Question

`2h-3\ \textgreater \ 15`

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{ (2h)/(2) > (18)/(2) }`

`\quad \quad \quad \quad\tt{ \frac{ \cancel{ \color{red}2}h}{ \cancel{ \color{red}2}} > (18)/(2) }`

`\quad \quad \quad \quad\tt{ h > (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{(2h)/(2)>(18)/(2)}`

`\large\text{CANCEL out: }\mathsf{(2)/(2)}\large\text{ because it gives you 1}`

`\large\text{KEEP: }\mathsf{(18)/(2)}\large\text{ because it gives you your h-value is being compared to}`

`\large\text{NEW EQUATION: }\mathsf{h > (18)/(2)}`

`\mathsf{(18)/(2)}`

`\mathsf{= 18/ 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:)}`