# Thread: Can anyone help me with algebra question

1. ## Can anyone help me with algebra question

Hi,
I am a mum who never learned algebra. My daughter is learning it at school and I am unable to help with homework. Is there someone out there kind enough to show me how to solve the below question so I can help my daughter out with similar problems.

2W + 3W = W
W+1 W-1

Thanks

2. 2W + 3W = W
W+1 W-1

$\frac{2w}{w + 1} + \frac{3w}{w - 1} = w$

Getting rid of the fractions would be a first step. You do that by multiplying everything by a common denominator. The common denominator for w + 1 and w - 1 is simply their product, (w + 1)(w - 1):

$(w + 1)(w - 1)\left(\frac{2w}{w + 1} + \frac{3w}{w - 1}\right) = (w + 1)(w - 1)w$

What happens now? When you multiply (w + 1)(w - 1) by the first fraction the w + 1 cancels out, and you're left with 2w(w - 1). Likewise, when you multiply (w + 1)(w - 1) by the second fraction the w - 1 cancels out, and you're left with 3w(w + 1).

$2w(w - 1) + 3w(w + 1) = (w + 1)(w - 1)w$

As for the right side, you'll have to multiply it out. (w + 1)(w - 1) is a special product; it equals $w^2 - 1$. Then distribute the w over the $w^2 - 1$:

$2w(w - 1) + 3w(w + 1) = (w^2 - 1)w$
$2w(w - 1) + 3w(w + 1) = w^3 - w$

Let's use the distributive property on the left side and start combining like terms:

\begin{aligned}
2w(w - 1) + 3w(w + 1) &= w^3 - w \\
2w^2 - 2w + 3w^2 + 3w &= w^3 - w \\
5w^2 + w &= w^3 - w \\
0 &= w^3 - 5w^2 - 2w
\end{aligned}

Factor out a w:
\begin{aligned}
w^3 - 5w^2 - 2w &= 0 \\
w(w^2 - 5w - 2) &= 0
\end{aligned}

The quadratic is not factorable. Set each factor equal to zero. First you have w = 0. Then, for the quadratic, use the quadratic formula:
\begin{aligned}
w &= \frac{5 \pm \sqrt{(-5)^2 - 4(1)(-2)}}{2(1)} \\
w &= \frac{5 \pm \sqrt{25 + 8}}{2} \\
w &= \frac{5 \pm \sqrt{33}}{2} \\
\end{aligned}

\begin{aligned}