What are the solutions to the equation (x+7}{12}-\frac{x^2+1)/4 =1-2/3 ?

The solutions to the equation (x+7}{12}-\frac{x^2+1)/4 =1-2/3 are x=0,x= 1/3

Find the zeros of (x+7}{12}-\frac{x^2+1)/4 =1-2/3

The zeros of (x+7}{12}-\frac{x^2+1)/4 =1-2/3 are x=0,x= 1/3

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\frac{x + 7}{12} - \frac{x ^{2} + 1}{4} = 1 - \frac{2}{3}

x = 0, x = \frac{1}{3}

Decimal Notation

x = 0, x = 0.33333 \dots

Solution steps

\frac{x + 7}{12} - \frac{x ^{2} + 1}{4} = 1 - \frac{2}{3}

Find Least Common Multiplier of

12, 4, 3 : 12

Multiply by LCM=

12

\frac{x + 7}{12} \cdot 12 - \frac{x ^{2} + 1}{4} \cdot 12 = 1 \cdot 12 - \frac{2}{3} \cdot 12

Simplify

x + 7 - 3 (x^{2} + 1) = 12 - 8

x + 7 - 3 (x^{2} + 1) = 4

Expand

x + 7 - 3 (x^{2} + 1) : - 3 x^{2} + x + 4

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

Move

4

to the left side

- 3 x^{2} + x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 3 ) \cdot 0}{2 ( - 3 )}

1^{2} - 4 (- 3) \cdot 0 = 1

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

Separate the solutions

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

x = \frac{- 1 + 1}{2 ( - 3 )} : 0

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

The solutions to the quadratic equation are:

x = 0, x = \frac{1}{3}

(x + 3)^{2} - 1 = 35

\frac{10 m ^{6} n ^{3}}{5 m ^{2} n ^{7}}

3 x = - 9

2 ∣ x - 3 ∣ + 1 = 7

3 m^{2} - 27 = 0

What are the solutions to the equation (x+7}{12}-\frac{x^2+1)/4 =1-2/3 ?
The solutions to the equation (x+7}{12}-\frac{x^2+1)/4 =1-2/3 are x=0,x= 1/3
Find the zeros of (x+7}{12}-\frac{x^2+1)/4 =1-2/3
The zeros of (x+7}{12}-\frac{x^2+1)/4 =1-2/3 are x=0,x= 1/3