# Thread: Solving equations involving rational functions of linears.

1. ## Solving equations involving rational functions of linears.

i have 4 problems that my dad sent me. /=fraction ^=exponent
1. 2y/y-3=-1/y
so far i got
2y^2=-y+3

2. x/2=3x+8/x
i got
x^2=6x+16
then i dont know what to do next?

3. x-2/x+6= x/2
so far
2x-4=x^2+6

4. y+2/-y=2/y+3
I DONT KNOW WHAT TO DO FOR THIS ONE?!

My questions on proportions.

1. is it possible for a portion to have two different answers for the same variable?

2. Is it true you cant add percents unless same?

Someone help

i have 4 problems that my dad sent me. /=fraction ^=exponent
1. 2y/y-3=-1/y
so far i got
2y^2=-y+3
And why call there "proportions"?

To start, your equation needs a set of brackets:
2y / (y - 3) = -1 / y

Your 2y^2 = -y + 3 is correct; go 1 step further:
2y^2 + y - 3 = 0

Now solve that for y, by factoring or the quadratic formula.

i have 4 problems that my dad sent me. /=fraction ^=exponent
1. 2y/y-3=-1/y
so far i got
2y^2=-y+3

2. x/2=3x+8/x
i got
x^2=6x+16
then i dont know what to do next?

3. x-2/x+6= x/2
so far
2x-4=x^2+6

4. y+2/-y=2/y+3
I DONT KNOW WHAT TO DO FOR THIS ONE?!

My questions on proportions.

1. is it possible for a portion to have two different answers for the same variable?

2. Is it true you cant add percents unless same?

Someone help

4. 1. is it possible for a portion to have two different answers for the same variable?
As wilmer said your terminology seems a bit unusual. The equations you posted all involve fractions (portions?), but can be re-arranged into quadratic equations. Quadratic equations can have up to 2 solutions.

for example: x= 1/x has two solutions: x=1, x=-1. You could have got this by rearranging to get $\displaystyle x^2 = 1$

4. y+2/-y=2/y+3
to avoid ambiguity, I assume this is

for this one, try multiplying both sides by (y+3)(y), then expand the brackets on both sides to get a standard quadratic equation