NJSLA Algebra II 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 $7\textsf{.}$

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

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

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 $\left(-1,\dfrac{1}{2}\right)\textsf{.}$

From $3x = 21$, $x = 7$. Substitute into the second equation: $14 + y = 19$, so $y = 5$. Then $y - x = 5 - 7 = -2$.

Substitute $y = -x + 8$ into the second equation: $4x - x + 8 = 14$, so $3x = 6$ and $x = 2$. Then $y = 6$, and $x + y = 8$.

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