Question

❌❌DONT SPAMM❌❌
PLEASE SOLVE THIS QUESTION

SOLVE ANY TWO QUESTION​

Answer 1

Explanation:

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Answer 2

Answer:

1. a ,b , c are in continued proportion ...

[ Given...]

a/b = c/a

Let a/b = c/ d = k

a = bk and c = dk

11a² + 9ac / 11b² + 9bd = a² + 3ac / b² + 3bd

From L.H.S

11a² + 9ac / 11b² + 9bd

=11 ( bk )² + 9 ( bk ) ( dk )/ 11b² + 9bd ...From (i)..

= 11b²k² + 9bdk² / 11b² + 9bd

= k² ( 11b² + 9bd ) / 11b² + 9d = k

From R.H.S

a² + 3ac / b² + 3bd

( bk )² + 3( bk ) ( dk ) / b² + 3bd....From (I)....

= b²k² + 3bdk² / b² + 3bd

= k² ( b² + 3bd ) / b² + 3bd = k

L.H.S. = R. H.S

Hence proved ....

2. we have given

2/x + 3/y = 14 ....(1)

5/x - 4/y = -2 ......(2)

On multiplying equation (1) by 5 and equation (2) by 2 to make the coefficient of x equal , we get the equation

10/x + 15/y = 65 .....(3)

10/x - 8/y = -4 ........(4)

On subtracting. equation (3) from (4) we get ,

10/x - 8/y - 10/x - 15/y = -4-65

-8/y - 15/y = -69

-23/y = -69

= 23/69 = y

= y = 1/2

on putting y = 1/2 in equation (3) we get

5/x - 4/y = -2

5/x + 4/1/2 = 7

= 5/x = 8 = 7

= 5/x = 15

= x = 5/15

= x = 1/3

Hence , x = 1/3 and y = 1/2 , which is required solution.

$hope \: it \: helps \: you$