For what value of k the equation has a unique solution?

hello friends the question is find the value of k for which the following system of equations has a unique solution equation is X + 2 Y is equals to 35 X + K Y + 7 is equal to zero net start solution equation is X + 2 Y is equal to 3 and the second is 5 x + K Y + 7 is equal to zero the value of 1857 is to be too is kt13 the sea to is 7:00 now for unique solution the equation must have a one by two

Main To is not equal to given by V2 so by this one is 182 is 5 not equals to given is to find the B2 is ke ke is not equals to 10 that means the value of k can be any real number but not 10 that is our answer thank you

For what value of k, the system of equations
x + 2y = 3,
5x + ky + 7 = 0
Have (i) a unique solution, (ii) no solution?
Also, show that there is no value of k for which the given system of equation has infinitely namely solutions

The given system of equations:
x + 2y = 3
⇒ x + 2y - 3 = 0                           ….(i)
And, 5x + ky + 7 = 0                   …(ii)
These equations are of the following form:
`a_1x+b_1y+c_1 = 0, a_2x+b_2y+c_2 = 0`
where, `a_1 = 1, b_1= 2, c_1= -3 and a_2 = 5, b_2 = k, c_2 = 7`
(i) For a unique solution, we must have:
∴ `(a_1)/(a_2) ≠ (b_1)/(b_2) i.e., 1/5 ≠ 2/k ⇒ k ≠ 10`
Thus for all real values of k other than 10, the given system of equations will have a unique solution.
(ii) In order that the given system of equations has no solution, we must have:
`(a_1)/(a_2) = (b_1)/(b_2 )≠ (c_1)/(c_2)`
`⇒ 1/5 ≠ 2/k ≠ (−3)/7`
`⇒ 1/5 ≠ 2/k and 2/k ≠ (−3)/7`
`⇒k = 10, k ≠ 14/(−3)`
Hence, the required value of k is 10.
There is no value of k for which the given system of equations has an infinite number of solutions.

GIVEN:

To find: To determine for what value of k the system of equation has

(1) Unique solution

(2) No solution

(3) Infinitely many solution

We know that the system of equations

(1) For Unique solution

Here,

Hence for

(2) For no solution

Here,

Hence for

(3) For infinitely many solution

Here,

But since here

Hence the system does not have infinitely many solutions.

While DonAntonio's answer is certainly correct, it is likely that your question comes from a class where determinants have not yet been discussed, so you may need a different perspective.

In that case, recall that your system will be inconsistent if, after row reduction, you have a row of the form $( 0 \ 0 \ 0 \mid 1)$ since this row would correspond to the equation $0x+0y+0z=1$ which clearly has no solutions.

On the other hand, you also don't want a row of of the form $( 0 \ 0 \ 0 \mid 0)$, which would give you a free variable and hence infinitely many solutions.

Thus, you should find the values of $k$ for which $2-6k-4k^2 = 0$. By our discussion, we can see that as long as you avoid those values of $k$, your system will have a solution, and this solution will be unique.

You can find these "bad" values of $k$ by methods from high school algebra, e.g. the quadratic formula.

How do you find the value of k for a unique solution?

How do you know if an equation has a unique solution?

For what value of k pair of linear equations have unique solution?