A progressive collection of 20 practice problems on parametric equations, designed to teach not only how to find the solution, but above all how to correctly discuss all possible cases as the parameter varies.
In each exercise we will carefully analyse when division is permitted, which values of the parameter must be excluded, and under what circumstances the equation becomes determinate, inconsistent, or indeterminate.
Let us recall the fundamental idea: in parametric equations, the parameter must never be treated as a number that is certainly nonzero. Before dividing or simplifying an expression involving the parameter, one must always check for which values it can vanish.
For example, in the equation:
\[ (a-1)x=3 \]
we cannot immediately divide by \(a-1\), because:
\[ a-1=0 \quad \Longleftrightarrow \quad a=1 \]
and division by zero is not defined.
For this reason, solving a parametric equation almost always requires a case analysis.
Exercise 1 — level ★☆☆☆☆
Solve and discuss the equation:
\[ ax=6 \]
Answer
\[ \begin{cases} a\ne0 & \Rightarrow S=\left\{\frac{6}{a}\right\} \\ a=0 & \Rightarrow S=\varnothing \end{cases} \]
Solution
The unknown of the equation is \(x\), while \(a\) is a real parameter.
The equation is:
\[ ax=6 \]
The coefficient of the unknown \(x\) is \(a\).
We cannot immediately divide both sides by \(a\), because the parameter might take the value:
\[ a=0 \]
and division by zero is not defined.
We must therefore distinguish two cases.
Case \(a\ne0\)
If:
\[ a\ne0 \]
then we may divide both sides of the equation by \(a\):
\[ x=\frac{6}{a} \]
For every fixed value of the parameter \(a\) with \(a\ne0\), the equation has exactly one solution:
\[ S=\left\{\frac{6}{a}\right\} \]
Case \(a=0\)
If instead:
\[ a=0 \]
we substitute this value into the original equation:
\[ 0\cdot x=6 \]
that is:
\[ 0=6 \]
This is a false statement, since zero does not equal six.
There is therefore no real value of \(x\) that satisfies the equation.
The equation is thus inconsistent.
Therefore:
\[ S=\varnothing \]
Exercise 2 — level ★☆☆☆☆
Solve and discuss the equation:
\[ (a-2)x=4 \]
Answer
\[ \begin{cases} a\ne2 & \Rightarrow S=\left\{\frac{4}{a-2}\right\} \\ a=2 & \Rightarrow S=\varnothing \end{cases} \]
Solution
The unknown of the equation is \(x\), while \(a\) is a real parameter.
The equation is:
\[ (a-2)x=4 \]
The coefficient of the unknown \(x\) is:
\[ a-2 \]
Before dividing both sides by \(a-2\), we must check when this quantity vanishes.
We solve:
\[ a-2=0 \]
obtaining:
\[ a=2 \]
We must therefore discuss the two cases separately.
Case \(a\ne2\)
If:
\[ a\ne2 \]
then:
\[ a-2\ne0 \]
We may therefore divide both sides by \(a-2\):
\[ x=\frac{4}{a-2} \]
The equation thus has exactly one solution:
\[ S=\left\{\frac{4}{a-2}\right\} \]
Case \(a=2\)
If instead:
\[ a=2 \]
we substitute this value into the original equation:
\[ (2-2)x=4 \]
that is:
\[ 0\cdot x=4 \]
hence:
\[ 0=4 \]
This statement is false.
No real value of \(x\) satisfies the equation.
The equation is therefore inconsistent.
Therefore:
\[ S=\varnothing \]
Exercise 3 — level ★☆☆☆☆
Solve and discuss the equation:
\[ (a+1)x=0 \]
Answer
\[ \begin{cases} a\ne-1 & \Rightarrow S=\{0\} \\ a=-1 & \Rightarrow S=\mathbb{R} \end{cases} \]
Solution
The equation is:
\[ (a+1)x=0 \]
The coefficient of the unknown is:
\[ a+1 \]
Before dividing by this quantity, we must determine when it can vanish.
We solve:
\[ a+1=0 \]
obtaining:
\[ a=-1 \]
We distinguish two cases.
Case \(a\ne-1\)
If:
\[ a\ne-1 \]
then:
\[ a+1\ne0 \]
We may therefore divide both sides by \(a+1\):
\[ x=0 \]
The equation thus has exactly one solution:
\[ S=\{0\} \]
Case \(a=-1\)
If instead:
\[ a=-1 \]
we substitute this value into the original equation:
\[ (-1+1)x=0 \]
that is:
\[ 0\cdot x=0 \]
hence:
\[ 0=0 \]
This is always true, regardless of the value assigned to \(x\).
Every real number therefore satisfies the equation.
The equation is thus indeterminate.
Therefore:
\[ S=\mathbb{R} \]
Exercise 4 — level ★★☆☆☆
Solve and discuss the equation:
\[ (a-3)x=a+1 \]
Answer
\[ \begin{cases} a\ne3 & \Rightarrow S=\left\{\frac{a+1}{a-3}\right\} \\ a=3 & \Rightarrow S=\varnothing \end{cases} \]
Solution
Consider the equation:
\[ (a-3)x=a+1 \]
The coefficient of the unknown is:
\[ a-3 \]
Before dividing by this quantity, we must check when it vanishes.
We solve:
\[ a-3=0 \]
obtaining:
\[ a=3 \]
We must therefore discuss the two cases separately.
Case \(a\ne3\)
If:
\[ a\ne3 \]
then:
\[ a-3\ne0 \]
We may therefore divide both sides by \(a-3\):
\[ x=\frac{a+1}{a-3} \]
The equation thus has exactly one solution:
\[ S=\left\{\frac{a+1}{a-3}\right\} \]
Case \(a=3\)
If instead:
\[ a=3 \]
we substitute this value into the original equation:
\[ (3-3)x=3+1 \]
that is:
\[ 0\cdot x=4 \]
hence:
\[ 0=4 \]
This statement is false.
The equation is therefore inconsistent.
Therefore:
\[ S=\varnothing \]
Exercise 5 — level ★★☆☆☆
Solve and discuss the equation:
\[ (a-2)x=2a-4 \]
Answer
\[ \begin{cases} a\ne2 & \Rightarrow S=\{2\} \\ a=2 & \Rightarrow S=\mathbb{R} \end{cases} \]
Solution
Consider the equation:
\[ (a-2)x=2a-4 \]
The coefficient of the unknown is:
\[ a-2 \]
Before dividing by this quantity, we must check when it vanishes.
We solve:
\[ a-2=0 \]
giving:
\[ a=2 \]
We distinguish two cases.
Case \(a\ne2\)
If:
\[ a\ne2 \]
then we may divide both sides by \(a-2\):
\[ x=\frac{2a-4}{a-2} \]
We now factor out \(2\) from the numerator:
\[ 2a-4=2(a-2) \]
giving:
\[ x=\frac{2(a-2)}{a-2} \]
Since in this case:
\[ a-2\ne0 \]
we may cancel the common factor:
\[ x=2 \]
The equation thus has exactly one solution:
\[ S=\{2\} \]
Case \(a=2\)
If instead:
\[ a=2 \]
we substitute this value into the original equation:
\[ (2-2)x=2\cdot2-4 \]
that is:
\[ 0\cdot x=0 \]
hence:
\[ 0=0 \]
This is always true.
Every real number therefore satisfies the equation.
The equation is thus indeterminate.
Therefore:
\[ S=\mathbb{R} \]
Exercise 6 — level ★★☆☆☆
Solve and discuss the equation:
\[ (a^2-1)x=a+1 \]
Answer
\[ \begin{cases} a \ne \pm 1 & \Rightarrow S=\left\{\frac{1}{a-1}\right\} \\ a=1 & \Rightarrow S=\varnothing \\ a=-1 & \Rightarrow S=\mathbb{R} \end{cases} \]
Solution
Consider the equation:
\[ (a^2-1)x=a+1 \]
The coefficient of the unknown is:
\[ a^2-1 \]
Before dividing by this quantity, we must check when it vanishes.
We solve:
\[ a^2-1=0 \]
This is a difference of squares:
\[ a^2-1=(a-1)(a+1) \]
We therefore obtain:
\[ (a-1)(a+1)=0 \]
giving:
\[ a=1 \]
or:
\[ a=-1 \]
We must therefore discuss three distinct cases.
Case \(a\ne\pm1\)
If:
\[ a\ne1 \qquad \text{and} \qquad a\ne-1 \]
then:
\[ a^2-1\ne0 \]
We may therefore divide both sides by \(a^2-1\):
\[ x=\frac{a+1}{a^2-1} \]
We factor the denominator:
\[ a^2-1=(a-1)(a+1) \]
obtaining:
\[ x=\frac{a+1}{(a-1)(a+1)} \]
Since in this case:
\[ a+1\ne0 \]
we may cancel the factor \(a+1\):
\[ x=\frac{1}{a-1} \]
The equation therefore has exactly one solution:
\[ S=\left\{\frac{1}{a-1}\right\} \]
Case \(a=1\)
If:
\[ a=1 \]
we substitute into the original equation:
\[ (1^2-1)x=1+1 \]
that is:
\[ 0\cdot x=2 \]
hence:
\[ 0=2 \]
This is a contradiction.
Therefore:
\[ S=\varnothing \]
Case \(a=-1\)
If instead:
\[ a=-1 \]
we substitute into the equation:
\[ ((-1)^2-1)x=-1+1 \]
that is:
\[ 0\cdot x=0 \]
hence:
\[ 0=0 \]
This is always true.
Every real number satisfies the equation.
Therefore:
\[ S=\mathbb{R} \]
Exercise 7 — level ★★★☆☆
Solve and discuss the equation:
\[ (a-1)x+2=a \]
Answer
\[ \begin{cases} a\ne1 & \Rightarrow S=\left\{\frac{a-2}{a-1}\right\} \\ a=1 & \Rightarrow S=\varnothing \end{cases} \]
Solution
Consider the equation:
\[ (a-1)x+2=a \]
We first isolate the term containing the unknown.
Subtracting \(2\) from both sides:
\[ (a-1)x=a-2 \]
The coefficient of the unknown is:
\[ a-1 \]
We must check when this coefficient vanishes.
We solve:
\[ a-1=0 \]
obtaining:
\[ a=1 \]
We therefore distinguish two cases.
Case \(a\ne1\)
If:
\[ a\ne1 \]
then we may divide both sides by \(a-1\):
\[ x=\frac{a-2}{a-1} \]
The equation thus has exactly one solution:
\[ S=\left\{\frac{a-2}{a-1}\right\} \]
Case \(a=1\)
If instead:
\[ a=1 \]
we substitute this value into the original equation:
\[ (1-1)x+2=1 \]
that is:
\[ 0\cdot x+2=1 \]
hence:
\[ 2=1 \]
This statement is false.
The equation is therefore inconsistent.
Therefore:
\[ S=\varnothing \]
Exercise 8 — level ★★★☆☆
Solve and discuss the equation:
\[ (a+2)x=a(x+1) \]
Answer
For every real value of \( a \): \[ S=\left\{\frac{a}{2}\right\} \]
Solution
Consider the equation:
\[ (a+2)x=a(x+1) \]
The right-hand side contains a product. Expanding by the distributive property:
\[ a(x+1)=ax+a \]
The equation becomes:
\[ (a+2)x=ax+a \]
Expanding the left-hand side:
\[ ax+2x=ax+a \]
We now collect all terms containing \(x\) on the left-hand side.
Subtracting \(ax\) from both sides:
\[ ax+2x-ax=a \]
The terms \(ax\) cancel:
\[ 2x=a \]
Dividing both sides by \(2\):
\[ x=\frac{a}{2} \]
In this exercise no condition on the parameter arises, because after simplification the coefficient of the unknown is the number \(2\), which never vanishes.
The equation therefore always has exactly one real solution.
Therefore:
\[ S=\left\{\frac{a}{2}\right\} \]
Exercise 9 — level ★★★☆☆
Solve and discuss the equation:
\[ (a-4)x=2a-8 \]
Answer
\[ \begin{cases} a\ne4 & \Rightarrow S=\{2\} \\ a=4 & \Rightarrow S=\mathbb{R} \end{cases} \]
Solution
Consider the equation:
\[ (a-4)x=2a-8 \]
The coefficient of the unknown \(x\) is:
\[ a-4 \]
Before dividing by \(a-4\), we must check when this quantity vanishes.
We solve:
\[ a-4=0 \]
obtaining:
\[ a=4 \]
We must therefore distinguish two cases.
Case \(a\ne4\)
If:
\[ a\ne4 \]
then:
\[ a-4\ne0 \]
We may therefore divide both sides by \(a-4\):
\[ x=\frac{2a-8}{a-4} \]
Factoring out \(2\) from the numerator:
\[ 2a-8=2(a-4) \]
hence:
\[ x=\frac{2(a-4)}{a-4} \]
Since we are working in the case \(a\ne4\), we may cancel the factor \(a-4\):
\[ x=2 \]
Therefore:
\[ S=\{2\} \]
Case \(a=4\)
If instead:
\[ a=4 \]
we substitute this value into the original equation:
\[ (4-4)x=2\cdot4-8 \]
that is:
\[ 0\cdot x=8-8 \]
hence:
\[ 0\cdot x=0 \]
The statement:
\[ 0=0 \]
is always true, regardless of the value of \(x\).
The equation is therefore indeterminate and every real number is a solution:
\[ S=\mathbb{R} \]
Exercise 10 — level ★★★☆☆
Solve and discuss the equation:
\[ (a+3)x=a^2-9 \]
Answer
\[ \begin{cases} a\ne-3 & \Rightarrow S=\{a-3\} \\ a=-3 & \Rightarrow S=\mathbb{R} \end{cases} \]
Solution
Consider the equation:
\[ (a+3)x=a^2-9 \]
The coefficient of the unknown \(x\) is:
\[ a+3 \]
Before dividing by \(a+3\), we determine when this coefficient vanishes:
\[ a+3=0 \]
giving:
\[ a=-3 \]
We therefore distinguish the cases.
Case \(a\ne-3\)
If:
\[ a\ne-3 \]
then:
\[ a+3\ne0 \]
We may divide both sides by \(a+3\):
\[ x=\frac{a^2-9}{a+3} \]
We factor the numerator as a difference of squares:
\[ a^2-9=(a-3)(a+3) \]
obtaining:
\[ x=\frac{(a-3)(a+3)}{a+3} \]
Since in the case \(a\ne-3\) we have \(a+3\ne0\), we may cancel:
\[ x=a-3 \]
Therefore:
\[ S=\{a-3\} \]
Case \(a=-3\)
If instead:
\[ a=-3 \]
we substitute into the original equation:
\[ (-3+3)x=(-3)^2-9 \]
that is:
\[ 0\cdot x=9-9 \]
hence:
\[ 0\cdot x=0 \]
This is true for every real value of \(x\).
The equation is therefore indeterminate:
\[ S=\mathbb{R} \]
Exercise 11 — level ★★★★☆
Solve and discuss the equation:
\[ (a^2-4)x=a-2 \]
Answer
\[ \begin{cases} a \ne 2 \ \text{and}\ a \ne -2 & \Rightarrow S=\left\{\frac{1}{a+2}\right\} \\ a=2 & \Rightarrow S=\mathbb{R} \\ a=-2 & \Rightarrow S=\varnothing \end{cases} \]
Solution
Consider the equation:
\[ (a^2-4)x=a-2 \]
The coefficient of the unknown is:
\[ a^2-4 \]
Before dividing by this quantity, we must check when it vanishes.
We solve:
\[ a^2-4=0 \]
Factoring as a difference of squares:
\[ a^2-4=(a-2)(a+2) \]
We obtain:
\[ (a-2)(a+2)=0 \]
giving:
\[ a=2 \]
or:
\[ a=-2 \]
We must therefore discuss three distinct cases.
Case \(a\ne2\) and \(a\ne-2\)
If:
\[ a\ne2 \qquad \text{and} \qquad a\ne-2 \]
then:
\[ a^2-4\ne0 \]
We may divide both sides by \(a^2-4\):
\[ x=\frac{a-2}{a^2-4} \]
Substituting the factored form of the denominator:
\[ x=\frac{a-2}{(a-2)(a+2)} \]
Since in this case:
\[ a-2\ne0 \]
we may cancel the factor \(a-2\):
\[ x=\frac{1}{a+2} \]
The equation therefore has exactly one solution:
\[ S=\left\{\frac{1}{a+2}\right\} \]
Case \(a=2\)
If:
\[ a=2 \]
we substitute into the original equation:
\[ (2^2-4)x=2-2 \]
that is:
\[ 0\cdot x=0 \]
This is always true.
The equation is therefore indeterminate and every real number is a solution:
\[ S=\mathbb{R} \]
Case \(a=-2\)
If instead:
\[ a=-2 \]
we substitute into the equation:
\[ ((-2)^2-4)x=-2-2 \]
that is:
\[ 0\cdot x=-4 \]
hence:
\[ 0=-4 \]
This is a contradiction.
The equation has no solution.
Therefore:
\[ S=\varnothing \]
Exercise 12 — level ★★★★☆
Solve and discuss the equation:
\[ (a-1)(x-2)=0 \]
Answer
\[ \begin{cases} a\ne1 & \Rightarrow S=\{2\} \\ a=1 & \Rightarrow S=\mathbb{R} \end{cases} \]
Solution
Consider the equation:
\[ (a-1)(x-2)=0 \]
This is a product equal to zero.
Recall the zero-product property:
\[ A\cdot B=0 \quad \Longleftrightarrow \quad A=0 \ \text{or} \ B=0 \]
In our case the two factors are:
\[ a-1 \]
and:
\[ x-2 \]
However, care is needed: the parameter \(a\) is not the unknown of the equation. The only unknown is \(x\).
For this reason we must analyse the values of the parameter.
Case \(a\ne1\)
If:
\[ a\ne1 \]
then:
\[ a-1\ne0 \]
The first factor cannot therefore vanish.
For the product to be zero, the second factor must vanish:
\[ x-2=0 \]
giving:
\[ x=2 \]
Therefore:
\[ S=\{2\} \]
Case \(a=1\)
If instead:
\[ a=1 \]
the first factor becomes:
\[ a-1=0 \]
The equation therefore takes the form:
\[ 0\cdot(x-2)=0 \]
that is:
\[ 0=0 \]
This is always true, regardless of the value of \(x\).
Every real number therefore satisfies the equation.
Therefore:
\[ S=\mathbb{R} \]
Exercise 13 — level ★★★★☆
Solve and discuss the equation:
\[ (a+2)(x-1)=a+2 \]
Answer
\[ \begin{cases} a\ne-2 & \Rightarrow S=\{2\} \\ a=-2 & \Rightarrow S=\mathbb{R} \end{cases} \]
Solution
Consider the equation:
\[ (a+2)(x-1)=a+2 \]
The factor \(a+2\) appears on both sides, but we cannot cancel it without first discussing the case in which it vanishes.
We examine:
\[ a+2=0 \]
giving:
\[ a=-2 \]
We distinguish two cases.
Case \(a\ne-2\)
If:
\[ a\ne-2 \]
then:
\[ a+2\ne0 \]
We may divide both sides by \(a+2\):
\[ x-1=1 \]
Adding \(1\) to both sides:
\[ x=2 \]
Therefore:
\[ S=\{2\} \]
Case \(a=-2\)
If instead:
\[ a=-2 \]
we substitute into the original equation:
\[ (-2+2)(x-1)=-2+2 \]
that is:
\[ 0\cdot(x-1)=0 \]
hence:
\[ 0=0 \]
This is always true, regardless of the value of \(x\).
The equation is therefore indeterminate and every real number is a solution:
\[ S=\mathbb{R} \]
Exercise 14 — level ★★★★☆
Solve and discuss the equation:
\[ (a-1)x=a(x-1) \]
Answer
\[ S=\{a\} \]
Solution
Consider the equation:
\[ (a-1)x=a(x-1) \]
We expand both sides.
The left-hand side is:
\[ (a-1)x=ax-x \]
The right-hand side is:
\[ a(x-1)=ax-a \]
The equation becomes:
\[ ax-x=ax-a \]
Subtracting \(ax\) from both sides:
\[ ax-x-ax=ax-a-ax \]
that is:
\[ -x=-a \]
Multiplying both sides by \(-1\):
\[ x=a \]
In this exercise no special cases need to be distinguished, because after simplification the coefficient of the unknown is \(-1\), which never vanishes.
Therefore, for every real value of the parameter \(a\), the equation has exactly one solution:
\[ S=\{a\} \]
Exercise 15 — level ★★★★☆
Solve and discuss the equation:
\[ \frac{x}{a-1}=2 \]
Answer
\[ a\ne1 \quad \Rightarrow \quad S=\{2a-2\} \]
For \(a=1\) the equation is undefined.
Solution
In this equation the parameter appears in the denominator:
\[ \frac{x}{a-1}=2 \]
Before solving, we must impose the condition for the denominator to be defined.
The denominator cannot be zero:
\[ a-1\ne0 \]
hence:
\[ a\ne1 \]
If \(a=1\), the equation is meaningless, since it would involve division by zero.
For \(a\ne1\), we may multiply both sides by \(a-1\):
\[ x=2(a-1) \]
Expanding:
\[ x=2a-2 \]
Therefore:
\[ a\ne1 \quad \Rightarrow \quad S=\{2a-2\} \]
Exercise 16 — level ★★★★☆
Solve and discuss the equation:
\[ \frac{x-1}{a+2}=3 \]
Answer
\[ a\ne-2 \quad \Rightarrow \quad S=\{3a+7\} \]
For \(a=-2\) the equation is undefined.
Solution
In this equation the parameter appears in the denominator:
\[ \frac{x-1}{a+2}=3 \]
Before solving, we must determine for which values of the parameter the equation is defined.
The denominator cannot be zero:
\[ a+2\ne0 \]
hence:
\[ a\ne-2 \]
If \(a=-2\), the equation is undefined, since it would involve division by zero.
Assuming therefore:
\[ a\ne-2 \]
we may multiply both sides by \(a+2\):
\[ x-1=3(a+2) \]
Expanding the right-hand side:
\[ x-1=3a+6 \]
Adding \(1\) to both sides:
\[ x=3a+7 \]
Therefore:
\[ a\ne-2 \quad \Rightarrow \quad S=\{3a+7\} \]
Exercise 17 — level ★★★★★
Solve and discuss the equation:
\[ (a-1)x=2a-2 \]
Answer
\[ \begin{cases} a\ne1 & \Rightarrow S=\{2\} \\ a=1 & \Rightarrow S=\mathbb{R} \end{cases} \]
Solution
Consider the equation:
\[ (a-1)x=2a-2 \]
The coefficient of the unknown is:
\[ a-1 \]
Before dividing by \(a-1\), we must check when this coefficient vanishes:
\[ a-1=0 \]
giving:
\[ a=1 \]
We distinguish two cases.
Case \(a\ne1\)
If:
\[ a\ne1 \]
then:
\[ a-1\ne0 \]
We may therefore divide both sides by \(a-1\):
\[ x=\frac{2a-2}{a-1} \]
Factoring out \(2\) from the numerator:
\[ 2a-2=2(a-1) \]
hence:
\[ x=\frac{2(a-1)}{a-1} \]
Since in this case \(a-1\ne0\), we may cancel:
\[ x=2 \]
Therefore:
\[ S=\{2\} \]
Case \(a=1\)
If instead:
\[ a=1 \]
we substitute this value into the original equation:
\[ (1-1)x=2\cdot1-2 \]
that is:
\[ 0\cdot x=0 \]
hence:
\[ 0=0 \]
This is true for every real value of \(x\).
The equation is therefore indeterminate:
\[ S=\mathbb{R} \]
Exercise 18 — level ★★★★★
Solve and discuss the equation:
\[ (a+1)x=a^2-1 \]
Answer
\[ \begin{cases} a\ne-1 & \Rightarrow S=\{a-1\} \\ a=-1 & \Rightarrow S=\mathbb{R} \end{cases} \]
Solution
Consider the equation:
\[ (a+1)x=a^2-1 \]
The coefficient of the unknown \(x\) is:
\[ a+1 \]
Before dividing by \(a+1\), we must check when this coefficient vanishes:
\[ a+1=0 \]
giving:
\[ a=-1 \]
We distinguish two cases.
Case \(a\ne-1\)
If:
\[ a\ne-1 \]
then:
\[ a+1\ne0 \]
We may divide both sides by \(a+1\):
\[ x=\frac{a^2-1}{a+1} \]
Factoring the numerator as a difference of squares:
\[ a^2-1=(a-1)(a+1) \]
obtaining:
\[ x=\frac{(a-1)(a+1)}{a+1} \]
Since in this case \(a+1\ne0\), we may cancel the factor \(a+1\):
\[ x=a-1 \]
Therefore:
\[ S=\{a-1\} \]
Case \(a=-1\)
If instead:
\[ a=-1 \]
we substitute this value into the original equation:
\[ (-1+1)x=(-1)^2-1 \]
that is:
\[ 0\cdot x=1-1 \]
hence:
\[ 0\cdot x=0 \]
This is true for every real value of \(x\).
The equation is therefore indeterminate:
\[ S=\mathbb{R} \]
Exercise 19 — level ★★★★★
Solve and discuss the equation:
\[ ax+a=2x+2 \]
Answer
\[ \begin{cases} a\ne2 & \Rightarrow S=\{-1\} \\ a=2 & \Rightarrow S=\mathbb{R} \end{cases} \]
Solution
Consider the equation:
\[ ax+a=2x+2 \]
We bring all terms containing \(x\) to the left-hand side and all remaining terms to the right-hand side.
Subtracting \(2x\) from both sides:
\[ ax-2x+a=2 \]
Subtracting \(a\) from both sides:
\[ ax-2x=2-a \]
Factoring out \(x\) on the left-hand side:
\[ x(a-2)=2-a \]
We observe that:
\[ 2-a=-(a-2) \]
so the equation becomes:
\[ x(a-2)=-(a-2) \]
The coefficient of the unknown is:
\[ a-2 \]
We must therefore distinguish the case \(a-2\ne0\) from the case \(a-2=0\).
Case \(a\ne2\)
If:
\[ a\ne2 \]
then:
\[ a-2\ne0 \]
We may divide both sides by \(a-2\):
\[ x=-1 \]
Therefore:
\[ S=\{-1\} \]
Case \(a=2\)
If instead:
\[ a=2 \]
we substitute into the original equation:
\[ 2x+2=2x+2 \]
This is true for every real value of \(x\).
The equation is therefore indeterminate:
\[ S=\mathbb{R} \]
Exercise 20 — level ★★★★★
Solve and discuss the equation:
\[ (a-1)x=a+3 \]
Answer
\[ \begin{cases} a\ne1 & \Rightarrow S=\left\{\dfrac{a+3}{a-1}\right\} \\ a=1 & \Rightarrow S=\varnothing \end{cases} \]
Solution
Consider the equation:
\[ (a-1)x=a+3 \]
The coefficient of the unknown \(x\) depends on the parameter \(a\). For this reason we cannot immediately solve the equation by dividing by \(a-1\): we must first check when this coefficient vanishes.
We examine the condition:
\[ a-1=0 \]
from which:
\[ a=1 \]
We must therefore distinguish two cases.
Case \(a\ne1\)
If:
\[ a\ne1 \]
then:
\[ a-1\ne0 \]
We may therefore divide both sides by \(a-1\):
\[ x=\frac{a+3}{a-1} \]
The equation thus has exactly one solution.
Therefore:
\[ S=\left\{\dfrac{a+3}{a-1}\right\} \]
Case \(a=1\)
If instead:
\[ a=1 \]
we substitute this value into the original equation:
\[ (1-1)x=1+3 \]
that is:
\[ 0\cdot x=4 \]
hence:
\[ 0=4 \]
This is a contradiction, since zero cannot equal four.
No real number satisfies the equation.
The equation is therefore inconsistent.
Therefore:
\[ S=\varnothing \]
Concluding remarks
First-degree parametric equations illustrate how the behaviour of an equation changes as a parameter varies. Unlike ordinary numerical equations, it is not enough to carry out the calculations mechanically: one must carefully analyse the values of the parameter that can make the coefficient of the unknown vanish.
In many of the exercises above we have seen that:
\[ ax+b=0 \]
behaves differently depending on the value of the coefficient \(a\):
- if \(a\ne0\), the equation has exactly one solution;
- if \(a=0\), the equation may become inconsistent or indeterminate.
For this reason, when studying parametric equations it is essential to:
- identify the coefficient of the unknown;
- determine when that coefficient vanishes;
- discuss all possible cases separately.
This line of reasoning is not confined to first-degree equations: it is a fundamental technique that arises throughout algebra, in the study of systems of equations, inequalities, functions, and more advanced topics.