Solving for L

C

cmsmith

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Feb 14, 2007
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4
It's been a long time, please be patient...(the answer [L>=160])

solve for L
U = 1 over 1 + 2 (80/L) >=.5

step 1 (Hopefully this first step is correct?)
.5 <= 1 over 1 + 160/L

step 2
.5 <= 1/1 + 1 over 160/L

step 3
.5 <= 1 + L/160

step 4 (I'm close but the ".5" is not part of the answer, why?)
160 <= .5 + L
 
cmsmith said:
It's been a long time, please be patient...(the answer [L>=160])
solve for L
U = 1 over 1 + 2 (80/L) >=.5
Click to expand...
WHY have you got 2 equal signs? That's not an equation, sorry.
 
So you mean:\(\displaystyle \L \;\frac{1}{1\,+\,2(\frac{80}{L})}\,\ge\,.5\)
 
actually, the solution set is L < -160 or L > 160.

from your given solution, I'll assume that L is a positive value.

1/[1 + 2(80/L)] > 1/2

1/(1 + 160/L) > 1/2

multiply numerator and denominator of the left side by L ...

L/(L + 160) > 1/2

multiply both sides of the inequality by (L + 160) ...

L > L/2 + 80

subtract L/2 from both sides ...

L/2 > 180

multiply both sides by 2 ...

L > 160
 
This is very helpful and I think I understand. Let me expand a little, is the below correct?

1/[1 + 2(80/L)] >= 1/2

1/[1+160/L] >= 1/2

L/L+160 >= 1/2

[L+160/1]*[L/L+160] >= [1/2]*[L+160/1]

L(L+160)/L+160 >= L+160/2

L >= L/2 + 160/2

L - L/2 >= 160/2

2L/2 - 1L/2 >= 80

L/2 >= 80

(2/1)*(L/2) >= 80 * (2/1)

L >= 160
 
Let me try again...is the below correct?

1/[1 + 2(80/L)] >= 1/2

1/[1+160/L] >= 1/2

L/L+160 >= 1/2

[L+160/1]*[L/L+160] >= [1/2]*[L+160/1]

L >= (L+160)/2

(2/1)*L >= [(L+160)/2]*(2/1)

2L >= L + 160

2L - L >= L + 160 - L

L >= 160
 
Answer your own question ... did you arrive at the desired result using correct algebra?
 
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