Solve for \( x \).

\[
\frac{1}{x-1} - \frac{3}{x+2} = \frac{1}{4}
\]

x − 1 1 ​ − x + 2 3 ​ = 4 1 ​ ( x − 1 ) ( x + 2 ) x + 2 − 3 ( x − 1 ) ​ = 4 1 ​ ( x − 1 ) ( x + 2 ) x + 2 − 3 x + 3 ​ = 4 1 ​ x 2 + x − 2 5 − 2 x ​ = 4 1 ​ 4 ( 5 − 2 x ) = x 2 + x − 2 20 − 8 x = x 2 + x − 2 x 2 + 9 x − 22 = 0 ​
Solvin this quadratic equation with Bhaskara formula we find:
x = -11 or x = 2

Answered by Ryan2 | 2024-06-10

x − 1 1 ​ − x + 2 3 ​ = 4 1 ​ / ⋅ 4 ( x − 1 ) ( x + 2 ) x − 1 4 ( x − 1 ) ( x + 2 ) ​ − x + 2 3 ⋅ 4 ( x − 1 ) ( x + 2 ) ​ = 4 4 ( x − 1 ) ( x + 2 ) ​ 4 ( x + 2 ) − 12 ( x − 1 ) = ( x − 1 ) ( x + 2 ) 4 x + 8 − 12 x + 12 = x 2 + 2 x − x − 2 − 8 x + 20 = x 2 + x − 2 − x 2 − 9 x + 22 = 0 / ⋅ ( − 1 ) x 2 + 9 x − 22 = 0 ⇒ x 2 + 11 x − 2 x − 22 = 0 x ( x + 11 ) − 2 ( x + 11 ) = 0 ( x + 11 ) ( x − 2 ) = 0 \ ⇔ ( x + 11 = 0 or x − 2 = 0 ) x = − 11 or x = 2

Answered by kate200468 | 2024-06-10

The solutions for x in the equation x − 1 1 ​ − x + 2 3 ​ = 4 1 ​ are x = 2 and x = − 11 . This was achieved by manipulating the fractions, simplifying, and applying the quadratic formula. As a result, we ended with a quadratic equation that provided the two solutions after solving it.
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Answered by Ryan2 | 2024-12-26