Question

How do you solve \[\left| 3x+6 \right|>15\]?

Answer 1

Hint: This type of question is based on the concept of inequality. To solve \[\left| 3x+6 \right|>15\], we first have to consider the modulus function. According to the definition of modulus function, we get 3x+6>15 and –(3x+6)>15. First consider 3x+6>15. Then, take 3 common from the LHS. Also, we can express 15 as the product of 3 and 5. Divide the whole inequality by 3 and cancel the common terms. Then, subtract the inequality by 2. Now, consider –(3x+6)>15. Then, take 3 common from the LHS. Also, we can express 15 as the product of 3 and 5. Divide the whole equation by 3 and cancel the common terms. Then, add the inequality by 2.

Complete step by step answer:
According to the question, we are asked to solve \[\left| 3x+6 \right|>15\].
We have been given the equation is \[\left| 3x+6 \right|>15\]. --------(1)
First, we have to consider the modulus function.
According to the definition of modulus function, we know that
\[\left| x \right|=\left\{ \begin{align}
  & \text{x , x > 0} \\
 & \text{-x , x < 0} \\
\end{align} \right.\]
Here, when x>-2, we get 3x+6>15.
And when x<-2, we get 3x+6<15.
Case (i)
Consider x>-2.
Therefore, \[\left| 3x+6 \right|=3x+6\].
Therefore, the inequality is \[3x+6>15\].
We find that 3 are common in both the terms of the LHS.
On taking 3 common out of the bracket, we get
\[3\left( x+2 \right)>15\]
We can write the expression as
\[3\left( x+2 \right)>3\times 5\].
Let us divide the whole inequality by 3.
\[\Rightarrow \dfrac{3\left( x+2 \right)}{3}>\dfrac{3\times 5}{3}\]
We find that 3 are common in both the numerator and the denominator of LHS and RHS. On cancelling 3, we get
\[x+2>5\]
Now, we have to subtract 2 from both the sides of the inequality. We get
\[x+2-2>5-2\]
We know that terms with the same magnitude and opposite signs cancel out.
\[\Rightarrow x>5-2\]
On further simplifications, we get
\[x>3\].
We can express the values of x as \[\left( 3,\infty \right)\].
Case (ii)
Consider x<-2.
Therefore, \[\left| 3x+6 \right|=-\left( 3x+6 \right)\].
Therefore, the inequality is \[-\left( 3x+6 \right)>15\].
We find that 3 are common in both the terms of the LHS.
On taking 3 common out of the bracket, we get
\[-3\left( x+2 \right)>15\]
We can write the expression as
\[-3\left( x+2 \right)>3\times 5\].
Let us divide the whole inequality by 3.
\[\Rightarrow \dfrac{-3\left( x+2 \right)}{3}>\dfrac{3\times 5}{3}\]
We find that 3 are common in both the numerator and the denominator of LHS and RHS. On cancelling 3, we get
\[-\left( x+2 \right)>5\]
On taking the negative sign inside the bracket, we get
\[-x-2>5\]
Now, we have to add 2 from both the sides of the inequality. We get
\[-x-2+2>5+2\]
We know that terms with the same magnitude and opposite signs cancel out.
\[\Rightarrow -x>5+2\]
On further simplifications, we get
\[-x>7\].
But we need to find the value of x.
Using the rule of inequality, if \[-x>a\], then \[xHere, a=7.
We get x<-7.
We can express the values of x as \[\left( -\infty ,-7 \right)\].
Therefore, the values of x considering the two cases are \[\left( -\infty ,-7 \right)\cup \left( 3,\infty \right)\].

Hence, the values of x in the inequality \[\left| 3x+6 \right|>15\] are \[\left( -\infty ,-7 \right)\cup \left( 3,\infty \right)\].

Note: We can also write the solution as x>3. Whenever we get such types of questions, we have to make certain calculations so that we get the variable alone on the LHS. Avoid calculation mistakes based on sign conventions. We should not express the values of x in a closed bracket, that is \[\left[ 3,\infty \right]\]and \[\left[ -\infty ,-7 \right]\]. Then, the solution is completely wrong.