Linear equations in two variables
Given that x = y = 0 is not the only solution to the following system of linear equations, determine all the possible values of k : x + 2 y 2 x + y = k x = k y .
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8 solutions
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nice solution... my logic was:
2y=x(k-1) (1) 2x=y(k-1) (2)
Adding (1) and (2) gives 2(y+x)=x+y(k-1) This implies k-1=2 Which gives k=3, however it doesn't give the negative solution. What's wrong with my logic? Thanks
Generally a quadratic equation have two sol. Bt you did it without seeing that can it be qurad. So failed to see
x=+or - y, if x=y, then 3x=kx=> k = 3, and your ans is k=3 how it solve, i dont understand, so please help me for this solution please
brilliant give brilliant idea.................
well done @Jubayer Nirjhor ;)
nice it is super and amazing
It was simply easy
If a linear system of equations has more than one solution, then that means there are infinitely many solutions. Hence, these two equations are linearly dependent, and their coefficients are in constant proportion:
1 − k 2 = 2 1 − k
Solving gives k = 1 ± 2 .
Smartest and conceptual of all solutions.........
x + 2y = kx .............. (1)
2x + y = ky ............... (2)
Adding (1) and (2) :
3( x + y ) = k( x + y )
Therefore, k = 3
Subtracting (1) from (2) :
1( x - y ) = k( y - x )
k( y - x ) = -1( y -x )
Therefore, k = -1
Hence value of k are − 1 and 3
Note that 3 ( x + y ) = k ( x + y ) does not necessarily imply that k = 3 . Why, or why not?
yes, if (x + y) = 0, then k may not necessarily be equal to 3.
Similarly, you can use augmented matrix and solve for K
why soo broad
Yes, this was how I solved!
Re-arranged as: ( 1 − k ) x + 2 y = 0 , 2 x + ( 1 − k ) y = 0 , the determinant of the coefficient matrix is ( 1 − k ) 2 − 4 = k 2 − 2 k − 3 = ( k − 1 ) ( k + 3 ) . The system of equations has an infinite number of solutions when this determinant is zero. (Since the "constants column" has zero for both entries, the system of equations is dependent. This can be confirmed by "row-reducing" the system.) So there will be more than one solution for k = 1 and k = -3 .
Hello,
given that 2y + x = kx , 2x + y = ky ,
as for 2y + x = kx,
kx - x= 2y
x(k -1) = 2y
x = 2y / (k-1)----> (1st)
as for 2x + y = ky,
ky - y = 2x
y(k -1) = 2x---->(2nd),
substitute (1st) into (2nd),
y(k-1) = 2[ 2y / (k-1) ]
y(k-1)(k-1) = 4y
k^2 - 2k + 1 = 4
k^2 -2k -3 = 0
(k + 1)( k-3) = 0
k = -1 , k=3,
therefore , k= -1,3....
thanks....
x + 2y = kx ..............(i)
2x + y = ky ...............(ii)
Adding (i) and (ii):
3( x + y ) = k( x + y )
Therefore, k = 3
Subtracting (i) from (ii):
1( x - y ) = k( y - x )
k( y - x ) = -1( y -x )
Therefore, k = -1
Hence value of k are -1 and 3
x+2y=kx gives y=x(k-1)/2 . . . (1)
2x+y=ky gives y=2x/k-1 . . . (2). Equating (1) nd (2) (k-1)^2=4 or, k-1=+2 or -2. Giving k=-1 or 3.
Good question
divide the first equation on x and use the result to replace y/x by (k-1)/2 you will get k^2 -2k-3=0 So you will get k=3 or k=-1
Rewriting the equations: 2 y = ( k − 1 ) x , 2 x = ( k − 1 ) y
If k = 1 then the solution is x = y = 0 so we assume x , y = 0 and k = 1 . Dividing the rewritten equations gives: 2 x 2 y = ( k − 1 ) y ( k − 1 ) x ⟹ x y = y x ⟹ x 2 = y 2 ⟹ x = ± y
If x = y then substituting in the equation gives: 3 x = k x ⟹ k = 3
If x = − y , substituting gives: y = − k y ⟹ k = - 1
So the answer is: - 1 , 3 .