
Solve the inequality. Write the solution set in interval notation.
\[\]\[\left| { - 8x + 9} \right| < 6\]
A. \[\left( { - \infty , - \dfrac{{17}}{8}} \right)\bigcup {\left( { - \dfrac{{15}}{8}, - \infty } \right)} \]
B. \[\left( {\dfrac{3}{8}, - \dfrac{{15}}{8}} \right)\]
C. \[\left( {\dfrac{3}{8},\dfrac{{15}}{8}} \right)\]
D. \[\left( { - \infty ,\dfrac{3}{8}} \right)\bigcup {\left( {\dfrac{{15}}{8},\infty } \right)} \]
Answer
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Hint: We are given linear inequality \[| - 8x + 9| < 6\]and we have to solve it in a set of interval notation. We will apply an absolute rule to solve this inequality
If \[|u| < a\],
\[a > 0\]
then, \[ - a < u < + a\]
We know that the Absolute value of a number is the distance of value from the origin regardless of the direction Absolute value is denoted by two vertical lines enclosing the number or expression.
The absolute value of \[x\]is express as \[|x| = a\]which implies,
\[x = + a\] and \[x = - a\]
Complete step-by-step answer:
First of all, we need to isolate the absolute value expression on the left side of an inequality
\[| - 8x + 9| < 6\]
But, in our problem, it is already isolated on the left side of an inequality
Now, according to absolute rule If \[|u| < a\], \[a > 0\]
Then, \[ - a < |u| < + a\]
Now we will remove the absolute value bars by setting up a compound inequality Since our absolute value is less than a number then we have to set up three-part compound inequality which is below
\[ - 6| - 8x + 9|| < 6\]
Now, subtract\[\;9\]both side and inside the absolute value
\[ - 6 - 9 < - 8x + 9 - 9 < 6 - 9\]
\[ = - 15 < - 8x < - 3\]
Now, divide by \[ - 8\]to isolate the absolute value
\[\dfrac{{ - 15}}{{ - 8}} < \dfrac{{ - 8x}}{{ - 8}} < \dfrac{{ - 3}}{{ - 8}}\]
\[ = \dfrac{{ - 15}}{{ - 8}} < \dfrac{{ - 8x}}{{ - 8}} < \dfrac{{ - 3}}{{ - 8}}\]
So,\[x \in \left( {\dfrac{3}{8},\dfrac{{15}}{8}} \right)\] the interval notation is \[\left( {\dfrac{3}{8},\dfrac{{15}}{8}} \right)\]
So, the correct answer is “Option C”.
Note: When you multiply or divide by a negative number reverse the inequality sign.
Now, it should be taken care that whenever we divide or multiply by a negative number to both sides of the inequality, the inequality sign must be reversed to keep a true statement.
While taking out the absolute values, always remember to observe every step to avoid mistakes because a single wrong inequality leads to the wrong solution.
If \[|u| < a\],
\[a > 0\]
then, \[ - a < u < + a\]
We know that the Absolute value of a number is the distance of value from the origin regardless of the direction Absolute value is denoted by two vertical lines enclosing the number or expression.
The absolute value of \[x\]is express as \[|x| = a\]which implies,
\[x = + a\] and \[x = - a\]
Complete step-by-step answer:
First of all, we need to isolate the absolute value expression on the left side of an inequality
\[| - 8x + 9| < 6\]
But, in our problem, it is already isolated on the left side of an inequality
Now, according to absolute rule If \[|u| < a\], \[a > 0\]
Then, \[ - a < |u| < + a\]
Now we will remove the absolute value bars by setting up a compound inequality Since our absolute value is less than a number then we have to set up three-part compound inequality which is below
\[ - 6| - 8x + 9|| < 6\]
Now, subtract\[\;9\]both side and inside the absolute value
\[ - 6 - 9 < - 8x + 9 - 9 < 6 - 9\]
\[ = - 15 < - 8x < - 3\]
Now, divide by \[ - 8\]to isolate the absolute value
\[\dfrac{{ - 15}}{{ - 8}} < \dfrac{{ - 8x}}{{ - 8}} < \dfrac{{ - 3}}{{ - 8}}\]
\[ = \dfrac{{ - 15}}{{ - 8}} < \dfrac{{ - 8x}}{{ - 8}} < \dfrac{{ - 3}}{{ - 8}}\]
So,\[x \in \left( {\dfrac{3}{8},\dfrac{{15}}{8}} \right)\] the interval notation is \[\left( {\dfrac{3}{8},\dfrac{{15}}{8}} \right)\]
So, the correct answer is “Option C”.
Note: When you multiply or divide by a negative number reverse the inequality sign.
Now, it should be taken care that whenever we divide or multiply by a negative number to both sides of the inequality, the inequality sign must be reversed to keep a true statement.
While taking out the absolute values, always remember to observe every step to avoid mistakes because a single wrong inequality leads to the wrong solution.
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