What are the solutions to the equation ((x-1)*(x+2)}{12}-\frac{(x+1)*(x-2))/6-1=(x-3)/3 ?

The solutions to the equation ((x-1)*(x+2)}{12}-\frac{(x+1)*(x-2))/6-1=(x-3)/3 are x=-2,x=1

Find the zeros of ((x-1)*(x+2)}{12}-\frac{(x+1)*(x-2))/6-1=(x-3)/3

The zeros of ((x-1)*(x+2)}{12}-\frac{(x+1)*(x-2))/6-1=(x-3)/3 are x=-2,x=1

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\frac{( x - 1 ) \cdot ( x + 2 )}{12} - \frac{( x + 1 ) \cdot ( x - 2 )}{6} - 1 = \frac{x - 3}{3}

x = - 2, x = 1

Solution steps

\frac{( x - 1 ) ( x + 2 )}{12} - \frac{( x + 1 ) ( x - 2 )}{6} - 1 = \frac{x - 3}{3}

Find Least Common Multiplier of

12, 6, 3 : 12

Multiply by LCM=

12

\frac{( x - 1 ) ( x + 2 )}{12} \cdot 12 - \frac{( x + 1 ) ( x - 2 )}{6} \cdot 12 - 1 \cdot 12 = \frac{x - 3}{3} \cdot 12

Simplify

(x - 1) (x + 2) - 2 (x + 1) (x - 2) - 12 = 4 (x - 3)

Expand

(x - 1) (x + 2) - 2 (x + 1) (x - 2) - 12 : - x^{2} + 3 x - 10

Expand

4 (x - 3) : 4 x - 12

- x^{2} + 3 x - 10 = 4 x - 12

Move

12

to the left side

- x^{2} + 3 x + 2 = 4 x

Move

4 x

to the left side

- x^{2} - x + 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 1 ) \cdot 2}{2 ( - 1 )}

(- 1)^{2} - 4 (- 1) \cdot 2 = 3

x_{1, 2} = \frac{- ( - 1 ) \pm 3}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 3}{2 ( - 1 )}, x_{2} = \frac{- ( - 1 ) - 3}{2 ( - 1 )}

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

x = \frac{- ( - 1 ) - 3}{2 ( - 1 )} : 1

The solutions to the quadratic equation are:

x = - 2, x = 1

\frac{1}{12 m ^{3} n} + \frac{8}{22 m ^{2} n ^{2}}

3 (4 c + 1) - 4 (4 c - 3)

(\frac{4}{9})^{\frac{1}{2}}

x (a - 7) - y (a - 7)

x = 0.6 \cdot (0.18 + 0.6 - x)

What are the solutions to the equation ((x-1)*(x+2)}{12}-\frac{(x+1)*(x-2))/6-1=(x-3)/3 ?
The solutions to the equation ((x-1)*(x+2)}{12}-\frac{(x+1)*(x-2))/6-1=(x-3)/3 are x=-2,x=1
Find the zeros of ((x-1)*(x+2)}{12}-\frac{(x+1)*(x-2))/6-1=(x-3)/3
The zeros of ((x-1)*(x+2)}{12}-\frac{(x+1)*(x-2))/6-1=(x-3)/3 are x=-2,x=1