SOLUTION: In the word problem , three consecutive numbers where 3times the first is 4 more than 2 times the second. The answer I came up with is 3z=4+2z+1+z+2.

Question 176427: In the word problem , three consecutive numbers where 3times the first is 4 more than 2 times the second. The answer I came up with is 3z=4+2z+1+z+2.
Found 3 solutions by jim_thompson5910, Fombitz, MathTherapy:
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
Let
x=first number
x+1=second number
x+2=third number

"3times the first is 4 more than 2 times the second." translates to

Start with the given equation.

Distribute.

Combine like terms on the right side.

Subtract from both sides.

Combine like terms on the left side.

----------------------------------------------------------------------

Answer:

So the answer is which means that the numbers are: 6, 7, and 8
Answer by Fombitz(32388)   (Show Source): You can put this solution on YOUR website!
First: z
Second: z+1
Third: z+2
3*(First)=4+2*(Second)
Now substitute,

Remember, the second number is (z+1), 2 times that is 2(z+1) or 2z+2.
Also, the third number wasn't mentioned but you had it in your equation???

Now take this equation and solve for z.
Re-post if you get stuck.
Answer by MathTherapy(10557)   (Show Source): You can put this solution on YOUR website!
In the word problem , three consecutive numbers where 3times the first is 4 more than 2 times the second. The answer I came up with is 3z=4+2z+1+z+2.

The real solution to the problem is:
1st #: z
2nd #: z + 1
3rd #: z + 2
3(z) = 2(z + 1) + 4
3z = 2z + 2 + 4
z = 6
Numbers are 6, 7, & 8
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