On solving 25x+y−3x−y=1,40x+y+2x−y=5, we get

25x+y−3x−y=1 40x+y+2x−y=5 Let 1x+y=a;1x−y=b25a−3b=1→(1)40a+2b=5→(2) (1)×(2)⇒50a−6b=2(2)×3⇒120a+6b=15170a =17 a=17170a=11025×110−3b=1−3b=11−52−3b=−32b=−32×1−3 b=12∴1x+y=110x+y=101x−y=12x−y=2x+y=10x−y=22x=12x=6x+y=106+y=10y=4∴x+6&y=4

On solving 25x+y−3x−y=1,40x+y+2x−y=5, we get

25x+y−3x−y=1 40x+y+2x−y=5 Let 1x+y=a;1x−y=b25a−3b=1→(1)40a+2b=5→(2) (1)×(2)⇒50a−6b=2(2)×3⇒120a+6b=15170a =17 a=17170a=11025×110−3b=1−3b=11−52−3b=−32b=−32×1−3 b=12∴1x+y=110x+y=101x−y=12x−y=2x+y=10x−y=22x=12x=6x+y=106+y=10y=4∴x+6&y=4

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

On solving $\frac{25}{x + y} - \frac{3}{x - y} = 1, \frac{40}{x + y} + \frac{2}{x - y} = 5$ , we get

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$x =4, y =6$

$x = 6, y = 4$

$x = 3, y = 1$

$x = 5, y = 2$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\frac{25}{x + y} - \frac{3}{x - y} = 1 \begin{array}{l} \frac{40}{x + y} + \frac{2}{x - y} = 5 \\ Let \frac{1}{x + y} = a; \frac{1}{x - y} = b \\ 25 a - 3 b = 1 \to (1) \\ 40 a + 2 b = 5 \to (2) \end{array} \begin{aligned} (1) \times (2) \Rightarrow 50 a - 6 b = 2 \\ (2) \times 3 \Rightarrow 120 a + 6 b = 15 \\ 170 a = 17 \end{aligned} \begin{array}{l} a = \frac{17}{170} \\ a = \frac{1}{10} \\ 25 \times \frac{1}{10} - 3 b = 1 \\ - 3 b = \frac{1}{1} - \frac{5}{2} \\ - 3 b = \frac{- 3}{2} \\ b = - \frac{3}{2} \times \frac{1}{- 3} \end{array} \begin{array}{l} b = \frac{1}{2} \\ ∴ \frac{1}{x + y} = \frac{1}{10} \\ x + y = 10 \\ \frac{1}{x - y} = \frac{1}{2} \\ x - y = 2 \\ \begin{matrix} x + y = 10 \\ x - y = 2 \end{matrix} \\ \begin{array}{l} 2 x = 12 \\ x = 6 \\ x + y = 10 \\ 6 + y = 10 \\ y = 4 \\ ∴ x + 6 & y = 4 \end{array} \end{array}$