What is the answer to (2(x+1))/7+(3(2x-5))/8 =0 ?

The answer to (2(x+1))/7+(3(2x-5))/8 =0 is x= 89/58

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(2(x+1))/7+(3(2x-5))/8 =0

\frac{2 ( x + 1 )}{7} + \frac{3 ( 2 x - 5 )}{8} = 0

x = \frac{89}{58}

Solution steps

\frac{2 ( x + 1 )}{7} + \frac{3 ( 2 x - 5 )}{8} = 0

Multiply by LCM

16 (x + 1) + 21 (2 x - 5) = 0

Expand

16 (x + 1) + 21 (2 x - 5) : 58 x - 89

58 x - 89 = 0

Move

89

to the right side

58 x = 89

Divide both sides by

58

x = \frac{89}{58}

x^{2} - 11 x + 24

x^{2} - 7 x = - 12

(x + y)^{2} + 9 (x + y) - 10

4 x^{2} + 9 x + 4 = 0

(4 + i)^{2}

What is the answer to (2(x+1))/7+(3(2x-5))/8 =0 ?
The answer to (2(x+1))/7+(3(2x-5))/8 =0 is x= 89/58
Solve for x: (2(x+1))/7+(3(2x-5))/8 =0
The solution is x= 89/58