Help with Algebra problem, please!

[PiCO]

My brain is not working tonight.

 

Solve for y:

 

6[5y-(3y-1)] = 4(3y-7)

 

It's actually a greater-than or equal in the middle, but I can't figure out how to do that on my keyboard.

 

Thanks for any help!

[Jacquelyn in NC]

6[5y-(3y-1)] >= 4(3y-7)

 

Use the distributive property on both sides:

6[5y-3y+1] >= 12y - 28

 

Add like terms:

6[2y + 1] >= 12y - 28

 

Distributive property again:

12y + 6 >= 12y - 28

 

6>= -28

ALWAYS TRUE, so y is any real number.

 

Hope that helps.

[Firestar Academy]

6[5y-(3y-1)] >= 4(3y-7)

 

Use the distributive property on both sides:

6[5y-3y+1] >= 12y - 28

 

Add like terms:

6[2y + 1] >= 12y - 28

 

Distributive property again:

12y + 6 >= 12y - 28

 

6>= -28

ALWAYS TRUE, so y is any real number.

 

Hope that helps.

 

 

:iagree:

[kiana]

Jacquelyn is right.

[Gailmegan]

That's what I got too, well except I had divided 2 out earlier, so I had 3>= -14. Same answer.

[PiCO]

6[5y-(3y-1)] >= 4(3y-7)

 

Use the distributive property on both sides:

6[5y-3y+1] >= 12y - 28

 

Add like terms:

6[2y + 1] >= 12y - 28

 

Distributive property again:

12y + 6 >= 12y - 28

 

6>= -28

ALWAYS TRUE, so y is any real number.

 

Hope that helps.

 

Thanks, that helps a lot! For some reason not coming up with an actual number was not on my radar. I thought I was doing something wrong when I eliminated all the ys.

[kiana]

Thanks, that helps a lot! For some reason not coming up with an actual number was not on my radar. I thought I was doing something wrong when I eliminated all the ys.

 

That messes with people's heads all the time. :)