An algebra problem by Sunny Dhondkar

Algebra Level 3

{ x + y = 2 k x + y = 4 x + k y = 5 \begin{cases} x+y=2 \\ kx + y = 4 \\ x+ky = 5 \end{cases}

Find the number of possible values of k k such that the system of equations above has at least one solution.

1 2 Infinitely many 0

Sunny Dhondkar by Sunny Dhondkar , 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.

2 solutions

Ali Jan
Oct 12, 2016

adding equations 2 and 3 you get

(K+1)X + (K+1)Y = 9.

Factorising the left side we get (K+1)(X+Y) = 9.

From equation 1 we have X+Y = 2

Substituting in for X+Y we have 2(K+1) = 9

solving this for the K clearly gives 1 value.

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Luiz Claudio
Mar 2, 2019

x + y + k x + y + x + k y = 11 x+y+kx+y+x+ky = 11

( k ( x + y ) + 2 ( x + y ) = 11 k(x+y)+2(x+y)=11

k ( 2 ) + 2 ( 2 ) = 11 k(2)+2(2) =11

2 k = 7 ⟹ k = 3 , 5 2k = 7 \implies k= 3,5

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...