Calculus Examples

$0.095 x - e^{- 2 x^{2} + y} = (\frac{1}{12}) - e^{- 2 x^{2} + y}$

Step 1

Simplify $\frac{1}{12} - e^{- 2 x^{2} + y}$ .

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Step 1.1

To write $- e^{- 2 x^{2} + y}$ as a fraction with a common denominator, multiply by $\frac{12}{12}$ .

$0.095 x - e^{- 2 x^{2} + y} = \frac{1}{12} - e^{- 2 x^{2} + y} \cdot \frac{12}{12}$

Step 1.2

Combine $- e^{- 2 x^{2} + y}$ and $\frac{12}{12}$ .

$0.095 x - e^{- 2 x^{2} + y} = \frac{1}{12} + \frac{- e^{- 2 x^{2} + y} \cdot 12}{12}$

Step 1.3

Combine the numerators over the common denominator.

$0.095 x - e^{- 2 x^{2} + y} = \frac{1 - e^{- 2 x^{2} + y} \cdot 12}{12}$

Step 1.4

Multiply $12$ by $- 1$ .

$0.095 x - e^{- 2 x^{2} + y} = \frac{1 - 12 e^{- 2 x^{2} + y}}{12}$

Step 2

Move all terms containing $y$ to the left side of the equation.

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Step 2.1

Subtract $\frac{1 - 12 e^{- 2 x^{2} + y}}{12}$ from both sides of the equation.

$0.095 x - e^{- 2 x^{2} + y} - \frac{1 - 12 e^{- 2 x^{2} + y}}{12} = 0$

Step 2.2

To write $- e^{- 2 x^{2} + y}$ as a fraction with a common denominator, multiply by $\frac{12}{12}$ .

$0.095 x - e^{- 2 x^{2} + y} \cdot \frac{12}{12} - \frac{1 - 12 e^{- 2 x^{2} + y}}{12} = 0$

Step 2.3

Combine $- e^{- 2 x^{2} + y}$ and $\frac{12}{12}$ .

$0.095 x + \frac{- e^{- 2 x^{2} + y} \cdot 12}{12} - \frac{1 - 12 e^{- 2 x^{2} + y}}{12} = 0$

Step 2.4

Combine the numerators over the common denominator.

$0.095 x + \frac{- e^{- 2 x^{2} + y} \cdot 12 - (1 - 12 e^{- 2 x^{2} + y})}{12} = 0$

Step 2.5

Simplify each term.

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Step 2.5.1

Simplify the numerator.

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Step 2.5.1.1

Multiply $12$ by $- 1$ .

$0.095 x + \frac{- 12 e^{- 2 x^{2} + y} - (1 - 12 e^{- 2 x^{2} + y})}{12} = 0$

Step 2.5.1.2

Apply the distributive property.

$0.095 x + \frac{- 12 e^{- 2 x^{2} + y} - 1 \cdot 1 - (- 12 e^{- 2 x^{2} + y})}{12} = 0$

Step 2.5.1.3

Multiply $- 1$ by $1$ .

$0.095 x + \frac{- 12 e^{- 2 x^{2} + y} - 1 - (- 12 e^{- 2 x^{2} + y})}{12} = 0$

Step 2.5.1.4

Multiply $- 12$ by $- 1$ .

$0.095 x + \frac{- 12 e^{- 2 x^{2} + y} - 1 + 12 e^{- 2 x^{2} + y}}{12} = 0$

Step 2.5.1.5

Add $- 12 e^{- 2 x^{2} + y}$ and $12 e^{- 2 x^{2} + y}$ .

$0.095 x + \frac{0 - 1}{12} = 0$

Step 2.5.1.6

Subtract $1$ from $0$ .

$0.095 x + \frac{- 1}{12} = 0$

Step 2.5.2

Move the negative in front of the fraction.

$0.095 x - \frac{1}{12} = 0$

Step 3

Add $\frac{1}{12}$ to both sides of the equation.

$0.095 x = \frac{1}{12}$

Step 4

Divide each term in $0.095 x = \frac{1}{12}$ by $0.095$ and simplify.

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Step 4.1

Divide each term in $0.095 x = \frac{1}{12}$ by $0.095$ .

$\frac{0.095 x}{0.095} = \frac{\frac{1}{12}}{0.095}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $0.095$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{0.095 x}{0.095} = \frac{\frac{1}{12}}{0.095}$

Step 4.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{1}{12}}{0.095}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{1}{12} \cdot \frac{1}{0.095}$

Step 4.3.2

Divide $1$ by $0.095$ .

$x = \frac{1}{12} \cdot 10.52631578$

Step 4.3.3

Combine $\frac{1}{12}$ and $10.52631578$ .

$x = \frac{10.52631578}{12}$

Step 4.3.4

Divide $10.52631578$ by $12$ .

$x = 0.87719298$

Step 5

The variable $y$ got canceled.

All real numbers

Step 6

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$