Solve the inequality $2y - 1 < \frac{11y}{4} < \frac{1}{4}$. Answer: ... Given -25 ≤ x ≤ 50 and -15 ≤ y ≤ -5, what is the smallest possible value for x + y^2? Answer: ...

1. Split the compound inequality into two inequalities.

$2y-1<\frac{11y}{4}\quad \text{and}\quad \frac{11y}{4}<\frac14$

2. Solve the left inequality.

$2y-1<\frac{11y}{4}$
Multiply by $4$:
$8y-4<11y$
$-4<3y$
$y>-\frac43$

3. Solve the right inequality and combine both conditions.

$\frac{11y}{4}<\frac14$
Multiply by $4$:
$11y<1$
$y<\frac{1}{11}$

Combining with $y>-\frac43$ gives
$-\frac43<y<\frac{1}{11}.$

1. Find the minimum possible values contributing to $x+y^2$.

Since $-25\leq x\leq 50$, the smallest possible value of $x$ is $-25$.

Since $-15\leq y\leq -5$, the smallest possible value of $y^2$ occurs at $y=-5$:
$y^2=(-5)^2=25.$

2. Calculate the smallest possible value of the expression.

Smallest possible value:
$x+y^2=-25+25=0.$