1. The problem involves solving and understanding the expressions related to $y^2$ given by the equations: - $-6 = y^2$ - $3 = y^2$ - $7 - y^2$ 2. First, recall that $y^2$ represents the square of $y$, which is always greater than or equal to zero for real numbers. 3. Analyze each equation: - For $-6 = y^2$, since $y^2 \geq 0$ for all real $y$, this equation has no real solution because $-6$ is negative. - For $3 = y^2$, solve for $y$: $$y = \pm \sqrt{3}$$ - For the expression $7 - y^2$, this is not an equation but an expression that depends on $y$. 4. To understand $7 - y^2$, note that since $y^2 \geq 0$, the maximum value of $7 - y^2$ is when $y^2$ is minimum (i.e., 0), so: $$7 - y^2 \leq 7$$ and it decreases as $|y|$ increases. 5. Summary: - No real solutions for $-6 = y^2$. - Two real solutions for $3 = y^2$ are $y = \pm \sqrt{3}$. - $7 - y^2$ is an expression that decreases as $|y|$ increases, with a maximum value of 7 at $y=0$. Final answers: - $-6 = y^2$ has no real solutions. - $3 = y^2$ solutions: $y = \pm \sqrt{3}$. - $7 - y^2$ is an expression with maximum 7 at $y=0$.