## Help with Solving an Equation Containing Radicals...

Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.

### Help with Solving an Equation Containing Radicals...

Hi,
I've got the following problem:
$y\sqrt{3} + 1 = 3 - y$

I add y to both sides to rid y on the right. Then I subtract 1 from both sides to get this:

$2y\sqrt{3}=2$

Divide both sides by 2 to get:
$y\sqrt{3} = 1$

then divide by $\sqrt{3}$ to get:
$y = \frac{1}{\sqrt{3}}$

multiply the fraction by $\frac{\sqrt{3}}{\sqrt{3}}$ to get rid of the radical and that leaves me with:
$y = \frac{\sqrt{3}}{3}$

The book gives the answer as:
$\sqrt{3} - 1$

I can't tell where I've taken a wrong step.

Ideas?

mb
maroonblazer

Posts: 48
Joined: Thu Aug 12, 2010 11:16 am

### Re: Help with Solving an Equation Containing Radicals...

maroonblazer wrote:Hi,
I've got the following problem:
$y\sqrt{3} + 1 = 3 - y$

I add y to both sides to rid y on the right. Then I subtract 1 from both sides to get this:

$2y\sqrt{3}=2$

this is not correct

Martingale

Posts: 363
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA

### Re: Help with Solving an Equation Containing Radicals...

Martingale wrote:
maroonblazer wrote:Hi,
I've got the following problem:
$y\sqrt{3} + 1 = 3 - y$

I add y to both sides to rid y on the right. Then I subtract 1 from both sides to get this:

$2y\sqrt{3}=2$

this is not correct

Ah, ok, thanks. Instead, I think I should have done:
$y + y\sqrt{3} = 2$
$y(1 + \sqrt{3}) = 2$

Multiply the numerator and denominator by the conjugate, $1-\sqrt{3}$, to get:
$\frac{2-\sqrt{3}}{-2}$

which simplifies to:
$-1+\sqrt{3}$

Thanks again!
mb
maroonblazer

Posts: 48
Joined: Thu Aug 12, 2010 11:16 am

maroonblazer wrote:Instead, I think I should have done:
$y + y\sqrt{3} = 2$
$y(1 + \sqrt{3}) = 2$

Multiply the numerator and denominator by the conjugate, $1-\sqrt{3}$...

What "numerator and denominator"? Or do you mean that you first divided through to get $y$ by itself? If so, then:

. . . . .$y\left(1\, +\, \sqrt{3}\right)\, =\, 2$

. . . . .$y\, =\, \frac{2}{1\, +\, \sqrt{3}}$

Then you did the multiplication by $1\, -\, sqrt{3}$ to get:

. . . . .$y\, =\, \left(\frac{2}{1\, +\, \sqrt{3}}\right)\left(\frac{1\, -\, \sqrt{3}}{1\, -\, \sqrt{3}}\right)$

maroonblazer wrote:$\frac{2-\sqrt{3}}{-2}$

How did the $2$ multiply only on the $1$ and not also the $\sqrt{3}\,?$

stapel_eliz

Posts: 1705
Joined: Mon Dec 08, 2008 4:22 pm

### Re:

stapel_eliz wrote:
maroonblazer wrote:Instead, I think I should have done:
$y + y\sqrt{3} = 2$
$y(1 + \sqrt{3}) = 2$

Multiply the numerator and denominator by the conjugate, $1-\sqrt{3}$...

What "numerator and denominator"? Or do you mean that you first divided through to get $y$ by itself? If so, then:

. . . . .$y\left(1\, +\, \sqrt{3}\right)\, =\, 2$

. . . . .$y\, =\, \frac{2}{1\, +\, \sqrt{3}}$

Then you did the multiplication by $1\, -\, sqrt{3}$ to get:

. . . . .$y\, =\, \left(\frac{2}{1\, +\, \sqrt{3}}\right)\left(\frac{1\, -\, \sqrt{3}}{1\, -\, \sqrt{3}}\right)$

maroonblazer wrote:$\frac{2-\sqrt{3}}{-2}$

How did the $2$ multiply only on the $1$ and not also the $\sqrt{3}\,?$

Sorry, yes, I skipped a step above. And you're right - I should've multiplied on the $\sqrt{3}$ too.
maroonblazer

Posts: 48
Joined: Thu Aug 12, 2010 11:16 am