New to Algrebra, please help. 2(y-3) = y-6?

Question

I get to 2y + y
-- --
y y then what?

Answer 1

2(y-3) = y-6

This is what you are supposed to calculate : a value for the y, such that the equation holds.

for instance try y = 1

2(1-3) = 1-6
2 - 6 = 1- 6
and you see that this is not so, thus y=1 is not a solution.

the standard way to solve is : move all y''s to the left of the=, and all numbers to the right of the =

2(y-3) = y - 6
2y - 6 = y - 6
2y -y - 6 = -6 ( move left y to the right, sign changes )
2y - y = -6 + 6 moved left -6 to right, change sign

y = 0. ( adding the 2y - y together, and the -6 + 6

so the solution is y = o

Answer 2

Ok...

2(y-3) = y - 6
distribute...
2y - 2*3 = y - 6
2y - 6 = y - 6
Add 6 to both sides...
2y = y
Subtract y from both sides...
y = 0

Answer 3

2(y-3)=y-6

sol'n: checking;

2(y-3)=y-6 2(y-3)=y-6
2y-6=y-6 2(0-3)=0-6
2y-y=6-6 2(-3)= -6
y=0 -6 = -6

use distributive property on the left side of the equation.
then, you combine similar terms in order to get the answer.
if you know the answer, try to check it by simply
substituting your answer to the given equation.
if both equations are equal then your answer is correct bu if not, then try another solution.

Answer 4

No. 2(y-3) = 2y-6. Multiplication is distributive over *all* of the terms inside the parenthesis ☺


Doug

Answer 5

Well, 2y + y= 3y, and is 2y + y = 0, then 3y=0. Divide both sides by 3: y= 0/3. or y=0.

Answer 6

2y = y so y = 0

Answer 7

2(y-3)=1y-6
distribute the 2 to (y-3)
2y-6=1y-6
now subtract the 1y from both sides
1y-6=(-6)
add 6 to both sides to get y alone
1y=0

Answer 8

2(y-3)=y-6 /-(y-6)
2(y-3)-y+6=0
2y - 6 -y +6=0
y=0

Answer 9

2y-6=y-6
2y-y=-6+6
y=0.
this is the answer.
ur given equation is satisfied only when y=0.

Answer 10

2y - 6 = y - 6
2y - y = -6 + 6
y = 0

hehe. i'm kinda rusty at math. it's been a semester without math in college. damn. haha.