Is it possible for the solution set of a system of linear inequalities to be all real numbers? Explain your reasoning.

Question

JustSaiyan · Answer

@Arcadiouse

JustSaiyan · Answer

@Angle

Angle · Answer

x + 3y = 2
2x + 6y = 4

JustSaiyan · Answer

Thank you.

Angle · Answer

what do you get when you try solving for x and y?

JustSaiyan · Answer

A head ache?

Angle · Answer

LOL
yes
but I tried making the example easy ;)
you gonna try? :3

Arcadiouse · Answer

Yeah lol.

M: Mental
A: Abuse
T: To
H: Humans

JustSaiyan · Answer

This is algebra.

Arcadiouse · Answer

Still math.

JustSaiyan · Answer

x + 7y = 9
5x + 9y = 16

Angle · Answer

ewwww
what are you doing? x'D

JustSaiyan · Answer

Pretending to know how this works.

JustSaiyan · Answer

But could you answer the actual question?

Angle · Answer

x + 3y = 2
2x + 6y = 4

you can divide by 2 through the entire second equation and get
x + 3y = 2
but that's just the first equation

Angle · Answer

I'm saying that you can have a system of equations where it's just the same equation

and the same equation = that same equation = solutions are all real numbers

Angle · Answer

since the equation overlaps itself perfectly, all real numbers are the answer

JustSaiyan · Answer

Thank you.

JustSaiyan · Answer

It seems this site is more helpful than PA, OS, and brainly.

Angle · Answer

but do you understand how I explained it?

JustSaiyan · Answer

Yes.

Angle · Answer

these sorts of equations are really easy to confuse with the "no solution" type of system of equations, so just trying to solve for x and y is your best bet of figuring such equations out ^_^