Math Question?

Question

Solve: 2y/y+4 – 2y/y-4=y^2+16/y^2-16
I came up with no solution

Answer 1

First, look at what will make your denoms = 0 and throws those out as possible answers. y can't be 4 or -4.

Now, clear fractions completely by multiplying both sides by what would be the common denom. In this case, if we did want to put everything over the same denom, it would be (y+4)(y-4). So, the trick is to multiply both sides by that and get rid of the fractions completely.

Multiply the first term by (y+4)(y-4) and the (y+4) will go away leaving 2y(y-4)

In the second term, the (y-4) factors reduce and you get
-2y(y+4)

And, on the right hand side, (I'm assuming (y^2 + 16) is the numerator), you just get left with the top
y^2 + 16.

Resulting equation is: 2y(y-4) - 2y(y+4) = y^2 + 16

Expand: 2y^2 - 8y -2y^2 -8y = y^2 + 16.

Take everything to one side and you get:
y^2 + 16y +16 = 0.

You use the Quadratic Formula and get: y=

-16 +/- sqrt(256-4*1*16)
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2

Answer 2

You want to get rid of the fractions. You can do that by multiplying the whole equation with each of the denominators. But wait a second -- something special here:

[1] ... (y + 4) * (y - 4) = y^2 - 16

So if we multiply with (y + 4) and with (y - 4), all fractions should be gone. The new formula becomes

[2] ... 2y * (y - 4) - 2y * (y + 4) = y^2 + 16

Keep in mind, however, that y cannot be -4 or +4, for that would lead to division by zero in the original equation.

Expand:

[3] ... (2y^2 - 8y) - (2y^2 + 8y) = y^2 + 16

[4] ... -16y = y^2 + 16
[5] ... y^2 + 16y + 16 = 0

Use quadratic formula or complete the square:

[6] ... (y + 8)^2 - 48 = 0
[7] ... y + 8 = +/- sqrt(48) = +/- 4 sqrt 3
[8] ... y = -8 +/- 4 sqrt 3

Answer 3

Asume you left out some parentheses and the problem is

2y/(y + 4) - 2y/(y - 4) = (y^2 + 16)/(y^2 - 16)

Note that y^2 - 16, referred to as a differenct of two squares, factors inro (y + 4)(y - 4) and that this is a common denominator for the fractions in the equation. We can clear the denominators by multiplying both sides by (y + 4)(y - 4)

2y(y - 4) - 2y(y + 4) = (y^2 + 16)

The factor in (y + 4)(y - 4) that is not in the denominator of a fraction now shows up in what was the numerator

2y^2 - 8y - 2y^2 - 8y = y^2 + 16
-16y = y^2 + 16
0 = y^2 + 16y + 16

which is usually written as

y^2 + 16y + 16 = 0

If I haven't made an error, this is a quadratic equation that doesn't factor easily, and can be solved using the quadratic formula

[-b +- sqrt(b^2 - 4ac)]/2a

and here a = 1, b = 16, c = 16

Answer 4

None of the above are correct.

If you multiply both sides by (y^2 - 16), you end up with a polynomial of order 4:

y^4-16y^2+16y+16 = 0

This can be rewritten as:

(y^3+2y^2-12y-8)(y-2) = 0

So you know that one solution is y=2.

Now you can try to find other roots but I think you may run into some problems unless you use Newton's root finding method. However the remaining roots may be complex.

DutchProf: He has made an error because he forgot to multiply y^2 by (y-4)(y+4). This equation is not a quadratic. It is a polynomial of order 4.

Hold on: Am I reading this correctly?

2y/(y+4) - 2y/(y-4) = y^2 + 16/(y^2-16)

OR is it:

2y/(y+4) - 2y/(y-4) = (y^2 + 16)/(y^2-16)

Answer 5

2y/y+4 – 2y/y-4=y^2+16/y^2-16
2 + 4 - 2 = y^2+16/y^2-16
y^2+16/y^2 + 20 = 0
y^4 + 20y^2 + 16 = 0
y^2 = {-20 -V(...)}/2 <0 no solution
or
y^2 = {-20 -V(400 - 64)}/2
y^2 = -10 + V84 < 0 So no solution
You are right: 2y/y+4 – 2y/y-4=y^2+16/y^2-16 has no solution.
Th

Answer 6

2y/y+4 - 2y/y-4 = y^2+16/y^2-16

multiply the equation by (y+4)(y-4) and we get

2y(y-4) -2y(y+4) = y^2+16 {(y^2-16) = (y+4)(y-4)}

2y^2-8y-2y^2-8y = y^2+16

-16y = y^2+16

y^2+16y+16 = 0

y = {-16+sqrt(265-64)}/2 or{ -16-sqrt(256-64)}/2

y= (-8+4sqrt3) or (-8-4sqrt3)

=

Answer 7

DutchProf and Kenyai have given you the correct answer. There are two solutions. -8+/-4sqrt3.

Answer 8

(2y)/(y+4) - (2y)/(y-4) = (y2+16)/(y2-16)
(2y2 - 8y) - (2y2 + 8y) = (y2+16)
-16y = y2+16
y2 + 16y + 16 = 0

Use the quadratic formula.

y = -8 +/- 4sqrt(3)

Answer 9

the = sign tells you that both sides are equal...this means that is either the answer or depending on the instructions your probly supposed to do something else

Answer 10

Isn't it strange that the denominators on the elft side are factors of those on the right?

Now you do it......