SAT Algebra. Practice it free.

Slope-intercept form is $y = mx + b$. Substituting $(4, 1)$ with $m = \dfrac{3}{2}$ gives $1 = \dfrac{3}{2}(4) + b = 6 + b$, so $b = -5$. The equation is $y = \dfrac{3}{2}x - 5\textsf{.}$

The equation means $2x + 10 = 16$ or $2x + 10 = -16$. So $2x = 6$ or $2x = -26$, and $x = 3$ or $x = -13$. Both solutions are $-13$ and $3\textsf{.}$

Rewrite $|5x + 10| = |5(x + 2)| = |5|\,|x + 2| = 5|x + 2|$. Since $5|x + 2| = 35$, divide by $5$ to get $|x + 2| = 7$. The expression $x + 2$ equals $7$ or $-7$; the positive value is $7\textsf{.}$

Divide both sides by $12$: $x = 5$. Then $5x = 5 \cdot 5 = 25\textsf{.}$

Divide both sides by 2: $x < \dfrac{5}{2}$, so $k = \dfrac{5}{2}$.

Substitute $q = 3$: $2p + 5(3) = 39$, so $2p + 15 = 39$. Then $2p = 24$ and $p = 12$.

Slope-intercept form is $f(x) = mx + b\textsf{.}$ The cost increases by $\text{\char36}4$ per mile, so $m = 4\textsf{.}$

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