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If xy different from zero and x^2y^2=6
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26 Jan 2024, 00:34
26 Jan 2024, 00:34
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60% (01:58) correct
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If \(xy \neq 0 \) and \(x^2y^2-xy=6\) which of the following could be y in terms of x?
Indicate all correct options.
A. \(\frac{6}{x}\)
B. \(- \frac{2}{x}\)
C. \(\frac{3}{x}\)
Indicate all correct options.
A. \(\frac{6}{x}\)
B. \(- \frac{2}{x}\)
C. \(\frac{3}{x}\)
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If xy different from zero and x^2y^2=6
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27 Jan 2024, 20:04
27 Jan 2024, 20:04
1
Simplify \(x^2y^2 - xy\) = 6 into \(xy(xy - 1) = 6\)
Plug in the options:
A: \(y = \frac{6}{x}\)
\(x(\frac{6}{x})(x(\frac{6}{x}) - 1) = 6(6 - 1) = 30\)
B: \(y = -\frac{2}{x}\)
\(x(-\frac{2}{x})(x(-\frac{2}{x}) - 1) = -2(-3) = 6\)
C: \(y = \frac{3}{x}\)
\(x(\frac{3}{x})(x(\frac{3}{x}) - 1) = 3(3 - 1) = 6\)
Plug in the options:
A: \(y = \frac{6}{x}\)
\(x(\frac{6}{x})(x(\frac{6}{x}) - 1) = 6(6 - 1) = 30\)
B: \(y = -\frac{2}{x}\)
\(x(-\frac{2}{x})(x(-\frac{2}{x}) - 1) = -2(-3) = 6\)
C: \(y = \frac{3}{x}\)
\(x(\frac{3}{x})(x(\frac{3}{x}) - 1) = 3(3 - 1) = 6\)
Re: If xy different from zero and x^2y^2=6
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22 Apr 2024, 13:27
22 Apr 2024, 13:27
1
x^2y^2 - xy = 6
subtract 6 from both sides
x^2y^2 - xy - 6 = 0
x^2y^2 - 3xy + 2xy - 6 = 0
xy(xy - 3) + 2(xy - 3) = 0
xy - 3 = 0 xy + 2 = 0
y = 3/x , y = -2/x
subtract 6 from both sides
x^2y^2 - xy - 6 = 0
x^2y^2 - 3xy + 2xy - 6 = 0
xy(xy - 3) + 2(xy - 3) = 0
xy - 3 = 0 xy + 2 = 0
y = 3/x , y = -2/x
