Parameters in Systems of Equations

[DualDog321]

I generally struggle with these parameter problems. By "one unique solution," I know it mean that the lines would intersect at one point and no real numbers work for both of the equations. So far, I have tried putting the equations into slope-intercept form, but I am not sure what to do from there.
Thanks!

 

[MarkFL]

Take a look at this live graph:

Desmos | Graphing Calculator

www.desmos.com

Move the slider to let \(s\) take on a variety of values.

Looking at that, what would you say about the value of \(s\) for when there is no solution for the given system?

 

[DualDog321]

Ah, so when s=2/3 there is no solutions because they are parallel, so that means s≠2/3 in order for the function to have one unique solutions.
Thank you

 

[MarkFL]

Yes, suppose we multiply the second equation by -2/3 so that the system becomes:

[MATH]2x-sy=1[/MATH]
[MATH]-2x+\frac{2}{3}y=-\frac{4}{3}[/MATH]
Adding these, we get:

[MATH]\frac{2}{3}y-sy=-\frac{1}{3}[/MATH]
Multiply by 3:

[MATH]2y-3sy=-1[/MATH]
Factor:

[MATH](2-3s)y=-1[/MATH]
Divide by \(2-3s\):

[MATH]y=\frac{1}{3s-2}[/MATH]
We can now see that in order for \(y\) to be defined, we require:

[MATH]3s-2\ne0[/MATH]
This implies, as you correctly found:

[MATH]s\ne\frac{2}{3}[/MATH]

 

[pka]

View attachment 18124
I generally struggle with these parameter problems. By "one unique solution," I know it mean that the lines would intersect at one point and no real numbers work for both of the equations. So far, I have tried putting the equations into slope-intercept form, but I am not sure what to do from there.

If you understand determinate what value of \(s\) makes \(\left| {\begin{array}{*{20}{c}} 2&{ - s} \\ 3&{ - 1} \end{array}} \right| \ne 0\)

 

[Steven G]

I know it mean that the lines would intersect at one point and no real numbers work for both of the equations.

I agree that the lines will intersect at one point. But what does it mean for no real numbers work for both of the equations??? What does it mean for a real number to work for both equations.
Your goal is to find a pair of numbers (x,y) that work for both equations.

Just pretend that s is a real number (like 7 for example) and solve the system of equations. Or even better, use pka's hint.

2x-sy= 1 ==> 6x - 3sy = 3
3x-y = 2 ==> 6x - 2y = 4

Subtracting yields (2-3s)y = -1. So y = -1/(2-3s). y is only defined when 2-3s is not 0 or s is not 2/3

 

[DualDog321]

Yes, thanks for pointing that out, I made a mistake in the original post as I looked at my notes incorrectly. It was meant to say "a unique pair of real numbers that satisfy both equations."

 

[HallsofIvy]

I am a bit annoyed at the wording of the problem! "Find the value of the real number parameter s such that the system of equations has a unique solution."

The use of the singular "the value" implies that there is one such value which is not true. Better wording would have been
"Find all values of the real number parameter s such that the system of equations has a unique solution."

 

[JeffM]

I am a bit annoyed at the wording of the problem! "Find the value of the real number parameter s such that the system of equations has a unique solution."

The use of the singular "the value" implies that there is one such value which is not true. Better wording would have been
"Find all values of the real number parameter s such that the system of equations has a unique solution."

Even better would be "Find all real values of the parameter s that permit a unique solution to the resulting system of equations."

My point is that there is a different system of equations for each value of s. I am being fussy, but problems should give as little scope for confusion as possible.