Solving equation

• Jan 11th 2010, 11:18 AM
Mukilab
Solving 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 AM
e^(i*pi)
Quote:

Originally Posted by Mukilab
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 $\displaystyle 4=y^{2}-2y+1$

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 AM
Mukilab
Quote:

Originally Posted by e^(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

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

to

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

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

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

to

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

o__O

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

Multiplying by $\displaystyle 2(y-1)^2$ is to clear the denominators
• Jan 11th 2010, 12:11 PM
bigwave
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 PM
Quote:

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

to

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

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 $\displaystyle e^{i*pi}$ meant when he mentioned + and - in relation to the square roots?
• Jan 11th 2010, 12:58 PM
Mukilab
/facepalm
• Jan 11th 2010, 01:08 PM
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 PM
e^(i*pi)
In this case

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

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