Question

"2h-3\ \textgreater \ 15" align="absmiddle" class="latex-formula">

Answer 1

Problem:

$\quad \quad \quad \quad\tt{2h - 3 > 15}$

Let's solve!

$\quad \quad \quad \quad\tt{2h - 3 > 15}$

Add both sides by 3.

$\quad \quad \quad \quad\tt{2h - 3 + 3 > 15 + 3}$

$\quad \quad \quad \quad\tt{2h \: \: \cancel{ \color{red}- 3 + 3} \: > 15 + 3}$

$\quad \quad \quad \quad\tt{2h > 18}$

Divide it by 2.

$\quad \quad \quad \quad\tt{ \frac{2h}{2} > \frac{18}{2} }$

$\quad \quad \quad \quad\tt{ \frac{ \cancel{ \color{red}2}h}{ \cancel{ \color{red}2}} > \frac{18}{2} }$

$\quad \quad \quad \quad\tt{ h > \frac{18}{2} }$

$\quad \quad \quad \quad\tt{ h > 9}$

Hence the answer is:

$\quad \quad \quad \quad \boxed{\tt { \color{green}h > 9}}$

________

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Answer 2

$\huge\text{Hey there!}$

$\large\textsf{2h - 3 }\mathsf{>}\large\textsf{ 15}$

$\large\text{ADD 3 to BOTH SIDES}$

$\large\textsf{2h - 3 + 3 }>\large\textsf{ 15 + 3}$

$\large\text{CANCEL out: \textsf{-3 + 3} because that gives you 0}$

$\large\text{KEEP: \textsf{15 + 3} because it helps you solve for your h-value}$

$\large\textsf{15 + 3 = \boxed{\large\textsf{\bf 18}}}$

$\large\text{NEW EQUATION: \textsf{2h} }\mathsf{>}\large\textsf{ 18}$

$\large\text{DIVIDE 2 to BOTH SIDES}$

$\mathsf{\dfrac{2h}{2}>\dfrac{18}{2}}$

$\large\text{CANCEL out: }\mathsf{\dfrac{2}{2}}\large\text{ because it gives you 1}$

$\large\text{KEEP: }\mathsf{\dfrac{18}{2}}\large\text{ because it gives you your h-value is being compared to}$

$\large\text{NEW EQUATION: }\mathsf{h > \dfrac{18}{2}}$

$\mathsf{\dfrac{18}{2}}$

$\mathsf{= 18\div 2}$

$=\boxed{\mathsf{\large\textsf {\bf 9}}}$

$\boxed{\boxed{\large\textsf{Answer: }\mathsf{\bf h > 9}}}\huge\checkmark\\\\\boxed{\boxed{\rm \bf It\ it\ an\ O P E N E D\ circle\ shaded}}\\\boxed{\boxed{\rm \bf to\ the\ right\ starting\ at\ point\ 9}}\checkmark$

$\large\text{Good luck on your assignment and enjoy your day!}$

~$\frak{Amphitrite1040:)}$