Question

(c - 3a, 3b - 2c) is a solution of the equation x + y — 3 = 0, If a = 2k, b = 5k, c = 7k; k ≠ 0 and k ∈ R, then find the value of k.

Answer 1

x=c-3a
y=3b-2c

Now,x+y-3=0
→(c-3a)+(3b-2c)-3=0
c=7k
a=2k
b=5k

So the equation becomes

(7k-3(2k))+(3(5k)-2(7k))-3=0
→(7k-6k)+(15k-14k)-3=0
→1k+1k-3=0
→2k-3=0
→2k=3
→k=3/2

And Given k is a real number.

So k is 3/2. and it satisfies all conditions.

Hope it helps...

Answer 2

Hi ,

It is given that ,

1 ) ( c - 3a , 3b - 2c ) is a solution of

the equation x + y - 3 = 0

and

a = 2k , b = 5k , c = 7k

Now ,

( c - 3a , 3b - 2c )

= ( 7k - 3×2k , 3×5k - 2×7k )

= ( 7k - 6k , 15k - 14k )

= ( k , k )

Therefore ,

Substitute ( k , k ) in the equation

x + y - 3 = 0, we get

k + k - 3 = 0

=>2k = 3

k = 3/2

I hope this helps you.

: )