SOL Algebra I Equations and Inequalities. Practice it free.

Subtract the fifth-year salary from the first: $\text{\char36}60{,}800 - \text{\char36}52{,}000 = \text{\char36}8{,}800$. Over five years there are four equal raises, so divide by $5 - 1 = 4$ to get $x = \text{\char36}2{,}200$. Add $\text{\char36}2{,}200$ each year after the first: $\text{\char36}52{,}000$, $\text{\char36}54{,}200$, $\text{\char36}56{,}400$, $\text{\char36}58{,}600$, $\text{\char36}60{,}800$. Sum all five years: $\text{\char36}282{,}000$.

The equation means $3x - 12 = 24$ or $3x - 12 = -24$. Adding $12$ gives $3x = 36$ or $3x = -12$. Dividing by $3$ gives $x = 12$ or $x = -4$. Both solutions are $-4$ and $12\textsf{.}$

Rewrite $|6x + 12| = |6(x + 2)| = |6|\,|x + 2| = 6|x + 2|$. Since $6|x + 2| = 48$, divide by $6$ to get $|x + 2| = 8$. The expression $x + 2$ equals $8$ or $-8$; the positive value is $8\textsf{.}$

At least $2$ gives $d \ge 2$. Less than $7$ gives $d < 7$, so $2 \le d < 7$.

Subtract $4$ from both sides: $8x = 32$, so $x = 4$. Then $2x = 2 \cdot 4 = 8\textsf{.}$

In slope-intercept form $y = mx + b$, the constant term $b$ is the $y$-intercept. Here $b = 7$.

The slope is $m = \dfrac{4 - 1}{6 - 2} = \dfrac{3}{4}$. Substitute $(2, 1)$: $1 = \dfrac{3}{4}(2) + b$, so $b = -\dfrac{1}{2}$. The equation is $y = \dfrac{3}{4}x - \dfrac{1}{2}$.

Add 2: $3x \ge 9$. Divide by 3: $x \ge 3$, so $k = 3$.