The solution set of the inequality for R is given by [tex]-\frac{1172}{85} \leq x\leq \frac{1228}{55}[/tex] .
An inequality in mathematics is a relation that compares two numbers or other mathematical expressions in an unequal way. It is frequently used to compare two numbers on the number line based on their sizes. Different types of inequalities are represented by a variety of notations such as < ,> ,≤ and ≥.
The modulus function takes the absolute value of a function.
To solve the given inequality we have simplify the inequality first.
[tex]x-\frac{7}{9} |(\frac{12}{5} -6x)|+50\ge -30[/tex]
[tex]or, x-\frac{7}{9}\times \frac{6}{5} |(2-5x)|+50\ge -30\\or,x-\frac{14}{15}|2-5x|\ge-80\\[/tex]
Now we separate the inequalities:
[tex]x-\frac{14}{15}(2-5x)\ge-80[/tex] for 2x-5 ≥ 0
[tex]x-\frac{14}{15}(-(2-5x))\ge-80[/tex] for 2x-5 ≤ 0
Now we will solve the two inequalities separate and find the intersection.
[tex]x-\frac{14}{15}(2-5x)\ge-80\\or,15x-28+70x\ge-1200\\or,x\ge -\frac{1172}{85}[/tex]
Again
2x-5 ≥ 0
or, [tex]x\le\frac{2}{5}[/tex]
hence [tex]-\frac{1172}{85}\le x\le\frac{2}{5}[/tex]
Similarly solving the other inequality we get
[tex]\frac{2}{5}\le x\le\frac{1228}{55}[/tex]
Now we find the intersection of the two inequalities:
[tex]-\frac{1172}{85}\le x\le\frac{1228}{55}[/tex]
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