Why I Get X & Y Mixed Up When Plotting Graphs

[Ziggletooth]

Ok so I don't know what is wrong with me, but when it comes to graphs I often get x and y mixed up and the whole thing becomes very confusing.

I had this question

plot a graph using y = 2/3x

The answer is:

if x = 3
y = (2/3 * 3) = 2
x = 3, y = 2

and these are integers which can be plotted very easily on the graph.

I got confused and did this

y = 2/3x
(y * 3) = (2/3x * 3)
3y = 2x
(3y / 2) = (2x / 2)
x = 1.5y

Obviously this is very wrong (if x = 3, y = 4.5 which is != 2 as above), but I'm not sure why. This is an equation isn't it? and those are the sort of balancing things you do to solve equations, so why has it gone so horribly wrong?

 

[evinda]

Ok so I don't know what is wrong with me, but when it comes to graphs I often get x and y mixed up and the whole thing becomes very confusing.

I had this question

plot a graph using y = 2/3x

The answer is:

if x = 3
y = (2/3 * 3) = 2
x = 3, y = 2

and these are integers which can be plotted very easily on the graph.

I got confused and did this

y = 2/3x
(y * 3) = (2/3x * 3)
3y = 2x
(3y / 2) = (2x / 2)
x = 1.5y

Obviously this is very wrong (if x = 3, y = 4.5 which is != 2 as above), but I'm not sure why. This is an equation isn't it? and those are the sort of balancing things you do to solve equations, so why has it gone so horribly wrong?

When you want to plot the graph of a line $y=ax$, you pick a value of $x$ that you want, say $x_1$, and you find the corresponding $y$, say $y_1=ax_1$ and since $(0,0)$ satisfies the equation of the line, you draw a line that passes through $(0,0)$ and $(x_1,y_1)$.

 

[Ziggletooth]

When you want to plot the graph of a line $y=ax$, you pick a value of $x$ that you want, say $x_1$, and you find the corresponding $y$, say $y_1=ax_1$ and since $(0,0)$ satisfies the equation of the line, you draw a line that passes through $(0,0)$ and $(x_1,y_1)$.

I'm sorry, I don't think I was clear. I understand that much but I want to know what was wrong with my method, did I do an operation wrong or can you not do it that way, and if so why not? Because it looks like an equation to me and that's how one would usually simplify an equation.

I don't wish to barrage this forum but in addition to that I do have another question. I said at the beginning I seem to get the coordinates x and y mixed up all the time and I can't reliably tell which is which. Here are two questions and both times I got the coordinates mixed up perhaps you can help me understand which is which.

So the first question is to plot

y = 1 2/3x

I do this
y = 5/3x
3y = 5x

and then I plot the coordinates 5,3 because I read it as 'for every 3y there is 5x' I go up 3 places on the y-axis and then along 5 places on the x axis.

The correct answer is 3,5

My brain doesn't seem to jump to this conclusion, so I figure I'll remember that (a * y = b * x) = (a * x = b * y ) that is to say, switch the coefficients because it seems like that works for some reason.

Anyway the next question

Graph the line that represents a proportional relationship between d and t with the property that an increase of 5 units in t corresponds to an increase of 2 units in d.

What is the unit rate of change of d with respect to t? (That is, a change of 1 unit in t will correspond to a change of how many units in d?)

I figure it's 5t = 2d so d = 2/5

Now I go to graph it, but I remember the rule I came up with about switching the coefficients, so I move 2 places on the t axis and 5 places on the d axis... and what a surprise, I got it wrong again it's the other way. I just can't seem to win here.

Can someone help me clear the mist surrounding this concept so I can place the coordinates correctly.

 

[mrtwhs]

y = 2/3x
(y * 3) = (2/3x * 3)
3y = 2x
(3y / 2) = (2x / 2)
x = 1.5y

Obviously this is very wrong (if x = 3, y = 4.5 which is != 2 as above), but I'm not sure why. This is an equation isn't it? and those are the sort of balancing things you do to solve equations, so why has it gone so horribly wrong?

When you say \(\displaystyle x=3\), that means you put the 3 where the \(\displaystyle x\) is!

\(\displaystyle 3=1.5y\).

Now try it.