Step by stepplease

Solve

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

I got to

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

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- Jan 11th 2010, 11:18 AMMukilabSolving equation
**Step by step**please

Solve

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

I got to

$\displaystyle 4=(y-1)^{2}$ or $\displaystyle 4=y^{2}-2y+1$ - Jan 11th 2010, 11:35 AMe^(i*pi)
Multiply throughout by $\displaystyle 2(y-1)^2$ to clear the denominator

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

Simplify

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

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

Divide by 4

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

Take the square root of both sides and don't forget the $\displaystyle \pm$ sign - Jan 11th 2010, 11:55 AMMukilab
- Jan 11th 2010, 12:04 PMe^(i*pi)
- Jan 11th 2010, 12:11 PMbigwave
another way

x every term by 2 you have

$\displaystyle

\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

$\displaystyle

y-1 = \frac{3}{2}\Rightarrow

y = \frac{3}{2} + \frac{2}{2} = \frac{5}{2}

$ - Jan 11th 2010, 12:23 PMArchie Meade
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 $\displaystyle e^{i*pi}$ meant when he mentioned + and - in relation to the square roots? - Jan 11th 2010, 12:58 PMMukilab
/facepalm

- Jan 11th 2010, 01:08 PMArchie Meade
There's no problem,

no-one is judging,

but you learn from seeing your errors and correcting them.

Now, there are 2 answers,

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. - Jan 11th 2010, 01:14 PMe^(i*pi)
In this case

$\displaystyle

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

$

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