Need help with a simple algebra problem

tboo

Diamond Member
Jun 25, 2000
7,626
1
81
I havent done anything with math in over 14 years & my son asked me to help him with a few of his problems. One really stumped me(im sure its really easy).

Here is the problem: simplify the equation

x+y/x+y-4 - y/y-4

Any help with a walkthough on how you solved it is appreciated

Thanks
 

gorcorps

aka Brandon
Jul 18, 2004
30,741
456
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Man I really need some parenthesis in there:
is it [(x+y)/(x+y-4)] - [y/y-4]?
 

SagaLore

Elite Member
Dec 18, 2001
24,036
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Originally posted by: tboo
x+y/x+y-4 - y/y-4

Woohoo, easy stuff. Okay, the rule is that you solve multiplication/division first, then add/subtract, from left to right. Treat anything inside parenthesis as separate and solve those initially.

But the way you formatted the equation is really weird. I can't tell how stuff is grouped. Can you try and format that a little better, so it looks exactly like it does on his homework?
 

hdeck

Lifer
Sep 26, 2002
14,530
1
0
Originally posted by: gorcorps
Man I really need some parenthesis in there:
is it [(x+y)/(x+y-4)] - [y/y-4]?

i'm assuming so based on how he put a space around the subtraction sign in the middle, but omitted it in the other places.
 

SagaLore

Elite Member
Dec 18, 2001
24,036
21
81
Originally posted by: gorcorps
Man I really need some parenthesis in there:
is it [(x+y)/(x+y-4)] - [y/y-4]?

Thats the thing - if they aren't there to begin with, you can't assume the addition/subtraction is supposed to have them, or it completely changes the answer. :shocked:
 

tboo

Diamond Member
Jun 25, 2000
7,626
1
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Originally posted by: SagaLore
Originally posted by: tboo
x+y/x+y-4 - y/y-4

Woohoo, easy stuff. Okay, the rule is that you solve multiplication/division first, then add/subtract, from left to right. Treat anything inside parenthesis as separate and solve those initially.

But the way you formatted the equation is really weird. I can't tell how stuff is grouped. Can you try and format that a little better, so it looks exactly like it does on his homework?

Ok lets see if I can do this on the forum:

x+y
------
x+y-4

minus

y
----
y-4



is that better?

Thanks for your help
 

gorcorps

aka Brandon
Jul 18, 2004
30,741
456
126
Well if it's grouped like I said then what slackwarelinux posted is right. How I did it:

You have to start by getting a common denominator of (x+y-4)(y-4), on the left of the minus sign you'll multiply by (y-4)/(y-4) and on the right you'll multiply by (x+y-4)/(x+y-4). Now properly FOIL and carry through all the numerator operations, then cancel out where you can, and you'll end up with:

-4x/(x+y-4)(y-4)
 

PlasmaBomb

Lifer
Nov 19, 2004
11,636
2
81
Done out long hand...

(x+y)(y-4) - y(x+y-4)
________________

(x+y-4)(y-4)



Goes to (multiply out the brackets)-

xy - 4x + y^2 - 4y -xy -y^2 + 4y
____________________________

(x+y-4)(y-4)



Cancel out terms -

-4x
______________
(x+y-4)(y-4)


Then do the same for the bottom.
 
Oct 27, 2007
17,009
5
0
Originally posted by: tboo
Originally posted by: SagaLore
Originally posted by: tboo
x+y/x+y-4 - y/y-4

Woohoo, easy stuff. Okay, the rule is that you solve multiplication/division first, then add/subtract, from left to right. Treat anything inside parenthesis as separate and solve those initially.

But the way you formatted the equation is really weird. I can't tell how stuff is grouped. Can you try and format that a little better, so it looks exactly like it does on his homework?

Ok lets see if I can do this on the forum:

x+y
------
x+y-4

minus

y
----
y-4



is that better?

Thanks for your help

yeah that's not going to work on Fusetalk :( But I can see what you mean.

(x+y)/(x+y-4) - y/(y-4)

It's still not an equation, but try this.

I probably made a mistake there, I rushed through it.

Also note this isn't an equation, it's an expression.

Edit - ok, yeah I made a hash of that judging by other people's answers. Oops.
 

PlasmaBomb

Lifer
Nov 19, 2004
11,636
2
81
Originally posted by: GodlessAstronomer

yeah that's not going to work on Fusetalk :( But I can see what you mean.

(x+y)/(x+y-4) - y/(y-4)

It's still not an equation, but try this.

I probably made a mistake there, I rushed through it.

Also note this isn't an equation, it's an expression.

Edit - ok, yeah I made a hash of that judging by other people's answers. Oops.

Looks good :)
 

tboo

Diamond Member
Jun 25, 2000
7,626
1
81
Ok, one more question.

The next step in the problem is to multiply the answer by 1/x
 
Oct 27, 2007
17,009
5
0
You need to divide the entire expression by x. Happily this expression has only one term in the numerator and that term is a multiple of x, so just can just get rid of the x, making the numerator -4.

When you're helping your son with this, make sure he knows for sure why he is doing these things. Don't just tell him to drop the x in front of the -4, ask him why that's what he should do. If he doesn't know, go back through this stuff in his text book.
 

tboo

Diamond Member
Jun 25, 2000
7,626
1
81
Originally posted by: GodlessAstronomer
You need to divide the entire expression by x. Happily this expression has only one term in the numerator and that term is a multiple of x, so just can just get rid of the x, making the numerator -4.

When you're helping your son with this, make sure he knows for sure why he is doing these things. Don't just tell him to drop the x in front of the -4, ask him why that's what he should do. If he doesn't know, go back through this stuff in his text book.

You lost me on the answer to the last part
 
Oct 27, 2007
17,009
5
0
Multiplying by 1/x is identical to dividing by x (they are inverse operations). You can think of the answer to the first part of the problem as being "negative four, multiplied by x, divided by (the denominator)". You can divide that whole thing by x by removing the "multiply by x" part (remember, division is the inverse of multiplication). Or you could think of it as multiplying x by 1/x. If you draw that on paper it should be immediately obvious that this is x/x, which is equal to 1. So that expression from earlier multiplied by 1/x becomes "-4, multiplied by 1, divided by (the denominator)".
 

tboo

Diamond Member
Jun 25, 2000
7,626
1
81
Originally posted by: GodlessAstronomer
Multiplying by 1/x is identical to dividing by x (they are inverse operations). You can think of the answer to the first part of the problem as being "negative four, multiplied by x, divided by (the denominator)". You can divide that whole thing by x by removing the "multiply by x" part (remember, division is the inverse of multiplication). Or you could think of it as multiplying x by 1/x. If you draw that on paper it should be immediately obvious that this is x/x, which is equal to 1. So that expression from earlier multiplied by 1/x becomes "-4, multiplied by 1, divided by (the denominator)".

Doesnt the denominator get multiplied by x as well?
 
Oct 27, 2007
17,009
5
0
Originally posted by: tboo
Originally posted by: GodlessAstronomer
Multiplying by 1/x is identical to dividing by x (they are inverse operations). You can think of the answer to the first part of the problem as being "negative four, multiplied by x, divided by (the denominator)". You can divide that whole thing by x by removing the "multiply by x" part (remember, division is the inverse of multiplication). Or you could think of it as multiplying x by 1/x. If you draw that on paper it should be immediately obvious that this is x/x, which is equal to 1. So that expression from earlier multiplied by 1/x becomes "-4, multiplied by 1, divided by (the denominator)".

Doesnt the denominator get multiplied by x as well?

Multiplying the denominator by x is identical to dividing the numerator by x.