Question

Solve the inequality \[\left| {3x - 4} \right| > \left| {2x + 1} \right|\]?

Answer 1

Hint: Use the concept of absolute value with inequality and open the inequality for the given equation. Bring the right hand side to the denominator of the left hand side and make it as one single modulus function with a fraction in it. Find the value of x between two numbers by adding or subtracting values to the inequality.
Absolute value of an inequality \[\left| x \right| > k\] can be written as \[ - k > x > k\]

Complete step by step answer:
We are given the inequality \[\left| {3x - 4} \right| > \left| {2x + 1} \right|\]
Divide both sides of the inequality by \[\left| {2x + 1} \right|\]
\[\dfrac{{\left| {3x - 4} \right|}}{{\left| {2x + 1} \right|}} > \dfrac{{\left| {2x + 1} \right|}}{{\left| {2x + 1} \right|}}\]
Cancel same terms from numerator and denominator of right hand side of the inequality
 \[\dfrac{{\left| {3x - 4} \right|}}{{\left| {2x + 1} \right|}} > 1\]
We can use the formula \[\dfrac{{\left| p \right|}}{{\left| q \right|}} = \left| {\dfrac{p}{q}} \right|\]
\[\left| {\dfrac{{3x - 4}}{{2x + 1}}} \right| > 1\]
We know the inequality of greater than type with modulus i.e. \[\left| x \right| > k\] can be written as \[ - k > x > k\]
So we can write the inequality in equation (1) as \[ - 1 > \dfrac{{3x - 4}}{{2x + 1}} > 1\]
Multiply all the three terms in inequality by \[2x + 1\]
\[ \Rightarrow - 1 \times (2x + 1) > \dfrac{{3x - 4}}{{2x + 1}} \times (2x + 1) > 1 \times (2x + 1)\]
\[ \Rightarrow - 2x - 1 > 3x - 4 > 2x + 1\]
Add 4 to all the terms in inequality
\[ \Rightarrow - 2x - 1 + 4 > 3x - 4 + 4 > 2x + 1 + 4\]
\[ \Rightarrow - 2x + 3 > 3x > 2x + 5\]
Now we solve first inequality: \[ - 2x + 3 > 3x\]
\[ \Rightarrow 3 > 3x + 2x\]
\[ \Rightarrow 3 > 5x\]
\[ \Rightarrow \dfrac{3}{5} > x\] … (2)
Also, we solve the second inequality: \[3x > 2x + 5\]
\[ \Rightarrow 3x - 2x > 5\]
\[ \Rightarrow x > 5\] … (3)
So, from equations (2) and (3) we can write that x lies in intervals \[\left( { - \infty ,\dfrac{3}{5}} \right]\]and \[\left[ {5,\infty } \right)\]
So, the solution of the inequality \[\left| {3x - 4} \right| > \left| {2x + 1} \right|\] is \[\left( { - \infty ,\dfrac{3}{5}} \right] \cup \left[ {5,\infty } \right)\]
\[\therefore \]The solution of the inequality \[\left| {3x - 4} \right| > \left| {2x + 1} \right|\]is \[\left( { - \infty ,\dfrac{3}{5}} \right] \cup \left[ {5,\infty } \right)\]

Note: Do not make the mistake of opening all the 4 possibilities of signs of the two inequalities i.e. both positive, both negative and two for one positive and one negative. Keep in mind we have the expansion of inequality for the case where one of the sides of inequality has absolute terms and the other as constant value, so we make one side as constant and then apply the formula.