6(10y+2 / 6) +4 = 6(-8y+4 / 6)

10y+2 = -8y+4

2y/2 -2 = 4-2

2y/2 = 2/2

y = 1

**However, the correct answer is -1**

What am i missing?

I've spent a good 3 months getting to elementary algebra.. how long will it take to get to calculus?

- Thread starter bobisaka
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6(10y+2 / 6) +4 = 6(-8y+4 / 6)

10y+2 = -8y+4

2y/2 -2 = 4-2

2y/2 = 2/2

y = 1

What am i missing?

I've spent a good 3 months getting to elementary algebra.. how long will it take to get to calculus?

If, as I suspect the problem is

\(\displaystyle \dfrac{5y - 1}{3} + 4 = \dfrac{-\ 8y + 4}{6}\),

you did not write it down correctly. It should be

(5y - 1)/3 + 4 = (- 8y + 4)/6.

Review PEMDAS.

Now getting on to substance. You were

\(\displaystyle \dfrac{5y - 1}{3} + 4 = \dfrac{-\ 8y + 4}{6} \implies 6 * \left ( \dfrac{5y - 1}{3} + 4 \right ) = 6 \left (\dfrac{-\ 8y + 4}{6} \right ) \implies 2(5y - 1) + 6 * 4 = 4 - 8y.\)

Do you see why that is the right way?

- Joined
- Dec 30, 2014

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For your first mistake lets look at an example.

2 + 5 = 7

2*3 +5 = 7*3 means 6+5 = 21 or 11=21 which is not true. That is you can't multiply the whole right side by 3 and just part of the left side by 3 and expect to get equality.

In your 2nd line you clearly denoted that you are multiply the rhs by 6, but you only denoted that you are multiplying part of the lhs by 6. Why are you not multiply the 4 by the 6 as well?!!

You wrote in your 1st line 5y-1/3, this means \(\displaystyle 5y-\dfrac{1}{3}\). Was that what you meant or did you mean \(\displaystyle \dfrac{5y-1}{3}??\)

In your 2nd line -1/3 became +2/6 which is not true? Do you see why?

Your 4th line is a disaster. Please fix the earlier results and post back. Thanks.

2 + 5 = 7

2*3 +5 = 7*3 means 6+5 = 21 or 11=21 which is not true. That is you can't multiply the whole right side by 3 and just part of the left side by 3 and expect to get equality.

In your 2nd line you clearly denoted that you are multiply the rhs by 6, but you only denoted that you are multiplying part of the lhs by 6. Why are you not multiply the 4 by the 6 as well?!!

You wrote in your 1st line 5y-1/3, this means \(\displaystyle 5y-\dfrac{1}{3}\). Was that what you meant or did you mean \(\displaystyle \dfrac{5y-1}{3}??\)

In your 2nd line -1/3 became +2/6 which is not true? Do you see why?

Your 4th line is a disaster. Please fix the earlier results and post back. Thanks.

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