# Math Help - Equation

1. ## Equation

I have y=x+4/x-3

For some reason I have to change it to :

y = 1 + 7/x-3

I'm not sure how to do this, but mostly I have no idea "Why" we change it to the second form. It may have to do with graphing but I have no idea.

Please shed some light on this..

Thanks!

2. Originally Posted by phgao
I have y=x+4/x-3

For some reason I have to change it to :

y = 1 + 7/x-3

I'm not sure how to do this, but mostly I have no idea "Why" we change it to the second form. It may have to do with graphing but I have no idea.

Please shed some light on this..

Thanks!
Are you sure that you have included all necessary parentheses?

$y=(x+4)/x-3$

would expand to:

$y=1+4/x-3$,

which is closer to the form you are asking for, but we would normally
collect the numeric terms together to:

$y=4/x-2$.

RonL

3. Um this is how i was shown...
y = (x+4)/(x-3)
y+4 = 1(x-3)+7
y+4 = (x-3)[1+(7/(x-3))]

And this last step I don't understand

y = 1 + [7/(x-3)]

Also I was wondering if you know why you would change the equation into this form? Thx.

4. Originally Posted by phgao
Um this is how i was shown...
y = (x+4)/(x-3)
y+4 = 1(x-3)+7
y+4 = (x-3)[1+(7/(x-3))]

And this last step I don't understand

y = 1 + [7/(x-3)]

Also I was wondering if you know why you would change the equation into this form? Thx.
To be honest I don't understand the first step.

If:

$y=\frac{x+4}{x-3}$

then when $x=0, \ y=-4/3$, while if

$y+4 = 1(x-3)+7$,

then when $x=0, \ y=0$,

RonL

5. Originally Posted by phgao
Um this is how i was shown...
y = (x+4)/(x-3)
.
.
.
y = 1 + [7/(x-3)]

Also I was wondering if you know why you would change the equation into this form? Thx.
Hello,

if you do the polynomial division:
$(x+4) : (x-3)=1$
The remainder is 7. So you get
$\frac{x+4}{x-3}=1+\frac{7}{x-3}$

If you have to calculate the equation of the symptotic line it's much easier to use the 2nd form of the equation:
$\lim _{\csub {x \rightarrow \infty}}{ \frac {x+4}{x-3}}=\lim_{\csub {x \rightarrow \infty}}{\left(1+\frac{7}{x-3} \right)}= 1$

I've attached the graph of your function to demonstrate that the y-values get closer and closer to 1.

Bye

6. Originally Posted by phgao
I have y=x+4/x-3

For some reason I have to change it to :

y = 1 + 7/x-3

I'm not sure how to do this, but mostly I have no idea "Why" we change it to the second form. It may have to do with graphing but I have no idea.

Please shed some light on this..

Thanks!
(x +4)/(x-3)
You do that by long division and you'd get
1, remainder 7
which is
1 +7/(x-3).

Why change the (x+4)/(x-3) into 1 +7/(x-3) ?
Maybe you are asked to integrate [(x+4)/(x-3)]dx.

7. Hi:

As I read through this thread, I am drawn back in time to Professor Frugale's pre-calculus class and origin of the two points I'd like to bring to the discussion.

First, for a simple method of determining the quotient of two polynomials of like degree, consider the equation, y = (x+2)/(x+5). Before tackling the division, however, recall the following property of fractions: (a+b)/c = a/c + b/c. We learned this in grade school, without the variables, of course. Mrs. Nevel's echoing echoes would echo ad infinitum: "If the denominators are alike, then just add (subtract) the numerators." Okay, on with the problem...

(x+2)/(x+5) = (x+2+3-3)/(x+5) = (x+5-3)/(x+5) = (x+5)/(x+5) - 3/(x+5) [Nevel's Prop] = 1 - 3/(x+5). Hence y = 1 - 3/(x+5).

In practice, the division can be done mentally in less time than it takes for one to wonder why. Note that y, a function of x, has both vertical and horizontal asymptotes. This new form is intended to facilitate visualization of asymptotic behavior for the upcoming student. By examination, it is intuitively apparent that, as x becomes increasingly large, 3/(x+5) diminishes toward zero. Restated: As x tends toward infinity, y=1-3/(x+5) becomes, virtually, y=1, the graph of which is a horizontal line one unit above the x-axis. Hence we have a horizontal asymptote at y=1. The technique is constructive as the student begins to form the concept of a limit - fundamental to the calculus at nearly every stop. Moreover, as one member previously noted, the "divided" form is more readily integrated in some cases. For instance, it is less than obvious that [(x+2)/(x+5)]dx has integral x - 3ln[abs.val.(x+5)] + constant. Rewriting as above, however, renders integration into cake. The tool indeed has its rightful place among the arsenal of the formidable mathematical enthusiast.

Incidentally, do you know which "very" popular property of mathematics (a+b)/c = a/c + b/c is an example of?

Regards,

Rich B.

8. Are you saying that : 1-(3/(X+5))

Is that same as: 1+( 7/X-3))

Because they're not. Right?

Incidentally, do you know which "very" popular property of mathematics (a+b)/c = a/c + b/c is an example of?
What is this? Isnt it just adding...

9. Originally Posted by phgao
Are you saying that : 1-(3/(X+5))

Is that same as: 1+( 7/X-3))

Because they're not. Right?

What is this? Isnt it just adding...
No, they are not. I was just giving another example.

It is adding indeed (along with another operation). But (2^3)(2^5) = 2^8 is just multiplying. Nevertheless, it is multiplication via one of the "special" properties of exponents. Likewise, (a+b)/c = a/c + b/c utilizes a special property, the name of which shall remain incognito until others have had an opportunity to guess (answer).

Enjoy the day!

Rich B.