Re arranging an equation.

**el123** · August 27th 2009, 05:45 PM

Ok so i have this equation which I am having difficulty in figuring out how it was derived.

this is the original equation

$\frac{M_s}{P} = L_0 +L_yY +L_rr$

this is the equation i cant figure out how to get to.

$r = \frac{1}{L_r}[\frac{M_s}{P}-L_0] - \frac{L_y}{L_r} Y$

If someone has the time to do a step by step explanation of how to rearrange it to get r, it would help me immensely.

Thanks.

**QM deFuturo** · August 27th 2009, 06:20 PM

Originally Posted by el123

Ok so i have this equation which I am having difficulty in figuring out how it was derived.

this is the original equation

$\frac{M_s}{P} = L_0 +L_yY +L_rr$

this is the equation i cant figure out how to get to.

$r = \frac{1}{L_r}[\frac{M_s}{P}-L_0] - \frac{L_y}{L_r} Y$

If someone has the time to do a step by step explanation of how to rearrange it to get r, it would help me immensely.

Thanks.

(1) $r + \frac{L_yY}{L_r} = \frac{1}{L_r}[\frac{M_s}{P} - L_0]$ First we added to $\frac{L_yY}{L_r}$ both sides

(2) $rL_r + \frac{L_r L_yY}{L_r} = \frac{M_s}{P} - L_0$ Then we multiply both sides by $L_r$ to clear it on the RHS. Note we can cancel these in the second term of the LHS

(3) $rL_r + L_yY + L_0 = \frac{M_s}{P}$ Adding $L_0$ to both sides

You can then rearrange terms to make it look like the answer above.

**el123** · August 27th 2009, 06:26 PM

thanks!

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