AK STAR Grade 9 Algebra. Practice it free.

The equation means $x + 8 = 21$ or $x + 8 = -21$. So $x = 13$ or $x = -29$. Both solutions are $-29$ and $13\textsf{.}$

Rewrite $|2x + 10| = |2(x + 5)| = |2|\,|x + 5| = 2|x + 5|$. Since $2|x + 5| = 26$, divide by $2$ to get $|x + 5| = 13$. The expression $x + 5$ equals $13$ or $-13$; the positive value is $13\textsf{.}$

The greatest common factor of $5x^{2}$ and $10x$ is $5x\textsf{.}$ Factoring gives $5x(x + 2)$.

More than $70$ means $s > 70$. At most $100$ means $s \le 100$, so $70 < s \le 100$.

Check both constraints: $12a + 8c \le 48$ and $c \ge 2\textsf{.}$ The pair $(2, 3)$ costs $12(2) + 8(3) = 48$ and has $3$ child tickets, so both constraints are met.

$81 - x^{2} = 9^{2} - x^{2} = (9 - x)(9 + x)$.

Here $a = 1$, $b = 0$, and $c = -25$. Substituting into $b^{2} - 4ac$ gives $0^{2} - 4(1)(-25) = 100\textsf{,}$ so there are $2$ distinct real roots.

Here $a = 1$, $b = 6$, and $c = 9$. Substituting into $b^{2} - 4ac$ gives $6^{2} - 4(1)(9) = 0$, so there is exactly $1$ distinct real solution.