First, subtract 7 to both sides.

\(\displaystyle 3y-xy+7=2\)

\(\displaystyle 3y-xy=-5\)

Factor the y on the left hand side.

\(\displaystyle y(3-x)=-5\)

Divide both sides by 3 – x.

\(\displaystyle y=\frac{-5}{3-x}\)

Take an extra step by factoring the minus sign on the denominator.

\(\displaystyle y=\frac{-5}{-(x-3)}\)

Cancel the minus signs.

\(\displaystyle y=\frac{5}{x-3}\)

Let \(\displaystyle P_{1} = (1, \; -2) \; \; and \; \; P_{2} = (-3, \; 6)\).

First, calculate the slope between the two points.

\(\displaystyle m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{6 - (-2)}{-3 - 1} = \frac{8}{-4} = -2\)

Next, use the slope-intercept form to calculate the intercept. We are able to plug in our value for the slope, as well the the values for \(\displaystyle P_1\).

\(\displaystyle y = mx + b\)

\(\displaystyle -2 = -2(1) + b\)

\(\displaystyle b=0\)

Using slope-intercept form, where we know \(\displaystyle m=-2\) and \(\displaystyle b=0\), we can see that the equation for this line is \(\displaystyle y = -2x\).

To calculate a line passing through two points, we first need to calculate the slope, \(\displaystyle \small m\).

\(\displaystyle \small m=\frac{\Delta y}{\Delta x} =\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{8-2}{1-(-3)} = \frac{6}{4} = \frac{3}{2}\)

Now that we have the slope, we can plug it into our equation for a line in slope intercept form.

\(\displaystyle \small y=\frac{3}{2}x+b\)

To solve for \(\displaystyle \small b\), we can plug in one of the points we were given. For the sake of this example, let's use \(\displaystyle \small (-3,2)\), but realize either point will give use the same answer.

\(\displaystyle \small 2=\frac{3}{2}(-3)+b\)

\(\displaystyle \small 2=\frac{-9}{2}+b\)

\(\displaystyle \small 2+\frac{9}{2} = b\)

\(\displaystyle \small \frac{13}{2} = b\)

Now that we have solved for b, we can plug that into our slope intercept form and produce and the answer

\(\displaystyle \small y=\frac{3}{2}x+\frac{13}{2}\)

We are looking for a point, \(\displaystyle \small (x,y)\), where these two lines intersect. While there are many ways to solve for \(\displaystyle \small x\) and \(\displaystyle \small y\) given two equations, the simplest way I see is to use the elimination method since by adding the two equations together, we can eliminate the \(\displaystyle \small x\) variable.

\(\displaystyle 7y=7\)

Dividing both sides by 7, we isolate y.

\(\displaystyle \small y=1\)

Now, we can plug y back into either equation and solve for x.

\(\displaystyle \small 2x+(5)(1)=8\)

\(\displaystyle \small 2x+5=8\)

Next, we can solve for x.

\(\displaystyle \small 2x=3\)

\(\displaystyle \small x=(3/2)\)

Therefore, the point where these two lines intersect is \(\displaystyle \left(\frac{3}{2}, 1\right)\).

The first thing is to plug in the given value so you equation is 

\(\displaystyle 2=\frac{1}{3}x+9\).  

Then you must subtract \(\displaystyle 9\) from both sides in order to get \(\displaystyle x\) by itself.  

You now have 

\(\displaystyle -7=\frac{1}{3}x\) 

and you must multiply by \(\displaystyle 3\) for each side and you get \(\displaystyle x=-21\).

Add \(\displaystyle 7x\) on both sides.

\(\displaystyle -7x-16 +(7x)= 3x+2+(7x)\)

\(\displaystyle -16 = 10x+2\)

Subtract 2 from both sides.

\(\displaystyle -16 -2= 10x+2-2\)

\(\displaystyle -18 = 10x\)

Divide by 10 on both sides.

\(\displaystyle \frac{-18}{10} =\frac{ 10x}{10}\)

Reduce the fractions.

The answer is:  \(\displaystyle -\frac{9}{5}\)

Distribute the right side.

\(\displaystyle -2(2-3x) = -4 +6x\)

Rewrite the equation.

\(\displaystyle 3x+5 = -4+6x\)

Subtract \(\displaystyle 3x\) on both sides.

\(\displaystyle 3x+5 -3x= -4+6x-3x\)

\(\displaystyle 5=-4+3x\)

Add 4 on both sides.

\(\displaystyle 5+4=-4+3x+4\)

\(\displaystyle 9=3x\)

Divide by three on both sides.

\(\displaystyle \frac{9}{3}=\frac{3x}{3}\)

The answer is:  \(\displaystyle 3\)

You must get \(\displaystyle x\) by itself so you must add \(\displaystyle 5\) to both side which results in 

\(\displaystyle 16=x^{2}\).  

You must get the square root of both side to undue the exponent.  

This leaves you with \(\displaystyle x=4\).  

But since you square the \(\displaystyle x\) in the equation, the original value you plug can also be its negative value since squaring it will make it positive anyway.  

This means your answer can be \(\displaystyle x=4\) or \(\displaystyle x=-4\).

The asymptote of this equation can be found by observing that \(\displaystyle e^{x-3} > 0\) regardless of \(\displaystyle x\). We are thus solving for the value of \(\displaystyle y\) as \(\displaystyle e^{x-3}\) approaches zero.

\(\displaystyle -2e^{x-3} < -2\cdot 0\)

\(\displaystyle -2e^{x-3} < 0\)

\(\displaystyle -2e^{x-3} -5 < 0-5\)

\(\displaystyle -2e^{x-3} -5 < -5\)

\(\displaystyle y< -5\)

So the value that \(\displaystyle y\) cannot exceed is \(\displaystyle -5\), and the line \(\displaystyle y = -5\) is the asymptote.

An exponential equation of the form \(\displaystyle f(x) = ae^{x-h} +k\) has only one asymptote - a horizontal one at \(\displaystyle y = k\). In the given function, \(\displaystyle k = -2\), so its one and only asymptote is \(\displaystyle y = -2\).