There are two ways to solve this:

-The 1st equation can be mutliplied by $\displaystyle 2$ while the 2nd equation can be multiplied by $\displaystyle -3$ and added to the 1st equation to make it a single variable equation where

$\displaystyle 22y=-66$  

$\displaystyle y=-3$.

This can be plugged into either equation to get $\displaystyle x=2$

or

-The 2nd equation can be simplified to,

$\displaystyle x-2y=8$  

$\displaystyle x=8+2y$.

This value for $\displaystyle x$ can then be substituted into the first equation to make the equation single variable in $\displaystyle y, 3(8+2y)+5y=-9$.

Solving, gives $\displaystyle y=-3$, which can be plugged into either original equation to get $\displaystyle x=2$

Plug in $\displaystyle -14$ for $\displaystyle x$ in the $\displaystyle f(x)$ equation to get

 $\displaystyle f(-14)=7+(\frac{42}{-14})=7-3=4$

Given the equation,

 $\displaystyle f(x)=5x^3+2x-24$ and  $\displaystyle x=2$

Plug in $\displaystyle 2$ for $\displaystyle x$ to the equation, $\displaystyle f(2)=5(2)^3+2(2)-24$ 

Solve and simplify. 

$\displaystyle f(2)=5(8)+4-24$

$\displaystyle f(2)=40+4-24$

$\displaystyle f(2)=44-24$

$\displaystyle f(2)=20$

$\displaystyle f(x)=\frac{4x-2}{x+3}, x\neq-3$

Plug in the $\displaystyle x$ value for $\displaystyle x$.

$\displaystyle f(0)=\frac{4(0)-2}{(0)+3}$

Simplify

$\displaystyle f(0)=\frac{0-2}{3}$

Subtract

$\displaystyle f(0)=\frac{-2}{3}$

$\displaystyle x+y=32$ rearranges to

$\displaystyle x=32-y$

and

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

$\displaystyle 32-y=\frac{2}{3}y-2$

$\displaystyle 32-1\frac{2}{3}y=-2$

$\displaystyle -1\frac{2}{3}y=-34$

$\displaystyle -\frac{5}{3}y=-34$

$\displaystyle y=\frac{-34}{1}\cdot\frac{-3}{5}=\frac{102}{5}=20\frac{2}{5}$

$\displaystyle x=32-20\frac{2}{5}=11\frac{3}{5}$

In the second equation, you can substitute $\displaystyle 3x + 4$ for $\displaystyle y$ from the first.

$\displaystyle 2x + 3y = 34$

$\displaystyle 2x + 3 (3x + 4) = 34$

$\displaystyle 2x + 3 (3x) + 3 (4) = 34$

$\displaystyle 2x + 9x + 12 = 34$

$\displaystyle 11x + 12 = 34$

$\displaystyle 11x = 22$

$\displaystyle x = 2$

Now, substitute 2 for $\displaystyle x$ in the first equation:

$\displaystyle y = 3x + 4$

$\displaystyle y = 3 (2) + 4$

$\displaystyle y = 6 + 4$ 

$\displaystyle y = 10$

The solution is $\displaystyle (2, 10)$

To solve for $\displaystyle x$, you must isolate it from the other variables. Start by adding $\displaystyle b$ to both sides to give you $\displaystyle ax=cd+b$. Now, you need only to divide $\displaystyle a$ from both sides to completely isolate $\displaystyle x$. This gives you a solution of $\displaystyle x=\frac{cd+b}{a}$.

On the equation, replace x with 2 and then simplify.

$\displaystyle y=x^3-4x+9$

$\displaystyle y=(2)^3-4(2)+9$

$\displaystyle y=8-8+9$

$\displaystyle y=9$

$\displaystyle x^2-5x-36$

This is a quadratic equation. We can solve  for $\displaystyle \small x$ either by factoring or using the quadratic formula. Since this equation is factorable, I will present the factoring method here.

The factored form of our equation should be in the format $\displaystyle (A+B)(C+D)=0$.

To yield the first value in our original equation ($\displaystyle \small x^2$), $\displaystyle \small A=x$ and $\displaystyle \small C=x$.

$\displaystyle (x+B)(x+D)=0$

To yield the final term in our original equation ($\displaystyle \small -36$), we can set $\displaystyle \small B=-9$ and $\displaystyle \small D=4$.

$\displaystyle -9*4=-36\ \text{and}\ -9+4=-5$

$\displaystyle (x-9)(x+4)=0$

Now that the equation has been factored, we can evaluate $\displaystyle \small x$. We set each factored term equal to zero and solve.

$\displaystyle \begin{matrix} x-9=0 &x+4=0 \\ x-9+9=0+9&x+4-4=0-4 \\ x=9 & x=-4 \end{matrix}$

To find the point where these two lines intersect, set the equations equal to each other, such that $\displaystyle y$ is substituted with the $\displaystyle x$ side of the second equation. Solving this new equation for $\displaystyle x$ will give the $\displaystyle x$-coordinate of the point of intersection.

$\displaystyle 3x = x - 2$

Subtract $\displaystyle x$ from both sides.

$\displaystyle (3x) - x = (x - 2) - x$

$\displaystyle 2x = - 2$

Divide both sides by 2.

$\displaystyle \frac{2x}{2} = \frac{-2}{2}$

$\displaystyle x = - 1$

Now substitute $\displaystyle -1$ into either equation to find the $\displaystyle y$-coordinate of the point of intersection.

$\displaystyle y = 3x$

$\displaystyle y= 3(-1)$

$\displaystyle y = -3$

With both coordinates, we know the point of intersection is $\displaystyle (-1,-3)$. One can plug in $\displaystyle -1$ for $\displaystyle x$ and $\displaystyle -3$ for $\displaystyle y$ in both equations to verify that this is correct.