A kosarad üres
Inequalities
It happens that we do not examine in which cases do certain quantities take equal values, but on the contrary, the question is when is one of the quantities greater than or less than the other one. The problems formulated this way lead to inequalities, which can be defined similarly as the equations.
Example 1
For which real numbers is it fulfilled that
\latex{2x-3\lt\frac{x}{3}+2?}
Solution
We can consider both sides of the inequality as the assignment rules of functions, and as a solution we have to give those real numbers of the fundamental set of the inequality for which the substitution values of the functions are in the given relation.
In this case if
\latex{f (x) = 2x - 3}
and
\latex{g(x)=\frac{x}{3}+2},
then we are looking for those values \latex{ x }, for which
\latex{f (x)\lt g(x)}.
According to the other definition we consider the inequality as a logic function for which the solutions change its logic value to true. In both cases we try to give this solution set. For this all of the methods listed for the equations can be used, but we have to be careful with some steps when applying them.
If we plot the graphs of the functions on the two sides of the inequality in a common coordinate system, then we get the solution graphically.
This method has its limits, since we cannot always read the solution in question precisely enough.
This method has its limits, since we cannot always read the solution in question precisely enough.
Let us plot the functions on the two sides of example \latex{ 1 }. (Figure 8)
Figure 8
\latex{y=2x-3}
\latex{y=\frac{x}{3}+2}
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ y }
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ x }
\latex{ -1 }
\latex{ -2 }
\latex{ -3 }
\latex{ -1 }
\latex{ -2 }
\latex{ -3 }
In Figure 8 it can be seen that the two graphs intersect each other at point \latex{ x = 3 }. For smaller values the graph of function \latex{ f } is below the graph of function \latex{ g }, therefore the solution set of the inequality will be interval \latex{] –\infty; 3[}.
We can give the answer in the following form too: \latex{ x \lt 3 }.
Example 2
Let us solve the following inequality:
\latex{-2(x-3)-1\leq\frac{x-1}{2}+5}.
Solution
We give the solution the algebraic way now.
\latex{-2x+6-1\leq\frac{x-1}{2}+5}
\latex{-2x+5\leq\frac{x-1}{2}+5}
Multiplying both sides by \latex{ 2 }, and then applying the balance method:
\latex{-4x+10\leq x-1+10};
\latex{-4x+1\leq x};
\latex{1\leq 5x};
\latex{\frac{1}{5}\leq x}.
\latex{-4x+1\leq x};
\latex{1\leq 5x};
\latex{\frac{1}{5}\leq x}.
So the solution set of the inequality is \latex{\left[\frac{1}{5};\;\infty\right[}, a left-closed, right-open interval, i.e. \latex{\frac{1}{5}\leq x}. (Figure 9)
⯁ ⯁ ⯁
Following this method we can do the following steps:
- We can add or subtract the same constant to or from both sides of the inequality.
- We can also add or subtract an expression containing an unknown if it does not change the domain of the inequality.
- We can multiply or divide both sides of the inequality by a positive constant.
Figure 9
\latex{\frac{1}{5}}
\latex{ 0 }
\latex{ 1 }
We have to be careful when doing operations with negative numbers. Namely if for example we multiply both sides of the correct inequality \latex{ 1 \lt 2 } by a negative number, for example by \latex{ -1 }, then the correct inequality becomes this: \latex{ -1 \gt -2 }.
It means the following:
If we multiply or divide the inequality by a negative number, then we have to reverse the direction of the inequality that the solution set of the inequality stays unchanged.
It implies that if we multiply or divide by an expression containing an unknown, then we have to consider that it can be both positive and negative; therefore these cases should be examined separately.
We also have to be careful when taking the reciprocal of the two sides of the inequality, since \latex{2\lt3}, but \latex{\frac{1}{2}\gt \frac{1}{3}} . At the same time \latex{–2\lt3}, and \latex{-\frac{1}{2}\lt\frac{1}{3}}. It means that we have to pay special attention to this step, and that we have to examine the possible signs of values the expressions on the two sides of the inequality can take.
Example 3
Let us solve inequality \latex{\frac{6-2x}{x+5}\leq0}.
Solution I
Because of the denominator \latex{x\neq -5}. We can give the solution with taking it into consideration that the signs of the factors in the fractional expression on the left side determine the sign of the fraction.
Let us mark the zeros of the factors on a number line:
\latex{6-2x = 0}, i.e. \latex{x = 3};
\latex{x + 5 = 0}, i.e. \latex{x = –5}.
\latex{x + 5 = 0}, i.e. \latex{x = –5}.
Let us first consider the factor \latex{ 6 - 2x }. Above the number line let us mark the range, in which factor \latex{ 6 - 2x } is negative, with a dashed line, and the range, in which it is positive, with a solid line. Since this factor is in the numerator, its value can be equal to \latex{ 0 }. It is denoted by a filled circle at the intersection point.
Similarly we can examine the factor in the denominator too, where an empty circle denotes that it cannot take the value of \latex{ 0 }, since it does not have a meaning there. (Figure 10)
Figure 10
\latex{ 0 \lt x + 5 }
\latex{ 0 ≤ 6 - 2x }
\latex{ -5 }
\latex{ 0 }
\latex{ 3 }
\latex{ 3 }
\latex{ 3 }
\latex{ 0 }
\latex{ 0 }
\latex{ -5 }
\latex{ -5 }
On the red straight line we can see the solution. According to this the value of the fraction will be less than or equal to zero if
\latex{x\lt –5} or \latex{x\geq3}.
We can also give the solution with the help of intervals:
\latex{] –\infty;–5[\;\cup\; [3;\infty[}.
This method becomes especially efficient when we are examining inequalities containing products with several factors.
Solution II
If we give the solution by multiplying by expression \latex{ x + 5 }, then we can distinguish two cases. (Figure 11)
- If \latex{x+5\gt0}, i.e. \latex{x\gt-5}, then we get inequality \latex{6-2x\leq 0} the solution of which is \latex{x\geq3.}
This interval is a part of the interval given in the initial condition, thus these numbers will be solutions of the inequality.
- If \latex{x+5\lt0}, i.e. \latex{x\lt-5}, then we get inequality \latex{6-2x\geq0} the solution of which is \latex{x\leq3.} In this case the real numbers satisfying the inequality formulated as the condition will be the solutions, i.e. \latex{\lt-5}.
Summarised: the solutions of the inequality based on the values obtained in the two cases
\latex{x\lt-5} or \latex{x\geq 3}.
\latex{x+5\gt0}
\latex{6-2x\leq0}
Figure 11
\latex{x+5\lt0}
\latex{6-2x\geq0}
\latex{ -5 }
\latex{ 3 }
\latex{ 3 }
\latex{ -5 }
b)
a)
Example 4
Let us solve the following inequality:
\latex{\frac{(x-2)(x+5)}{(x+4)(x-6)}+\frac{(x-1)(x+5)}{(x+2)(x-6)}\geq\frac{2(x+5)}{x-6}}.
Solution
Because of the denominators,
\latex{x\neq–4}; \latex{x\neq6}; \latex{x\neq–2}.
To proceed we first find and use the common denominator which will be
\latex{(x + 4)(x-6)(x + 2)},
then we rearrange the inequality so that one side is equal to zero.
\latex{\frac{(x-2)(x+5)(x+2)+(x-1)(x+5)(x+4)-2(x+5)(x+4)(x+2)}{(x+4)(x-6)(x+2)}\geq 0}.
After that we factorise by factoring out in the numerator.
\latex{\frac{(x+5)(x^{2}-4+x^{2}+3x-4-2x^{2}-12x-16)}{(x+4)(x-4)(x+2)}\geq 0},
\latex{\frac{(x+5)(-9x-24)}{(x+4)(x-6)(x+2)}\leq 0}.
Figure 12
\latex{x+5\geq0}
\latex{3x+8\geq0}
\latex{x+4\gt0}
\latex{x-6\gt0}
\latex{x+2\gt0}
\latex{-5}
\latex{5}
\latex{0}
\latex{-\frac{8}{3}}
Dividing both sides by \latex{ -3 } we get the following inequality:
\latex{\frac{(x+5)(3x+8)}{(x+4)(x-6)(x+2)}\leq 0}.
Then we apply the method reviewed in example \latex{ 3 } by drawing the straight lines belonging to each factor on which we mark the set of signs. (Figure 12)
While examining the fraction we can state the following:
- An expression with several factors is equal to zero if and only if any of the factors in the numerator is equal to zero.
- An expression with several factors is positive if and only if none of its factors is equal to zero and the number of negative factors is even.
- An expression with several factors is negative if and only if none of its factors is equal to zero and the number of negative factors is odd.
Considering these we can see the solution set in the figure and we can represent it below the number line. The solution of the inequality will be the
following:
following:
\latex{x\leq-5}, or \latex{-4\lt x\leq-\frac{8}{3}}, or \latex{-2\lt x \lt6}.
Giving the solution with the help of intervals:
\latex{]-\infty;–5]\;\cup\;]–4;-\frac{8}{3}]\;\cup\;]-2; 6[}.
Exercises
{{exercise_number}}. Solve the following inequalities graphically.
- \latex{2x-3\lt x+1}
- \latex{\frac{x}{2}-1\geq-x+1}
- \latex{x^{2}\leq-3x+4}
- \latex{\mid x\mid\leq\frac{x}{2}+1}
{{exercise_number}}. Solve the following inequalities.
- \latex{\frac{x-1}{2}-\frac{x}{3}\gt0}
- \latex{\frac{x}{2}-\frac{x-2}{3}\gt2x-3}
- \latex{\frac{x-1}{2}-\frac{x}{3}\gt\frac{2x-2}{5}}
- \latex{\frac{3x-1}{2}+\frac{x}{3}\leq\frac{2x+3}{4}-\frac{x-1}{6}}
{{exercise_number}}. For which real numbers are the following inequalities satisfied?
- \latex{(2x+1)(x-1)\leq0}
- \latex{(2x+1)(x-1)(3x-6)\leq0}
- \latex{(2x-3)(x+2)(-x+2)\gt0}
- \latex{(2x-3)(x+2)( -x+2)(3-x)\gt0}
{{exercise_number}}. Solve the following inequalities.
- \latex{\frac{x-1}{x+1}\leq0}
- \latex{\frac{x-1}{x+1}\leq1}
- \latex{\frac{(x-1)(x+2)}{x+1}\leq0}
{{exercise_number}}. For which real numbers are the following inequalities satisfied?
- \latex{\frac{x+1}{x}\lt\frac{x}{x+1}}
- \latex{\frac{x+1}{x-1}\geq\frac{x-2}{x+1}}
- \latex{\frac{x+1}{x(x+2)}+\frac{x-1}{(x+3)(x+2)}\lt\frac{2}{x+2}}
