Asked by Selina
{y-2}/(a^2-5a+6)=(a+1)/(a-3)-(a+2)/(a-2)
How can I solve for y in terms of the other variables???
I tried many times, but I just get y=6,can't express y in terms of variables.
Can anyone please correct me or teach me?THANKS A LOT!
How can I solve for y in terms of the other variables???
I tried many times, but I just get y=6,can't express y in terms of variables.
Can anyone please correct me or teach me?THANKS A LOT!
Answers
Answered by
Reiny
The denominator of the left side factors to
(a-2)(a-3) which just happen to be found as denominators on the right side.
so...
if we multiply both sides by (a-2)(a-3) we get
y - 2 = (a+1)(a-2) - (a+2)(a-3)
expand the right sided, take the -2 over to the right side, and collect your like terms.
You are done!
(a-2)(a-3) which just happen to be found as denominators on the right side.
so...
if we multiply both sides by (a-2)(a-3) we get
y - 2 = (a+1)(a-2) - (a+2)(a-3)
expand the right sided, take the -2 over to the right side, and collect your like terms.
You are done!
Answered by
Selina
that means that still equal to 6 right?
Answered by
Reiny
gee, I guess you are right!
both the a^2 and the a term drop out.
I should have finished my solution, :)
y = 6 !!!!
both the a^2 and the a term drop out.
I should have finished my solution, :)
y = 6 !!!!
Answered by
Selina
however,6 is not a variable..the question is asking to solve for y in terms of other variables
Answered by
Reiny
why don't you state in your solution that the variables canceled out?
Trust your math!
In this case you were not able to solve in terms of the other variable.
In other words y was not a function of a
I suppose one could get silly and write
y = 6 + 0a
Trust your math!
In this case you were not able to solve in terms of the other variable.
In other words y was not a function of a
I suppose one could get silly and write
y = 6 + 0a
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.