PARCC Algebra I Reasoning with Equations and Inequalities. Practice it free.

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 $7\textsf{.}$

Here $a = 1$, $b = 0$, and $c = 9$. Substituting into $b^{2} - 4ac$ gives $0^{2} - 4(1)(9) = -36\textsf{,}$ so there are $0$ distinct real roots.

Here $a = 1$, $b = -4$, and $c = 4$. Substituting into $b^{2} - 4ac$ gives $(-4)^{2} - 4(1)(4) = 0$, so there is exactly $1$ distinct real solution.

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 $y = 6$. From $x + y \le 10$, $x \le 4$. From $x \ge y - 2$, $x \ge 4$. The only listed value that satisfies both is $x = 4$.

Set $p(x)=q(x)$: $\dfrac{1}{2}x+1=x^{2}-\dfrac{1}{2}x-1$, so $0=x^{2}-x-2$. Then $(x-2)(x+1)=0$, and $x=2$ or $x=-1$. For $x=-1$, $p(-1)=\dfrac{1}{2}$, so one solution is $\left(-1,\dfrac{1}{2}\right)\textsf{.}$