Question

2a+4b=d
solve for b

Answer 1

I just tried to solve 2a + 4b = d so I subtracted 2 from both sides and divided both sides by four. So now I'm stuck with d-2 divided by 4 and I can't get the answer

Answer 2

So this is what you did?

\[ \large {2a + 4b = d \\ 2a + 4b - 2 = d - 2 \\ {2\over4}a + b -{2\over4} = {d\over4}-2} \]

Not sure how that helps you find b. ;)

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Move the 2a over to the other side:

\[ \large {2a+4b\ \ \ \ \ \ \ \ \ =d \\ 2a+4b-2a=d-2a \\ \ \ \ \ \ \ \ \ \ \ 4b \ \ \ \ \ \ \ \ = d - 2a}\]

Now that we have only 4b on one side, just divide by 4 and we're done:

\[ \large {4b = d - 2a \\ \ \\ \frac{4b}{4} = \frac{d - 2a}{4} \\ \ \\ b = \frac{d - 2a}{4}}\]

Answer 3

4b=d-2a is what you should have
Now divide by 4, getting this answer:

Answer 4

Well the problem is 2a+4b=d ; b so I know I'm supposed to solve for d. So this is what I did:
2a + 4b = d
-2. -2. (2a-2 cancels outs)
4b= d-2
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4. 4. ( 4b over 4 cancels out) so that leaves me with. D-2 over 4

Answer 5

2a-2 does not cancel out.

Only 2a-2a = 0 cancels out.

Or 2-2 = 0.

2a-2 does not = 0.