
Solve the following inequation and represent the solution set on a number line.
\[-8\dfrac{1}{2}<-\dfrac{1}{2}-4x\le 7\dfrac{1}{2},x\in R\]
Answer
588.9k+ views
Hint: At first, break the given inequality into two inequalities and then do basic operations keeping the rules of the inequality to find the ranges of x and finally find the intersection of the range.
Complete step-by-step answer:
In this question, we are given an inequation and we have to find the solution set of x and represent it on a number line. The inequation given in the question is,
\[-8\dfrac{1}{2}<-\dfrac{1}{2}-4x\le 7\dfrac{1}{2},\text{ where }x\in R\]
So, at first, let’s break the inequation into two in equations as,
\[-8\dfrac{1}{2}<-\dfrac{1}{2}-4x\text{ and }\dfrac{-1}{2}-4x\le 7\dfrac{1}{2}\]
We will solve the inequality one by one. So, the first inequality is,
\[-8\dfrac{1}{2}<-\dfrac{1}{2}-4x\]
We will add (4x + 9) to both sides of the inequality, the sides of the inequality will not change. So, we get,
\[4x+9-8\dfrac{1}{2}<-\dfrac{1}{2}-4x+4x+9\]
\[\Rightarrow 4x+\dfrac{1}{2}<8\dfrac{1}{2}\]
Now, we will subtract \[\dfrac{1}{2}\] from both the sides of the inequality, the sides of the inequality will not change, so we get,
\[4x+\dfrac{1}{2}-\dfrac{1}{2}<8\dfrac{1}{2}-\dfrac{1}{2}\]
\[\Rightarrow 4x<8\]
Now, we will divide by 4 to the sides, so we get,
\[x<\dfrac{8}{4}\]
\[\Rightarrow x<2\]
Now, we will go for the second inequality,
\[\text{ }\dfrac{-1}{2}-4x\le 7\dfrac{1}{2}\]
So, we will first add \[\dfrac{1}{2}\] to both the sides, so we get,
\[\dfrac{-1}{2}-4x+\dfrac{1}{2}\le 7\dfrac{1}{2}+\dfrac{1}{2}\]
\[\Rightarrow -4x\le 8\]
Now, we will divide by – 4 to both sides, here the side of the inequality changes. So, we get,
\[x\ge -2\]
In the earlier inequality, we got x < 2 and in the latter one, we get \[x\ge -2\] which can be written as, \[-2\le x<2.\] So, in the number line it will be,
Note: The sides of the inequality only change in two cases. The first one is when the inequality is multiplied by a negative number and the second one is when divided by a negative number.
Complete step-by-step answer:
In this question, we are given an inequation and we have to find the solution set of x and represent it on a number line. The inequation given in the question is,
\[-8\dfrac{1}{2}<-\dfrac{1}{2}-4x\le 7\dfrac{1}{2},\text{ where }x\in R\]
So, at first, let’s break the inequation into two in equations as,
\[-8\dfrac{1}{2}<-\dfrac{1}{2}-4x\text{ and }\dfrac{-1}{2}-4x\le 7\dfrac{1}{2}\]
We will solve the inequality one by one. So, the first inequality is,
\[-8\dfrac{1}{2}<-\dfrac{1}{2}-4x\]
We will add (4x + 9) to both sides of the inequality, the sides of the inequality will not change. So, we get,
\[4x+9-8\dfrac{1}{2}<-\dfrac{1}{2}-4x+4x+9\]
\[\Rightarrow 4x+\dfrac{1}{2}<8\dfrac{1}{2}\]
Now, we will subtract \[\dfrac{1}{2}\] from both the sides of the inequality, the sides of the inequality will not change, so we get,
\[4x+\dfrac{1}{2}-\dfrac{1}{2}<8\dfrac{1}{2}-\dfrac{1}{2}\]
\[\Rightarrow 4x<8\]
Now, we will divide by 4 to the sides, so we get,
\[x<\dfrac{8}{4}\]
\[\Rightarrow x<2\]
Now, we will go for the second inequality,
\[\text{ }\dfrac{-1}{2}-4x\le 7\dfrac{1}{2}\]
So, we will first add \[\dfrac{1}{2}\] to both the sides, so we get,
\[\dfrac{-1}{2}-4x+\dfrac{1}{2}\le 7\dfrac{1}{2}+\dfrac{1}{2}\]
\[\Rightarrow -4x\le 8\]
Now, we will divide by – 4 to both sides, here the side of the inequality changes. So, we get,
\[x\ge -2\]
In the earlier inequality, we got x < 2 and in the latter one, we get \[x\ge -2\] which can be written as, \[-2\le x<2.\] So, in the number line it will be,
Note: The sides of the inequality only change in two cases. The first one is when the inequality is multiplied by a negative number and the second one is when divided by a negative number.
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