Help! I’m confused!!!?

3/2y – y = 4 + 1/2y

Multiply by 2y:

3 – 2y^2 = 8y + 1

2y^2 + 8y – 2 = 0

2(y^2 + 4y – 1) = 0

y = ( – 4 +/- sqrt(4^2 + 4) ) / 2

= -2 +/- sqrt(20)/2

= -2 +/- sqrt(5).

First, we have to get all the terms with “Y” on one side of the equation and all terms that only have a number on the other side. Let’s move the 1/2Y from the right side to the left by subtracting it from both sides. That gives us:

3/2Y -Y -1/2Y = 4

Now we combine all the coefficients of Y (those are the numbers that are multiplying the “Y”). That gives us:

(3/2 -1 -1/2)Y = 4 or,

(0)Y = 4

Any number times zero = zero. That would set up the equation:

0 = 4

This is an impossibility as written. Either the equation was mis-copied or the teacher is trying to see what happens when you are presented with a contradiction.

3/(2y) – y = 4 + 1/(2y)

okay…

FIRST multiply by the whole equation by “2y” so that you no longer are dealing with fractions with 2y in the denominator….

When you do that (multiply through by “2y”), you should get…

3 -2y^2 = 8y + 1

now… move everything to the right side of the = sign so that you are left with 0 on the left side…. and you get…

0 = 2y^2 + 8y -2

which is the same as if you said…

2y^2 + 8y – 2 = 0

now you have to use the quadratic equation to find you roots “y”…

The Quadratic formula is

y = -b +/- sqrt(b^2 – 4ac)

…….————————-

………………..2a

where a = 2, b = 8 and c = -2

You are going to have two answers for “y” using the quadratic equation… y1 and y2

Let’s figure out y1:

y1 = -b + sqrt(b^2 – 4ac)

…….————————- and now substitue a=2, b=8, and c = -2

………………..2a

y1 = -8 + sqrt(8^2 – 4*2*-2)

…….————————-

………………..2(2)

y1 = -8 + sqrt(64 +16)

…….———————– = 0.236 So…. y1 = 0.236

………………4

Now…let’s figure out y2:

y2 = -b – sqrt(b^2 – 4ac)

…….————————- and now substitue a=2, b=8, and c = -2

………………..2a

y2 = -8 – sqrt(8^2 – 4*2*-2)

…….————————-

………………..2(2)

y2 = -8 – sqrt(64 +16)

…….———————– = -4.236 So…. y2 = -4.236

………………4

3/2y – y = 4 + 1/2y

3/2y=4+1/2y+y

3/2y=4+3/2y

3/2y-3/2y=4

0=4

This algebraic problem is false.

i bet you’ve done all the possible. so this makes it just “impossible solution”. not all problems can work

you probably ended up with 0=4…so it is just not possible to solve..

i think that’s what my teacher told me