What is expand f(x)((2(x-1))/3+(3(2x+1))/9 =9) ?

The solution to expand f(x)((2(x-1))/3+(3(2x+1))/9 =9) is (4xf(x))/3-(f(x))/3 =9

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expand f(x)((2(x-1))/3+(3(2x+1))/9 =9)

expand

f (x) (\frac{2 ( x - 1 )}{3} + \frac{3 ( 2 x + 1 )}{9} = 9)

\frac{4 x f ( x )}{3} - \frac{f ( x )}{3} = 9

Solution steps

f (x) (\frac{2 ( x - 1 )}{3} + \frac{3 ( 2 x + 1 )}{9} = 9)

= f (x) (\frac{2 ( x - 1 )}{3} + 9 = \frac{3 ( 2 x + 1 )}{9})

Apply the distributive law:

a (b + c) = ab + a c

a = f (x), b = \frac{2 ( x - 1 )}{3}, c = \frac{3 ( 2 x + 1 )}{9} = 9

f (x) \frac{2 ( x - 1 )}{3} + f (x) \frac{3 ( 2 x + 1 )}{9} = 9

\frac{2 ( x - 1 )}{3} f (x) + \frac{3 ( 2 x + 1 )}{9} f (x) = 9

Expand

\frac{2 ( x - 1 )}{3} f (x) + \frac{3 ( 2 x + 1 )}{9} f (x) : \frac{4 x f ( x )}{3} - \frac{f ( x )}{3}

\frac{4 x f ( x )}{3} - \frac{f ( x )}{3} = 9

- \frac{1}{x ( x + 1 )}

(3 x - y^{2} + 1) d x

V (- 4, - 3) \cdot P (- 7, - 6)

\frac{t ^{2} - 1}{( t ^{2} + 1 ) ^{2}}

Q (- 5, 3) (3, - 2)

What is expand f(x)((2(x-1))/3+(3(2x+1))/9 =9) ?
The solution to expand f(x)((2(x-1))/3+(3(2x+1))/9 =9) is (4xf(x))/3-(f(x))/3 =9