In order to solve the equation for $\displaystyle x$, we need to isolate $\displaystyle x$ on one side of the equation:

$\displaystyle \frac{1}{3}x+\frac{2}{3}y=\frac{7}{9}$

$\displaystyle \frac{1}{3}x=\frac{7}{9}-\frac{2}{3}y$

$\displaystyle x=\frac{21}{9}-2y$

Reducing the fraction,

$\displaystyle x=\frac{7}{3}-2y$

$\displaystyle 3d + ad + y = 10$

$\displaystyle 3d + ad + y - y= 10 - y$

$\displaystyle 3d + ad = 10 - y$

$\displaystyle \left (3 + a \right )d = 10 - y$

$\displaystyle \frac{\left (3 + a \right )d }{3+a}= \frac{10 - y }{3+a}$

$\displaystyle d= \frac{10 - y }{3+a}$

$\displaystyle v = 1+ \sqrt[3]{w-7}$

$\displaystyle v- 1 = 1+ \sqrt[3]{w-7} - 1$

$\displaystyle \sqrt[3]{w-7} = v- 1$

$\displaystyle \left (\sqrt[3]{w-7}\right )^{3} = \left (v- 1 \right )^{3}$

$\displaystyle w-7= v^{3} -3v^{2}+3v-1$

$\displaystyle w-7+ 7= v^{3} -3v^{2}+3v-1 + 7$

$\displaystyle w = v^{3} -3v^{2}+3v+6$

$\displaystyle a^{3}- y + 1= c^{2}$

$\displaystyle a^{3}- y + 1+ y - 1= c^{2} + y - 1$

$\displaystyle a^{3} = c^{2} + y - 1$

$\displaystyle \sqrt[3]{ a^{3} } =\sqrt[3]{ c^{2} + y - 1}$

$\displaystyle a=\sqrt[3]{ c^{2} + y - 1}$

 $\displaystyle v^{2}+ 4vt + 4t^{2}$ is a perfect square trinomial:

$\displaystyle v^{2}+ 4vt + 4t^{2} = v^{2}+ 2 \cdot 2 t + (2t)^{2} = (v+2t)^{2}$

The equation can be rewritten as 

$\displaystyle (v+2t)^{2} = h + 30$

By the square-root property, since no assumption was made about the sign of any variable, 

$\displaystyle v+2t = \pm\sqrt{ h + 30 }$

$\displaystyle v+2t -v = \pm \sqrt{ h + 30 }-v$

$\displaystyle 2t= -v\pm \sqrt{ h + 30 }$

$\displaystyle \frac{2t}{2}=\frac{- v\pm \sqrt{ h + 30 } }{2}$

$\displaystyle t=\frac{- v\pm \sqrt{ h + 30 }}{2}$

Therefore, 

$\displaystyle t=\frac{- v + \sqrt{ h + 30 }}{2}$ or $\displaystyle t=\frac{- v - \sqrt{ h + 30 } }{2}$.

The statement is a quadratic equation in $\displaystyle y$, so it can be solved using the quadratic formula, 

$\displaystyle y = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

where $\displaystyle a = 1,b = 6, c = Q$

$\displaystyle y = \frac{-6 \pm \sqrt{6^{2}-4(1)(Q)}}{2 (1)}$

$\displaystyle y = \frac{-6 \pm \sqrt{36-4Q}}{2}$

$\displaystyle y = \frac{-6 \pm \sqrt{4(9-Q)}}{2}$

$\displaystyle y = \frac{-6 \pm 2 \sqrt{9-Q}}{2}$

$\displaystyle y = \frac{2\left (-3 \pm \sqrt{9-Q} \right )}{2}$

$\displaystyle y = -3 \pm \sqrt{9-Q}$

We are given a classic quadratic equation, but we aren't asked for the solutions, just how many distinct solutions there are. Remember, distinct solutions are different solutions. If we get two solutions that are the same numbers, they do not count. 

The quickest way to solve this involves some factoring. 

Start by pulling out a 3

$\displaystyle \small 3(x^2-9)=0$

Now, within our parentheses, we have a classic difference of squares. The interior factors further to look like this.

$\displaystyle \small \small 3{(x+3)(x-3)}=0$

From here we can either solve the equation and count our solutions, or we can recognize that the two factors are different and therefore will give different solutions. Let's solve it by using the Zero Product Property

Solution 1

$\displaystyle \small x+3=0$

$\displaystyle \small x=-3$

Solution 2

$\displaystyle \small x-3=0$

$\displaystyle \small x=3$

Thus, we have two distinct solutions!

$\displaystyle h = f+ \sqrt{3g+ 4}$

$\displaystyle h- f = f+ \sqrt{3g+ 4} - f$

$\displaystyle \sqrt{3g+ 4} = h- f$

$\displaystyle \left (\sqrt{3g+ 4} \right )^{2}= \left (h- f \right )^{2}$

$\displaystyle 3g+4= h^{2}-2hf+f^{2}$

$\displaystyle 3g+4- 4= h^{2}-2hf+f^{2} - 4$

$\displaystyle 3g = h^{2}-2hf+f^{2} - 4$

$\displaystyle \frac{3g }{3}= \frac{h^{2}-2hf+f^{2} - 4}{3}$

$\displaystyle g= \frac{h^{2}-2hf+f^{2} - 4}{3}$

$\displaystyle K$ is 44% of $\displaystyle P$, so $\displaystyle 2K$ is $\displaystyle 2 \times 44 \% = 88 \%$ of $\displaystyle P$.

Also, $\displaystyle 3P$ is 300% of $\displaystyle P$.

Add these:

$\displaystyle 2K+3P$ is $\displaystyle 88 \% + 300 \% = 388 \%$ of $\displaystyle P$

To solve the equation, we first group the $\displaystyle x$ terms on one side and the constants on the other side:

$\displaystyle -3x+18=7x-2$

$\displaystyle -10x=-20$

Now we can simply divide both sides by $\displaystyle -10$ to solve for $\displaystyle x$:

$\displaystyle -10x=-20$

$\displaystyle x=2$