3 equations ,3 variables..

Algebra Level 4

If a,b,c are real numbers such that a b + 1 b \frac{ab+1}{b} = 7 3 \frac{7}{3} b c + 1 c \frac{bc+1}{c} =4 a c + 1 a \frac{ac+1}{a} =1

Find a×b×c

Also try my sets Algebra Geometry Combinatorics Logical problems


The answer is 1.

Naitik Sanghavi by naitik sanghavi , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Siddharth Singh
Nov 4, 2015

Multiplying all the 3 equations:

( a + 1 b ) ( b + 1 c ) ( c + 1 a ) = 28 3 (a+\frac{1}{b})(b+\frac{1}{c})(c+\frac{1}{a})=\frac{28}{3}

a b c + 1 a b c + a + b + c + 1 a + 1 b + 1 c = 28 3 abc+\frac{1}{abc}+a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{28}{3}

putting the values from the given equations we get:

a b c + 1 a b c = 2 abc+\frac{1}{abc}=2

Solving the equation gives

Hence a b c = 1 abc=1

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...