An algebra problem by A Former Brilliant Member

Algebra Level 2

{ x y = − 8 z y = − 2 z x = 4 \begin{cases} xy=-8 \\ zy=-2 \\ zx=4\end{cases}

Let x , y , z x,y,z be numbers satisfying the system of equations above. Find the values of y y .

Try some of my problems: Math Problems - Set 1
± 4 \pm 4 ± 3 \pm 3 ± 1 \pm 1 ± 2 \pm 2

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A Former Brilliant Member by A Former Brilliant Member

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1 solution

Multiply the three equations.

( x y ) ( z y ) ( z x ) = ( − 8 ) ( − 2 ) ( 4 = ) ( 64 ) (xy)(zy)(zx)=(-8)(-2)(4=)(64)

x 2 y 2 z 2 = 64 x^2y^2z^2=64

( x y z ) 2 = 64 (xyz)^2=64

x y z = ± 8 xyz=±8

Divide the result by z x zx ,

x y z z x = \frac{xyz}{zx}= ± 8 4 \frac{±8}{4}

y = ± 2 y=±2

Since we manipulated the equations, we have to verify that our solutions do indeed satisfy the original conditions. Observe that y = ± 2 y = \pm 2 , x = ∓ 4 x = \mp4 , z = ∓ 1 z = \mp1 work.

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Note: You have demonstrated that y = ± 2 y = \pm 2 is a necessary condition to solve the system of equations. But is it sufficient ? IE Are there solutions to the original system of equations that allow for y = 2 , − 2 y = 2, -2 ?

Calvin Lin Staff - 4 years, 5 months ago

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Yes it is,substituting the values of y=+_2 satisfies all the equations

genis dude - 4 years, 5 months ago

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