What is the answer to solvefor x,z=f^1(y+2x)+f^2(y-3x) ?

The answer to solvefor x,z=f^1(y+2x)+f^2(y-3x) is x=(z-yf-yf^2)/(f(2-3f));\quad f\ne 0,f\ne 2/3

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solvefor x,z=f^1(y+2x)+f^2(y-3x)

solve for

x, z = f^{1} (y + 2 x) + f^{2} (y - 3 x)

x = \frac{z - y f - y f ^{2}}{f ( 2 - 3 f )}; f \neq = 0, f \neq = \frac{2}{3}

Solution steps

z = f^{1} (y + 2 x) + f^{2} (y - 3 x)

Switch sides

f^{1} (y + 2 x) + f^{2} (y - 3 x) = z

Apply rule

a^{1} = a

f^{1} = f

f (y + 2 x) + f^{2} (y - 3 x) = z

Expand

f (y + 2 x) : y f + 2 x f

y f + 2 x f + f^{2} (y - 3 x) = z

Expand

f^{2} (y - 3 x) : y f^{2} - 3 x f^{2}

y f + 2 x f + y f^{2} - 3 x f^{2} = z

Move

y f

to the right side

2 x f + y f^{2} - 3 x f^{2} = z - y f

Move

y f^{2}

to the right side

2 x f - 3 x f^{2} = z - y f - y f^{2}

Factor

2 x f - 3 x f^{2} : x f (2 - 3 f)

x f (2 - 3 f) = z - y f - y f^{2}

Divide both sides by

f (2 - 3 f); f \neq = 0, f \neq = \frac{2}{3}

x = \frac{z - y f - y f ^{2}}{f ( 2 - 3 f )}; f \neq = 0, f \neq = \frac{2}{3}

8 x = - 7

\frac{6 - x}{4} = \frac{x}{8}

x, m (5 - x) = 2 - y

x, 3 x - y + 2 z = 12

\frac{x}{( 2 c + 1 )} + \frac{( 2 c + 1 )}{( 2 c )} = \frac{21}{12}

What is the answer to solvefor x,z=f^1(y+2x)+f^2(y-3x) ?
The answer to solvefor x,z=f^1(y+2x)+f^2(y-3x) is x=(z-yf-yf^2)/(f(2-3f));\quad f\ne 0,f\ne 2/3