NJSLA Algebra II Creating Equations. Practice it free.

More than $2$ means $w > 2$. At most $10$ means $w \le 10$, so $2 < w \le 10$.

In slope-intercept form $y = mx + b$, the coefficient of $x$ is the slope. Here $m = -5$.

The slope is $m = \dfrac{1 - (-3)}{6 - 2} = 1$. Substitute $(2, -3)$: $-3 = 1(2) + b$, so $b = -5$. The equation is $y = x - 5$.

A maximum means $c$ can be at most that value, so use $\le\textsf{.}$ Seven times $n$ plus $15$ is $7n + 15\textsf{,}$ giving $c \le 7n + 15\textsf{.}$

Subtract the third-year salary from the first: $\text{\char36}36{,}400 - \text{\char36}32{,}500 = \text{\char36}3{,}900$. Over three years there are two equal raises, so divide by $3 - 1 = 2$ to get $x = \text{\char36}1{,}950$. Add $\text{\char36}1{,}950$ each year after the first: $\text{\char36}32{,}500$, $\text{\char36}34{,}450$, $\text{\char36}36{,}400$. Sum all three years: $\text{\char36}103{,}350$.

The system is $2L + 3S = 54$ and $L + 5S = 55\textsf{.}$ Double the second equation: $2L + 10S = 110\textsf{.}$ Subtract the first: $7S = 56\textsf{,}$ so $S = 8\textsf{.}$

Amira has $5$ times as many stickers as Ben, so $a = 5b$. Together they have $90$ stickers, so $a + b = 90\textsf{.}$

The factors $l$, $w$, and $h$ are multiplied, so $h$ is isolated by using the inverse operation (division) on both sides of the equation $\dfrac{V}{\textcolor{red}{lw}} = \dfrac{lwh}{\textcolor{red}{lw}}$. Simplifying gives $h = \dfrac{V}{lw}$.