KSA Grade 10 Algebra. Practice it free.

The equation means $x + 16 = 27$ or $x + 16 = -27$. So $x = 11$ or $x = -43$. Both solutions are $-43$ and $11\textsf{.}$

Either $x - 2 = 11$, giving $x = 13$ and $2x = 26$, or $x - 2 = -11$, giving $x = -9$ and $2x = -18$. The positive value of $2x$ is $26\textsf{.}$

The greatest common factor of $12x^{2}$ and $8x$ is $4x\textsf{.}$ Factoring gives $4x(3x + 2)$.

More than $6$ means $p > 6$. At most $8$ means $p \le 8$, so $6 < p \le 8$.

The money equation is $4x + 6y = 22\textsf{.}$ Ticket counts must be whole numbers that are zero or greater. The pair $(4, 1)$ gives $4(4) + 6(1) = 22$ and both counts are whole and nonnegative.

$49 - x^{2} = 7^{2} - x^{2} = (7 - x)(7 + x)$.

Here $a = 1$, $b = -10$, and $c = 25$. Substituting into $b^{2} - 4ac$ gives $(-10)^{2} - 4(1)(25) = 0\textsf{,}$ so there is exactly $1$ distinct real root.

Here $a = 2$, $b = -7$, and $c = 3$. Substituting into $b^{2} - 4ac$ gives $(-7)^{2} - 4(2)(3) = 49 - 24 = 25$.