(a^2)x+(a-1)=(a+1)x, solve for x

tammie

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Joined
Sep 8, 2007
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3
Hello! : )

This is my first post, so not quite sure if syntax is correct or if this should be in beginning algebra or intermediate/advanced algebra. I'm taking the course College Algebra.

Book says answer is x=(1-a)/(a^2-a-1)

I've tried 3 different ways, but think I may be distributing prematurely when there would be an easier way:

1) (a^2)x+(a-1)=(a+1)x
(a^2)x+(a-1)=ax+x
-x+(a^2)+(a-1)=ax
(a^2)-x+a-1=(ax)/x
-x=(-a^2)-a+1+((ax)/x)
x=(a^2)+a-1-((ax)/x)
I don't know if this is correct or how to get my answer into book answer form from here.

2) (a^2)x+(a-1)=(a+1)x
(a^2)x=-a+1+(a+1)x
x=(-a+1+(a+1)x)/a^2
x=(-a+1+ax+x)/a^2
x=(1-a+ax+x)/a^2
Again, don't know if this is correct or how to get into book form.

3) (a^2)x+(a-1)=(a+1)x if I divide both sides by a^2, do I do it like this:
x+(a-1)=((a+1)x)/a^2 or like this:
x+((a-1)/a^2)=((a+1)x)/a^2 ???

Please help, it's early in the morning and I'm all cracked out on coffee!!!! : P

Thanks so much,
Tammie : )
 

galactus

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Staff member
Joined
Sep 28, 2005
Messages
7,216
tammie said:
Hello! : )

This is my first post, so not quite sure if syntax is correct or if this should be in beginning algebra or intermediate/advanced algebra. I'm taking the course College Algebra.

Book says answer is x=(1-a)/(a^2-a-1)

I've tried 3 different ways, but think I may be distributing prematurely when there would be an easier way:

1) (a^2)x+(a-1)=(a+1)x
(a^2)x+(a-1)=ax+x
-x+(a^2)+(a-1)=ax
(a^2)-x+a-1=(ax)/x
-x=(-a^2)-a+1+((ax)/x)
x=(a^2)+a-1-((ax)/x)
I don't know if this is correct or how to get my answer into book answer form from here.

2) (a^2)x+(a-1)=(a+1)x
(a^2)x=-a+1+(a+1)x
x=(-a+1+(a+1)x)/a^2
x=(-a+1+ax+x)/a^2
x=(1-a+ax+x)/a^2
Again, don't know if this is correct or how to get into book form.

3) (a^2)x+(a-1)=(a+1)x if I divide both sides by a^2, do I do it like this:
x+(a-1)=((a+1)x)/a^2 or like this:
x+((a-1)/a^2)=((a+1)x)/a^2 ???

Please help, it's early in the morning and I'm all cracked out on coffee!!!! : P

Thanks so much,
Tammie : )
Hello Tammie:

\(\displaystyle \L\\a^{2}x+(a-1)=(a+1)x\)

\(\displaystyle \L\\a^{2}x-(a+1)x=1-a\)

Common factor of x on the left:

\(\displaystyle \L\\x(a^{2}-a-1)=1-a\)

\(\displaystyle \L\\x=\frac{1-a}{a^{2}-a-1}\)
 

tammie

New member
Joined
Sep 8, 2007
Messages
3
Thank you!!!!

You Rock!!!! : )
 

tammie

New member
Joined
Sep 8, 2007
Messages
3
PS....

How do you get the math to show up like you did? For example, you don't have to type "a^2", it just shows up as A squared.
 

galactus

Super Moderator
Staff member
Joined
Sep 28, 2005
Messages
7,216
That's LaTex, Tammie. You can do it too. Click on 'quote' at the upper right corner of my post to see the code I typed to make it display that way.
 
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