## This is a question that requires the use of algebra.

In order to solve this problem, we must first find the value of y when x is 8. This can be done by substituting 8 in for x and solving for y:

y = __

__x + y _

__72. The answer is: 16. We can then use our algebra skills to solve for the value of x when y is 16:

-16(__)__+_yz__=0_(?)+. First, we subtract -16 from both sides and combine like terms so that it looks like this: __-(_-16)=__(-38). Next, divide by 38 on each side of the equation in order to find a quotient and remainder. __ (?/38)=(-)15/(18), which means that there are 15 parts with an 18 left over. To get rid of the leftover 18, take 18 away from both sides of the equation. This leaves us with __(0/38)=__15, which tells us that x is equal to 15 when y is 16.

### Numbering: not numbered but can be formatted as such for ease of reading if desired

Bullet Point: If we know something about how one variable changes in relation to another, then we may need some algebra skills to solve a related problem. In this example, we found out that y varies directly as x (so it needs an exponent), and so solving for “y” was done using variables like “x”. We were able to find the answer by substituting different values into our equation until the correct value appeared; and once solved, use more algebra skills to figure out the answer.

Bullet Point: Solving for y in terms of x is a skill that can be used with more complex equations, such as when we need to find V=Px+Q using algebra skills; all because one variable changes directly as another.

See how this solved problem relates back to some key ideas about algebra and math? You should always be aware of what you are solving for (or what it might relate back to) before trying any methods or formulas!

Bullet Point: In this example, we found out that y varies directly as x so they needed an exponent.

Bullet Point: You can also use this to find the value of y when x is a number other than 72, just in case you have not yet memorized that equation. Just input different values into our formula until we get one that satisfies what we are looking for. For example, if I were solving for y with an x of 12, then my first equation would be solved like so: y=x^0+72=$12$+72 =84 . If it’s still no good or goes over your head, don’t worry! We’ll talk about how to solve equations more later on in this post.

### Let me know if any part needs clarification and I will try my best to help you out.

-Bullet Point: We also talked about how to solve for y when x is an exponent, but that example was a little more complicated so we left it as homework! For those of you who are not quite up to the challenge, I will give the answer in this section for your convenience. Don’t worry if you don’t want to try solving it yourself because there’s another way of finding values without algebra and math formulas that we’ll talk about below.

A solution would be worked out like this:

y=x^0+72=$12$+72 =84. If we were looking for y with an x of 81 instead then our equation would be solved like so: y=(81^0+72)=$12$+81*80 =79.

Bullet Point: The equation for the general shape of a line is y=mx+b.

Bullet Point: In order to find the slope, which is how much “change” there will be in y as x changes by one unit, we take two points on our graph that are close together and use them to calculate the change in Y (y) between those two points over an increment of X (x).

We can draw this with your favorite drawing tool or free program like Microsoft paint or paint.net if you want! Drawing it out helps me visualize what’s happening better than just looking at numbers. I have included some examples below so you know what they should look like! You may need to zoom way into the image to make things more clear.

There are a few ways we can calculate the slope, but they all involve taking two points on our graph that are close together and using them to calculate the change in Y (y) between those two points over an increment of X (x).

The first is by finding what’s called “the point-slope formula” which requires us to take both coordinates of any one point from the line equation: y=mx+b and plugging it into this form where m is for slope. This would look like \(m=-\frac{(Y_0)(X_0)-(Y_n)(X_n)} {(\Delta x)(\Delta y)}\)

The second is by simply taking the average of both coordinates for any two points on the line equation: y=mx+b and plugging it into this form \(m=-\frac{(Y_0)(X_0)+(Y_n)(X_n)} {(\Delta x)(\Delta y)}\)

### Both methods are equally accurate, but I would recommend point-slope because it’s a little more straightforward.

For example, take these coordinate pairs from the graph below which correspond to Point A: (x=72,y=-) and Point B:(x=12,y=18). Plug them into our formula above using \(m=-\frac{(Y_0)(X_0)+(-18)} {(\Delta x)(-72)}\). The slope of the line is \(m=-\frac{ -12}{ -72}\) which gives us a value for m. If we plug that into our formula, we get an answer of -19/35. An easy way to find out if it was correct would be place both values on either side of the equation and see if they match (ie: y=mx+b). In this example, (-18)=(-12)*x + b or y=(y)+(x)*(-) => 18-(12)/(-36)=(y)-((x))

This means that y=-19/35 when x is -36.

y)}\) *Both methods are equally accurate, but I would recommend point-slope because it’s a little more straightforward. For example, take these coordinate pairs from the graph below which correspond to Point A: (x=72,y=-) and Point B:(x=12,y=18). Plug them into our formula above using \(m=-\frac{(Y_0)(X_0)+(-18)} {(\Delta x)(-72)}\). The slope of the line is \(m=-\frac{ -12}{ -72}\) which gives us a value for m. If we plug that into the point-slope formula, we get y=-19/35.

*Doing this for Point B gives us a slope of m=\( \frac{18}{12}\) and an equation of \(y=17/11 = 17.0927\). For both methods to be correct, the two values should match up (ie: Y_0=(x)(y)). In our example with Point A (- 18) equals -36 when plugged in on either side of the equation so it is also valid that \(\Delta x = -72\) because they are equivalent numbers.

If we take the equation \(y=17.0927\) and solve for x, it equals 12.0577 which is equivalent to Point B (x = 12). This means that point A can also be correctly solved by \(\Delta x=-\frac{12}{ -72}\) as if solving for y: \[y=-(\frac{\Delta x}{m}) -18 \\ Y_0=(X)(Y) == 17.0927(x)=(12)(17.0927)\] so our final answer would be y=13/23.

For both methods of finding the value at a different X coordinate, they should match up when plugged into either side of the equation because they are equivalent numbers. For example, if we take the equation \(x=17.0927\) and solve for y it equals 12.0577 which is equivalent to point B (y=-12). This means that Point A can also be correctly solved by \(\Delta x=-\frac{12}{-72}\) as if solving for x: \[Y_0=(X)(Y) == 17.0927(x)=(13)(17.0927)\] so our final answer would be Y=13/23.* *If you didn’t understand this last explanation about how points match up on both sides of an inequality symbol when plugging in different values, please click here In summary: There are two