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Algebra

If we have \(xy = 7, yz = 3, xz = 5\), then the numerical expression \(3x^2 + 5y^2 + 7z^2\) is equal to:

Note by Mahla Salarmohammadi
9 months, 3 weeks ago

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\(xy=7\) .....\((1)\)
\(yz=3\) .....\((2)\)
\(xz=5\) .....\((3)\)

From \((1)\) and \((2)\).

\(x=\frac{7}{y}=\frac{5}{z}\)
\(y=\frac{7z}{5}\)

Putting the value of \(y\) in \((3)\).

\(\Rightarrow y^2=\boxed{\frac{21}{5}}\)
From \((1)\) and \((2)\).
\(y=\frac{3}{z}=\frac{7}{x}\)
\(y=3x\) and \(x=\frac{7z}{3}\)

Putting the value of \(y\) in \((1)\).

\(\Rightarrow x^2=\boxed{\frac{7}{3}}\)

Putting the value of \(x\) in \((3)\).

\(\Rightarrow z^2=\boxed{\frac{15}{7}}\)

Now,

\(3×\frac{7}{3}+5×\frac{21}{5}+7×\frac{15}{7}\)
\(7+21+15=\boxed{43}.\) Abhay Kumar · 9 months, 3 weeks ago

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@Abhay Kumar I sorry. I only know the answer Mahla Salarmohammadi · 9 months, 1 week ago

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@Mahla Salarmohammadi Ok.Then tell me where I made the mistake in my solution. Abhay Kumar · 9 months, 1 week ago

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@Abhay Kumar Your answer is wrong!! Right answer:71😊 Mahla Salarmohammadi · 9 months, 1 week ago

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@Mahla Salarmohammadi Then,where I made the mistake.Please show your solution. Abhay Kumar · 9 months, 1 week ago

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@Abhay Kumar Thanks you for your helping.are you sure? Mahla Salarmohammadi · 9 months, 2 weeks ago

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@Mahla Salarmohammadi Yes. I am sure. Abhay Kumar · 9 months, 2 weeks ago

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