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MEDAL!!! Solve the expression...
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What expression?
\[\large \frac{ 2k }{ 4k - 4 } - \frac{ 2k }{ k-1 }\]
I got: \[\large \frac{ 3k(-k + 1) }{ 2(k-1)^2 }\]
\( \dfrac{ 2k }{ 4k - 4 } - \dfrac{ 2k }{ k-1 }\) \( \dfrac{ 2k }{ 4(k - 1) } - \dfrac{ 2k }{ k-1 }\) \( \dfrac{1}{4}\left(\dfrac{ 2k }{ k -1 } \right) -1\left(\dfrac{ 2k }{ k -1 } \right) \)
\( \left(\dfrac{1}{4}-1\right)\left(\dfrac{ 2k }{ k -1 } \right) \)
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\( \left(\dfrac{-3}{4}\right)\left(\dfrac{ 2k }{ k -1 } \right) \)
\( \dfrac{ -6k }{ 4k -4 } \)
simpler, no?
Yes but is my answer correct?
yes
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thank you!
and both our answers simplify to \( \dfrac{ -3k }{ 2k -2 } \)
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