How do you solve (x(x+1)/5) -(x+1)/6=1/3?

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x(x+1)/5 -(x+1)/6 =1/3
Multiplying both sides by 30
6x(x+1)-5(x+1)=10
6x^2+6x-5x-5-10=0
6x^2+x-15=0
6x^2+10x-9x-15=0
2x(3x+5)-3(3x+5)=0
(3x+5)(2x-3)=0
If 3x+5=0,then 3x= -5 or x= -5/3
If 2x-3=0,then 2x=3 or x=3/2
x=3/2 or -5/3 ans

6x (x + 1) - 5 (x + 1) = 10
6x² + 6x - 5x - 5 = 10
6x² + x - 15 = 0
(3x + 5)(2x - 3) = 0
x = - 5 / 3 , x = 3 / 2

The first thing to do is multiply through by 5*6 to clear the fractions

5*6*[x(x+1)]/5 = 6*[x(x+1)]

5*6*[(x+1)/6] = 5*(x+1)

5*6/3 = 10

The expression is

6*x(x+1) - 5*(x+1) = 10

multiply through

6x^2 + 6x - 5x - 5 = 0

6x^2 + 1x - 5 = 0

this factors as

(6x - 5)(x+1) = 0

6x - 5 = 0

x = 5/6 x+1 = 0 x = -1

x(x + 1)/5 - (x + 1)/6 = 1/3
(x^2 + x)/5 - (x + 1)/6 = 1/3
30[(x^2 + x)/5 - (x + 1)/6] = 30(1/3)
6(x^2 + x) - 5(x + 1) = 10
6x^2 + 6x - 5x - 5 = 10
6x^2 + x - 5 - 10 = 0
6x^2 + x - 15 = 0
(2x - 3)(3x + 5) = 0

2x - 3 = 0
2x = 3
x = 3/2
x = 1 1/3

3x + 5 = 0
3x = -5
x = -5/3
x = -1 2/3

∴ x = 1 1/3 , -1 2/3

Multiplay (distributiv law), bring all to one side (=subtract 1/3), then use quadratic formula.

(x(x+1)/5) -(x+1)/6 = 1/3
(x^2+x)/5 - (x+1)/6 = 1/3
(x^2+x) - (5x+5)/6 = 5/ 3 <multiply across by 5
6x^2+6x-5x-5 =10 <multiply across by 6
6x^2+x-15 =0
(3x + 5)(2x - 3) = 0
x = - 5 / 3 , x = 3 / 2