Exercises: Algebraic fractions

Simplify the following algebraic fraction by taking out common factors in the numerator and in the denominator: $\frac{4x^3 - 2x^2}{6x^2 - 2x}$

Check, by cross-multiplying, whether the algebraic fractions $\frac{x-3}{x+1}$ and $\frac{x^2-9}{x^2+4x+3}$ are equivalent.

Perform the following addition of algebraic fractions and simplify the result: $\frac{3x+2}{x-2} + \frac{-x-6}{x-2}$

Multiply the following algebraic fractions and give the result in its simplest form: $\frac{x}{x+2} \cdot \frac{x^2-4}{x^2}$

Divide the following algebraic fractions and simplify: $\frac{2x+2}{x-1} : \frac{4x+4}{x^2-1}$

Find the value of the algebraic fraction $\frac{x^2 - 3x + 2}{x^2 + 1}$ for $x = -2$.

Simplify the following algebraic fraction using algebraic identities: $\frac{x^2 - 25}{x^2 - 10x + 25}$

Perform the following operation and simplify the result: $\frac{x^2+5x}{x+3} - \frac{x^2-9}{x+3}$

Lucía has to simplify the algebraic fraction $\dfrac{x^2+4x}{x^2+6x+8}$ on an exam and writes the following on her paper:

1. a) Identify the conceptual mistake Lucía made in her first step and explain in detail why it is not mathematically valid to cancel out the $x^2$ term. Provide a simple numerical counterexample (using a fraction with concrete numbers and additions) that shows that cancelling out addends leads to an incorrect result.
2. b) Solve the simplification correctly by factoring the numerator and the denominator, and state which values of $x$ do not belong to the domain of the original fraction.

Find an algebraic fraction $\dfrac{P(x)}{Q(x)}$ that simultaneously satisfies the following three conditions:

1) Its simplified irreducible form is $\dfrac{2}{x-5}$.

2) The only values that do not belong to its domain (forbidden values) are $x = 5$ and $x = -3$.

3) The polynomials $P(x)$ and $Q(x)$ must be fully expanded in the final answer (without indicated products or factored parentheses).

Explain step by step how you deduced the numerator and the denominator from the given domain restrictions.

Simplify the following algebraic fraction by first factorising the numerator and the denominator: $\frac{x^3 - 4x^2 + 4x}{x^3 - 2x^2 - 4x + 8}$

Perform the following operations by finding a common denominator and simplify the result: $\frac{2}{x-1} + \frac{x}{x+1} - \frac{2x}{x^2-1}$

The base of a rectangular solar panel is $x+5$ metres and its height is $x-5$ metres. The side of a square solar panel is $x$ metres. Write down the algebraic fraction representing the quotient of the area of the rectangular panel and the area of the square panel. Calculate the value of this fraction if $x = 10$ metres.

A hot water tap takes $x$ hours to fill a bath, while the cold water tap takes $x+3$ hours. Write a simplified algebraic fraction representing the fraction of the bath that fills in one hour if both taps are turned on together.

Perform the following combined operation with algebraic fractions and simplify the result: $\left( \frac{1}{x} - \frac{2}{x+2} \right) : \frac{x-2}{x^2+2x}$

The total cost in euros of manufacturing $x$ bicycles is given by the polynomial $C(x) = 3x^2 + 150x$. The total revenue in euros from selling those $x$ bicycles is expressed by the polynomial $I(x) = 5x^2 + 200x$. Write down the irreducible algebraic fraction that represents the ratio of revenue to costs.

Perform the following subtraction of algebraic fractions and simplify as much as possible: $\frac{x^2+x}{x^2-1} - \frac{x}{x-1}$

A train travels a distance of $120$ km at a constant speed of $v$ km/h. On the return journey, due to rain, its speed decreases by $10$ km/h. Write down the algebraic fraction representing the total time for the round trip in terms of $v$, simplifying to a single fraction.

To calculate and simplify the subtraction $\dfrac{x}{x^2-9} - \dfrac{1}{2x+6}$, two 4th-year ESO students propose different strategies:

• Marcos proposes multiplying both denominators directly by each other to find a common denominator: $(x^2-9) \cdot (2x+6) = 2x^3+6x^2-18x-54$.

• Ainhoa proposes first factoring each denominator, obtaining $x^2-9 = (x-3)(x+3)$ and $2x+6 = 2(x+3)$, to calculate the least common multiple ($\text{lcm}$) from there.

1. a) Compare both procedures and justify which of the two is preferable to minimise the difficulty of the calculations and avoid unnecessarily high-degree polynomials.
2. b) Carry out the operation following the most convenient method, simplify the result until you obtain an irreducible fraction, and indicate which values of $x$ must be excluded from the expression.

Consider the algebraic expression $F(x) = \dfrac{3x^2 - 5x}{x^2 - 1} - 3$.

1. a) Estimate: before you start calculating with pencil and paper, mentally analyse what happens when the variable $x$ takes a huge positive value (for example, $x = 100\,000$). What value do you think the quotient $\dfrac{3x^2-5x}{x^2-1}$ approaches when ignoring the lower-degree terms? From that deduction, make a reasoned estimate of what value $F(x)$ will approach for very large numbers, and whether you expect the result to be slightly positive or slightly negative.
2. b) Check: bring the subtraction to a common denominator, carry out the operations in the numerator, and simplify $F(x)$ into a single irreducible algebraic fraction.
3. c) Substitute $x = 100$ into the simplified fraction obtained in part b), calculate its decimal value, and reason whether it confirms the estimate you made in part a).

Perform the operation and simplify the following expression as much as possible: $\frac{x^2-1}{x^2+2x+1} \cdot \left( \frac{2x}{x-1} - \frac{x+1}{x} \right)$

Determine the value of the parameter $k$ so that the algebraic fraction $\frac{x^2 - kx + 10}{x-5}$ can be simplified (that is, the numerator is divisible by the denominator). Once $k$ is found, write down the simplified expression.

Find the polynomial $P(x)$ that makes the following equality between algebraic fractions true: $\frac{x^2-9}{P(x)} = \frac{x-3}{x^2+2x}$

An open-topped box is to be made from a rectangular sheet of cardboard with dimensions $x+6$ and $x+4$ centimetres, by cutting squares of side $2$ cm from the four corners and folding up the flaps. Write down the irreducible algebraic fraction representing the ratio of the volume of the box to the area of its base.

Show, by manipulating and simplifying the algebraic fractions, that the following expression is equal to $2ab$: $\frac{a^3-b^3}{a-b} - \frac{a^3+b^3}{a+b}$

In a chemistry laboratory, $x$ litres of solution A, costing $\frac{20}{x}$ euros per litre, are mixed with $x+4$ litres of solution B, costing $\frac{30}{x+4}$ euros per litre. Write down the algebraic fraction representing the price per litre of the resulting solution, simplifying the expression as much as possible.

Simplify the following algebraic expression in the form of a complex fraction: $1 + \frac{1}{1 + \frac{1}{x-1}}$

An athlete runs on a circular track. During the first half of the distance, they run at a speed of $v$ m/s, and during the second half, they run at a speed of $v-2$ m/s. Given that the average speed for the whole distance is calculated as total distance divided by total time, show using algebraic fractions that the average speed is $\frac{2v(v-2)}{2v-2}$ m/s.

A medium-distance train covers a round-trip journey between two cities connected by a $120\text{ km}$ long track. On the outbound journey it travels at a constant speed $v$ (in $\text{km/h}$), while on the return journey it travels at a faster speed of $(v + 20)\text{ km/h}$.

1. a) Knowing that time is calculated as the quotient of distance and speed, write the algebraic fraction representing the total time of the round-trip journey, $T(v)$, adding the times of both legs using the least common multiple of their denominators.
2. b) The overall average speed of the entire journey is defined as the total distance travelled divided by the total time spent: $V_m(v) = \dfrac{240}{T(v)}$. Simplify this complex fraction and show that the overall average speed can be expressed as $V_m(v) = \dfrac{v(v+20)}{v+10}$.
3. c) A traveller claims: "Since on the way there we went at $v$ and on the way back at $v+20$, the average speed of the entire journey is exactly the ordinary arithmetic mean: $\dfrac{v + (v+20)}{2} = v + 10$". Prove algebraically, by calculating the difference $V_m(v) - (v+10)$ as a single irreducible fraction, that this claim is false, and deduce whether the true overall average speed is always strictly greater or strictly less than that arithmetic mean for any speed $v > 0$.

Elena solves the following combined operation on the board:

In her notebook she writes down the following steps:

• Step 1: Subtraction inside the parentheses: $\dfrac{x(x+1)-2(x-1)}{(x-1)(x+1)} = \dfrac{x^2-x+2}{x^2-1}$.

• Step 2: Convert the division into a product by inverting the second fraction: $\dfrac{x^2-x+2}{x^2-1} \cdot \dfrac{x^2-1}{x^2+x-2}$.

• Step 3: Cancel the block $(x^2-1)$ present in both the numerator and denominator: $\dfrac{x^2-x+2}{x^2+x-2}$.

• Step 4: Cancel $x^2$ with $x^2$, divide $\dfrac{-x}{x} = -1$ and $\dfrac{2}{-2} = -1$, deducing that the final result is $1$.

• Step 5: State that the expression can be evaluated at any real number except at $x = 1$ and $x = -1$.

1. a) Elena made two serious conceptual errors: one algebraic in Step 4 and another in domain analysis in Step 5 (relating to the division of fractions). Explain what each error consists of and why it invalidates her reasoning.
2. b) Solve the exercise correctly starting from Step 3 by factoring the polynomials where possible, and determine the complete list of all values of $x$ for which the original expression lacks mathematical meaning.