Question

$$(\frac{\sqrt{16}+2x)}{3}-8<-4$$

Answer

x&lt;-4/3

Solution


Since \(4\times 4=16\), the square root of \(16\) is \(4\).
\[(4+2x)\times 3-8<-4\]
Regroup terms.
\[3(4+2x)-8<-4\]
Add \(8\) to both sides.
\[3(4+2x)<-4+8\]
Simplify  \(-4+8\)  to  \(4\).
\[3(4+2x)<4\]
Divide both sides by \(3\).
\[4+2x<\frac{4}{3}\]
Subtract \(4\) from both sides.
\[2x<\frac{4}{3}-4\]
Simplify  \(\frac{4}{3}-4\)  to  \(-\frac{8}{3}\).
\[2x<-\frac{8}{3}\]
Divide both sides by \(2\).
\[x<-\frac{\frac{8}{3}}{2}\]
Simplify  \(\frac{\frac{8}{3}}{2}\)  to  \(\frac{8}{3\times 2}\).
\[x<-\frac{8}{3\times 2}\]
Simplify  \(3\times 2\)  to  \(6\).
\[x<-\frac{8}{6}\]
Simplify  \(\frac{8}{6}\)  to  \(\frac{4}{3}\).
\[x<-\frac{4}{3}\]

Decimal Form: -1.333333