Question

plz .....help me ...​

Answer 1

Answer:

(d) none of the above

Step-by-step explanation:

x+1/y+1=4/5 (let, x and y be the required

5x+5=4y+4 fraction)

5x+1=4y

y=5x+1/4 .... (1)

x-5/y-5=1/2

2x-10=y-5

y=2x-5 .... (2)

from (1) and (2),

5x+1/4=2x-5

5x+1=8x-20

3x=21

x=7

y=2x-5

=2×7-5

=9

hence, the fraction is x/y=7/9

Answer 2

$\sf In \: quadratic \: equation, \: f(x) =ax^2+bx+c \\\\\sf </p><p>Sum \: of \: roots \: (\alpha+\beta) = \frac{-b} {a} \\\\\sf </p><p>Product \: of \: roots \: ( \alpha\beta) = \frac{c} {a} \\\\\sf </p><p> Now,\: In \: the \: question. \\\\\sf </p><p>\frac{1}{a\alpha+b} + \frac{1}{a\beta+b} \\\\\sf </p><p>For \: ease, \\\\\sf </p><p>\: Let \: (a\alpha+b) \: be \: m \: and \: (a\beta+b) \: be \: n, \: then \\\\\sf </p><p>Taking, \: LCM \: of \: denominators.. \\\\\sf </p><p>We \: get, \\\\\sf </p><p>\frac{1}{m}+ \frac{1}{n}= \frac{m+n} {mn} \\\\\sf </p><p>Replace \: the \: values \: of \: m \: and \: n \\\\\sf </p><p>=\frac{(a\beta+b) + (a\alpha+b)} {(a\alpha+b) (a\beta+b)} \\\\\sf </p><p>=\frac{\orange{a}\beta+ \orange{a}\alpha+b+b} {a^2\alpha\beta + \green{ab}\alpha+\green{ab}\beta+b^2} \\\\\sf </p><p>=\frac{\orange{a}(\beta+\alpha)+2b} {a^2\alpha\beta + \green{ab}(\alpha+\beta)+b^2} \\\\\\\\\sf </p><p>Putting \: the \: values \: of \: \alpha+\beta \: and \: \alpha\beta,\: we \: get \\\\\sf </p><p>=\frac{a \: x \: (\frac{-b} {a} )+2b} {a^2 \:x \: \frac{c} {a} + ab \: x \: (\frac{-b} {a} )+b^2} \\\\\sf </p><p>=\frac{\cancel{a}\: x \: (\frac{-b} {\cancel{a}} )+2b} {a \:x \: \frac{c} {\cancel{a}} + \cancel{a}b\: x \: (\frac{-b} {\cancel{a}} )+b^2} \\\\\sf </p><p>=\frac{b} {ac-b^2+b^2}</p><p>=\frac{b} {ac- \cancel{b^2} + \cancel{b^2}}</p><p>=\frac{b} {ac}</p><p>$