I don't know what I should do with both of these equations: -3x+6y=1 & 9x+ky=-3. The question I have to answer is: what would k be so that both equations have a solution of no solution. I'm just really stuck on what to do. I've tried a lot of possibilities. Even making it into slope-intercept form or y=mx+b. But that doesn't work.

The question is:

Consider the following equations.

-3x + 6y = 1

9x + ky = -3.

What would k be so that both equations have a solution or no solution.

At k = -18, the equation has no solution, otherwise, there is a solution.

Step-by-step explanation:

First, solve the given equations

-3x + 6y = 1(1)

9x + ky = -3(2)

simultaneously by elimination, taking k as constant.

Multiply (1) by 3 and add to (2)

(k + 18)y = 0

y = 0/(k + 18)(3)

For there to be a solution, then

k ≠ 18

If k = 18, we have y = 0/0, and the equations have no solution.

Using y = 0 in (1)

-3x = 1

x = -1/3

Therefore, we have the following:

SOLUTION: For k ≠ -18, such that x = -1/3, y = 0.

NO SOLUTION: For k = -18

the sum of the lengths of any two sides of a triangle is greater than the length of the third side. if you take the three sides of a triangle and add them in pairs, the sum is greater than (not equal to) the third side. if that is not true, then it is not possible to construct a triangle with the given side lengths.

step-by-step explanation: