Question

Find integers x,y,z such that x^2z+y^2z+4xy=40 and x^2+y^2+xyz=20

Answer 1

Refer to my solution and please mark it as branliest

Answer 2

Therefore the values of x = ±√5, y = ±√5, and z = 2.

Given:

x²z+y²z+4xy=40 and x²+y²+xyz=20

To Find:

The values of x, y, and z.

Solution:

The given problem is very simple and can be solved by following this approach.

Given equations: x²z+y²z+4xy=40 and x²+y²+xyz=20

Let equation - 1 be x²z+y²z+4xy=40 and equation - 2 be x²+y²+xyz=20

Let us multiply the second equation with '2' on both sides.

⇒ 2x² + 2y² + 2xyz = 40

Now second equation is subtracted from the first equation.

⇒ ( 2x² + 2y² + 2xyz ) - ( x²z+y²z+4xy ) = 40 - 40

⇒ 2x² + 2y² + 2xyz - x²z - y²z - 4xy = 0

⇒ 2x² - x²z + 2y² - y²z + 2xyz - 4xy = 0

⇒ x² ( 2 - z ) + y² ( 2 - z ) + 2xy ( z - 2 ) = 0

⇒ x² ( 2 - z ) + y² ( 2 - z ) - 2xy ( 2 - z ) = 0

⇒ ( 2 - z ) ( x² + y² - 2xy ) = 0

⇒ ( 2 - z ) ( x - y )² = 0

⇒ ( 2 - z ) = 0 or ( x - y )² = 0

⇒ z = 2 or x = y

Substitute z = 2 and x = y in equation 1,

⇒ x²z+y²z+4xy=40

⇒ ( x² × 2 ) + ( x² × 2 ) + 4x.x=40

⇒ 2x² + 2x² + 4x² = 40

⇒ 8x² = 40 ⇒ x² = 40/8 = 5 ⇒ x = ±√5

∴ x = y = ±√5

Therefore the values of x = ±√5, y = ±√5, and z = 2.

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