A progressive collection of 20 practice problems on algebraic fractions, designed to develop a thorough understanding of domain conditions, simplification, operations, and the most common pitfalls to avoid.
In each exercise, the focus is not merely on carrying out the calculations, but on understanding why every step is valid. Algebraic fractions depend fundamentally on the denominator: before performing any transformation, one must always determine for which values the expression is defined.
Exercise 1 — level ★☆☆☆☆
Find the domain of the fraction:
\[ \frac{x+1}{x-3} \]
Answer
\[ x\neq 3 \]
Solution
An algebraic fraction is defined only when its denominator is non-zero. We therefore require:
\[ x-3\neq 0. \]
Solving this condition gives:
\[ x\neq 3. \]
This means the fraction is defined for all real numbers except \(3\).
The domain is therefore:
\[ \mathbb{R}\setminus\{3\}. \]
Exercise 2 — level ★☆☆☆☆
Find the domain of the fraction:
\[ \frac{2x-1}{x^2-9} \]
Answer
\[ x\neq -3, \qquad x\neq 3 \]
Solution
The denominator must be non-zero:
\[ x^2-9\neq 0. \]
To analyse this condition, we factor the polynomial:
\[ x^2-9=(x-3)(x+3). \]
The condition therefore becomes:
\[ (x-3)(x+3)\neq 0. \]
A product is non-zero if and only if none of its factors is zero. We must therefore require:
\[ x-3\neq 0 \qquad \text{and} \qquad x+3\neq 0. \]
This gives:
\[ x\neq 3, \qquad x\neq -3. \]
The domain of the fraction is:
\[ \mathbb{R}\setminus\{-3,3\}. \]
Exercise 3 — level ★☆☆☆☆
Simplify the fraction:
\[ \frac{x^2-4}{x^2+4x+4} \]
Answer
\[ \frac{x-2}{x+2}, \qquad x\neq -2 \]
Solution
To simplify an algebraic fraction, we must first factor both the numerator and the denominator.
The numerator is a difference of squares:
\[ x^2-4=(x-2)(x+2). \]
The denominator is a perfect square:
\[ x^2+4x+4=(x+2)^2. \]
The fraction therefore becomes:
\[ \frac{x^2-4}{x^2+4x+4} = \frac{(x-2)(x+2)}{(x+2)^2}. \]
Before simplifying, we must impose the domain condition:
\[ (x+2)^2\neq 0. \]
From which:
\[ x+2\neq 0, \qquad x\neq -2. \]
We can now cancel the common factor \(x+2\):
\[ \frac{(x-2)(x+2)}{(x+2)^2} = \frac{x-2}{x+2}. \]
Therefore:
\[ \frac{x^2-4}{x^2+4x+4} = \frac{x-2}{x+2}, \qquad x\neq -2. \]
The condition \(x\neq -2\) must be retained even after simplification.
Exercise 4 — level ★☆☆☆☆
Simplify:
\[ \frac{3x^2}{6x} \]
Answer
\[ \frac{x}{2}, \qquad x\neq 0 \]
Solution
Even when an algebraic fraction appears straightforward, the first step is always to examine the denominator. In this case the denominator is:
\[ 6x. \]
A fraction is defined only when its denominator is non-zero, so we require:
\[ 6x\neq 0. \]
Since \(6\) is non-zero, the product \(6x\) vanishes only when \(x=0\). Therefore:
\[ x\neq 0. \]
We can now proceed with the simplification. Writing the numerator in factored form:
\[ 3x^2=3\cdot x\cdot x. \]
The denominator can likewise be written as:
\[ 6x=6\cdot x. \]
Therefore:
\[ \frac{3x^2}{6x} = \frac{3\cdot x\cdot x}{6\cdot x}. \]
The factor \(x\) appears in both the numerator and the denominator. We may cancel it because we have already established that \(x\neq 0\). Were \(x\) equal to zero, the original denominator would vanish and the fraction would be undefined.
This leaves:
\[ \frac{3x}{6}. \]
We now simplify the numerical coefficient:
\[ \frac{3x}{6}=\frac{x}{2}. \]
Hence:
\[ \frac{3x^2}{6x} = \frac{x}{2}, \qquad x\neq 0. \]
The condition \(x\neq 0\) must be retained. The original fraction is not defined at \(x=0\), whereas the expression \(\frac{x}{2}\) would be defined there. Omitting the domain condition would therefore result in a loss of essential information.
Exercise 5 — level ★★☆☆☆
Simplify:
\[ \frac{x^2+5x}{x^2-25} \]
Answer
\[ \frac{x}{x-5}, \qquad x\neq -5,\ 5 \]
Solution
To simplify an algebraic fraction, we must express both the numerator and denominator as products. Only then can we identify any common factors.
Consider the numerator:
\[ x^2+5x. \]
Both terms share the factor \(x\). Factoring it out gives:
\[ x^2+5x=x(x+5). \]
Now consider the denominator:
\[ x^2-25. \]
This is a difference of squares, since:
\[ 25=5^2. \]
Therefore:
\[ x^2-25=x^2-5^2=(x-5)(x+5). \]
The fraction can therefore be rewritten as:
\[ \frac{x^2+5x}{x^2-25} = \frac{x(x+5)}{(x-5)(x+5)}. \]
Before cancelling the common factor \(x+5\), we must establish the domain conditions for the original fraction. The denominator must be non-zero:
\[ x^2-25\neq 0. \]
Using the factorisation just obtained, this condition becomes:
\[ (x-5)(x+5)\neq 0. \]
A product is non-zero if none of its factors is zero. Hence:
\[ x-5\neq 0 \qquad \text{and} \qquad x+5\neq 0. \]
From which:
\[ x\neq 5 \qquad \text{and} \qquad x\neq -5. \]
We can now cancel the common factor \(x+5\):
\[ \frac{x(x+5)}{(x-5)(x+5)} = \frac{x}{x-5}. \]
Therefore:
\[ \frac{x^2+5x}{x^2-25} = \frac{x}{x-5}, \qquad x\neq -5,\ 5. \]
The value \(x=-5\) must remain excluded even though the factor \(x+5\) no longer appears in the final expression. Simplification changes the form of the fraction, but does not alter the domain of the original.
Exercise 6 — level ★★☆☆☆
Simplify:
\[ \frac{x^2-3x}{x^2-6x+9} \]
Answer
\[ \frac{x}{x-3}, \qquad x\neq 3 \]
Solution
To simplify the fraction correctly, we must first factor the numerator and denominator.
Starting with the numerator:
\[ x^2-3x. \]
Both terms contain the factor \(x\). Factoring it out gives:
\[ x^2-3x=x(x-3). \]
Now consider the denominator:
\[ x^2-6x+9. \]
This trinomial is a perfect square. Indeed:
\[ (x-3)^2=x^2-6x+9. \]
Therefore:
\[ x^2-6x+9=(x-3)^2. \]
The fraction becomes:
\[ \frac{x^2-3x}{x^2-6x+9} = \frac{x(x-3)}{(x-3)^2}. \]
Before simplifying, we must impose the domain condition. The original denominator cannot equal zero:
\[ x^2-6x+9\neq 0. \]
Since:
\[ x^2-6x+9=(x-3)^2, \]
the condition becomes:
\[ (x-3)^2\neq 0. \]
A square is non-zero if and only if its base is non-zero. Hence:
\[ x-3\neq 0. \]
From which:
\[ x\neq 3. \]
We can now cancel one factor of \(x-3\):
\[ \frac{x(x-3)}{(x-3)^2} = \frac{x}{x-3}. \]
We thus obtain:
\[ \frac{x^2-3x}{x^2-6x+9} = \frac{x}{x-3}, \qquad x\neq 3. \]
The condition \(x\neq 3\) is not a minor detail: at \(x=3\) the original denominator equals zero, so that value cannot be admitted.
Exercise 7 — level ★★☆☆☆
Find a common denominator for:
\[ \frac{2}{x} \qquad \text{and} \qquad \frac{3}{x+1} \]
Answer
\[ \frac{2(x+1)}{x(x+1)} \qquad \text{and} \qquad \frac{3x}{x(x+1)}, \qquad x\neq -1,\ 0 \]
Solution
Reducing two fractions to a common denominator means rewriting them as equivalent fractions sharing a common denominator. This operation is essential, for instance, whenever one wishes to add or subtract algebraic fractions.
The denominators of the two fractions are:
\[ x \qquad \text{and} \qquad x+1. \]
Before constructing the common denominator, we determine the domain conditions. We must require:
\[ x\neq 0 \qquad \text{and} \qquad x+1\neq 0. \]
The second condition is equivalent to:
\[ x\neq -1. \]
The overall conditions are therefore:
\[ x\neq 0, \qquad x\neq -1. \]
Since the denominators \(x\) and \(x+1\) share no common factors, the lowest common denominator is their product:
\[ x(x+1). \]
Consider the first fraction:
\[ \frac{2}{x}. \]
To obtain the denominator \(x(x+1)\), we multiply both numerator and denominator by the missing factor \(x+1\):
\[ \frac{2}{x} = \frac{2(x+1)}{x(x+1)}. \]
Now consider the second fraction:
\[ \frac{3}{x+1}. \]
Here the missing factor is \(x\), so:
\[ \frac{3}{x+1} = \frac{3x}{x(x+1)}. \]
The two fractions reduced to a common denominator are therefore:
\[ \frac{2(x+1)}{x(x+1)} \qquad \text{and} \qquad \frac{3x}{x(x+1)}, \qquad x\neq -1,\ 0. \]
The domain conditions ensure that the factors appearing in the denominators are non-zero. It is for this reason that the transformations performed yield equivalent fractions on the common domain.
Exercise 8 — level ★★☆☆☆
Calculate:
\[ \frac{1}{x-2}+\frac{3}{x+2} \]
Answer
\[ \frac{4(x-1)}{(x-2)(x+2)}, \qquad x\neq -2,\ 2 \]
Solution
To add two algebraic fractions with different denominators, we must first find a common denominator. Before performing any transformation, however, we determine the domain conditions.
The denominators are:
\[ x-2 \qquad \text{and} \qquad x+2. \]
We must therefore require:
\[ x-2\neq 0 \qquad \text{and} \qquad x+2\neq 0. \]
From these conditions we obtain:
\[ x\neq 2 \qquad \text{and} \qquad x\neq -2. \]
The most convenient common denominator is the product of the two denominators:
\[ (x-2)(x+2). \]
The first fraction is missing the factor \(x+2\). We therefore multiply both numerator and denominator by \(x+2\):
\[ \frac{1}{x-2} = \frac{x+2}{(x-2)(x+2)}. \]
The second fraction is missing the factor \(x-2\). We multiply numerator and denominator by \(x-2\):
\[ \frac{3}{x+2} = \frac{3(x-2)}{(x-2)(x+2)}. \]
Now that both fractions share the same denominator, we can add the numerators and keep the common denominator:
\[ \frac{1}{x-2}+\frac{3}{x+2} = \frac{x+2+3(x-2)}{(x-2)(x+2)}. \]
Expanding the numerator:
\[ x+2+3(x-2)=x+2+3x-6. \]
Collecting like terms:
\[ x+2+3x-6=4x-4. \]
Therefore:
\[ \frac{1}{x-2}+\frac{3}{x+2} = \frac{4x-4}{(x-2)(x+2)}. \]
We can factor \(4\) from the numerator:
\[ 4x-4=4(x-1). \]
Hence:
\[ \frac{1}{x-2}+\frac{3}{x+2} = \frac{4(x-1)}{(x-2)(x+2)}, \qquad x\neq -2,\ 2. \]
No further simplification is possible, since the factor \(x-1\) does not appear in the denominator. The conditions \(x\neq -2\) and \(x\neq 2\) must nonetheless be stated, as they arise from the original denominators.
Exercise 9 — level ★★☆☆☆
Calculate:
\[ \frac{x}{x+1}-\frac{1}{x-1} \]
Answer
\[ \frac{x^2-2x-1}{(x+1)(x-1)}, \qquad x\neq -1,\ 1 \]
Solution
The expression involves a difference of algebraic fractions. As always, we begin with the domain conditions, identifying the values that cannot be assigned to the variable.
The denominators are:
\[ x+1 \qquad \text{and} \qquad x-1. \]
We must require:
\[ x+1\neq 0 \qquad \text{and} \qquad x-1\neq 0. \]
Therefore:
\[ x\neq -1 \qquad \text{and} \qquad x\neq 1. \]
The common denominator is:
\[ (x+1)(x-1). \]
The first fraction lacks the factor \(x-1\), so:
\[ \frac{x}{x+1} = \frac{x(x-1)}{(x+1)(x-1)}. \]
The second fraction lacks the factor \(x+1\), so:
\[ \frac{1}{x-1} = \frac{x+1}{(x+1)(x-1)}. \]
We can now subtract the numerators. It is important to keep the parentheses, since the minus sign distributes across the entire numerator of the second fraction:
\[ \frac{x}{x+1}-\frac{1}{x-1} = \frac{x(x-1)-(x+1)}{(x+1)(x-1)}. \]
Expanding the numerator:
\[ x(x-1)-(x+1)=x^2-x-x-1. \]
Collecting like terms:
\[ x^2-x-x-1=x^2-2x-1. \]
We therefore obtain:
\[ \frac{x}{x+1}-\frac{1}{x-1} = \frac{x^2-2x-1}{(x+1)(x-1)}. \]
The numerator contains neither the factor \(x+1\) nor the factor \(x-1\), so no further simplification is possible.
Therefore:
\[ \frac{x}{x+1}-\frac{1}{x-1} = \frac{x^2-2x-1}{(x+1)(x-1)}, \qquad x\neq -1,\ 1. \]
Exercise 10 — level ★★☆☆☆
Calculate:
\[ \frac{x^2-1}{x}\cdot \frac{2x}{x+1} \]
Answer
\[ 2(x-1), \qquad x\neq -1,\ 0 \]
Solution
When multiplying algebraic fractions, it is almost always advisable to factor the polynomials before multiplying. This allows us to identify common factors immediately and simplify the calculation.
First, however, we determine the domain conditions. The denominators present are:
\[ x \qquad \text{and} \qquad x+1. \]
We must therefore require:
\[ x\neq 0 \qquad \text{and} \qquad x+1\neq 0. \]
The second condition is equivalent to:
\[ x\neq -1. \]
The domain conditions are therefore:
\[ x\neq 0, \qquad x\neq -1. \]
Now consider the product:
\[ \frac{x^2-1}{x}\cdot \frac{2x}{x+1}. \]
The numerator \(x^2-1\) is a difference of squares:
\[ x^2-1=(x-1)(x+1). \]
Substituting this factorisation, we obtain:
\[ \frac{x^2-1}{x}\cdot \frac{2x}{x+1} = \frac{(x-1)(x+1)}{x}\cdot \frac{2x}{x+1}. \]
We can now cancel the factor \(x+1\), which appears in the numerator of the first fraction and the denominator of the second. This cancellation is valid because in the domain we have required \(x+1\neq 0\), i.e. \(x\neq -1\).
We can also cancel the factor \(x\), which appears in the denominator of the first fraction and the numerator of the second. This is likewise valid since we have required \(x\neq 0\).
After cancelling, we are left with:
\[ 2(x-1). \]
Therefore:
\[ \frac{x^2-1}{x}\cdot \frac{2x}{x+1} = 2(x-1), \qquad x\neq -1,\ 0. \]
Even though the final result is a polynomial, the domain conditions must not be forgotten: the original expression was not defined at \(x=0\) or \(x=-1\).
Exercise 11 — level ★★★☆☆
Calculate:
\[ \frac{x^2-4}{x^2-2x}:\frac{x+2}{x} \]
Answer
\[ 1, \qquad x\neq -2,\ 0,\ 2 \]
Solution
The expression involves a division of algebraic fractions. In such cases, we must ensure not only that the denominators are non-zero, but also that the fraction acting as divisor is itself non-zero.
The denominator of the first fraction is:
\[ x^2-2x. \]
Factoring out \(x\) gives:
\[ x^2-2x=x(x-2). \]
We must therefore require:
\[ x(x-2)\neq 0. \]
From which:
\[ x\neq 0, \qquad x\neq 2. \]
The denominator of the second fraction is also \(x\), giving us the same condition:
\[ x\neq 0. \]
We must now consider an additional condition: the fraction \(\dfrac{x+2}{x}\) is the divisor. Since division by zero is not permitted, we require:
\[ \frac{x+2}{x}\neq 0. \]
In the domain where \(x\neq 0\), a fraction is zero if and only if its numerator is zero. We must therefore require:
\[ x+2\neq 0. \]
From which:
\[ x\neq -2. \]
The overall conditions are:
\[ x\neq -2,\qquad x\neq 0,\qquad x\neq 2. \]
We can now convert the division into multiplication by the reciprocal:
\[ \frac{x^2-4}{x^2-2x}:\frac{x+2}{x} = \frac{x^2-4}{x^2-2x}\cdot \frac{x}{x+2}. \]
We factor the polynomials:
\[ x^2-4=(x-2)(x+2) \]
and
\[ x^2-2x=x(x-2). \]
Substituting, we obtain:
\[ \frac{x^2-4}{x^2-2x}\cdot \frac{x}{x+2} = \frac{(x-2)(x+2)}{x(x-2)}\cdot \frac{x}{x+2}. \]
We can now cancel the common factors. The factor \(x-2\) appears in both a numerator and a denominator, as do the factors \(x+2\) and \(x\). All these cancellations are valid because we have excluded the values that would make these factors zero:
\[ x\neq 2,\qquad x\neq -2,\qquad x\neq 0. \]
After cancellation, we are left with:
\[ 1. \]
Hence:
\[ \frac{x^2-4}{x^2-2x}:\frac{x+2}{x} = 1, \qquad x\neq -2,\ 0,\ 2. \]
The final result is a constant, but the original expression was not defined at \(x=-2\), \(x=0\), or \(x=2\). The domain conditions must therefore be stated.
Exercise 12 — level ★★★☆☆
Simplify:
\[ \frac{x^2+2x+1}{x^2-1} \]
Answer
\[ \frac{x+1}{x-1}, \qquad x\neq -1,\ 1 \]
Solution
To simplify an algebraic fraction, we must first factor the numerator and denominator. Only after this step can we identify common factors to cancel.
Consider the numerator:
\[ x^2+2x+1. \]
This trinomial is the square of the binomial \(x+1\), since:
\[ (x+1)^2=x^2+2x+1. \]
Therefore:
\[ x^2+2x+1=(x+1)^2. \]
Now consider the denominator:
\[ x^2-1. \]
This is a difference of squares:
\[ x^2-1=x^2-1^2. \]
Applying the difference of squares formula gives:
\[ x^2-1=(x-1)(x+1). \]
The fraction becomes:
\[ \frac{x^2+2x+1}{x^2-1} = \frac{(x+1)^2}{(x-1)(x+1)}. \]
Before cancelling the common factor \(x+1\), we must determine the domain conditions for the original fraction:
\[ x^2-1\neq 0. \]
Using the factorisation:
\[ x^2-1=(x-1)(x+1), \]
we obtain:
\[ (x-1)(x+1)\neq 0. \]
A product is non-zero if and only if none of its factors is zero. Therefore:
\[ x-1\neq 0 \qquad \text{and} \qquad x+1\neq 0. \]
From which:
\[ x\neq 1 \qquad \text{and} \qquad x\neq -1. \]
We can now cancel one factor of \(x+1\):
\[ \frac{(x+1)^2}{(x-1)(x+1)} = \frac{x+1}{x-1}. \]
Therefore:
\[ \frac{x^2+2x+1}{x^2-1} = \frac{x+1}{x-1}, \qquad x\neq -1,\ 1. \]
The value \(x=-1\) must remain excluded even though the factor \(x+1\) has been cancelled. The original fraction was not defined at \(x=-1\).
Exercise 13 — level ★★★☆☆
Calculate:
\[ \frac{2}{x-1}+\frac{x}{x^2-1} \]
Answer
\[ \frac{3x+2}{x^2-1}, \qquad x\neq -1,\ 1 \]
Solution
The expression involves a sum of algebraic fractions. To add fractions with different denominators, we must first bring them to a common denominator.
First, however, we determine the domain conditions. The denominators are:
\[ x-1 \qquad \text{and} \qquad x^2-1. \]
We must require:
\[ x-1\neq 0 \qquad \text{and} \qquad x^2-1\neq 0. \]
Factoring:
\[ x^2-1=(x-1)(x+1). \]
The conditions therefore become:
\[ x-1\neq 0 \qquad \text{and} \qquad (x-1)(x+1)\neq 0. \]
From this we obtain:
\[ x\neq 1, \qquad x\neq -1. \]
The most convenient common denominator is:
\[ x^2-1=(x-1)(x+1). \]
The second fraction already has this denominator:
\[ \frac{x}{x^2-1}. \]
The first fraction, however, has denominator \(x-1\). To obtain the denominator \((x-1)(x+1)\), we multiply both numerator and denominator by \(x+1\):
\[ \frac{2}{x-1} = \frac{2(x+1)}{(x-1)(x+1)}. \]
We can now add:
\[ \frac{2}{x-1}+\frac{x}{x^2-1} = \frac{2(x+1)+x}{(x-1)(x+1)}. \]
Expanding the numerator:
\[ 2(x+1)+x=2x+2+x. \]
Collecting like terms:
\[ 2x+2+x=3x+2. \]
Therefore:
\[ \frac{2}{x-1}+\frac{x}{x^2-1} = \frac{3x+2}{(x-1)(x+1)}. \]
Since \((x-1)(x+1)=x^2-1\), we may also write:
\[ \frac{3x+2}{(x-1)(x+1)} = \frac{3x+2}{x^2-1}. \]
Hence:
\[ \frac{2}{x-1}+\frac{x}{x^2-1} = \frac{3x+2}{x^2-1}, \qquad x\neq -1,\ 1. \]
No further simplification is possible, since the numerator \(3x+2\) shares no common factors with the denominator.
Exercise 14 — level ★★★☆☆
Simplify the expression:
\[ \left(\frac{x}{x+2}-\frac{2}{x}\right)\cdot \frac{x(x+2)}{x-2} \]
Answer
\[ \frac{x^2-2x-4}{x-2}, \qquad x\neq -2,\ 0,\ 2 \]
Solution
The expression contains a bracket involving a difference of algebraic fractions, followed by a product. Before carrying out the calculations, we must determine the domain conditions.
The denominators present are:
\[ x+2,\qquad x,\qquad x-2. \]
We must therefore require:
\[ x+2\neq 0,\qquad x\neq 0,\qquad x-2\neq 0. \]
From which:
\[ x\neq -2,\qquad x\neq 0,\qquad x\neq 2. \]
We first simplify the bracket:
\[ \frac{x}{x+2}-\frac{2}{x}. \]
The common denominator is:
\[ x(x+2). \]
The first fraction lacks the factor \(x\), so:
\[ \frac{x}{x+2} = \frac{x^2}{x(x+2)}. \]
The second fraction lacks the factor \(x+2\), so:
\[ \frac{2}{x} = \frac{2(x+2)}{x(x+2)}. \]
Subtracting the two fractions gives:
\[ \frac{x}{x+2}-\frac{2}{x} = \frac{x^2-2(x+2)}{x(x+2)}. \]
Expanding the numerator:
\[ x^2-2(x+2)=x^2-2x-4. \]
The expression in the bracket therefore equals:
\[ \frac{x^2-2x-4}{x(x+2)}. \]
The original expression becomes:
\[ \frac{x^2-2x-4}{x(x+2)}\cdot \frac{x(x+2)}{x-2}. \]
We can now cancel the common factor \(x(x+2)\), which appears in the denominator of the first fraction and the numerator of the second. This cancellation is valid because our domain conditions have already excluded \(x=0\) and \(x=-2\), the values that would make those factors zero.
This leaves:
\[ \frac{x^2-2x-4}{x-2}. \]
We cannot simplify further by cancelling \(x-2\), since the numerator \(x^2-2x-4\) does not have \(x-2\) as a factor. Indeed, substituting \(x=2\) gives:
\[ 2^2-2\cdot 2-4=4-4-4=-4\neq 0. \]
Hence:
\[ \left(\frac{x}{x+2}-\frac{2}{x}\right)\cdot \frac{x(x+2)}{x-2} = \frac{x^2-2x-4}{x-2}, \qquad x\neq -2,\ 0,\ 2. \]
Exercise 15 — level ★★★☆☆
Solve the rational equation:
\[ \frac{x+1}{x-3}=2 \]
Answer
\[ S=\{7\} \]
Solution
In a rational equation, the first step is always to determine the domain conditions. Values that make any denominator zero cannot be solutions of the equation.
In this case the denominator is:
\[ x-3. \]
We must therefore require:
\[ x-3\neq 0. \]
From which:
\[ x\neq 3. \]
We can now solve the equation:
\[ \frac{x+1}{x-3}=2. \]
We multiply both sides by \(x-3\). This is valid because we are working within the domain of the equation, where \(x-3\neq 0\).
This gives:
\[ x+1=2(x-3). \]
Expanding the right-hand side:
\[ x+1=2x-6. \]
Collecting the \(x\) terms on one side and the constant terms on the other:
\[ 1+6=2x-x. \]
Therefore:
\[ 7=x. \]
We have found \(x=7\). We must now verify that this value satisfies the domain condition.
Since:
\[ 7\neq 3, \]
the value found is acceptable.
Therefore:
\[ S=\{7\}. \]
Exercise 16 — level ★★★☆☆
Solve the rational equation:
\[ \frac{x}{x-1}=\frac{2}{x-1} \]
Answer
\[ S=\{2\} \]
Solution
Before solving the equation, we determine the domain conditions. The denominator is \(x-1\), which must be non-zero:
\[ x-1\neq 0. \]
From which:
\[ x\neq 1. \]
Within the domain of the equation, both sides share the same denominator: \(x-1\). Since this denominator is non-zero, we may equate the numerators:
\[ x=2. \]
We must now check whether the value found is consistent with the domain condition.
The condition requires: \(x\neq 1\).
Since \(2\neq 1\), the value \(x=2\) is acceptable.
The solution set is therefore:
\[ S=\{2\}. \]
Exercise 17 — level ★★★★☆
Solve the rational equation:
\[ \frac{x+2}{x-1}=\frac{3x}{x-1} \]
Answer
\[ S=\varnothing \]
Solution
Before solving the equation, we must determine the domain conditions. Even though both denominators are equal, this step cannot be skipped.
The common denominator is:
\[ x-1. \]
We must therefore require:
\[ x-1\neq 0. \]
From which:
\[ x\neq 1. \]
Within the domain of the equation, the denominator \(x-1\) is non-zero. We may therefore equate the numerators:
\[ x+2=3x. \]
Collecting the \(x\) terms on one side:
\[ 2=3x-x. \]
Therefore:
\[ 2=2x. \]
Dividing by \(2\), we obtain:
\[ x=1. \]
The value found must, however, be checked against the domain condition. We had required: \(x\neq 1\).
The value \(x=1\) is therefore not acceptable, since it makes the original denominator zero:
\[ 1-1=0. \]
Consequently, the value found must be rejected.
The equation has no solutions:
\[ S=\varnothing. \]
Exercise 18 — level ★★★★☆
Solve the rational equation:
\[ \frac{1}{x}+\frac{1}{x+1}=1 \]
Answer
\[ S= \left\{ \frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2} \right\} \]
Solution
We begin with the domain conditions. The denominators present in the equation are:
\[ x \qquad \text{and} \qquad x+1. \]
We must therefore require:
\[ x\neq 0 \qquad \text{and} \qquad x+1\neq 0. \]
The second condition is equivalent to:
\[ x\neq -1. \]
Hence:
\[ x\neq 0, \qquad x\neq -1. \]
We now solve the equation:
\[ \frac{1}{x}+\frac{1}{x+1}=1. \]
The lowest common denominator is:
\[ x(x+1). \]
Within the domain of the equation, this product is non-zero. We may therefore multiply both sides by \(x(x+1)\).
Multiplying the first term by \(x(x+1)\) gives:
\[ \frac{1}{x}\cdot x(x+1)=x+1. \]
Multiplying the second term by \(x(x+1)\) gives:
\[ \frac{1}{x+1}\cdot x(x+1)=x. \]
Multiplying the right-hand side by \(x(x+1)\) gives:
\[ 1\cdot x(x+1)=x(x+1). \]
The equation therefore becomes:
\[ x+1+x=x(x+1). \]
Collecting like terms on the left-hand side:
\[ 2x+1=x(x+1). \]
Expanding the right-hand side:
\[ 2x+1=x^2+x. \]
Bringing all terms to the right-hand side:
\[ 0=x^2+x-2x-1. \]
Collecting like terms:
\[ 0=x^2-x-1. \]
We must therefore solve:
\[ x^2-x-1=0. \]
Applying the quadratic formula:
\[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. \]
Here: \(a=1\), \(b=-1\), \(c=-1\). Therefore:
\[ x=\frac{-(-1)\pm\sqrt{(-1)^2-4\cdot 1\cdot(-1)}}{2\cdot 1}. \]
Simplifying:
\[ x=\frac{1\pm\sqrt{1+4}}{2}. \]
Hence:
\[ x=\frac{1\pm\sqrt{5}}{2}. \]
We must now verify that the values found are consistent with the domain conditions. The conditions were: \(x\neq 0\), \(x\neq -1\).
The two values
\[ \frac{1-\sqrt{5}}{2} \qquad \text{and} \qquad \frac{1+\sqrt{5}}{2} \]
are neither \(0\) nor \(-1\), so both are acceptable.
Therefore:
\[ S= \left\{ \frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2} \right\}. \]
Exercise 19 — level ★★★★☆
Simplify:
\[ \frac{x^2-5x+6}{x^2-4}\cdot \frac{x+2}{x-3} \]
Answer
\[ 1, \qquad x\neq -2,\ 2,\ 3 \]
Solution
The expression involves a product of algebraic fractions. Before carrying out the multiplication, it is advisable to factor all polynomials. The domain conditions, however, must be derived from the denominators of the original expression.
The denominators are:
\[ x^2-4 \qquad \text{and} \qquad x-3. \]
We must therefore require:
\[ x^2-4\neq 0 \qquad \text{and} \qquad x-3\neq 0. \]
Factoring the first denominator:
\[ x^2-4=(x-2)(x+2). \]
The condition \(x^2-4\neq 0\) therefore becomes:
\[ (x-2)(x+2)\neq 0. \]
A product is non-zero if none of its factors is zero. Hence:
\[ x-2\neq 0 \qquad \text{and} \qquad x+2\neq 0. \]
From which:
\[ x\neq 2 \qquad \text{and} \qquad x\neq -2. \]
From the second condition:
\[ x-3\neq 0 \qquad \Longrightarrow \qquad x\neq 3. \]
The overall domain conditions are therefore:
\[ x\neq -2,\qquad x\neq 2,\qquad x\neq 3. \]
We now proceed with the simplification. Factoring the numerator of the first fraction:
\[ x^2-5x+6. \]
We look for two numbers whose product is \(6\) and whose sum is \(-5\). These numbers are \(-2\) and \(-3\). Therefore:
\[ x^2-5x+6=(x-2)(x-3). \]
Moreover, as already established:
\[ x^2-4=(x-2)(x+2). \]
The expression becomes:
\[ \frac{(x-2)(x-3)}{(x-2)(x+2)}\cdot \frac{x+2}{x-3}. \]
We can now cancel the common factors.
The factor \(x-2\) appears in both the numerator and denominator of the first fraction. We may cancel it since we have excluded \(x=2\).
The factor \(x+2\) appears in the denominator of the first fraction and the numerator of the second. We may cancel it since we have excluded \(x=-2\).
The factor \(x-3\) appears in the numerator of the first fraction and the denominator of the second. We may cancel it since we have excluded \(x=3\).
After all these cancellations, no variable factor remains:
\[ 1. \]
Therefore:
\[ \frac{x^2-5x+6}{x^2-4}\cdot \frac{x+2}{x-3} = 1, \qquad x\neq -2,\ 2,\ 3. \]
Even though the final result is the constant \(1\), the domain conditions must not be forgotten. The original expression is not defined at \(x=-2\), \(x=2\), or \(x=3\).
Exercise 20 — level ★★★★★
Simplify the expression:
\[ \left(\frac{x+1}{x^2-1}-\frac{1}{x-1}\right):\frac{x}{x+1} \]
Answer
\[ 0, \qquad x\neq -1,\ 0,\ 1 \]
Solution
This expression first involves a difference of algebraic fractions, followed by division by another fraction. For this reason, the domain conditions must be determined with particular care.
The denominators present are:
\[ x^2-1,\qquad x-1,\qquad x+1. \]
We must therefore require:
\[ x^2-1\neq 0,\qquad x-1\neq 0,\qquad x+1\neq 0. \]
Factoring:
\[ x^2-1=(x-1)(x+1). \]
The conditions on the denominators therefore give:
\[ x\neq 1 \qquad \text{and} \qquad x\neq -1. \]
There is, however, an additional condition. The fraction
\[ \frac{x}{x+1} \]
is the divisor of the expression. Dividing by a fraction means multiplying by its reciprocal, but this is only valid if the divisor fraction is itself non-zero.
We must therefore require:
\[ \frac{x}{x+1}\neq 0. \]
In the domain where \(x+1\neq 0\), a fraction is zero if and only if its numerator is zero. We must therefore also exclude:
\[ x=0. \]
The overall conditions are:
\[ x\neq -1,\qquad x\neq 0,\qquad x\neq 1. \]
We now work on the bracket:
\[ \frac{x+1}{x^2-1}-\frac{1}{x-1}. \]
Factoring the denominator of the first fraction:
\[ x^2-1=(x-1)(x+1). \]
Therefore:
\[ \frac{x+1}{x^2-1} = \frac{x+1}{(x-1)(x+1)}. \]
Within the domain of the expression we know that \(x+1\neq 0\). We may therefore cancel the common factor \(x+1\):
\[ \frac{x+1}{(x-1)(x+1)} = \frac{1}{x-1}. \]
The bracket becomes:
\[ \frac{1}{x-1}-\frac{1}{x-1}. \]
The difference of two equal fractions is zero:
\[ \frac{1}{x-1}-\frac{1}{x-1}=0. \]
The original expression therefore reduces to:
\[ 0:\frac{x}{x+1}. \]
By our domain conditions, we have required that \(\dfrac{x}{x+1}\neq 0\). The division is therefore valid and the result is:
\[ 0. \]
Therefore:
\[ \left(\frac{x+1}{x^2-1}-\frac{1}{x-1}\right):\frac{x}{x+1} = 0, \qquad x\neq -1,\ 0,\ 1. \]
The final result is zero, but the original expression is not defined at \(x=-1\), \(x=0\), or \(x=1\). These values must therefore remain excluded.