What is the answer to 1/9 [3x-6-5((7x)/2-5)]+13(x-5)+1/4 =0 ?

The answer to 1/9 [3x-6-5((7x)/2-5)]+13(x-5)+1/4 =0 is x= 11/2

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1/9 [3x-6-5((7x)/2-5)]+13(x-5)+1/4 =0

\frac{1}{9} [3 x - 6 - 5 (\frac{7 x}{2} - 5)] + 13 (x - 5) + \frac{1}{4} = 0

x = \frac{11}{2}

Decimal Notation

x = 5.5

Solution steps

\frac{1}{9} [3 x - 6 - 5 (\frac{7 x}{2} - 5)] + 13 (x - 5) + \frac{1}{4} = 0

\frac{1}{9} (3 x - 6 - 5 (\frac{7 x}{2} - 5)) = \frac{3 x - 6 - 5 ( \frac{7 x}{2} - 5 )}{9}

\frac{3 x - 6 - 5 ( \frac{7 x}{2} - 5 )}{9} + 13 (x - 5) + \frac{1}{4} = 0

Move

\frac{1}{4}

to the right side

\frac{3 x - 6 - 5 ( \frac{7 x}{2} - 5 )}{9} + 13 (x - 5) = - \frac{1}{4}

Expand

\frac{3 x - 6 - 5 ( \frac{7 x}{2} - 5 )}{9} + 13 (x - 5) : \frac{205 x}{18} + \frac{19}{9} - 65

\frac{205 x}{18} + \frac{19}{9} - 65 = - \frac{1}{4}

Multiply both sides by

18

205 x - 1132 = - \frac{9}{2}

Move

1132

to the right side

205 x = \frac{2255}{2}

Divide both sides by

205

x = \frac{11}{2}

\frac{5 - x}{2} + 4 (x - 1) = 2 x - \frac{3 x + 1}{2}

c, z = 27 c i s (24 0^{\circ})

4 (\frac{x}{2} - \frac{1}{2}) = 9 (\frac{x}{3} + \frac{1}{3}) - 5 (\frac{x}{5} + \frac{4}{5})

x, P = y - x (y^{2} + 2)

\frac{2 v - 3}{4} = \frac{6 v + 7}{3}

What is the answer to 1/9 [3x-6-5((7x)/2-5)]+13(x-5)+1/4 =0 ?
The answer to 1/9 [3x-6-5((7x)/2-5)]+13(x-5)+1/4 =0 is x= 11/2
Solve for x: 1/9 [3x-6-5((7x)/2-5)]+13(x-5)+1/4 =0
The solution is x= 11/2