How do you solve \[\left| 1-2x \right|=9\]?

Hint: Remove the modulus sign by using the formula: - if \[\left| x \right|=a\] then \[x=\pm a\], where ‘a’ must be a positive real number. Consider the positive and negative sign one by one and solve the obtained linear equation to get the two values of x.

Complete step by step solution:
Here, we have been provided with the expression: - \[\left| 1-2x \right|=9\] and we are asked to solve it. That means we have to find the values of x.
Now, we have been provided with the modulus function, so we need to use its basic properties to solve the question. We know that modulus of any number, whether positive or negative, is always positive, so even if the number inside the modulus sign is negative its output will be positive.
\[\because \left| 1-2x \right|=9\]
Using the property of modulus given as: - \[\left| x \right|=a\] then \[x=\pm a\], where ‘a’ must be a positive number or can be equal to 0, we have,
\[\Rightarrow \left( 1-2x \right)=\pm 9\]
Let us consider both the signs one – by – one and find the values of x.
(i) Considering the positive sign.
\[\begin{align}
  & \Rightarrow 1-2x=9 \\
 & \Rightarrow -2x=9-1 \\
 & \Rightarrow -2x=8 \\
\end{align}\]
Dividing both the sides with -2, we get,
\[\Rightarrow x=-4\]
(ii) Considering the negative sign.
\[\begin{align}
  & \Rightarrow 1-2x=-9 \\
 & \Rightarrow -2x=-9-1 \\
 & \Rightarrow -2x=-10 \\
\end{align}\]
Dividing both the sides with -2, we get,
\[\Rightarrow x=5\]

Hence, the solutions of the given equation are: - x = 5 and x = -4.

Note: One can use a different approach also to solve this question. What we can do is, we will substitute the expression inside the modulus sign equal to 0 and find the values of x. In the next step we will mark these values of x on a number line and consider the intervals one by one to check the sign of expression in that interval. We will remove the modulus sign according to the sign of the function in a particular interval. In the above question, for \[x<\dfrac{1}{2}\] the value of \[\left| 1-2x \right|\] will be \[\left( 1-2x \right)\] and for \[x\ge \dfrac{1}{2}\] it will be \[\left( 2x-1 \right)\].