a+b=2c,a+b=2c,a + b = 2c, find aa−c" role="presentation">aa−caa−c\frac{a}{{a - c}} + cb−c" role="presentation">cb−ccb−c\frac{c}{{b - c}} = ?" /> a+b=2c,a+b=2c,a + b = 2c, find aa−c" role="presentation">aa−caa−c\frac{a}{{a - c}} + cb−c" role="presentation">cb−ccb−c\frac{c}{{b - c}} = ?" /> a+b=2c,a+b=2c,a + b = 2c, find aa−c" role="presentation">aa−caa−c\frac{a}{{a - c}} + cb−c" role="presentation">cb−ccb−c\frac{c}{{b - c}} = ?" />

India’s No.1 Educational Platform For UPSC,PSC And All Competitive Exam

Download Our App

If $a + b = 2 c,$ $a + b = 2c,$ find $\frac{a}{a - c}$ $\frac{a}{{a - c}}$ + $\frac{c}{b - c}$ $\frac{c}{{b - c}}$ = ?

Answer:- 1

Explanation:-

Solution:

\begin{aligned} a + b = 2 c \\ Taking a = 2, b = 4 and c = 3 \\ So, 2 + 4 = 2 \times 3 \\ 6 = 6 \\ Now, \frac{a}{a - c} + \frac{c}{b - c} \\ = \frac{2}{2 - 3} + \frac{3}{4 - 3} \\ = \frac{2}{- 1} + \frac{3}{1} \\ = 1 \end{aligned}

Workspace Report

Post your Comments

find

\frac{a}{a - c}

$\frac{a}{{a - c}}$ +

\frac{c}{b - c}

$\frac{c}{{b - c}}$ = ?", "text": "If

a + b = 2 c,

$a + b = 2c,$ find

\frac{a}{a - c}

$\frac{a}{{a - c}}$ +

\frac{c}{b - c}

$\frac{c}{{b - c}}$ = ?", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} a + b = 2 c \\ Taking a = 2, b = 4 and c = 3 \\ So, 2 + 4 = 2 \times 3 \\ 6 = 6 \\ Now, \frac{a}{a - c} + \frac{c}{b - c} \\ = \frac{2}{2 - 3} + \frac{3}{4 - 3} \\ = \frac{2}{- 1} + \frac{3}{1} \\ = 1 \end{aligned}

$\eqalign{ & a + b = 2c \cr & {\text{Taking }}a = 2,{\text{ }}b = 4{\text{ and }}c = 3 \cr & {\text{So,}}2 + 4 = 2 \times 3 \cr & \boxed{6 = 6} \cr & {\text{Now,}}\frac{a}{{a - c}} + {\text{ }}\frac{c}{{b - c}} \cr & = \frac{2}{{2 - 3}} + {\text{ }}\frac{3}{{4 - 3}} \cr & = \frac{2}{{ - 1}} + \frac{3}{1} \cr & = 1 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution: