit violating points via the coordinate formulas.

Checking only the boundary intersection would have missed these failures far out in the unbounded region.

A zero row gives determinant $0$.

A row with exactly one 1 lets us expand along that row: the remaining $2\times2$ determinant is a difference of two numbers in $\{0,1\}$, so its absolute value is at most $1$.

Now suppose a row is $(1,1,1)$.

Move it to the top, affecting at most the determinant’s sign.

Clear the first column below it by subtracting this top row from a lower row whenever that lower row starts with 1.

In the remaining two columns, each lower row has entries either both in $\{0,1\}$ or both in $\{0,-1\}$.

Multiplying a negative row by $-1$ affects only the sign and leaves a $2\times2$ zero-one matrix.

Expanding down the cleared first column therefore bounds the original absolute determinant by $1$.

The only case that can exceed $1$ has exactly two ones in each row.

If any rows repeat, the determinant is zero.

Otherwise the three rows must be precisely the three types listed.

Direct expansion for any one ordering gives absolute determinant $2$, and row reordering preserves its absolute value.

This proves both the bound and the complete equality classification.

For (b), adding appropriate pairs of equations and subtracting the third gives the three formulas in the answer.

The determinant has absolute value $2$, so the real solution is unique but may have halves.

All three numerators have the same parity as $u+v+w$, making the even-sum condition exactly the integrality condition.

Their nonnegativity is exactly the three triangle inequalities stated.

These conditions make the formulas nonnegative integers; direct addition recovers $u,v,w$, proving sufficiency.