Finite Math Examples

$y = 2 x - | 4 - x^{2} |$

Step 1

Rewrite the equation as $2 x - | 4 - x^{2} | = y$ .

$2 x - | 4 - x^{2} | = y$

Step 2

Subtract $2 x$ from both sides of the equation.

$- | 4 - x^{2} | = y - 2 x$

Step 3

Divide each term in $- | 4 - x^{2} | = y - 2 x$ by $- 1$ and simplify.

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Step 3.1

Divide each term in $- | 4 - x^{2} | = y - 2 x$ by $- 1$ .

$\frac{- | 4 - x^{2} |}{- 1} = \frac{y}{- 1} + \frac{- 2 x}{- 1}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Dividing two negative values results in a positive value.

$\frac{| 4 - x^{2} |}{1} = \frac{y}{- 1} + \frac{- 2 x}{- 1}$

Step 3.2.2

Divide $| 4 - x^{2} |$ by $1$ .

$| 4 - x^{2} | = \frac{y}{- 1} + \frac{- 2 x}{- 1}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Simplify each term.

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Step 3.3.1.1

Move the negative one from the denominator of $\frac{y}{- 1}$ .

$| 4 - x^{2} | = - 1 \cdot y + \frac{- 2 x}{- 1}$

Step 3.3.1.2

Rewrite $- 1 \cdot y$ as $- y$ .

$| 4 - x^{2} | = - y + \frac{- 2 x}{- 1}$

Step 3.3.1.3

Move the negative one from the denominator of $\frac{- 2 x}{- 1}$ .

$| 4 - x^{2} | = - y - 1 \cdot (- 2 x)$

Step 3.3.1.4

Rewrite $- 1 \cdot (- 2 x)$ as $- (- 2 x)$ .

$| 4 - x^{2} | = - y - (- 2 x)$

Step 3.3.1.5

Multiply $- 2$ by $- 1$ .

$| 4 - x^{2} | = - y + 2 x$

Step 4

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$4 - x^{2} = \pm (- y + 2 x)$

Step 5

The complete solution is the result of both the positive and negative portions of the solution.

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Step 5.1

First, use the positive value of the $\pm$ to find the first solution.

$4 - x^{2} = - y + 2 x$

Step 5.2

Subtract $2 x$ from both sides of the equation.

$4 - x^{2} - 2 x = - y$

Step 5.3

Add $y$ to both sides of the equation.

$4 - x^{2} - 2 x + y = 0$

Step 5.4

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 5.5

Substitute the values $a = - 1$ , $b = - 2$ , and $c = 4 + y$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (- 1 \cdot (4 + y))}}{2 \cdot - 1}$

Step 5.6

Simplify.

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Step 5.6.1

Simplify the numerator.

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Step 5.6.1.1

Raise $- 2$ to the power of $2$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 1 \cdot (4 + y)}}{2 \cdot - 1}$

Step 5.6.1.2

Multiply $- 4$ by $- 1$ .

$x = \frac{2 \pm \sqrt{4 + 4 \cdot (4 + y)}}{2 \cdot - 1}$

Step 5.6.1.3

Apply the distributive property.

$x = \frac{2 \pm \sqrt{4 + 4 \cdot 4 + 4 y}}{2 \cdot - 1}$

Step 5.6.1.4

Multiply $4$ by $4$ .

$x = \frac{2 \pm \sqrt{4 + 16 + 4 y}}{2 \cdot - 1}$

Step 5.6.1.5

Add $4$ and $16$ .

$x = \frac{2 \pm \sqrt{20 + 4 y}}{2 \cdot - 1}$

Step 5.6.1.6

Factor $4$ out of $20 + 4 y$ .

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Step 5.6.1.6.1

Factor $4$ out of $20$ .

$x = \frac{2 \pm \sqrt{4 \cdot 5 + 4 y}}{2 \cdot - 1}$

Step 5.6.1.6.2

Factor $4$ out of $4 \cdot 5 + 4 y$ .

$x = \frac{2 \pm \sqrt{4 (5 + y)}}{2 \cdot - 1}$

Step 5.6.1.7

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm \sqrt{2^{2} (5 + y)}}{2 \cdot - 1}$

Step 5.6.1.8

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{5 + y}}{2 \cdot - 1}$

Step 5.6.2

Multiply $2$ by $- 1$ .

$x = \frac{2 \pm 2 \sqrt{5 + y}}{- 2}$

Step 5.6.3

Simplify $\frac{2 \pm 2 \sqrt{5 + y}}{- 2}$ .

$x = \frac{1 \pm \sqrt{5 + y}}{- 1}$

Step 5.6.4

Move the negative one from the denominator of $\frac{1 \pm \sqrt{5 + y}}{- 1}$ .

$x = - 1 \cdot (1 \pm \sqrt{5 + y})$

Step 5.6.5

Rewrite $- 1 \cdot (1 \pm \sqrt{5 + y})$ as $- (1 \pm \sqrt{5 + y})$ .

$x = - (1 \pm \sqrt{5 + y})$

Step 5.7

The final answer is the combination of both solutions.

$x = - 1 - \sqrt{5 + y}$

$x = - 1 + \sqrt{5 + y}$

Step 5.8

Next, use the negative value of the $\pm$ to find the second solution.

$4 - x^{2} = - (- y + 2 x)$

Step 5.9

Simplify $- (- y + 2 x)$ .

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Step 5.9.1

Rewrite.

$4 - x^{2} = 0 + 0 - (- y + 2 x)$

Step 5.9.2

Simplify by adding zeros.

$4 - x^{2} = - (- y + 2 x)$

Step 5.9.3

Apply the distributive property.

$4 - x^{2} = - - y - (2 x)$

Step 5.9.4

Multiply $- - y$ .

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Step 5.9.4.1

Multiply $- 1$ by $- 1$ .

$4 - x^{2} = 1 y - (2 x)$

Step 5.9.4.2

Multiply $y$ by $1$ .

$4 - x^{2} = y - (2 x)$

Step 5.9.5

Multiply $2$ by $- 1$ .

$4 - x^{2} = y - 2 x$

Step 5.10

Add $2 x$ to both sides of the equation.

$4 - x^{2} + 2 x = y$

Step 5.11

Subtract $y$ from both sides of the equation.

$4 - x^{2} + 2 x - y = 0$

Step 5.12

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 5.13

Substitute the values $a = - 1$ , $b = 2$ , and $c = 4 - y$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (- 1 \cdot (4 - y))}}{2 \cdot - 1}$

Step 5.14

Simplify.

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Step 5.14.1

Simplify the numerator.

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Step 5.14.1.1

Raise $2$ to the power of $2$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 1 \cdot (4 - y)}}{2 \cdot - 1}$

Step 5.14.1.2

Multiply $- 4$ by $- 1$ .

$x = \frac{- 2 \pm \sqrt{4 + 4 \cdot (4 - y)}}{2 \cdot - 1}$

Step 5.14.1.3

Apply the distributive property.

$x = \frac{- 2 \pm \sqrt{4 + 4 \cdot 4 + 4 (- y)}}{2 \cdot - 1}$

Step 5.14.1.4

Multiply $4$ by $4$ .

$x = \frac{- 2 \pm \sqrt{4 + 16 + 4 (- y)}}{2 \cdot - 1}$

Step 5.14.1.5

Multiply $- 1$ by $4$ .

$x = \frac{- 2 \pm \sqrt{4 + 16 - 4 y}}{2 \cdot - 1}$

Step 5.14.1.6

Add $4$ and $16$ .

$x = \frac{- 2 \pm \sqrt{20 - 4 y}}{2 \cdot - 1}$

Step 5.14.1.7

Factor $4$ out of $20 - 4 y$ .

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Step 5.14.1.7.1

Factor $4$ out of $20$ .

$x = \frac{- 2 \pm \sqrt{4 (5) - 4 y}}{2 \cdot - 1}$

Step 5.14.1.7.2

Factor $4$ out of $- 4 y$ .

$x = \frac{- 2 \pm \sqrt{4 (5) + 4 (- y)}}{2 \cdot - 1}$

Step 5.14.1.7.3

Factor $4$ out of $4 (5) + 4 (- y)$ .

$x = \frac{- 2 \pm \sqrt{4 (5 - y)}}{2 \cdot - 1}$

Step 5.14.1.8

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm \sqrt{2^{2} (5 - y)}}{2 \cdot - 1}$

Step 5.14.1.9

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{5 - y}}{2 \cdot - 1}$

Step 5.14.2

Multiply $2$ by $- 1$ .

$x = \frac{- 2 \pm 2 \sqrt{5 - y}}{- 2}$

Step 5.14.3

Simplify $\frac{- 2 \pm 2 \sqrt{5 - y}}{- 2}$ .

$x = 1 \pm \sqrt{5 - y}$

Step 5.15

The final answer is the combination of both solutions.

$x = 1 + \sqrt{5 - y}$

$x = 1 - \sqrt{5 - y}$

Step 5.16

The complete solution is the result of both the positive and negative portions of the solution.

$x = - 1 - \sqrt{5 + y}$

$x = - 1 + \sqrt{5 + y}$

$x = 1 + \sqrt{5 - y}$

$x = 1 - \sqrt{5 - y}$

$x = - 1 - \sqrt{5 + y}$

$x = - 1 + \sqrt{5 + y}$

$x = 1 + \sqrt{5 - y}$

$x = 1 - \sqrt{5 - y}$