If 3-x-y-2-x-y-2-and-9-x-y-4-x-y-1 then a x-1-2-y-3-2 b x-5-2-y-1-2 cx-3-2-y-1-2 d x-1-2-y-5-2

3/x+y+2/x y=2 9/x+y 4/x y=1 Put 1/x+y = u and 1/x y=v 3u +2v = 2 —(1) 9u 4v = 1—(2) (1) × 3 9u+6v = 6— (3) (3) (2) 10v = 5 v = 5/10 = 1/2 ...

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Byju's Answer

Standard X

Mathematics

Elimination Method of Finding Solution of a Pair of Linear Equations

If 3/x+y+2/x-...

Question

If $\frac{3}{x + y} + \frac{2}{x - y} = 2 a n d \frac{9}{x + y} - \frac{4}{x - y} = 1$ then

(a) x= $\frac{1}{2}$ ,y= $\frac{3}{2}$ (b) x= $\frac{5}{2}$ ,y= $\frac{1}{2}$ (c)x= $\frac{3}{2}$ ,y= $\frac{1}{2}$ (d) x= $\frac{1}{2}$ ,y= $\frac{5}{2}$

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Solution

$\frac{3}{x + y} + \frac{2}{x - y} = 2 \frac{9}{x + y} - \frac{4}{x - y} = 1 P u t \frac{1}{x + y} = u a n d \frac{1}{x - y} = v 3 u + 2 v = 2 - - - (1) 9 u - 4 v = 1 - - - (2) (1) \times 3 9 u + 6 v = 6 - - - (3) (3) - (2) 10 v = 5 v = \frac{5}{10} = \frac{1}{2} 3 u + 2 v = 2 3 u = 2 - 2 u 2 - 2 \times \frac{1}{2} = 2 - 1 = 1 3 u = 1 u = \frac{1}{3} \frac{1}{x + y} = u = \frac{1}{3} x + y = 3 \frac{1}{x - y} = v = \frac{1}{2} x - y = 2 2 x = 5 x = \frac{5}{2} y = 3 - \frac{5}{2} = \frac{6 - 5}{2} = \frac{1}{2} x = \frac{5}{2} a n d y = \frac{1}{2}$

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If 3x+y+2x−y=2 and 9x+y−4x−y=1 then (a) x=12,y=32 (b) x=52,y=12 (c)x=32,y=12 (d) x=12,y=52

3x+y+2x−y=29x+y−4x−y=1Put 1x+y=u and 1x−y=v3u+2v=2−−−(1)9u−4v=1−−−(2)(1)×39u+6v=6−−−(3)(3)−(2)10v=5v=510=123u+2v=23u=2−2u2−2×12=2−1=13u=1u=131x+y=u=13x+y=31x−y=v=12x−y=22x=5x=52y=3−52=6−52=12 x=52 and y=12

If $\frac{3}{x + y} + \frac{2}{x - y} = 2 a n d \frac{9}{x + y} - \frac{4}{x - y} = 1$ then

(a) x= $\frac{1}{2}$ ,y= $\frac{3}{2}$ (b) x= $\frac{5}{2}$ ,y= $\frac{1}{2}$ (c)x= $\frac{3}{2}$ ,y= $\frac{1}{2}$ (d) x= $\frac{1}{2}$ ,y= $\frac{5}{2}$