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

T

tammie

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Sep 8, 2007
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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 : )
 
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 : )
Click to expand...

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:

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

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