Question

Q9. Solve |x + 1| ≥ 4 graphically.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: |x + 1| \geqslant 4$

We know,

$\boxed{ \rm{ \: |x| \geqslant y \: \rm\implies \:x \leqslant - y \: \: or \: \: x \geqslant y \: }}$

So, using this definition of Modulus inequality, we have

$\rm \: x + 1 \leqslant - 4 \: \: or \: \: x + 1 \geqslant 4$

$\rm \: x \leqslant - 4 - 1 \: \: or \: \: x \geqslant 4 - 1$

$\rm \: x \leqslant - 5 \: \: or \: \: x \geqslant 3$

$\rm\implies \:x \: \in \: ( - \infty , - 5] \: \cup \: [3, \infty )$

$\rule{190pt}{2pt}$

Additional information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ |x| < y\rm\implies \: - y < x < y}\\ \\ \bigstar \: \bf{ |x| \leqslant y\rm\implies \: - y \leqslant x \leqslant y}\\ \\ \bigstar \: \bf{ |x| > y\rm\implies \: x < - y \: or \: x > y} \: \\ \\ \bigstar \: \bf{ |x| \geqslant y\rm\implies \:x \leqslant - y \: or \: x \geqslant y}\\ \\ \bigstar \: \bf{ |x - a| < y\rm\implies \:a - y < x < a + y}\\ \\ \bigstar \: \bf{ |x - a| \leqslant y\rm\implies \:a - y \leqslant x \leqslant a + y}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Answer:

Given 

x≤4,xεN  

The solution set ={1,2,3,4} 

These four numbers are shown indicating with thick dot on the number line.

hope it will help you