# Page 45

## 1. $$3y-5<7y+9$$

<details> <summary> Example: </summary>

$$5y-4<3y+8$$

We will solve for $y$ step by step:

\begin{align}
5y - 4 &< 3y + 8 \\
5y - 3y &< 8 + 4 \\
2y &< 12 \\
y &< \frac{12}{2} \\
y &< 6
\end{align}

So, the solution is $y < 6$.

In interval notation this is the interval $(-\infty,6)$.

</details>

## 2. $$5(2y+1)+4 \geq 4(3y-3)-7$$

<details> <summary> Example: </summary>

We will solve for $y$ step by step:

\begin{align}
4(3y + 1) + 5 &\geq 3(2y - 2) - 6 \\
12y + 4 + 5 &\geq 6y - 6 - 6 \\
12y + 9 &\geq 6y - 12 \\
12y - 6y &\geq -12 - 9 \\
6y &\geq -21 \\
y &\geq \frac{-21}{6} \\
y &\geq -\frac{7}{2}
\end{align}

So, the solution is $y \geq -\frac{7}{2}$.

In interval notation this is the interval $\left[-\dfrac{7}{2},\infty\right)$.

</details>

## 3. Find the domain and put in interval notation: $g(x)=\sqrt{6-7x}$

<details> <summary> Example: </summary>

$g(x)=\sqrt{5-3x}$

For the square root to be defined, the expression inside the square root must be non-negative:

\begin{align}
5 - 3x &\geq 0 \\
-3x &\geq -5 \\
x &\leq \frac{5}{3}
\end{align}

So, the domain of $g(x)$ is all $x$ values such that $x \leq \frac{5}{3}$.

In interval notation, the domain is:

$$\left(-\infty, \frac{5}{3}\right]$$

</details>

## 4. Find the domain and put in interval notation: $h(x)=\dfrac{x-3}{\sqrt{2x+5}}$

<details> <summary> Example: </summary>

$h(x)=\dfrac{x-2}{\sqrt{8x+9}}$

The square root in the denominator requires the expression inside the square root to be positive. Therefore, we need:

\begin{align}
8x + 9 &> 0 \\
8x &> -9 \\
x &> -\frac{9}{8}
\end{align}

So, $x$ must be greater than $\frac{-9}{8}$ for the square root to be defined, and the denominator cannot be zero.

In interval notation, the domain of $h(x)$ is:

$$\left( -\frac{9}{8}, \infty \right)$$
</details>