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Linear Equations: Star problems

Ten optional challenges to stretch your reasoning. Work on paper, use hints when you need them, and check the answer or full solution when you are ready. You can skip these problems and continue the course.

1 of 3 stars: Stretch
2 of 3 stars: Challenge
3 of 3 stars: Deep challenge

Stars indicate difficulty within this set.

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Problem 1 of 10

Problem 1 An equation that cannot have just one answer

Difficulty: 1 of 3 stars, Stretch

A real constant $k$ is fixed. Consider
$\frac{x+1}{x-2}=1+\frac{k}{x-2}.$
Find every $k$ for which the equation has at least one real solution. For each such $k$ , describe all solutions. Explain why no choice of $k$ produces exactly one solution.

Builds on Linear Equations in Disguise, Algebraic Fractions
1. Hint 1
  Rewrite the numerator $x+1$ using the denominator $x-2$ .
2. Hint 2
  On the domain $x\ne2$ , subtract $1$ and multiply by $x-2$ . What remains?
Answer

$k=3$ gives every real $x\ne2$ ; every other $k$ gives no solution.

Full solution

The original equation requires $x\ne2$ .
Since $x+1=(x-2)+3$ , the left side is $1+3/(x-2)$ .
Subtracting $1$ from both sides leaves
$3/(x-2)=k/(x-2).$

Multiplication by the nonzero denominator gives the condition $3=k$ , with no $x$ remaining.
If $k=3$ , this condition is true for every admissible $x$ , and substitution into the original equation confirms that both sides are identical.
If $k\ne3$ , the condition is false regardless of $x$ , so there are no solutions.

Thus the equation either accepts its entire domain or rejects it.
A variable appearing in an equation does not guarantee that the equation determines that variable.
Here the parameter decides whether the two expressions are the same.

Answer

$k=3$ gives every real $x\ne2$ ; every other $k$ gives no solution.

Key idea

After legitimate cancellation, a statement with no remaining variable signals either an identity or an inconsistency.
Problem 1: Mark complete
Problem 2 Four shifts, one shared center

Difficulty: 1 of 3 stars, Stretch

Solve
$\frac{x-2}{3}+\frac{x-5}{6}=\frac{x-8}{9}+\frac{x-11}{12}.$
Find a method that avoids distributing a common denominator across all four original numerators, and explain why the solution is unique.

Builds on Solving Linear Equations, Arithmetic with Expressions
1. Hint 1
  Each numerator differs from its denominator by the same expression.
2. Hint 2
  For example, $(x-2)/3=(x+1)/3-1$ . Make the same rewrite in all four terms.
Answer

$x=-1$ .

Full solution

The four fractions can be rewritten as $(x+1)/3-1$ , $(x+1)/6-1$ , $(x+1)/9-1$ , and $(x+1)/12-1$ .
The two $-1$ terms on each side cancel, leaving
$\left(\frac13+\frac16\right)(x+1)=\left(\frac19+\frac1{12}\right)(x+1).$

The coefficients are $1/2$ and $7/36$ , so subtraction gives
$(11/36)(x+1)=0.$
Since $11/36$ is nonzero, the only possible solution is $x=-1$ .
At this input every original fraction equals $-1$ , so both sides equal $-2$ .
This also verifies the answer without further calculation.

The rewrite exposes a shared center at $x=-1$ .
Once that common factor is visible, the equation becomes a comparison between two different multiples of the same quantity.

Answer

$x=-1$ .

Key idea

If each numerator has the same offset from its denominator, shifting the variable can expose a common factor.
Problem 2: Mark complete
Problem 3 Irrational coefficients with a simple structure

Difficulty: 1 of 3 stars, Stretch

Solve exactly
$\bigl(\sqrt2+\sqrt3\bigr)(x-1)=5+2\sqrt6.$
Explain how to obtain the answer without decimal approximations or rationalizing a denominator, and prove that there is exactly one real solution.

Builds on Solving Linear Equations, Fractional Exponents and Radicals
1. Hint 1
  Look for the right side inside the square of a sum.
2. Hint 2
  Expand $(\sqrt2+\sqrt3)^2$ . The coefficient of $x-1$ is positive.
Answer

$x=1+\sqrt2+\sqrt3$ .

Full solution

Expanding the square gives
$(\sqrt2+\sqrt3)^2=2+2\sqrt2\sqrt3+3=5+2\sqrt6.$

Write $u=\sqrt2+\sqrt3$ , which is a positive real number.
The equation becomes
$u(x-1)=u^2.$
Division by $u$ is valid because $u\ne0$ , so $x-1=u$ and
$x=1+\sqrt2+\sqrt3.$

Substituting this value returns $u^2=5+2\sqrt6$ , so it satisfies the original equation.
The division by a nonzero coefficient is reversible and forces exactly one value of $x$ ; there are no other solutions.
The irrational-looking coefficient is not an obstacle once the matching structure on the other side is recognized.

Answer

$x=1+\sqrt2+\sqrt3$ .

Key idea

Before dividing by a complicated coefficient, check whether the other side contains a recognizable multiple or power of it.
Problem 3: Mark complete
Problem 4 A parameter can force a forbidden answer

Difficulty: 2 of 3 stars, Challenge

For each real value of $a$ , solve
$\frac{a-1}{x-2}=\frac{a+1}{x+2}.$
Give a complete classification, including every parameter value for which no solution exists.

Builds on Linear Equations in Disguise, Solving for a Variable
1. Hint 1
  Record the excluded inputs before cross-multiplying.
2. Hint 2
  After clearing the nonzero denominators, the equation becomes $2x=4a$ . Check when that candidate is excluded.
Answer

For $a\notin\{-1,1\}$ , the unique solution is $x=2a$ . For $a=-1$ or $a=1$ , there is no solution.

Full solution

The original domain is $x\ne2,-2$ .
For an input in this domain, clearing denominators is reversible and gives
$(a-1)(x+2)=(a+1)(x-2).$

Expansion and cancellation yield $-2x+4a=0$ , so every solution must be $x=2a$ .
If $a\ne1,-1$ , then $2a$ is in the original domain, and reversing the algebra proves that it works.
Hence the solution is unique.

When $a=1$ , the candidate is $x=2$ , which is excluded; the original equation would instead demand $0=2/(x+2)$ at an admissible input, impossible.
When $a=-1$ , the candidate is $x=-2$ , also excluded, and the original equation would demand $-2/(x-2)=0$ , again impossible.
These two cases show why checking only the cleared equation would give an incorrect classification.

Answer

For $a\notin\{-1,1\}$ , the unique solution is $x=2a$ . For $a=-1$ or $a=1$ , there is no solution.

Key idea

A parameter may leave the algebraic candidate well defined while pushing it outside the original domain.
Problem 4: Mark complete
Problem 5 Inverting a formula with an exceptional branch

Difficulty: 2 of 3 stars, Challenge

The real numbers $p$ and $r$ are fixed, and
$p=\frac{x-r}{x+r}.$
Solve for $x$ and give all cases: which pairs $(p,r)$ yield exactly one real solution, no real solutions, or infinitely many real solutions? Remember the original denominator.

Builds on Solving for a Variable, Linear Equations in Disguise
1. Hint 1
  Multiplication by $x+r$ gives a linear equation, but do not divide by $1-p$ until you have checked whether it is zero.
2. Hint 2
  Treat $r=0$ separately. If $r\ne0$ and $p\ne1$ , check $x+r$ after substituting your candidate.
Answer

If $r\ne0$ and $p\ne1$ , exactly one solution: $x=\frac{r(1+p)}{1-p}$ . If $r=0,p=1$ , every $x\ne0$ works. All other pairs give no solution.

Full solution

The domain condition is $x+r\ne0$ .
If $r=0$ , the original fraction is $x/x=1$ for every $x\ne0$ .
Thus $p=1$ gives infinitely many solutions, namely all nonzero real $x$ , while $p\ne1$ gives none.

Now suppose $r\ne0$ .
Clearing the denominator yields $px+pr=x-r$ , or
$(1-p)x=r(1+p).$
If $p=1$ , this would require $0=2r$ , which is impossible.
If $p\ne1$ , the only candidate is
$x=\frac{r(1+p)}{1-p}.$

For that candidate, $x+r=2r/(1-p)\ne0$ , so it lies in the original domain.
Reversing the algebra verifies it and establishes uniqueness.
These cases cover every real pair $(p,r)$ , including the branch that would be lost by dividing by $1-p$ at the start.

Answer

If $r\ne0$ and $p\ne1$ , exactly one solution: $x=\frac{r(1+p)}{1-p}$ . If $r=0,p=1$ , every $x\ne0$ works. All other pairs give no solution.

Key idea

When solving a formula, every potentially zero divisor creates a case that deserves its own analysis.
Problem 5: Mark complete
Problem 6 The missing size of a data set

Difficulty: 2 of 3 stars, Challenge

A collection of $n$ real numbers, where $n\ge3$ , has mean $20$ . Removing its largest entry lowers the mean to $18$ . Removing its smallest entry from the original collection instead raises the mean to $21$ . The largest and smallest entries differ by $45$ .

Find $n$ and both extreme entries. Then exhibit a collection showing that all the conditions can actually hold.

Builds on Word Problems with Linear Equations
1. Hint 1
  Use the original total $20n$ to express each removed entry in terms of $n$ .
2. Hint 2
  The largest entry is $20n-18(n-1)$ ; obtain a corresponding expression for the smallest and use their difference.
Answer

$n=16$ , largest entry $50$ , smallest entry $5$ . One collection is $5,18$ , thirteen copies of $19$ , and $50$ .

Full solution

The original sum is $20n$ .
After removing the largest entry $L$ , the remaining $n-1$ numbers total $18(n-1)$ , so
$L=20n-18(n-1)=2n+18.$
Similarly, removing the smallest entry $S$ leaves total $21(n-1)$ , so
$S=20n-21(n-1)=21-n.$

The difference condition gives $(2n+18)-(21-n)=45$ , hence $3n-3=45$ and $n=16$ .
Substitution gives $L=50$ and $S=5$ .

These forced values still need a feasibility check.
Take the collection consisting of $5$ , $18$ , thirteen copies of $19$ , and $50$ .
It contains $16$ numbers with total
$5+18+247+50=320.$
Its mean is $20$ ; deleting $50$ leaves $270/15=18$ , and deleting $5$ leaves $315/15=21$ .
Its extreme entries really are $5$ and $50$ , whose difference is $45$ .

Answer

$n=16$ , largest entry $50$ , smallest entry $5$ . One collection is $5,18$ , thirteen copies of $19$ , and $50$ .

Key idea

Express each changed total using the same unknown count, and check that algebraically forced data can actually be realized.
Problem 6: Mark complete
Problem 7 Two equations with integer answers

Difficulty: 2 of 3 stars, Challenge

Find every integer $a$ for which each of the equations
$(a-1)x=a+5,\qquad (a+1)y=a-3$
has a unique integer solution. Give the corresponding pair $(x,y)$ for each $a$ , and prove completeness.

Builds on Solving Linear Equations
1. Hint 1
  First exclude parameter values that make one of the coefficients zero.
2. Hint 2
  For the remaining values, rewrite $x=1+6/(a-1)$ and $y=1-4/(a+1)$ . Both denominators must divide fixed integers.
Answer

$(a,x,y)=(-5,0,2),\ (-2,-1,5),\ (0,-5,-3),\ (3,4,0)$ .

Full solution

At $a=1$ , the first equation becomes $0x=6$ , and at $a=-1$ , the second becomes $0y=-4$ .
Neither parameter works.
For all other $a$ , both equations have unique real solutions, given by
$x=1+\frac6{a-1},\qquad y=1-\frac4{a+1}.$

The first value is an integer exactly when $a-1$ is a positive or negative divisor of $6$ .
Thus its candidate set is $\{-5,-2,-1,0,2,3,4,7\}$ .
The second is an integer exactly when $a+1$ is a positive or negative divisor of $4$ , giving $\{-5,-3,-2,0,1,3\}$ .

The intersection is $\{-5,-2,0,3\}$ .
Substitution produces the four triples in the answer.
Their coefficients $a-1$ and $a+1$ are nonzero, so every listed equation has a unique solution.
Conversely, any qualifying parameter must appear in both complete divisor lists, so no additional integers can work.

Answer

$(a,x,y)=(-5,0,2),\ (-2,-1,5),\ (0,-5,-3),\ (3,4,0)$ .

Key idea

An integer-valued linear-equation answer can impose a divisibility condition on the coefficient; simultaneous conditions become an intersection of finite lists.
Problem 7: Mark complete
Problem 8 Two borders with the same area

Difficulty: 3 of 3 stars, Deep challenge

A square courtyard has side length $L$ meters. In one design, a walkway of uniform width $2$ meters is added outside the courtyard. In another design, a walkway of uniform width $b$ meters lies inside the original boundary. Each walkway includes its four corner squares. The inside design must leave a central square of positive side length.

(a) If $b=3$ , find $L$ for which the two walkways have equal area, and find that area.

(b) Keep the outside width at $2$ , but allow $b$ to be any positive integer. Find every $b$ for which equal walkway areas are possible, and give the corresponding $L$ . The courtyard side $L$ need not be an integer.

Schematic diagrams; all lengths are in meters.

Text description of this figure

Two schematic drawings side by side. The left one, labeled Outside border, shows a square courtyard of side L with a shaded walkway of uniform width 2 running around the outside of it, including the four corner squares. The right one, labeled Inside border, shows a square courtyard of side L whose outer band, of uniform width b, is shaded as the walkway, leaving an unshaded central square. The drawings are schematic, and all lengths are in meters.

Builds on Word Problems with Linear Equations
1. Hint 1
  Compute each walkway as the difference of two square areas. Record the requirement $L>2b$ .
2. Hint 2
  Equality reduces to $(b-2)L=b^2+4$ . First decide whether $b\le2$ can work; then combine the formula for $L$ with $L>2b$ .
Answer

(a) $L=13$ meters and each walkway has area $120$ square meters. (b) Exactly $b=3$ with $L=13$ , and $b=4$ with $L=10$ .

Full solution

The outside walkway has area
$(L+4)^2-L^2=8L+16.$
The inside walkway has area $L^2-(L-2b)^2=4bL-4b^2$ , with the geometric restriction $L>2b$ .
Equal areas therefore require
$(b-2)L=b^2+4.$

For $b=3$ , this gives $L=13$ .
The central square has side $13-6=7>0$ , so the design is valid.
The outside area is $8\cdot13+16=120$ , agreeing with the inside area
$13^2-7^2=120.$

For the classification, the right side $b^2+4$ is positive and $L>0$ , so necessarily $b>2$ .
In that case
$L=(b^2+4)/(b-2).$
The central-square requirement becomes $b^2+4>2b(b-2)$ , or $(b-2)^2<8$ .
Because $b-2$ is a positive integer, it is $1$ or $2$ .
Thus $b=3$ or $4$ .

The first case was checked above.
The second gives $L=20/2=10$ , leaving central side $10-8=2>0$ ; both areas are $96$ .
Every algebraic and geometric restriction has now been checked.

Answer

(a) $L=13$ meters and each walkway has area $120$ square meters. (b) Exactly $b=3$ with $L=13$ , and $b=4$ with $L=10$ .

Key idea

In geometric modeling, the equation determines a candidate while the shape's existence conditions decide whether that candidate is meaningful.
Problem 8: Mark complete
Problem 9 Canceled games without double counting

Difficulty: 3 of 3 stars, Deep challenge

A tournament was planned so that every pair of players would play exactly one game. Before any games were played, six players withdrew, and exactly $87$ planned games were canceled.

(a) Find the original number of players. Explain why dividing $87$ by $6$ does not correctly count how many opponents each withdrawing player had.

(b) Exactly five original players were designated as seeds, and exactly two of those seeds withdrew. How many canceled games involved at least one seed?

Builds on Word Problems with Linear Equations
1. Hint 1
  Separate canceled games between two withdrawing players from canceled games with only one withdrawing player.
2. Hint 2
  For part (b), first count games involving either of the two withdrawing seeds. Then add games between the four withdrawing nonseeds and the three remaining seeds.
Answer

(a) $18$ original players. (b) $45$ canceled games involved at least one seed.

Full solution

Let $N$ be the original number of players.
The six withdrawing players had $6(N-6)$ games against players who stayed.
Among themselves, they had $6\cdot5/2=15$ games.
These are disjoint categories containing every canceled game, so
$6(N-6)+15=87.$
Thus $6N-21=87$ and $N=18$ .

Dividing $87$ by $6$ does not count opponents per withdrawing player because a canceled game between two withdrawing players belongs to both of their opponent lists.
Indeed, the six individual lists contain $6\cdot17=102$ entries; subtracting the $15$ doubly counted games gives $87$ .

For part (b), each of the two withdrawing seeds had $17$ planned games.
Their mutual game occurs in both lists, so the number involving at least one of these two seeds is
$17+17-1=33.$
The remaining canceled games involving a seed must pair one of the four withdrawing nonseeds with one of the three remaining seeds.
There are $4\cdot3=12$ such games, and none was counted among the $33$ .

Therefore the requested total is $33+12=45$ .
Separating the two disjoint categories also proves that every canceled game involving a seed has been counted exactly once.

Answer

(a) $18$ original players. (b) $45$ canceled games involved at least one seed.

Key idea

Build a linear equation from disjoint categories; when counting through overlapping lists, identify exactly which objects appear twice.
Problem 9: Mark complete
Problem 10 Design a linear equation behind three fractions

Difficulty: 3 of 3 stars, Deep challenge

Consider
$\frac1{x-1}+\frac2{x-2}+\frac3{x-3}=\frac{kx+h}{(x-1)(x-2)},$
where $k,h$ are real parameters.

(a) Find the unique value of $k$ for which the $x^2$ terms cancel after all denominators are cleared, regardless of $h$ .

(b) Using that value of $k$ , find all pairs of integers $(h,x)$ with $x>0$ satisfying the original equation. Prove completeness, including the excluded inputs.

Builds on Linear Equations in Disguise
1. Hint 1
  The original inputs exclude $1,2,3$ . Multiply by $(x-1)(x-2)(x-3)$ and compare the $x^2$ coefficients.
2. Hint 2
  With the correct $k$ , the equation becomes $(h+4)x=3h+18$ . Separate $h=-4$ , then write $x=3+6/(h+4)$ .
Answer

(a) $k=6$ . (b) $(h,x)=(-3,9),(-2,6),(-1,5),(2,4)$ .

Full solution

The original domain is $x\notin\{1,2,3\}$ .
Clearing denominators gives
$(x-2)(x-3)+2(x-1)(x-3)+3(x-1)(x-2)=(kx+h)(x-3).$

The left side expands to $6x^2-22x+18$ ; the right side is $kx^2+(h-3k)x-3h$ .
Hence the quadratic terms cancel exactly when $k=6$ .
With this value, the equation reduces to
$(h+4)x=3h+18.$
If $h=-4$ , it says $0=6$ , impossible.
Otherwise
$x=3+6/(h+4).$

For integer $h,x$ , the nonzero integer $t=h+4$ must divide $6$ .
Positive divisors $1,2,3,6$ give $x=9,6,5,4$ , all admissible, and yield the four pairs stated.
Negative divisors $-1,-2,-3,-6$ give
$x=-3,0,1,2.$
The first two violate positivity, and the last two violate the original domain.

This exhausts every divisor and the exceptional case $h=-4$ .
All retained inputs avoid $1,2,3$ , so the denominator-clearing steps reverse and verify each original equation.

Answer

(a) $k=6$ . (b) $(h,x)=(-3,9),(-2,6),(-1,5),(2,4)$ .

Key idea

Choose coefficients to control an equation's degree, then combine algebraic reduction with the original domain to classify integer solutions.
Problem 10: Mark complete