What am i missing?

bobisaka

Junior Member
Joined
Dec 25, 2019
Messages
70
(5y-1 / 3) +4 = -8y+4 / 6

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?
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
5,087
Quite a few months more, I fear. You need to be really good at algebra to be comfortable in 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 right to think of CLEARING FRACTIONS as a sensible first step, but you did it wrong. Here is the right way.

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

Do you see why that is the right way?
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
8,085
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.
 
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