NJSLA Algebra I Creating Equations. Practice it free.

Above $32$ and below $40$ are both strict, so $32 < t < 40$.

Set $y = 0$: $2x = 12$, so $x = 6$.

The slope is $m = \dfrac{5 - (-1)}{2 - (-4)} = 1$. Substitute $(-4, -1)$: $-1 = 1(-4) + b$, so $b = 3$. The equation is $y = x + 3$.

A maximum means $x$ can be at most that value, so use $\le\textsf{.}$ Four times $y$ minus $7$ is $4y - 7\textsf{,}$ giving $x \le 4y - 7\textsf{.}$

Subtract the third-year base from the first: $\text{\char36}35{,}200 - \text{\char36}31{,}000 = \text{\char36}4{,}200$. Over three years there are two equal increases, so divide by $3 - 1 = 2$ to get $x = \text{\char36}2{,}100$. Add $\text{\char36}2{,}100$ each year after the first: $\text{\char36}31{,}000$, $\text{\char36}33{,}100$, $\text{\char36}35{,}200$. Sum all three years: $\text{\char36}99{,}300$.

The system is $2l + 3s = 49$ and $4l + s = 53\textsf{.}$ Double the first equation: $4l + 6s = 98\textsf{.}$ Subtract the second: $5s = 45\textsf{,}$ so $s = 9\textsf{.}$ Then $4l + 9 = 53\textsf{,}$ $4l = 44\textsf{,}$ and $l = 11\textsf{.}$

The recipe uses $3$ times as much flour as sugar, so $f = 3s$. Together they measure $8$ cups, so $f + s = 8\textsf{.}$

The sum $b + c$ is divided by $2$, so $b + c$ is isolated by using the inverse operation (multiplication) on both sides, $\textcolor{red}{2}a = b + c$. Then, $b$ is added, so $c$ is isolated by using the inverse operation (subtraction) on both sides, $2a - \textcolor{red}{b} = c$. Simplifying gives $c = 2a - b$.