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Posted by **Rachale** on Tuesday, August 4, 2009 at 8:58am.

Please solve and check all proposed solutions

(x-2)/(x-1)+(2)/(x^2-1)= 0

- Algebra -
**nicholas**, Tuesday, August 4, 2009 at 9:03amyou're correct, x=1 and x=0

- Algebra -
**Rachale**, Tuesday, August 4, 2009 at 9:34amSo when we check the problem, would this be acceptable? Does all of this check out? Thanks!

(x-2)/(x-1) + (2)/(x^2-1) = 0

x-2 / x-1+2/(x-1)(x-2) = 0

To check:

0-2 = -2

0-1 = -1

-1/-2 = 0.5

02 – 1 = -1

-1/2 = -0.5

0.5 + -0.5 = 0

1-2 = -1

1-1 = 0

0/-1 = 0

12 – 1 = 0

0/2 = 0

0+0 = 0

x = 0, 1

- Algebra -
**Rachale**, Tuesday, August 4, 2009 at 9:36am12 – 1 = 0 suppose to be 1^2 - 1 = 0

- Algebra -
**Rachale**, Tuesday, August 4, 2009 at 9:37am02 – 1 = -1 suppose to be 0^2 - 1 = 0

- Algebra -
**MathMate**, Tuesday, August 4, 2009 at 10:59amYou're right in suspecting the zero of x=1.

If you subsitiute x=1 into the equation, you will get two indefinite values, hence x=1 has to be rejected.

Note the hint given by the question:

"...and check all proposed solutions ".

All you need to do is to substitute each of the proposed solution into the original equation, if they work out (i.e. LHS=RHS), then the solution is acceptable.

See also:

http://www.jiskha.com/display.cgi?id=1249326304

- Algebra -
**Rachale**, Tuesday, August 4, 2009 at 1:06pmWill this check work? Is this correct?

0-2 = -2

0-1 = -1

-1/-2 = 0.5

0^2 – 1 = -1

-1/2 = -0.5

0.5 + -0.5 = 0

THEN

1-2 = -1

1-1 = 0

0/-1 = 0

1^2 – 1 = 0

0/2 = 0

0+0 = 0

x = 0, 1 Does this all check out?

- Algebra -
**MathMate**, Tuesday, August 4, 2009 at 3:12pmThe first check is OK, so 0 is admitted as a solution.

The second check is not OK because at two places, you have inverted the numerator and denominator, namely:

0/-1 = 0 should read -1/0 = ∞

and

0/2 = 0 should read 2/0 = ∞

Since the proposed solution x=1 gives rise to infinite values, x=1 is not admissible as a solution.

- Algebra -
**Rachale**, Tuesday, August 4, 2009 at 3:20pmSo 1 doesn't work in the solution of the equation at all?

- Algebra -
**MathMate**, Tuesday, August 4, 2009 at 3:26pmOne (1) is rejected as a solution because when substituted into the original equation

(x-2)/(x-1)+(2)/(x^2-1)= 0

it gives a zero as the denominator for each term, which means that each term becomes infinite. The sum of the two terms (left-hand-side) is thus not equal to zero, so 1 is not admitted as a solution to the given equation.

- Algebra -
**Rachale**, Tuesday, August 4, 2009 at 3:29pmThanks for the explanation!

- Algebra -
**MathMate**, Tuesday, August 4, 2009 at 4:57pmYou are very welcome!

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