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Solving Rational Equations: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A nested equation

Solve $\frac1{1-5/x}=\frac72$ over the real numbers, checking the original equation.
1. Hint 1
  Identify the inner denominator restriction and where the entire denominator is zero.
2. Hint 2
  Write the left side as one rational expression before clearing denominators.
Answer

$x=7$ .

Full solution

The inner fraction requires $x\ne0$ , and $1-5/x$ is zero at $5$ , so also $x\ne5$ .
On that domain, multiplying the numerator and denominator of the left side by $x$ gives the equation
$x/(x-5)=7/2.$
Clearing the nonzero denominators gives
$2x=7(x-5)$ $x=7$

The candidate $7$ is neither $0$ nor $5$ .
Checking the original: $1-5/7=2/7$ , whose reciprocal is $7/2$ .
Thus $7$ is the solution.

Answer

$x=7$ .

Key idea

Nested rational equations need restrictions from both the inner and outer denominators.
Problem 1: Mark complete
Problem 2 A ratio of quadratic values

Solve $\frac{x^2+1}{x^2+4}=\frac12$ over the real numbers.
1. Hint 1
  The denominator is positive, so no real input is excluded.
2. Hint 2
  Clear the denominator and solve the resulting equation in $x^2$ .
Answer

$x=\sqrt2$ or $x=-\sqrt2$ .

Full solution

The denominator $x^2+4$ is positive for all real $x$ .
Multiplying by $2(x^2+4)$ gives
$2x^2+2=x^2+4$ $x^2=2$

The two candidates are $\sqrt2$ and $-\sqrt2$ .
Both have square $2$ , so both make the original fraction $3/6=1/2$ .

Answer

$x=\sqrt2$ or $x=-\sqrt2$ .

Key idea

A rational equation with no real denominator zeros can still have more than one real solution.
Problem 2: Mark complete
Problem 3 A zero fraction

Solve $\frac0{2x-9}=x+7$ over the real numbers.
1. Hint 1
  On its domain, a zero numerator makes the left side zero.
2. Hint 2
  If you clear the denominator, filter every resulting candidate using the original restriction.
Answer

$x=-7$ .

Full solution

The original denominator excludes $\frac92$ .
Multiplying by $2x-9$ gives
$0=(x+7)(2x-9)$
The candidates are $-7$ and $\frac92$ .

The candidate $\frac92$ is excluded and is extraneous.
At $-7$ , the original left side is $0/(-23)=0$ and the right side is $0$ , so $-7$ is the solution.

Answer

$x=-7$ .

Key idea

Clearing even a zero fraction can create an excluded candidate that must be rejected.
Problem 3: Mark complete
Problem 4 An average charge

A service charges a fixed 36 dollars plus 2 dollars per item. For one order, the average charge is 5 dollars per item. Write and solve a rational equation for the number of items, and state the domain and the separate condition supplied by the situation.
1. Hint 1
  Average charge divides total charge by the number of items.
2. Hint 2
  The mathematical denominator condition and the counting condition are different.
Answer

$n=12$ items; equation $(36+2n)/n=5$ ; domain $n\ne0$ ; situation requires a positive integer.

Full solution

With $n$ items, the average equation is
$\frac{36+2n}{n}=5$
Its denominator requires $n\ne0$ , while the situation requires a positive integer.

Multiplying by $n$ gives $36+2n=5n$ , so $n=12$ .
This is allowed by both filters.
The total charge is $60$ dollars, and $60/12=5$ dollars per item checks the average.

Answer

$n=12$ items; equation $(36+2n)/n=5$ ; domain $n\ne0$ ; situation requires a positive integer.

Key idea

In this average-charge model, the denominator requires a nonzero count, and the situation requires a positive whole number.
Problem 4: Mark complete
Problem 5 Two rectangle measurements

A rectangle has area 20 square cm and perimeter 18 cm. Let $w$ be the length of its shorter side in cm. Write a rational equation for $w$ , solve it, and explain which filter rejects any unusable root.
1. Hint 1
  Express the other side as area divided by $w$ .
2. Hint 2
  Both positive roots can satisfy the equation, so also use the word shorter.
Answer

$w+20/w=9$ ; roots $4,5$ ; shorter side $4$ cm, longer side $5$ cm; $5$ is rejected by context.

Full solution

The other side is $20/w$ , so the perimeter gives
$w+\frac{20}{w}=9$
The domain requires $w\ne0$ .
Multiplying by $w$ gives
$w^2-9w+20=0$ $(w-4)(w-5)=0$

Both $4$ and $5$ are genuine roots of the rational equation.
The context says $w$ is the shorter side.
When $w=5$ , the other side is $4$ , so that assignment is rejected by context, not by the domain.
For $w=4$ , the other side is $5$ ; their area is $20$ and perimeter is $18$ .

Answer

$w+20/w=9$ ; roots $4,5$ ; shorter side $4$ cm, longer side $5$ cm; $5$ is rejected by context.

Key idea

A genuine positive root can still fail a model's stated role for the variable.
Problem 5: Mark complete
Problem 6 Diluting a solution

A solution contains 24 grams of dissolved salt in 200 mL of liquid. Water is added without changing the salt mass until the concentration is $3/50$ gram per mL. Find the added water volume using a rational equation. Assume volumes add.
1. Hint 1
  Concentration is salt mass divided by the new liquid volume.
2. Hint 2
  Keep the nonzero denominator condition separate from the requirement that added water has nonnegative volume.
Answer

$200$ mL of water; equation $24/(200+x)=3/50$ ; domain $x\ne-200$ ; context requires $x\ge0$ .

Full solution

For added volume $x$ in mL, the concentration equation is
$\frac{24}{200+x}=\frac3{50}$
The denominator requires $x\ne-200$ , while the situation requires $x\ge0$ .

Clearing the denominators gives
$1200=3(200+x)$ $x=200$
The new volume is $400$ mL, and $24/400=3/50$ , checking the concentration and the physical condition.

Answer

$200$ mL of water; equation $24/(200+x)=3/50$ ; domain $x\ne-200$ ; context requires $x\ge0$ .

Key idea

When only water is added and volumes add, the concentration is the unchanged salt mass divided by the new total volume.
Problem 6: Mark complete
Problem 7 A family of equations

For real $k$ , classify the real solution set of $\frac2{x-1}+\frac{k}{x+1}=\frac4{x^2-1}$ for every value of $k$ .
1. Hint 1
  Both excluded inputs must remain excluded, even when a numerator coefficient is zero.
2. Hint 2
  After clearing denominators, move all terms to one side and expose the two factors.
Answer

If $k=-2$ : every real $x\ne-1,1$ ; if $k\ne-2$ : no solution.

Full solution

The original denominators exclude $-1$ and $1$ .
Multiplying every term by $(x-1)(x+1)$ gives
$2(x+1)+k(x-1)=4$ $(k+2)(x-1)=0$

If $k=-2$ , the cleared equation is true at every input, leaving the entire original domain.
If $k\ne-2$ , the only candidate is $x=1$ , which is excluded.
Thus those equations have no solution.

Answer

If $k=-2$ : every real $x\ne-1,1$ ; if $k\ne-2$ : no solution.

Key idea

A parameter can turn a rational equation from an identity on its domain into an equation with no allowed solution.
Problem 7: Mark complete
Problem 8 Two routes, one solution set

Solve $\frac{x^2-4x-77}{x-11}=9$ over the real numbers twice: once by simplifying the left side first, and once by multiplying both sides by $x-11$ first. Explain why the two routes produce different candidates but the same solution set.
1. Hint 1
  Simplifying and clearing both change the equation, so each route needs the original domain.
2. Hint 2
  Factor the numerator for the first route; move every term to one side and factor for the second.
Answer

The solution set is $\{2\}$ . Simplifying first gives only the candidate $2$ ; clearing first gives the candidates $2$ and $11$ , and $11$ is excluded. Both routes, filtered by the original domain $x\ne11$ , give $\{2\}$ .

Full solution

The original denominator excludes $11$ .

Route one: the numerator factors as $(x-11)(x+7)$ .
For $x\ne11$ , canceling the common factor leaves $x+7=9$ , so the only candidate is $2$ .

Route two: multiplying both sides by $x-11$ gives
$x^2-4x-77=9x-99.$
Collecting terms gives
$x^2-13x+22=0$ $(x-11)(x-2)=0$
The candidates are $2$ and $11$ .

The candidate $11$ is excluded, so it is extraneous and is rejected.
At $2$ , the original left side is $(4-8-77)/(-9)=(-81)/(-9)=9$ , which checks.

The lists differ because canceling $x-11$ removed the factor that clearing kept.
Canceling is valid only where $x\ne11$ , so route one silently used the restriction, while route two used it explicitly at the end.
Either way the original domain decides, and the solution set is $\{2\}$ .

Answer

The solution set is $\{2\}$ . Simplifying first gives only the candidate $2$ ; clearing first gives the candidates $2$ and $11$ , and $11$ is excluded. Both routes, filtered by the original domain $x\ne11$ , give $\{2\}$ .

Key idea

Simplifying first and clearing by the LCD can give different candidates, but once the excluded candidates are removed, both routes give the same solution set.
Problem 8: Mark complete
Problem 9 A defined expression

Two students discuss $\frac{x^2-8x+19}{3x+7}=0$ over the real numbers. One says there is no solution. The other says that because the expression is defined at $x=0$ , the equation must have at least one real solution. Decide who is correct and justify the complete solution set.
1. Hint 1
  Ask what being defined at an input tells you about the value there, and what it does not.
2. Hint 2
  Examine whether the numerator can be zero at a real input.
Answer

The first student is correct; there is no real solution.

Full solution

The original domain is
$x\ne-\frac73.$
Multiplying by the nonzero denominator on this domain gives
$x^2-8x+19=0$
Completing the square,
$(x-4)^2+3=0$
Since $(x-4)^2\ge0$ , the left side is at least $3$ for every real $x$ .
No real number satisfies the cleared equation, so the original has no real solution.

At $x=0$ the expression is defined but equals $\frac{19}{7}$ , not $0$ .
Domain membership says an input may be evaluated, not that it solves the equation.

Answer

The first student is correct; there is no real solution.

Key idea

An equation can have a nonempty domain and still have no real solution.
Problem 9: Mark complete
Problem 10 A candidate inside the domain

A student solves $\frac7{x-4}=\frac3{x-4}$ by multiplying both sides by $(x-4)(x+9)$ . The student finds the candidate $x=-9$ , notes that it is not an excluded input, and concludes that it is a solution. Find the equation's real solution set, and explain exactly where the student's reasoning fails.
1. Hint 1
  Multiplying both sides of an equation by an expression can be undone only where that expression is nonzero.
2. Hint 2
  Find the allowed input at which the student's multiplier is zero, and test it in the original equation.
3. Hint 3
  Compare with clearing by the LCD, $x-4$ , alone.
Answer

There is no real solution. The candidate $-9$ is extraneous even though it is an allowed input.

Full solution

The original domain is $x\ne4$ .
Multiplying by the LCD $x-4$ , which is nonzero on the whole domain, gives $7=3$ , which is false.
So the original equation has no solution.

The student's multiplier gives $7(x+9)=3(x+9)$ , so
$4(x+9)=0$
and the only candidate is $-9$ .
At $-9$ the original sides are $\frac7{-13}$ and $\frac3{-13}$ , which are not equal, so $-9$ is extraneous even though it is allowed.

The failure is the step from allowed to genuine.
That step is valid for the LCD because the LCD is nonzero at every allowed input, so the multiplication can be undone there.
The factor $x+9$ is zero at the allowed input $-9$ , where both sides of the new equation collapse to $0=0$ .

Answer

There is no real solution. The candidate $-9$ is extraneous even though it is an allowed input.

Key idea

An allowed root of the multiplied equation solves the original whenever the multiplier is nonzero at that input; otherwise, check it in the original equation.
Problem 10: Mark complete

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