Solving equation

• January 11th 2010, 11:18 AM
Mukilab
Solving equation

Solve
$\frac{3}{(y-1)^{2}}+\frac{3}{2(y-1)^{2}}=2$

I got to

$4=(y-1)^{2}$ or $4=y^{2}-2y+1$
• January 11th 2010, 11:35 AM
e^(i*pi)
Quote:

Originally Posted by Mukilab

Solve
$\frac{3}{(y-1)^{2}}+\frac{3}{2(y-1)^{2}}=2$

I got to

$4=(y-1)^{2}$ or $4=y^{2}-2y+1$

Multiply throughout by $2(y-1)^2$ to clear the denominator

$2(y-1)^2 \cdot \frac{3}{(y-1)^{2}}+ 2(y-1)^2 \cdot \frac{3}{2(y-1)^{2}}=2 \cdot 2(y-1)^2$

Simplify

$6 + 3 = 4(y-1)^2$

You could expand and use the quadratic formula but completing the square is more appropriate here IMO

Divide by 4

$(y-1)^2 = \frac{9}{4} = \left(\frac{3}{2}\right)^2$

Take the square root of both sides and don't forget the $\pm$ sign
• January 11th 2010, 11:55 AM
Mukilab
Quote:

Originally Posted by e^(i*pi)
Multiply throughout by $2(y-1)^2$ to clear the denominator

$2(y-1)^2 \cdot \frac{3}{(y-1)^{2}}+ 2(y-1)^2 \cdot \frac{3}{2(y-1)^{2}}=2 \cdot 2(y-1)^2$

Simplify

$6 + 3 = 4(y-1)^2$

You could expand and use the quadratic formula but completing the square is more appropriate here IMO

Divide by 4

$(y-1)^2 = \frac{9}{4} = \left(\frac{3}{2}\right)^2$

Take the square root of both sides and don't forget the $\pm$ sign

I simplified this
$2(y-1)^2 \cdot \frac{3}{(y-1)^{2}}+ 2(y-1)^2 \cdot \frac{3}{2(y-1)^{2}}=2 \cdot 2(y-1)^2$

to

$2(y-1)^2\cdot \frac{3}{(y-1)^2}+3=2$

o__O
• January 11th 2010, 12:04 PM
e^(i*pi)
Quote:

Originally Posted by Mukilab
I simplified this
$2(y-1)^2 \cdot \frac{3}{(y-1)^{2}}+ 2(y-1)^2 \cdot \frac{3}{2(y-1)^{2}}=2 \cdot 2(y-1)^2$

to

$2(y-1)^2\cdot \frac{3}{(y-1)^2}+3=2$

o__O

I simplified that bit for you, the answer is $6+3 = 4(y-1)^2$

Multiplying by $2(y-1)^2$ is to clear the denominators
• January 11th 2010, 12:11 PM
bigwave
another way

x every term by 2 you have

$
\frac{9}{(y-1)^2} = 4
\Rightarrow
4(y-1)^2 = 9
\Rightarrow
(y-1)^2 = \frac{9}{4}
$

take the square of both sides
$
y-1 = \frac{3}{2}\Rightarrow
y = \frac{3}{2} + \frac{2}{2} = \frac{5}{2}
$
• January 11th 2010, 12:23 PM
Quote:

Originally Posted by Mukilab
I simplified this
$2(y-1)^2 \cdot \frac{3}{(y-1)^{2}}+ 2(y-1)^2 \cdot \frac{3}{2(y-1)^{2}}=2 \cdot 2(y-1)^2$

to

$2(y-1)^2\cdot \frac{3}{(y-1)^2}+3=2$

o__O

Do you see the error you made here, Mukilab?

You could have cancelled terms on the left, you did cancel terms
in the middle but you dropped a term on the right.

Did you understand what $e^{i*pi}$ meant when he mentioned + and - in relation to the square roots?
• January 11th 2010, 12:58 PM
Mukilab
/facepalm
• January 11th 2010, 01:08 PM
There's no problem,
no-one is judging,
but you learn from seeing your errors and correcting them.

since (3)(3)=9 and (-3)(-3)=9 for example.

Of course, they are the same thing really.

3(-3) means subtract 3 three times,
while -3(-3) means the opposite of that which is add 3 three times.

So, square roots have 2 answers.
• January 11th 2010, 01:14 PM
e^(i*pi)
In this case

$
(y-1)^2 = \frac{9}{4} = \left(\frac{3}{2}\right)^2
$

$(y-1) = \pm \frac{3}{2}$