LEAP 2025 Algebra I Modeling & Application. Practice it free.

Point-slope form is $y - y_1 = m(x - x_1)\textsf{.}$ The slope is the coefficient of $(x - x_1)\textsf{,}$ so $m = 2\textsf{.}$

Substitute $y = 2$. From $2x - y \ge 4$, $2x - 2 \ge 4$, so $x \ge 3$. From $x + y \le 7$, $x \le 5$. The value $x = 4$ satisfies $3 \le 4 \le 5$.

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: $\text{\char36}103{,}350$.

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

There are $2$ times as many student tickets as adult tickets, so $s = 2a\textsf{.}$ In all there are $27$ tickets, so $s + a = 27\textsf{.}$

Substitute the point: $5 = 3(1) + b$, so $b = 5 - 3 = 2$.

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 $c \le 7n + 15\textsf{.}$