Answer
The solution is $\Big(-\infty,1\Big)\cup\Big(\dfrac{11}{3},\infty\Big)$
Work Step by Step
$|7-3x|\gt4$
Solving this absolute value inequality is equivalent to solving two separate inequalities, which are:
$7-3x\gt4$ and $7-3x\lt-4$
$\textbf{Solve the first inequality:}$
$7-3x\gt4$
Take $7$ to the right side:
$-3x\gt4-7$
$-3x\gt-3$
Take $3$ to divide the right side and reverse the inequality sign:
$x\lt\dfrac{-3}{-3}$
$x\lt1$
$\textbf{Solve the second inequality:}$
$7-3x\lt-4$
Take $7$ to the right side:
$-3x\lt-4-7$
$-3x\lt-11$
Take $3$ to divide the right side and reverse the inequality sign:
$x\gt\dfrac{-11}{-3}$
$x\gt\dfrac{11}{3}$
Expressing the solution in interval notation:
$\Big(-\infty,1\Big)\cup\Big(\dfrac{11}{3},\infty\Big)$
