First, multiplying each side of the equality by \(\displaystyle 5\) gives \(\displaystyle 3x > 30\). Next, dividing each side of the inequality by \(\displaystyle 3\) will solve for \(\displaystyle x\); \(\displaystyle x > 10\).

We simplify this inequality similarly to how we would simplify an equation

\(\displaystyle \dpi{100} \small 3x+8-8< 35-8\)

\(\displaystyle \dpi{100} \small \frac{3x}{3}< \frac{27}{3}\)

Thus \(\displaystyle \dpi{100} \small x< 9\)

In order to find the solution set, we solve \(\displaystyle 2x+12>42\) as we would an equation:

\(\displaystyle 2x+12>42\)

\(\displaystyle 2x>30\)

\(\displaystyle x>15\)

Therefore, the solution set is any value of \(\displaystyle x>15\).

Substituting 3 in for x, you will get 0 in the denominator of the fraction. It is not possible to have 0 be the denominator for a fraction so there is no possible solution to this equation.

A fraction is considered undefined when the denominator equals 0. Set the denominator equal to zero and solve for the variable.

\(\displaystyle \small x+4=0\)

\(\displaystyle \small x=-4\)

Multiply the equation on both sides by LCM \(\displaystyle \frac{x(x-2)}{1}\):

\(\displaystyle \frac{x+9}{x}+\frac{x+3}{x-2} = \frac{4x+11}{x}\)

\(\displaystyle \frac{x(x-2)}{1} \cdot \frac{x+9}{x}+\frac{x(x-2)}{1} \cdot\frac{x+3}{x-2} = \frac{x(x-2)}{1} \cdot\frac{4x+11}{x}\)

\(\displaystyle (x-2) \left ( x+9 \right ) + x( x+3 )= (x-2) (4x+11)\)

\(\displaystyle \left ( x^{2}+7x -18\right ) + ( x^{2}+3x )= 4x^{2}+3x -22\)

\(\displaystyle 4x^{2}+3x -22 = x^{2}+7x -18 + x^{2}+3x\)

\(\displaystyle 4x^{2}+3x -22 = 2x^{2}+10x -18\)

\(\displaystyle 4x^{2}+3x -22 - 2x^{2}-10x +18 = 2x^{2}+10x -18 - 2x^{2}-10x +18\)

\(\displaystyle 2x^{2}-7x -4 = 0\)

\(\displaystyle (x-4)(2x+1)= 0\)

\(\displaystyle x-4 = 0\)

\(\displaystyle x= 4\)

or 

\(\displaystyle 2x+1 = 0\)

\(\displaystyle 2x = -1\)

\(\displaystyle x = -\frac{1}{2}\)

Substitution confirms that these are the solutions. 

There are two solutions of unlike sign.

The problem is basically asking for what value of \(\displaystyle A\) the equation 

\(\displaystyle 6- |x + A | = 8\)

has no solution.

We can simplify as folllows:

\(\displaystyle 6- |x + A | - 6 = 8 - 6\)

\(\displaystyle - |x + A | = 2\)

\(\displaystyle |x + A | = - 2\)

Since the absolute value of a number must be nonnegative, regardless of the value of \(\displaystyle A\), this equation can never have a solution. Therefore, the correct response is that none of the given equations has a solution.

Multiply both sides by LCD \(\displaystyle \frac{10(y+3)}{1}\):

\(\displaystyle \frac{4}{y+3} = \frac{y+6}{10}\)

\(\displaystyle \frac{4}{y+3}\cdot \frac{10(y+3)}{1} = \frac{y+6}{10} \cdot \frac{10(y+3)}{1}\)

\(\displaystyle 4 \cdot 10 = (y+6) (y+3)\)

\(\displaystyle y^{2}+9y+18 = 40\)

\(\displaystyle y^{2}+9y+18- 40 = 40 - 40\)

\(\displaystyle y^{2}+9y-22 = 0\)

\(\displaystyle (y+11)(y-2) = 0\)

\(\displaystyle y+11 \Rightarrow y = -11\)

or

\(\displaystyle y-2 = 0 \Rightarrow y = 2\)

There are two solutions of unlike sign.