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OpenStudy (anonymous):
OpenStudy (pfenn1):
what would be the LCD?
OpenStudy (anonymous):
2
OpenStudy (pfenn1):
Okay, let's multiply both sides by 2\[\frac22 +\left( 8 \over 2x \right)=\left( 2x+8 \over 10 \right)\]
OpenStudy (pfenn1):
Let's simplify it\[1+\frac 4 x=\left( x+4 \over 5 \right)\] Now what is the LCD?
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OpenStudy (anonymous):
multiply both sides by 10 to get rid of the denominator on both sides. that's easier.
OpenStudy (pfenn1):
Okay let's try to multiply both sides by 10\[10+\left( 40 \over x \right)=2x+8\]
OpenStudy (pfenn1):
How about we multiply both sides by x\[10x +40=2x^2+8\]
OpenStudy (pfenn1):
We can rearrange and divide the whole thing by \[x^2 -x-20=0\]Now solve this?
OpenStudy (callisto):
Hmmm...
I do it in this way:
\[\frac{1}{2}+\frac{4}{2x}=\frac{x+4}{10}\]\[\frac{1}{2}+\frac{2}{x}=\frac{x+4}{10}\]\[\frac{1(x)}{2x}+\frac{2(2)}{(2)x}=\frac{x+4}{10}\]\[\frac{(x+4)}{2x}=\frac{x+4}{10}\]\[5(x+4)=(x+4)(x)\]\[5x+20=(x^2+4x)\]\[x^2-x-20=0\]\[(x-5)(x+4)=0\]So, x-5 =0 or x+4 =0
Can you solve it?
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