What is expand (x-y+3z=0,)(2x-3+z=5) ?

The solution to expand (x-y+3z=0,)(2x-3+z=5) is 2x^2-3x+xz=5

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expand (x-y+3z=0,)(2x-3+z=5)

expand

(x - y + 3 z = 0,) (2 x - 3 + z = 5)

2 x^{2} - 3 x + x z = 5

Solution steps

(x - y + 3 z = 0,) (2 x - 3 + z = 5)

= (x - y + 3 \cdot 0 =, z) (2 x - 3 + 5 = z)

Distribute parentheses

x \cdot 2 x + x (- 3) + x z = 5 + (- y) \cdot 2 x + (- y) (- 3) + (- y) z = 5 + 3 z = 0, 2 x + 3 z = 0, (- 3) + 3 z = 0, z = 5

Apply minus-plus rules

+ (- a) = - a, (- a) (- b) = ab

2 xx - 3 x + x z = 5 - 2 x y + 3 y - yz = 5 + 3 z = 0, 2 x - 3 z = 0, 3 + 3 z = 0, z = 5

Expand

2 xx - 3 x + x z : 2 x^{2} - 3 x + x z

2 x^{2} - 3 x + x z = 5

(\frac{1}{4} - 5 x)^{\frac{1}{2}}

\frac{x ^{4} + ( p - 3 ) x ^{2} + ( q + 3 )}{x ^{2} + x + 1}

\frac{3 ^{(n + 1) (n + 2)} + 65}{72}

(1 - e^{- a t}) sin (x)

\frac{( x - 1 ) ^{4 n} ( x ^{3} + 8 ) ( x - 4 )}{( x ^{2} - 2 x + 2 )}

What is expand (x-y+3z=0,)(2x-3+z=5) ?
The solution to expand (x-y+3z=0,)(2x-3+z=5) is 2x^2-3x+xz=5