Solving Rational Inequalities Steps & Examples

Rational Inequalities: A rational inequality is an inequality that contains a rational expression. Inequalities such as 3 2 x > 1 , 2 x x − 3 < 4 , 2 x − 3 x − 6 ≥ x , 3 2 x > 1 , 2 x x − 3 < 4 , 2 x − 3 x − 6 ≥ x , and 1 4 − 2 x 2 ≤ 3 x 1 4 − 2 x 2 ≤ 3 x are rational inequalities as they each contain a rational expression.

Some recommendations for solving rational inequalities

When solving linear inequalities, there is only one big thing: it is necessary to change the inequality sign when dividing (or multiplying) inequality by a negative number. To change the sign of inequality means to change the sign “less” to sign “more” or vice versa. At the same time, plus and minus signs, bypassing previously studied mathematical rules, should not be changed anywhere.

If we divide or multiply the inequality by a positive number, the sign of the inequality does not need to be changed. The rest of the solution of linear inequalities is completely identical to the solution of linear equations.

In linear and in any other rational inequalities in no case can one multiply or divide the left or right parts of the inequality by expressions containing a variable (except when this expression is positive or negative on the whole numerical axis, in this case when dividing by always negative expression the inequality sign needs to be changed, and when dividing by always a positive expression, the inequality sign needs to be preserved).

Solution of inequalities of the form:

It is carried out using the interval method , which consists of the following:

1. We depict the coordinate line on which we put all the numbers a _i . These numbers, arranged in ascending order, divide the coordinate line into ( n + 1) intervals of constancy of the function f ( x ).
2. Thus, having determined the sign of f ( x ) at any point of each interval (usually this point is chosen from the convenience of arithmetic operations), we determine the sign of the function on each interval. The main thing is not to substitute the gaps boundaries themselves into the function.
3. We write in response to all those intervals, the sign of the function on which correspond to the basic condition of inequality.

It should also be noted that it is not necessary to investigate the sign of the function on each interval by substituting some value from this interval. Thus, it is sufficient to define the sign of the function only on one interval (usually on the rightmost one), and then moving from this interval to the left along the numerical axis, the signs of the intervals can be alternated according to the principle:

• If the bracket from which the number was taken through which we are moving stands in an odd degree, then when passing through the corresponding point, the inequality sign changes .
• And if the corresponding bracket is in an even degree, then when passing through the corresponding point the inequality sign does not change .

In this case, the following remarks should be taken into account:

• In strict inequalities (signs “less” or “more”) the boundaries of the intervals never enter into the answer, and on the numerical axis they are represented by punctured dots.
• In non-strict inequalities (signs “less than or equal” or “greater than or equal”), the gap boundaries that are taken from the numerator always come in response and are depicted as filled dots (since at these points the function actually vanishes, which satisfies the condition).
• But the boundaries taken from the denominator in non-strict inequalities are always depicted as punctured points and never enter into response (since the denominator turns to zero at these points, which is unacceptable).
• In all inequalities, if the same bracket is in both the numerator and the denominator, then it cannot be reduced by this bracket. It is necessary to depict the point corresponding to it on the axis, and not to forget to exclude from the answer. In this case, when alternating the signs of the intervals, passing through this point, the sign does not need to be changed.

So once again the most important thing: when recording the final answer in inequalities, do not lose individual points that satisfy inequalities (these are the roots of the numerator in non-strict inequalities), and do not forget to exclude all roots of the denominator in all inequalities from the answer.

When solving rational inequalities of a more complex form than indicated above, it is necessary first to reduce them to this form with algebraic transformations, and then apply the interval method with all the subtleties already described. Thus, we can propose the following algorithm for solving rational inequalities :

1. All terms, fractions and other expressions must be moved to the left side of the inequality.
2. If necessary, bring the fraction to a common denominator.
3. Spread out the numerator and denominator of the resulting fraction into factors.
4. Solve the resulting inequality by the method of intervals.

Moreover, when solving rational inequalities, it is not allowed :

1. Multiply fractions “crosswise”.
2. As in the equations, it is impossible to reduce the factors with a variable on both sides of the inequality. If there are such factors, then after transferring all expressions to the left side of the inequality, they need to be put out of the brackets, and then take into account the points that they give after the final decomposition of the obtained expression into factors.
3. Separately consider the numerator and denominator of the fraction.

As in other topics in mathematics, when solving rational inequalities, you can use the method of replacing a variable . The main thing to remember is that after introducing a replacement, the new expression should become simpler and not contain the old variable. In addition, you must not forget to perform a reverse replacement.

When solving systems of rational inequalities, it is necessary to solve all inequalities included in the system in turn. The system requires the fulfillment of two or more conditions, and we are looking for those values ​​of an unknown quantity that satisfy all the conditions at once. Therefore, in the answer of the system of inequalities it is necessary to indicate the common parts of all solutions of individual inequalities (or the common parts of all shaded intervals representing the answers of each individual inequality).

When solving sets of rational inequalities, each of the inequalities is also solved in turn. The totality requires finding all the values ​​of a variable that satisfy at least one of the conditions. That is, any of the conditions, several conditions or all the conditions together. In the answer, the totality of inequalities indicate all parts of all solutions of individual inequalities (or all parts of all shaded intervals, representing the answers of each individual inequality).

The solution of some types of inequalities with modules

Inequalities with modules can and should be solved by consistently disclosing modules on intervals of their sign of constancy. Thus, it is necessary to act in the same way as when solving equations with modules (see below). But there are several relatively simple cases in which the solution of inequality with a module reduces to a simpler algorithm. For example, the solution of inequality of the form:

It is reduced to the solution of the system :

In particular, the inequality:

It can be replaced by an equivalent system :

Well, if in a similar inequality replace the sign “less” by “more”:

That his decision comes down to the solution of the totality :

In particular, the inequality:

May be replaced by an equivalent set :

Thus, it is necessary to remember that for the “module less” inequality, we get a system where both conditions must be met simultaneously, and for the “module more” inequality, we get the totality in which any of the conditions must be satisfied.

When solving rational inequalities with a module of the form:

It is advisable to move on to the following equivalent rational inequality without a module:

Such an inequality cannot be solved by extracting the root (if you honestly extract the root, then you need to put the modules again, and you return to the beginning, if you forget about the modules, this is equivalent to forgetting about them at the very beginning, and this is, of course, an error ). All brackets need to be moved to the left and, in any case without opening the brackets, apply the formula for the difference of squares.

Once again, to solve all other types of inequalities with modules other than those indicated above, it is necessary to disclose all modules included in the inequality at intervals of their sign-constancy and solve the resulting inequalities. Recall in more detail the general meaning of this algorithm:

• First we find the points on the numerical axis in which each of the expressions under the module vanishes.
• Next, we divide the entire numerical axis into the intervals between the obtained points and examine the sign of each of the submodular expressions on each interval. Note that to determine the sign of an expression, it is necessary to substitute in it any value of the variable from the interval, except for the boundary points. Choose those variable values ​​that are easy to substitute.
• Further, on each obtained interval, we reveal all the modules in the original inequality in accordance with their signs on this interval and solve the usual rational inequality obtained taking into account all the rules and subtleties of solving ordinary inequalities without modules.
• The solution of each of the inequalities obtained at a particular interval is combined into a system with the interval itself, and all such systems are combined into an aggregate. Thus, from the solutions of all inequalities, we select only those parts that are included in the interval on which this inequality was obtained and write all these parts into the final answer.

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