Algebra

If we have xy=7,yz=3,xz=5xy = 7, yz = 3, xz = 5, then the numerical expression 3x2+5y2+7z23x^2 + 5y^2 + 7z^2 is equal to:

#Algebra

Note by Mahla Salarmohammadi
5 years, 6 months ago

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Comments

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

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

x=7y=5zx=\frac{7}{y}=\frac{5}{z}
y=7z5y=\frac{7z}{5}

Putting the value of yy in (3)(3).

y2=215\Rightarrow y^2=\boxed{\frac{21}{5}}
From (1)(1) and (2)(2).
y=3z=7xy=\frac{3}{z}=\frac{7}{x}
y=3xy=3x and x=7z3x=\frac{7z}{3}

Putting the value of yy in (1)(1).

x2=73\Rightarrow x^2=\boxed{\frac{7}{3}}

Putting the value of xx in (3)(3).

z2=157\Rightarrow z^2=\boxed{\frac{15}{7}}

Now,

3×73+5×215+7×1573×\frac{7}{3}+5×\frac{21}{5}+7×\frac{15}{7}
7+21+15=43.7+21+15=\boxed{43}.

A Former Brilliant Member - 5 years, 6 months ago

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Thanks you for your helping.are you sure?

mahla salarmohammadi - 5 years, 6 months ago

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Yes. I am sure.

A Former Brilliant Member - 5 years, 5 months ago

Your answer is wrong!! Right answer:71😊

mahla salarmohammadi - 5 years, 5 months ago

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Then,where I made the mistake.Please show your solution.

A Former Brilliant Member - 5 years, 5 months ago

I sorry. I only know the answer

mahla salarmohammadi - 5 years, 5 months ago

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Ok.Then tell me where I made the mistake in my solution.

A Former Brilliant Member - 5 years, 5 months ago
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