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Algebra/Solving linear equations by substitution

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Question
QUESTION: 1. x/6 + y/6 = 1
  -1/2x - 1/3y = -5

2. 6x+ y = 5x +6
  4x-4/4 = 1

Pls. Be SURE to check Thanks!
14 sophomore

ANSWER: 1. I'm sure you mean -(1/2)x -(1/3)y = -5   multiply that by -6 to get 3x +2y = 30.
since the 1st eq. gives you x+y = 6, x = y-6 and substitute that into 3x+2y + 30
to get 3(6-y) +2y = 30 so y = -12 and x = 18
2. where is the "y" in your second equation?

---------- FOLLOW-UP ----------

QUESTION: Sorry it's like this:
2)  6x+ y = 5x +6
   4x-y/4 = 1

Additional:

3)  2x = 11-y
   1/2x+y =1/2

4)  0.5x = 3y + 4
   x = 2y + 4

5) x + y =1000
  0.05x = 57 - 0.06

Thanks again!!! I really really appreciate it!

Answer
2) re-write the first equation x+y = 6
since it leads to a simplier solution, I assume you mean (4x-y)/4 = 1
so multiply both sides by 4 to get 4x-y = 4 and then substitute y = 6-x from the 1st eq. to
get 4x-6+x = 4 so x = 2 and y = 4
3) Again, you should write it (1/2)x+y = 1/2, multiply by 2 to get x+2y = 1
substitute for y from eq. 1 to get x +2(11-2x) = 1 so x = 7 and y = -3
4) multiply the 1st eq. by 2 to get x = 6y+8 and since x = 2y+4
6y+8 = 2y+4, y = -1 and x = 2.
5) lost a "y" again.

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