(7)/(x-5)+(4)/(5-x) Which expression is equivalent to the sum for all x!=5 ? Choose 1 answer: (A) (11)/(x-5) (B) (11)/(5-x) (C) (3)/(x-5) (D) (3)/(5-x)

Rewrite fractions: We have two fractions: (7)/(x-5) and (4)/(5-x). To add them, we need a common denominator. Notice that (5-x) is the same as -(x-5). We can rewrite (4)/(5-x) as (4)/(-(x-5)) which is the same as (-4)/(x-5).Combine fractions: Now we have (7)/(x-5) + (-4)/(x-5). Since the denominators are the same, we can add the numerators directly.Add numerators: Adding the numerators: 7 + (-4) = 3.Final sum: So the sum of the two fractions is (3)/(x-5). This is equivalent to option (C).

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(7)/(x-5)+(4)/(5-x)
Which expression is equivalent to the sum for all
x!=5 ?
Choose 1 answer:
(A)
(11)/(x-5)
(B)
(11)/(5-x)
(C)
(3)/(x-5)
(D)
(3)/(5-x)

$\frac{7}{x-5}+\frac{4}{5-x}$ $\newline$ Which expression is equivalent to the sum for all $x \neq 5$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{11}{x-5}$ $\newline$ (B) $\frac{11}{5-x}$ $\newline$ (C) $\frac{3}{x-5}$ $\newline$ (D) $\frac{3}{5-x}$

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Full solution

\frac{7}{x-5}+\frac{4}{5-x}

\newline

Which expression is equivalent to the sum for all

x \neq 5

\newline

Choose

1

answer:

\newline

(A)

\frac{11}{x-5}

\newline

(B)

\frac{11}{5-x}

\newline

(C)

\frac{3}{x-5}

\newline

(D)

\frac{3}{5-x}

Rewrite fractions: We have two fractions: $(7)/(x-5)$ and $(4)/(5-x)$ . To add them, we need a common denominator. Notice that $(5-x)$ is the same as $- (x-5)$ . We can rewrite $(4)/(5-x)$ as $(4)/(-(x-5))$ which is the same as $(-4)/(x-5)$ .
Combine fractions: Now we have $(7)/(x-5) + (-4)/(x-5)$ . Since the denominators are the same, we can add the numerators directly.
Add numerators: Adding the numerators: $7 + (-4) = 3$ .
Final sum: So the sum of the two fractions is $(\frac{3}{x-5})$ . This is equivalent to option (C).