We are asked to solve three systems of equations and inequalities. Each system consists of an inequality and an equation, which must be satisfied simultaneously. a) $x+3 \neq 0$ and $x^2-9 = 0$ b) $x^2-x+1 \neq 0$ and $x^3+1 = 0$ c) $5x-6 \neq 0$ and $25x^2-36 = 0$

1. Problem Description

We are asked to solve three systems of equations and inequalities. Each system consists of an inequality and an equation, which must be satisfied simultaneously.

a) $x+3 \neq 0$ and $x^2-9 = 0$

b) $x^2-x+1 \neq 0$ and $x^3+1 = 0$

c) $5x-6 \neq 0$ and $25x^2-36 = 0$

2. Solution Steps

a)

First solve $x^2-9=0$.

$x^2 = 9$

$x = \pm 3$

Now check the inequality $x+3 \neq 0$.

$x \neq -3$

Therefore, the only solution is $x=3$.

b)

First solve $x^3+1=0$.

$x^3 = -1$

$x = -1$

Now check the inequality $x^2-x+1 \neq 0$.

$(-1)^2 - (-1) + 1 = 1 + 1 + 1 = 3 \neq 0$

Since the inequality is satisfied, the solution is $x = -1$.

c)

First solve $25x^2 - 36 = 0$.

$25x^2 = 36$

$x^2 = \frac{36}{25}$

$x = \pm \sqrt{\frac{36}{25}} = \pm \frac{6}{5}$

Now check the inequality $5x-6 \neq 0$.

$5x \neq 6$

$x \neq \frac{6}{5}$

Therefore, the only solution is $x = -\frac{6}{5}$.

3. Final Answer

a) $x=3$

b) $x=-1$

c) $x=-\frac{6}{5}$