- Example 1 - Solve the linear inequality \(1\frac{1}{4} \leq \frac{2}{3}x-1 \)
Solution - $$\begin{eqnarray} -\frac{2}{3}x &\leq& -1 - \frac{5}{4}/\cdot \left(-\frac{3}{2}\right)\\\nonumber x &\geq& \frac{3}{2} + \frac{15}{8} \\ \nonumber x &\geq& \frac{12+15}{8}\\ \nonumber x &\geq& \frac{27}{8}\\ x \in \Big[\frac{27}{8}, +\infty \Big\rangle\end{eqnarray}$$ - Example 2 Solve the linear inequality \(\frac{1}{2}\left(1-\frac{x+6}{6}\right) > \frac{1}{3}\left(\frac{x}{2} - \frac{2}{3}\right)\)
Solution \begin{eqnarray} \frac{1}{2}\left(1-\frac{x+6}{6}\right) &>& \frac{1}{3}\left(\frac{x}{2} - \frac{2}{3}\right)/\cdot 6 \\ 3\frac{x+6}{2} &>& x- \frac{4}{3}/\cdot 6 \\ 18-3x-18 &>& 6x-8 \\ -3x - 6x &>& -8\\ -9x &>& -8 /:(-9)\\x < \frac{8}{9} \Rightarrow x \in \Big\langle -\infty, \frac{8}{9}\Big\rangle\end{eqnarray} - Example 3 Solve the linear system of inequalities \begin{cases}\frac{x-3}{4} \geq x + 3 \\ \frac{4-x}{2} > x + 3\end{cases}
Solution \begin{eqnarray} \frac{x-3}{4} &\geq& x +3 /\cdot 4 \\x-3 &\geq& 4x + 12\\-3x &\geq& 15 /:(-3) \Rightarrow x \leq -5\end{eqnarray} \begin{eqnarray} \frac{4-x}{2} &>& x+3 /\cdot 2 \\4-x &>& 2x+6 \\ -3x &>& 2/:(-3) \Rightarrow x &<& -\frac{2}{3}\end{eqnarray} The soluton of the system are the numbers which are lower than or equal to -5 and lower than \(-\frac{2}{3}\). So the solution of the system is \(x\in \langle -\infty, -5]\) - Example 4 Solve the linear system of inequalities \begin{cases}\frac{5-2x}{2} \geq \frac{10-x}{4}\\ \frac{x+0.5}{3} > \frac{10-x}{4}\end{cases}
Solution\begin{eqnarray}\frac{5-2x}{2} &\geq& \frac{10-x}{4}/\cdot 4 \\10 -4x &\geq& 10-x \\ -3x &\geq& 0 /:(-3) \Rightarrow x \leq 0\end{eqnarray} \begin{eqnarray}\frac{x+0.5}{3} &>& \frac{10-x}{4}/\cdot 12 \\ 4x+2 &>& 30-3x\\ 7x &>& 28/:7 \Rightarrow x > 4 \end{eqnarray} The solution of the system are numbers that are lower than or equal to 0 and greater than 4 at the same time. Since this is imposible the system does not have any solutions. - Example 5 Solve the inequality \(\frac{4}{x-2} > 1.\)
Solution\begin{eqnarray}\frac{4}{x-2} &>& 1 \\\frac{4-x+2}{x-2}&>& 0 \\ \frac{6-x}{x-2} > 0\end{eqnarray} This inequlatiy can be solved by comparing positive and negative values of nominator and denominator. Here, two cases are observed - the frist case \begin{eqnarray} 6-x &>& 0 \Rightarrow x < 6 \\ x-2 &>& 0 \Rightarrow x > 2\\ x \in \langle 2, 6 \rangle \end{eqnarray} The second case \begin{eqnarray}6-x &<& 0 \Rightarrow x > 6 \\x-2 &<& 0 \Rightarrow x < 2 \\ x \in \oslash\end{eqnarray} - Example 6 Solve the inequality \((2x-3)(2-3x) \leq 0\)
Solution\begin{eqnarray} 2x-3 &\leq& 0 \Rightarrow x \leq \frac{3}{2}\\ 2-3x &\geq& 0 \Rightarrow x \leq \frac{2}{3}\\ x &\in& \Big\langle -\infty, -\frac{2}{3}\Big]\end{eqnarray} The second case. \begin{eqnarray}2x-3 &\geq& 0 \Rightarrow x \geq \frac{3}{2} \\ 2-3x &\leq& 0 \Rightarrow x \geq \frac{2}{3} \\ x &\in& \Big[\frac{3}{2}, +\infty \Big\rangle\end{eqnarray} The solution of the system of inequalites is $$x \in \Big\langle -\infty, \frac{2}{3}\Big]\cup \Big[\frac{3}{2}, +\infty\Big\rangle $$ - Example 7 - Solve the inequality \(\frac{x^2 +4}{2-x} \leq 0.\)
Solution The inequlity consits of quadratic function which is located in numerator and linear function in denominator. The entire system has to be less than or equal to 0. The numerator will be always positive for every real number \(x\) so it can be concluded that denominator must be negative so the entire inequality would be negative. In this case the expressio can not be equal to zero. So the enitre problem is reduced to solve the one inequlity. \begin{eqnarray}2-x &<& 0 \Rightarrow x > 2 \\x &\in & \langle 2, +\infty \rangle\end{eqnarray} - Example 8 Solve the inequality \(\frac{3-x}{2x+5} > 1 \)
SolutionBefore solving the inequality the inequality must be reduced. \begin{eqnarray}\frac{3-x}{2x+5} -1 &>& 0 \\ \frac{3-x-2x-5}{2x+5} &>& 0\\ \frac{-3x-2}{2x+5}&>& 0\end{eqnarray} The next step is to find the zero points for nominator and denominator. \begin{eqnarray}-3x-2 &=& 0 \Rightarrow x = -\frac{2}{3} \\ 2x +5 &=& 0 \Rightarrow x = -\frac{5}{2}\end{eqnarray} These zero points are determined to form a table and see how nominator and denomiator behave i.e. are they positive or negative in specific intervlas.\(-\infty \quad-\frac{5}{2}\) \(-\frac{5}{2} \quad -\frac{2}{3}\) \(-\frac{2}{3}\quad +\infty\) \(-3x-2\) + + - \(2x+5\) - + + \(\frac{-3x-2}{2x+5}\) - + - - Example 9 Solve the inequality \(5x-2 \geq 0 \)
Solution \begin{eqnarray}5x-2 &\geq& 0 \\ 5x &\geq& 2 /:5 \\ x &\geq& \frac{2}{5}\Rightarrow x\in \Big[\frac{2}{5}, +\infty\Big\rangle\end{eqnarray} - Example 10 Solve the inequality \(2x-5 \geq 0\)
Solution\begin{eqnarray}2x-5 &\geq& 0 \\2x &\geq& 5/:2 \\x &\geq& \frac{5}{2}\Rightarrow x \in \Big[\frac{5}{2}, +\infty\Big\rangle\end{eqnarray} - Example 11 Solve the inequality \(5x -2 < 0\)
Solution\begin{eqnarray}5x -2 &<& 0 \\ 5x &<& 2 /: 5 \\ x &<& \frac{5}{2} \Rightarrow x \in \Big\langle -\infty, \frac{5}{2}\Big\rangle\end{eqnarray} - Example 12 Solve the inequality \(2x-5>0\)
Solution\begin{eqnarray}2x-5 &>& 0\\ 2x &>& 5/:2 \\ x &>& \frac{5}{2} \Rightarrow x \in \Big\langle \frac{5}{2}, +\infty\Big\rangle \end{eqnarray} - Example 13 Solve the inequality \(2 \leq \frac{x+5}{2} < 4\)
Solution In this example we have two inequalities that will be solved separately. The solution of the first inequality. \begin{eqnarray}2 &\leq& \frac{x+5}{2}/\cdot 2 \\ x+5 &\geq& 4 \\ x &\geq& -1 \Rightarrow x \in [-1, +\infty\rangle\end{eqnarray} The solution of the second inequality \begin{eqnarray}\frac{x+5}{2} &<& 4 /\cdot 2 \\x+5 &<& 8 \\x &<& 3 \Rightarrow x \in \langle -\infty, 3 \rangle\end{eqnarray} Combining those two solution the soltuion of entire inequality is: \(x \in [-1,3\rangle\) - Example 13 Solve the inequality \(3-2x < 0 \)
Solution \begin{eqnarray}3-2x &<& 0 \\ -2x &<& -3 /:(-2) \\ x &>& \frac{3}{2}\Rightarrow x \in \Big\langle \frac{3}{2}, +\infty\Big\rangle\end{eqnarray} - Example 14 Solve the inequality \(3(2+x) > 2\)
Solution \begin{eqnarray}3(2+x) &>& 2 \\ 6 + 3x &>& 2 \\3x &>& -4/:3 \\ x &>& -\frac{4}{3}\end{eqnarray} - Example 15 Determine the solution of x in interval form of the inequality \(1-2x < 3\)
Solution \begin{eqnarray} 1-2x &<& 3 \\-2x &<& 2 /:(-2) \\ x &<& -1 \Rightarrow x\in \langle -\infty, -1\rangle \end{eqnarray} - Example 16 Sove the inequality \( \frac{1-x}{x} > 0 \)
Solution \begin{eqnarray} \frac{1-x}{x} &>& 0 \\ 1-x &>& 0 \Rightarrow x < 1 \\ x &>& 0 \\ x &\in& \langle 0, 1\rangle\end{eqnarray} - Example 17 Solve the inequality \(\frac{2x-4}{x+3} > 0 \)
Solution \begin{eqnarray}\frac{2x-4}{x+3} &>& 0 \\ 2x-4 &>& 0 \Rightarrow 2x > 4/:2 \Rightarrow x > 2\\ x +3 &<& 0 \Rightarrow x < -3 \\ \end{eqnarray} - Example 18 Solve the inequality \(\frac{x^2 +11}{(x-3)(x+1)} > 0 \)
Solution \begin{eqnarray}\frac{x^2 +11}{(x-3)(x+1)} &>& 0 \\ x - 3 &>& 0 \Rightarrow x > 3 \\ x+1 &<& 0 \Rightarrow x < -1 \\ x &\in& \langle -\infty, -1\rangle \cup \langle 3, +\infty\rangle \end{eqnarray} - Example 19 Solve the inequaltiy \(1 < \frac{2-x}{x+1} < 2\)
Solution \begin{eqnarray} x+1 &<& 2-x \\ 2x &<& 1/:2 \Rightarrow x < \frac{1}{2} \\ 2-x &<& 2x +2 \\ -3x &<& 0 /:(-3) \\ x &>& 0 \\ x &\in& \Big\langle 0, \frac{1}{2} \Big\rangle \end{eqnarray} - Example 2 Solve the inequality \(\frac{x+1}{x-3} < -1 \)
Solution \begin{eqnarray}\frac{x+1}{x-3} &<& -1 \\ \frac{x+1+x-3}{x-3} &<& 0 \\ \frac{2x-2}{x-3} &<& 0 /:2 \\\frac{x-1}{x-3} &<& 0\\x &<& 1 \\ x &>& 3 \\x &\in& \langle 1,3\rangle \end{eqnarray} - Example 2 Solve the inequality \(\frac{1-x}{x+1} \leq 0\)
Solution \begin{eqnarray}\frac{1-x}{x+1} &\leq& 0 \\ 1-x &\leq& 0 \\ x &\geq& 1 \\ x+1 &<& 0 \\ x&<& -1 \\ x &\in& \langle -\infty, -1\rangle \cup [1, +\infty \rangle\end{eqnarray} - Example 2 Solve the inequality \(\frac{1}{(x+2)(x+4)} < 0 \)
Solution\begin{eqnarray}\frac{1}{(x+2)(x+4)} &<& 0\\ x+2 &>& 0 \Rightarrow x > -2 \\ x+4 &<& 0 \Rightarrow x < -4\\ x+2 &<& 0\Rightarrow x < -2 \\ x + 4 &>& 0 \Rightarrow x> -4\\ x &\in& \langle -4, -2 \rangle \end{eqnarray} - Example 2 Solve the inequality \(\frac{4-x}{x-2} \geq 0\)
Solution \begin{eqnarray} 4-x &\geq& 0 \Rightarrow x \leq 4\\ x-2 &>& 0 \Rightarrow x > 2\\ x &\in& \langle 2, 4]\end{eqnarray} - Example 2 Solve the system of inequalities \begin{cases}\frac{5+2x}{2} \geq \frac{10 + x}{4}\\ \frac{x+0.5}{3} > \frac{10 + x}{4}\end{cases}
Solution Solution of the first inequality \begin{eqnarray}\frac{5+2x}{2} &\geq& \frac{10 + x}{4}/\cdot 4 \\ 10 + 4x &\geq& 10 + x \\ 3x &\geq& 0 /:3 \\ x &\geq& 0\Rightarrow x \in [0, +\infty\rangle\end{eqnarray} The solution of the second inequality \begin{eqnarray}\frac{x+0.5}{3} &>& \frac{10 - x}{4}/\cdot 12 \\ 4x + 2 &>& 30 - 3x \\ 7x &>& 28 \\ x &>& 4 \Rightarrow x \in \langle 4, +\infty \rangle \end{eqnarray} The solution of the system of inequalities is achieved by combining both solutions i.e. \(x\in \langle 4, +\infty\rangle - Example 2 - Solve the system of inequalites \begin{cases} 3-\frac{1}{2}x < 0 \\\frac{1}{2}x + 1 \geq 0 \end{cases}
Solution The solution of the first inequality \begin{eqnarray}3-\frac{1}{2}x &<& 0 \\ -\frac{1}{2}x &<& -3 /\cdot (-2) \\ x &>& 6 \Rightarrow x \in \langle 6, +\infty \rangle \end{eqnarray} The solution of the second inequality \begin{eqnarray}\frac{1}{2}x + 1 &\geq& 0\\ \frac{1}{2}x &\geq& -1 /\cdot 2 \\ x &\geq& -2 \Rightarrow x \in [-2, +infty\rangle \end{eqnarray} The solution of the system of inequalities is obtained by combining both solutions i.e. \(x \in \langle 6, +\infty \rangle\) - Example 2 Solve the system of inequalities \begin{cases}\frac{x-3}{4} \geq x+3 \\ \frac{4-x}{2} > x+3 \end{cases}
Solution The solution of the first inequality \begin{eqnarray}\frac{x-3}{4} &\geq& x+3/\cdot 4 \\ x-3 &\geq& 4x + 12 \\ -3x &\geq& 15 /:(-3) \\ x &\leq& -5 \Rightarrow x \in \langle -\infty, -5\rangle\end{eqnarray} The solution of the second inequality \begin{eqnarray}\frac{4-x}{2} &>& x+3/\cdot 2\\ 4-x &>& 2x + 6 \\ -3x &>& 2 /:(-3)\\ x &<& -\frac{2}{3} \Rightarrow x \in \Big\langle -\infty, -\frac{2}{3}\Big\rangle\end{eqnarray} The solution of the entire system of inequalities is obtained by combining the soltions of each inequality i.e. the solution is \(x \in \langle -\infty, -5\rangle.\) - Example 2 Solve the inequality \(\frac{3}{x-2} > 1\)
Solution Before we solve inequaltiy the initial inequaltiy must be simplifed. \begin{eqnarray}\frac{3}{x-2} &>& 1 \\ \frac{3-x+2}{x-2} &>& 0 \frac{5-x}{x-2} &>& 0\end{eqnarray} Now we have system of two inequalities one in the nominator and one in denominator. This system has two different cases. \begin{eqnarray}5-x &>& 0 \Rightarrow -x > -5 \Rightarrow x < 5\\ x-2 &>& 0 \Rightarrow x > 2\\ x &\in& \langle 2, 5 \rangle \end{eqnarray} The second case \begin{eqnarray} 5-x &<& 0 \Rightarrow x > 5 \\x-2 &<& 0 \Rightarrow x < 2 \\ x \in \langle -\infty, -2 \rangle \cup \langle 5, +\infty \rangle \end{eqnarray} The second case do not have any solutions - the intervals of both inequalities do not overlap. So the soltuon of the equation is \(x \in \langle 2, 5 \). - Example 2 Solve the inequality \(\frac{2x+6}{7x-9} < 1\)
Solution Before solving the inequality the inequality must bs simplifed. \begin{eqnarray}\frac{2x+6}{7x-9} &<& 1 \\ \frac{2x+6 - 7x + 9}{7x-9} < 0 \\ \frac{15-5x}{7x -9 } &<& 0 \end{eqnarray} \begin{eqnarray}15-5x &<& 0 \\-5x &<& -15 /:(-5) \\ x &>& 3 \\ 7x -9 &<& 0 \\ 7x &<& 9 /: 7 \\ x &<& \frac{9}{7}\\x&\in& \Big\langle -\infty, -\frac{9}{7}\Big\rangle \cup \langle 3, +\infty\rangle \end{eqnarray} - Example 2 Solve the inequality \(\frac{2x-5}{6x-9} < 2\)
Solution \begin{eqnarray}\frac{2x-5}{6x-9} &<& 2 \\ \frac{2x-5 - 12x + 18}{6x-9} &<& 0 \\ \frac{13-10x}{6x-9} &<& 0 \\ 13-10x &<& 0 \Rightarrow -10 x &<& -13 /:(-10) \Rightarrow x &>& \frac{13}{10} \\ 6x-9 &<& 0 \Rightarrow 6x < 9/: 6 \Rightarrow x < \frac{3}{2}\end{eqnarray} - Example 2 Solve the inequality \(\frac{x^2}{3-x}>0\)
Solution The nominator will always be greater than zero. The only condition that must be tested is the expression in denominator. \begin{eqnarray}3-x &>&0\\-x&>& -3 /\cdot(-1) \\x &<& 3 \Rightarrow x \in \langle -\infty, 3 \rangle\end{eqnarray} - Example 2 Solve the inequality \(\frac{2}{x-4} \geq 1\)
Solution Before we investigate inequalities in numinator and denominator the expressions must be simplified. \begin{eqnarray} \frac{2}{x-4} &\geq& 1 \\ \frac{2-x+4}{x-4} &\geq& 0 \\ \frac{6-x}{x-4} &\geq& 0\end{eqnarray} \begin{eqnarray} 6-x &\geq& 0 \\ -x &\geq& -6 /\cdot(-1) \\ x&\leq& 6 \end{eqnarray} The denominator must be greater than zero it can not be equal to 0. \begin{eqnarray} x-4 &>& 0 \\ x &>& 4 \end{eqnarray} The solution of the intial inequality can be written as \( x \in \langle 4, 6]\) - Example 2 Solve the inequality \(\frac{x-1}{2x-1} \leq \frac{1}{3}\)
Solution \begin{eqnarray} \frac{x-1}{2x-1} &\leq& \frac{1}{3}\\ \frac{x-1}{2x-1} - \frac{1}{3} &\leq& 0 \\ \frac{3x-3 - 2x +1}{6x-3} &\leq& 0 \\ \frac{x - 2}{6x-3} &\leq& 0\\ x-2 &\leq& 0 \Rightarrow x \leq 2 \\ 6x-3 &>& 0 \Rightarrow 6x &>& 3 /: 6 \Rightarrow x &>& \frac{1}{2} \\ x &\in& \Big\langle \frac{1}{2}, 2\Big] \end{eqnarray} - Example 2 Solve the inequality \(\frac{x}{x+1} - \frac{x^2+1}{x^2-1} >\frac{1}{2}\)
Solution \begin{eqnarray}\frac{x}{x+1} - \frac{x^2+1}{x^2-1} &>& \frac{1}{2} \\ \frac{x^2 - x - x^2 - 1}{(x-1)(x+1)} - \frac{1}{2} &>& 0 \\ \frac{-2x-2 -x^2 +1}{2(x^2 -1)} &>& 0 \\ \frac{-x^2 - 2x -1 }{2(x^2 -1)} &>& 0/\cdot 2 \\ -\frac{(x+1)^2}{(x-1)(x+1)} &>& 0 /\cdot (-1) \\ \frac{x+1}{x-1} &<& 0\\ x+1 &<& 0 \Rightarrow x <-1 \\ x-1 & >& 0 \Rightarrow x > 1 \\ x + 1 &>& 0 \Rightarrow x > -1 \\ x-1 &<& 0 \Rightarrow x < 1 \\ x &\in& \langle -1, 1 \rangle \end{eqnarray} - Example 2 Solve the inequality \(x-4[\frac{1}{4}x - 7[\frac{9}{7}x - 3 (\frac{7}{3}x +1)]] \leq 0 \)
Solution \begin{eqnarray} x-4\Big[\frac{1}{4}x - 7\Big[\frac{9}{7}x - 3 \Big(\frac{7}{3}x +1\Big)\Big]\Big] &\leq& 0 \\-84 - 160x &\leq& 0 \\ 160x &\geq& -84/:160 \Rightarrow x \geq -\frac{84}{160} \\x &\geq& -\frac{21}{40} \Rightarrow x \in \Big[-\frac{21}{40}, +\infty\Big\rangle \end{eqnarray} - Example 2 Solve the inequality \(x^2 (x-1) > 0\)
SolutionThe \(x^2\) will always be greater than 0 so the x-1 part of inequality must be investigated.\begin{eqnarray}x-1 &>& 0 \\x &>& 1 \Rightarrow x &\in& \langle 1, +\infty\rangle \end{eqnarray} - Example 2 Solve the inequality \(x^2 (x-1) < 0\)
Solution The main problem in this example is the \(x^2\) can not be lower than 0. It will alway be greater than 0 so the other part \(x-1\) must be less then 0.\begin{eqnarray} x-1 &<& 0 \\ x&<& 1 \Rightarrow x \in \langle -\infty, 1 \rangle \end{eqnarray}. The solution of entire inequaltiy can be wirtten as: \(x\in \langle -\infty, 1 \rangle \ {0}. - Example 2 Solve the inequality \(-3x +7 \leq -x + 5\)
Solution \begin{eqnarray}-3x +7 &\geq& -x + 5 \\ -2x &\geq& -2 /: (-2) \\ x &\leq& 1 \end{eqnarray}
How to solve linear inequalities ?
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