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If m is different from zero
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07 Sep 2021, 07:03
07 Sep 2021, 07:03
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Question Stats:
69% (01:08) correct
30% (01:44) wrong based on 36 sessions
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If \(m \neq 0 \)and \(m - \frac{1+2m^2}{2m}=\frac{y}{m}\), then y is =
A. \(-1\)
B. \(- \frac{1}{2}\)
C. \(-(1+4m^2)\)
D. \(- \frac{1+4m^2}{2}\)
E. \((2m+1)^2\)
A. \(-1\)
B. \(- \frac{1}{2}\)
C. \(-(1+4m^2)\)
D. \(- \frac{1+4m^2}{2}\)
E. \((2m+1)^2\)
If m is different from zero
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07 Sep 2021, 08:30
07 Sep 2021, 08:30
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\(m - \frac{1+2m^2}{2m}=\frac{y}{m}\)
Take the LCM
\(2m^2 - 1 - 2m^2 = 2y\)
\(y =\frac{-1}{2}\)
Answer B
Take the LCM
\(2m^2 - 1 - 2m^2 = 2y\)
\(y =\frac{-1}{2}\)
Answer B
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Re: If m is different from zero
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10 Sep 2021, 07:15
10 Sep 2021, 07:15
1
Carcass wrote:
If \(m \neq 0 \)and \(m - \frac{1+2m^2}{2m}=\frac{y}{m}\), then y is =
A. \(-1\)
B. \(- \frac{1}{2}\)
C. \(-(1+4m^2)\)
D. \(- \frac{1+4m^2}{2}\)
E. \((2m+1)^2\)
A. \(-1\)
B. \(- \frac{1}{2}\)
C. \(-(1+4m^2)\)
D. \(- \frac{1+4m^2}{2}\)
E. \((2m+1)^2\)
Given: \(m - \frac{1+2m^2}{2m}=\frac{y}{m}\)
To eliminate all fractions, we can multiply both sides of the equation by the least common multiple (LCM) of the denominators (2m and m).
The LCM of 2m and m is 2m, so we will multiply both sides by 2m to get: \(2m^2 - (1+2m^2)=2y\)
Simplify the left side to get: \(-1=2y\)
Divide both sides by 2 to get: to get: \(-\frac{1}{2}=y\)
Answer: B
