To prove a biconditional statement we have to prove two conditionals.
[tex]\begin{gathered} x\ge4\rightarrow3x\ge12 \\ 3x\ge12\rightarrow x\ge4 \end{gathered}[/tex]Let's prove the first one.
[tex]x\ge4[/tex]We multiply the inequality by 3.
[tex]\begin{gathered} 3x\ge4\cdot3 \\ 3x\ge12 \end{gathered}[/tex]Which proves the first conditional.
Let's prove the second conditional.
[tex]3x\ge12[/tex]We divide by 3 the inequality.
[tex]\begin{gathered} \frac{3x}{3}\ge\frac{12}{3} \\ x\ge4 \end{gathered}[/tex]Which proves the second conditional.
The inverse of the biconditional statement would be
[tex]x<4\leftrightarrow3x<12[/tex]To prove the inverse, we demonstrate each conditional.
[tex]\begin{gathered} x<4\rightarrow3x<12 \\ 3x<12\rightarrow x<4 \end{gathered}[/tex]Let's prove the first conditional.
[tex]x<4[/tex]We multiply by 3 the inequality.
[tex]3x<12[/tex]Which proves the first inequality.
Let's prove the second conditional.
[tex]3x<12\rightarrow x<4[/tex]We divide the inequality by 3.
[tex]\begin{gathered} \frac{3x}{3}<\frac{12}{3} \\ x<4 \end{gathered}[/tex]Which proves the second inequality.
At last, the contrapositive would be
[tex]3x<12\leftrightarrow x<4[/tex]Let's prove it using the same method as the previous ones.
The conditionals we have to prove are
[tex]\begin{gathered} 3x<12\rightarrow x<4 \\ x<4\rightarrow3x<12 \end{gathered}[/tex]First conditional proof.
[tex]\begin{gathered} \frac{3x}{3}<\frac{12}{3} \\ x<4 \end{gathered}[/tex]Second conditional proof.
[tex]\begin{gathered} x<4 \\ 3x<4\cdot3 \\ 3x<12 \end{gathered}[/tex]