What are the solutions to the equation (y(2y-1))/2 =(1-y)/3 ?

The solutions to the equation (y(2y-1))/2 =(1-y)/3 are y= 2/3 ,y=-1/2

Find the zeros of (y(2y-1))/2 =(1-y)/3

The zeros of (y(2y-1))/2 =(1-y)/3 are y= 2/3 ,y=-1/2

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\frac{y ( 2 y - 1 )}{2} = \frac{1 - y}{3}

y = \frac{2}{3}, y = - \frac{1}{2}

Decimal Notation

y = 0.66666 \dots, y = - 0.5

Solution steps

\frac{y ( 2 y - 1 )}{2} = \frac{1 - y}{3}

Find Least Common Multiplier of

2, 3 : 6

Multiply by LCM=

6

\frac{y ( 2 y - 1 )}{2} \cdot 6 = \frac{1 - y}{3} \cdot 6

Simplify

3 y (2 y - 1) = 2 (- y + 1)

Expand

3 y (2 y - 1) : 6 y^{2} - 3 y

Expand

2 (- y + 1) : - 2 y + 2

6 y^{2} - 3 y = - 2 y + 2

Move

2

to the left side

6 y^{2} - 3 y - 2 = - 2 y

Move

2 y

to the left side

6 y^{2} - y - 2 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 6 ( - 2 )}{2 \cdot 6}

(- 1)^{2} - 4 \cdot 6 (- 2) = 7

y_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 \cdot 6}

Separate the solutions

y_{1} = \frac{- ( - 1 ) + 7}{2 \cdot 6}, y_{2} = \frac{- ( - 1 ) - 7}{2 \cdot 6}

y = \frac{- ( - 1 ) + 7}{2 \cdot 6} : \frac{2}{3}

y = \frac{- ( - 1 ) - 7}{2 \cdot 6} : - \frac{1}{2}

The solutions to the quadratic equation are:

y = \frac{2}{3}, y = - \frac{1}{2}

4 x^{2} + 8 x + 3 = 8

8 - 2 x - x^{2} = 0

20 x^{2} - 30 x = 2 x + 45

3 (x - 2) (x + 1) - 2 (x - 1)^{2} = 4

64 x^{2} + 144 x + 81 = 25

What are the solutions to the equation (y(2y-1))/2 =(1-y)/3 ?
The solutions to the equation (y(2y-1))/2 =(1-y)/3 are y= 2/3 ,y=-1/2
Find the zeros of (y(2y-1))/2 =(1-y)/3
The zeros of (y(2y-1))/2 =(1-y)/3 are y= 2/3 ,y=-1/2