Question

AB/A+B=2 AC/A+C=5 BC/B+C=4 THEN FIND VALUE OF -C\10

Answer 1

"We have 3 equations. Taking a power of -1 for individual equations, taking a common denominator and adding.

2 ( A2BC2+ AB2C2 + A2B2C ) / A2B2C2 = 19 / 20

2 A2B2C2 ( 1/A +1/B + 1/C ) / A2B2C2 = 19 / 20

1/A + 1/B + 1/C = 19/40

From the first equation we have,

1/A + 1/B = 1 / 2

1/C = 19/40 – 1/2

C = -40

-C/10 = 4

"