PSAT 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 height must be at least $48.0$ inches. The minimum increase is $48.0 - 45.5 = 2.5\textsf{.}$

Add $3$ to both sides: $9x = 45$, so $x = 5$. Then $x - 1 = 5 - 1 = 4\textsf{.}$

Subtract $x$ from both sides: $3x + 1 > 10$. Subtract 1: $3x > 9$. Divide by 3: $x > 3$, so $k = 3$.

Substitute $y = 5$: $8x + 3(5) = 55$, so $8x + 15 = 55$. Then $8x = 40$ and $x = 5$.

Point-slope form is $y - y_1 = m(x - x_1)$. With $m = -\dfrac{1}{2}$ and $(x_1, y_1) = (6, 4)$, the equation is $y - 4 = -\dfrac{1}{2}(x - 6)\textsf{.}$

In $y = mx + b$, the $y$-intercept is $b = -7$.