Question

First, write the value(s) that make the denominator(s) zero. Then solve the equation. \frac{5}{x+1} + \frac{2}{x-1} = \frac{4}{(x+1)(x-1)} x \neq -1; \{1\} x \neq -1, 1; \{2\} No restrictions; \{1\} x \neq -1, 1; \emptyset

          First, write the value(s) that make the denominator(s) zero. Then solve the equation.
\frac{5}{x+1} + \frac{2}{x-1} = \frac{4}{(x+1)(x-1)}
x \neq -1; \{1\}
x \neq -1, 1; \{2\}
No restrictions; \{1\}
x \neq -1, 1; \emptyset

First, write the value(s) that make the denominator(s) zero. Then solve the equation.
(5)/(x+1) + (2)/(x-1) = (4)/((x+1)(x-1))
x ≠-1; {1}
x ≠-1, 1; {2}
No restrictions; {1}
x ≠-1, 1; ∅

Added by Gabriella M.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Charles Machakwa

08/25/2021

Step 1

For the first equation: x ≠ -1 (denominator x+1) x ≠ 1 (denominator x-1) For the second equation: There are no denominators in this equation, so there are no restrictions on x. Show more…

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First, write the value(s) that make the denominator(s) zero. Then solve the equation. (5)/(x+1) + (2)/(x-1) = (4)/((x+1)(x-1)) x ≠ -1; {1} x ≠ -1,1; {2} No restrictions; {1} x ≠ -1,1; (∅)/() First, write the value(s) that make the denominator(s) zero. Then solve the equation. x^2 + x - 1 = -1x - 1 ∅x - 1; {1} ∅x - 1,1; {2} No restrictions; {1} ∅x ≠ -1,1;