What is the answer to solvefor x,z=f(y-2x)+g(2y-x) ?

The answer to solvefor x,z=f(y-2x)+g(2y-x) is x=-(z-yf-2yg)/(g+2f);\quad-(2f+g)\ne 0

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solvefor x,z=f(y-2x)+g(2y-x)

solve for

x, z = f (y - 2 x) + g (2 y - x)

x = - \frac{z - y f - 2 y g}{g + 2 f}; - (2 f + g) \neq = 0

Solution steps

z = f (y - 2 x) + g (2 y - x)

Switch sides

f (y - 2 x) + g (2 y - x) = z

Expand

f (y - 2 x) : y f - 2 x f

y f - 2 x f + g (2 y - x) = z

Expand

g (2 y - x) : 2 y g - xg

y f - 2 x f + 2 y g - xg = z

Move

y f

to the right side

- 2 x f + 2 y g - xg = z - y f

Move

2 y g

to the right side

- 2 x f - xg = z - y f - 2 y g

Factor

- 2 x f - xg : - x (2 f + g)

- x (2 f + g) = z - y f - 2 y g

Divide both sides by

- (2 f + g); - (2 f + g) \neq = 0

x = - \frac{z - y f - 2 y g}{g + 2 f}; - (2 f + g) \neq = 0

x, x y + x - 2 y = 7

f = 24 + 66 - 21 - 66

x, 10 x + 12 y = 40

(5 - 2) r = 8

x, 2 x - \frac{1}{3} - y - \frac{2}{5} = 0

What is the answer to solvefor x,z=f(y-2x)+g(2y-x) ?
The answer to solvefor x,z=f(y-2x)+g(2y-x) is x=-(z-yf-2yg)/(g+2f);\quad-(2f+g)\ne 0