CMAS High school Algebra. Practice it free.

From $500$ through $1{,}200$, inclusive, means both endpoints are included: $500 \le e \le 1{,}200$.

Check both constraints: $5s + 3d \le 24$ and $d \ge 2\textsf{.}$ The pair $(3, 3)$ costs $5(3) + 3(3) = 24$ and has $3$ drinks, so both constraints are met.

Solve for $y$: $2y = -4x + 10$, so $y = -2x + 5$ and $m = -2$.

The slope is $m = \dfrac{5 - 2}{7 - 1} = \dfrac{1}{2}$. Substitute $(1, 2)$: $2 = \dfrac{1}{2}(1) + b$, so $b = \dfrac{3}{2}$. The equation is $y = \dfrac{1}{2}x + \dfrac{3}{2}$.

A maximum means $h$ can be at most that value, so use $\le\textsf{.}$ Five times $d$ minus $2$ is $5d - 2\textsf{,}$ giving $h \le 5d - 2\textsf{.}$

Subtract the fifth-year salary from the first: $\text{\char36}46{,}700 - \text{\char36}39{,}500 = \text{\char36}7{,}200$. Over five years there are four equal raises, so divide by $5 - 1 = 4$ to get $x = \text{\char36}1{,}800$. Add $\text{\char36}1{,}800$ each year after the first: $\text{\char36}39{,}500$, $\text{\char36}41{,}300$, $\text{\char36}43{,}100$, $\text{\char36}44{,}900$, $\text{\char36}46{,}700$. Sum all five years: $\text{\char36}215{,}500$.

The system is $3m + 5c = 22$ and $2m + 2c = 12\textsf{.}$ Divide the second equation by $2\textsf{:}$ $m + c = 6\textsf{,}$ so $c = 6 - m\textsf{.}$ Substitute into the first: $3m + 5(6 - m) = 22\textsf{,}$ so $3m + 30 - 5m = 22\textsf{,}$ $-2m = -8\textsf{,}$ and $m = 4\textsf{.}$

Lena is $2$ times as old as her brother, so $\ell = 2b$. The sum of their ages is $36$, so $\ell + b = 36\textsf{.}$