A system of inequalities consists of two or more inequalities that must hold simultaneously.
Solving a system of inequalities means finding all and only the values of the variable that satisfy every condition imposed by the system at the same time.
From a set-theoretic standpoint, the solution of a system is obtained by intersecting the solution sets of the individual inequalities.
Table of Contents
- Definition of a system of inequalities
- Fundamental principle
- Intersection of solution sets
- Graphical interpretation on the real number line
- Systems of linear inequalities
- Systems involving quadratic inequalities
- Systems involving rational inequalities
- Complete example using sign analysis
- Inconsistent systems and universally satisfied systems
- Most common mistakes
- General solution procedure
Definition of a system of inequalities
A system of inequalities is a collection of conditions, each expressed as an inequality, that must be satisfied simultaneously by the same variable.
For example:
\[ \begin{cases} x-1>0,\\ 2x+3\le 7. \end{cases} \]
A solution of the system is any real number that makes both inequalities true.
It is therefore not enough to satisfy just one condition: every inequality in the system must hold at the same time.
If we denote by:
\[ S_1 \]
the solution set of the first inequality, and by:
\[ S_2 \]
the solution set of the second, then the solution of the system is:
\[ S=S_1\cap S_2. \]
Fundamental principle
The fundamental principle of systems of inequalities states that:
the solution set of a system is the intersection of the solution sets of the individual inequalities.
This principle follows directly from the meaning of the word "system": all conditions must hold simultaneously.
To solve a system correctly, it is therefore best to proceed as follows:
- solve each inequality separately;
- determine the corresponding solution sets;
- compute the final intersection.
The most frequent mistake is precisely forgetting this last step.
Intersection of solution sets
Consider the system:
\[ \begin{cases} x>1,\\ x\le 4. \end{cases} \]
The first inequality has solution set:
\[ S_1=(1,+\infty). \]
The second inequality has solution set:
\[ S_2=(-\infty,4]. \]
The solution of the system is:
\[ S=S_1\cap S_2. \]
We therefore look for all numbers that belong to both sets simultaneously.
We obtain:
\[ S=(1,4]. \]
Indeed:
- the numbers must be strictly greater than \(1\);
- and at the same time less than or equal to \(4\).
The intersection thus represents the overlap of the two solution sets.
Graphical interpretation on the real number line
When working with systems of inequalities, it is very helpful to plot the solution sets on the real number line.
This makes it immediately clear which part of the line belongs to all solution sets simultaneously.
Consider:
\[ \begin{cases} x\ge -2,\\ x<3. \end{cases} \]
The first inequality describes all numbers greater than or equal to \(-2\).
The symbol:
\[ \ge \]
indicates that the endpoint belongs to the solution set.
The second inequality describes all numbers strictly less than \(3\).
The symbol:
\[ < \]
indicates that \(3\) itself is excluded from the solution.
Taking the intersection, we obtain:
\[ [-2,3). \]
Geometrically, this solution corresponds to the segment of the real line between \(-2\) and \(3\), including the left endpoint but excluding the right one.
Systems of linear inequalities
The simplest systems are those composed entirely of first-degree inequalities.
Consider:
\[ \begin{cases} 2x-1>3,\\ x+4\le 9. \end{cases} \]
We solve the first inequality:
\[ 2x-1>3. \]
Adding \(1\) to both sides:
\[ 2x>4. \]
Dividing by \(2\):
\[ x>2. \]
We now solve the second inequality:
\[ x+4\le 9. \]
Subtracting \(4\) from both sides:
\[ x\le 5. \]
We must therefore intersect:
\[ x>2 \]
with:
\[ x\le 5. \]
The result is:
\[ S=(2,5]. \]
The solution set consists of all numbers strictly greater than \(2\) and at the same time less than or equal to \(5\).
Systems involving quadratic inequalities
A system may also contain quadratic inequalities.
Consider:
\[ \begin{cases} x^2-4>0,\\ x-1\le 0. \end{cases} \]
We solve the first inequality:
\[ x^2-4>0. \]
Factoring the difference of squares:
\[ x^2-4=(x-2)(x+2). \]
The inequality becomes:
\[ (x-2)(x+2)>0. \]
A product is strictly positive when both factors share the same sign.
We therefore examine the sign of the product on each interval determined by the zeros of the factors.
- for \(x<-2\), both factors are negative, so their product is positive;
- for \(-2
2\), the="" factors="" have="" opposite="" signs,="" so="" their="" product="" is="" negative;="" ="" li=""> - for \(x>2\), both factors are positive, so their product is positive.
The solution set of the first inequality is therefore:
\[ (-\infty,-2)\cup(2,+\infty). \]
We now solve the second inequality:
\[ x-1\le 0. \]
This gives:
\[ x\le 1. \]
We now intersect the two solution sets:
\[ \left[(-\infty,-2)\cup(2,+\infty)\right]\cap(-\infty,1]. \]
The intersection is:
\[ S=(-\infty,-2). \]
Indeed, numbers greater than \(2\) cannot belong to the solution, since the second inequality requires them to be at most \(1\).
Systems involving rational inequalities
When a system contains rational inequalities, particular care must be taken with domain conditions.
Consider:
\[ \begin{cases} \dfrac{x+1}{x-2}>0,\\ x<5. \end{cases} \]
The fraction is undefined whenever the denominator equals zero.
We must therefore require:
\[ x\ne 2. \]
We now analyze the sign of the fraction:
\[ \frac{x+1}{x-2}>0. \]
The zeros of the numerator and denominator are:
\[ x=-1, \qquad x=2. \]
A fraction is strictly positive when the numerator and denominator are both positive or both negative.
This yields:
\[ x<-1 \qquad \text{or} \qquad x>2. \]
The second inequality imposes:
\[ x<5. \]
Intersecting:
\[ (-\infty,-1)\cup(2,+\infty) \]
with:
\[ (-\infty,5), \]
we obtain:
\[ S=(-\infty,-1)\cup(2,5). \]
The value:
\[ x=5 \]
is excluded because the second inequality is strict:
\[ x<5. \]
Likewise:
\[ x=2 \]
must be excluded, as it makes the denominator zero.
Complete example using sign analysis
Consider the system:
\[ \begin{cases} \dfrac{x^2-4}{x-1}\ge 0,\\ x<3. \end{cases} \]
We factor the numerator:
\[ x^2-4=(x-2)(x+2). \]
The inequality becomes:
\[ \frac{(x-2)(x+2)}{x-1}\ge 0. \]
The critical points are:
\[ x=-2, \qquad x=1, \qquad x=2. \]
The value:
\[ x=1 \]
must be excluded, as it makes the denominator zero.
We now examine the sign of the expression on each interval determined by the critical points.
- for \(x<-2\), the expression is negative;
- for \(-2
1\), the="" expression="" is="" positive;="" ="" li=""> - for \(1
2\), the="" expression="" is="" negative;="" ="" li=""> - for \(x>2\), the expression is positive.
Since the inequality is:
\[ \ge 0, \]
we must also include the zeros of the numerator:
\[ x=-2, \qquad x=2. \]
The solution set of the first inequality is therefore:
\[ [-2,1)\cup[2,+\infty). \]
The second inequality imposes:
\[ x<3. \]
Taking the intersection, we obtain:
\[ S=[-2,1)\cup[2,3). \]
Inconsistent systems and universally satisfied systems
A system of inequalities can be:
- inconsistent, if it has no solution;
- universally satisfied, if every real number is a solution.
Consider:
\[ \begin{cases} x>3,\\ x<1. \end{cases} \]
No real number can be simultaneously greater than \(3\) and less than \(1\).
The intersection is therefore empty:
\[ S=\varnothing. \]
Now consider:
\[ \begin{cases} x^2+1>0,\\ x^2+2>0. \end{cases} \]
Since:
\[ x^2\ge 0 \]
for every real number, both inequalities are always true.
The solution set is therefore:
\[ S=\mathbb{R}. \]
Most common mistakes
The most frequent errors when solving systems of inequalities are:
- forgetting to intersect the solution sets;
- taking the union of the solution sets instead of the intersection;
- incorrectly reversing the direction of an inequality;
- overlooking domain conditions in rational inequalities;
- making errors during sign analysis.
A particularly common slip is forgetting that the direction of an inequality reverses when both sides are multiplied or divided by a negative number.
For example:
\[ -2x>4. \]
Dividing both sides by \(-2\) reverses the inequality:
\[ x<-2. \]
Writing:
\[ x>-2 \]
would be incorrect.
General solution procedure
In general, the following step-by-step procedure leads to a correct solution for any system of inequalities:
- solve each inequality separately;
- determine the solution sets precisely;
- if helpful, plot the solution sets on the real number line;
- compute the intersection of the solution sets;
- express the final answer as an interval or a union of intervals.
This procedure is sufficient to handle correctly the vast majority of systems of inequalities encountered in a standard algebra course.