Precalculus Examples

{(x - 0.3)}^{2} + {(y - 3.2)}^{2} = 10 - 1

${(x - 0.3)}^{2} + {(y - 3.2)}^{2} = 10 - 1$

Step 1

Simplify the left side

{(x - 0.3)}^{2} + {(y - 3.2)}^{2}

${(x - 0.3)}^{2} + {(y - 3.2)}^{2}$ .

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Step 1.1

Simplify each term.

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Step 1.1.1

Rewrite

{(x - 0.3)}^{2}

${(x - 0.3)}^{2}$ as

(x - 0.3) (x - 0.3)

$(x - 0.3) (x - 0.3)$ .

(x - 0.3) (x - 0.3) + {(y - 3.2)}^{2} = 10 - 1

$(x - 0.3) (x - 0.3) + {(y - 3.2)}^{2} = 10 - 1$

Step 1.1.2

Expand

(x - 0.3) (x - 0.3)

$(x - 0.3) (x - 0.3)$ using the FOIL Method.

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Step 1.1.2.1

Apply the distributive property.

x (x - 0.3) - 0.3 (x - 0.3) + {(y - 3.2)}^{2} = 10 - 1

$x (x - 0.3) - 0.3 (x - 0.3) + {(y - 3.2)}^{2} = 10 - 1$

Step 1.1.2.2

Apply the distributive property.

x \cdot x + x \cdot - 0.3 - 0.3 (x - 0.3) + {(y - 3.2)}^{2} = 10 - 1

$x \cdot x + x \cdot - 0.3 - 0.3 (x - 0.3) + {(y - 3.2)}^{2} = 10 - 1$

Step 1.1.2.3

Apply the distributive property.

x \cdot x + x \cdot - 0.3 - 0.3 x - 0.3 \cdot - 0.3 + {(y - 3.2)}^{2} = 10 - 1

$x \cdot x + x \cdot - 0.3 - 0.3 x - 0.3 \cdot - 0.3 + {(y - 3.2)}^{2} = 10 - 1$

x \cdot x + x \cdot - 0.3 - 0.3 x - 0.3 \cdot - 0.3 + {(y - 3.2)}^{2} = 10 - 1

$x \cdot x + x \cdot - 0.3 - 0.3 x - 0.3 \cdot - 0.3 + {(y - 3.2)}^{2} = 10 - 1$

Step 1.1.3

Simplify and combine like terms.

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Step 1.1.3.1

Simplify each term.

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Step 1.1.3.1.1

Multiply

x

$x$ by

x

$x$ .

x^{2} + x \cdot - 0.3 - 0.3 x - 0.3 \cdot - 0.3 + {(y - 3.2)}^{2} = 10 - 1

$x^{2} + x \cdot - 0.3 - 0.3 x - 0.3 \cdot - 0.3 + {(y - 3.2)}^{2} = 10 - 1$

Step 1.1.3.1.2

Move

- 0.3

$- 0.3$ to the left of

x

$x$ .

x^{2} - 0.3 \cdot x - 0.3 x - 0.3 \cdot - 0.3 + {(y - 3.2)}^{2} = 10 - 1

$x^{2} - 0.3 \cdot x - 0.3 x - 0.3 \cdot - 0.3 + {(y - 3.2)}^{2} = 10 - 1$

Step 1.1.3.1.3

Multiply

- 0.3

$- 0.3$ by

- 0.3

$- 0.3$ .

x^{2} - 0.3 x - 0.3 x + 0.09 + {(y - 3.2)}^{2} = 10 - 1

$x^{2} - 0.3 x - 0.3 x + 0.09 + {(y - 3.2)}^{2} = 10 - 1$

x^{2} - 0.3 x - 0.3 x + 0.09 + {(y - 3.2)}^{2} = 10 - 1

$x^{2} - 0.3 x - 0.3 x + 0.09 + {(y - 3.2)}^{2} = 10 - 1$

Step 1.1.3.2

Subtract

0.3 x

$0.3 x$ from

- 0.3 x

$- 0.3 x$ .

x^{2} - 0.6 x + 0.09 + {(y - 3.2)}^{2} = 10 - 1

$x^{2} - 0.6 x + 0.09 + {(y - 3.2)}^{2} = 10 - 1$

x^{2} - 0.6 x + 0.09 + {(y - 3.2)}^{2} = 10 - 1

$x^{2} - 0.6 x + 0.09 + {(y - 3.2)}^{2} = 10 - 1$

Step 1.1.4

Rewrite

{(y - 3.2)}^{2}

${(y - 3.2)}^{2}$ as

(y - 3.2) (y - 3.2)

$(y - 3.2) (y - 3.2)$ .

x^{2} - 0.6 x + 0.09 + (y - 3.2) (y - 3.2) = 10 - 1

$x^{2} - 0.6 x + 0.09 + (y - 3.2) (y - 3.2) = 10 - 1$

Step 1.1.5

Expand

(y - 3.2) (y - 3.2)

$(y - 3.2) (y - 3.2)$ using the FOIL Method.

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Step 1.1.5.1

Apply the distributive property.

x^{2} - 0.6 x + 0.09 + y (y - 3.2) - 3.2 (y - 3.2) = 10 - 1

$x^{2} - 0.6 x + 0.09 + y (y - 3.2) - 3.2 (y - 3.2) = 10 - 1$

Step 1.1.5.2

Apply the distributive property.

x^{2} - 0.6 x + 0.09 + y \cdot y + y \cdot - 3.2 - 3.2 (y - 3.2) = 10 - 1

$x^{2} - 0.6 x + 0.09 + y \cdot y + y \cdot - 3.2 - 3.2 (y - 3.2) = 10 - 1$

Step 1.1.5.3

Apply the distributive property.

x^{2} - 0.6 x + 0.09 + y \cdot y + y \cdot - 3.2 - 3.2 y - 3.2 \cdot - 3.2 = 10 - 1

$x^{2} - 0.6 x + 0.09 + y \cdot y + y \cdot - 3.2 - 3.2 y - 3.2 \cdot - 3.2 = 10 - 1$

x^{2} - 0.6 x + 0.09 + y \cdot y + y \cdot - 3.2 - 3.2 y - 3.2 \cdot - 3.2 = 10 - 1

$x^{2} - 0.6 x + 0.09 + y \cdot y + y \cdot - 3.2 - 3.2 y - 3.2 \cdot - 3.2 = 10 - 1$

Step 1.1.6

Simplify and combine like terms.

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Step 1.1.6.1

Simplify each term.

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Step 1.1.6.1.1

Multiply

y

$y$ by

y

$y$ .

x^{2} - 0.6 x + 0.09 + y^{2} + y \cdot - 3.2 - 3.2 y - 3.2 \cdot - 3.2 = 10 - 1

$x^{2} - 0.6 x + 0.09 + y^{2} + y \cdot - 3.2 - 3.2 y - 3.2 \cdot - 3.2 = 10 - 1$

Step 1.1.6.1.2

Move

- 3.2

$- 3.2$ to the left of

y

$y$ .

x^{2} - 0.6 x + 0.09 + y^{2} - 3.2 \cdot y - 3.2 y - 3.2 \cdot - 3.2 = 10 - 1

$x^{2} - 0.6 x + 0.09 + y^{2} - 3.2 \cdot y - 3.2 y - 3.2 \cdot - 3.2 = 10 - 1$

Step 1.1.6.1.3

Multiply

- 3.2

$- 3.2$ by

- 3.2

$- 3.2$ .

x^{2} - 0.6 x + 0.09 + y^{2} - 3.2 y - 3.2 y + 10.24 = 10 - 1

$x^{2} - 0.6 x + 0.09 + y^{2} - 3.2 y - 3.2 y + 10.24 = 10 - 1$

x^{2} - 0.6 x + 0.09 + y^{2} - 3.2 y - 3.2 y + 10.24 = 10 - 1

$x^{2} - 0.6 x + 0.09 + y^{2} - 3.2 y - 3.2 y + 10.24 = 10 - 1$

Step 1.1.6.2

Subtract

3.2 y

$3.2 y$ from

- 3.2 y

$- 3.2 y$ .

x^{2} - 0.6 x + 0.09 + y^{2} - 6.4 y + 10.24 = 10 - 1

$x^{2} - 0.6 x + 0.09 + y^{2} - 6.4 y + 10.24 = 10 - 1$

x^{2} - 0.6 x + 0.09 + y^{2} - 6.4 y + 10.24 = 10 - 1

$x^{2} - 0.6 x + 0.09 + y^{2} - 6.4 y + 10.24 = 10 - 1$

x^{2} - 0.6 x + 0.09 + y^{2} - 6.4 y + 10.24 = 10 - 1

$x^{2} - 0.6 x + 0.09 + y^{2} - 6.4 y + 10.24 = 10 - 1$

Step 1.2

Simplify the expression.

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Step 1.2.1

Add

0.09

$0.09$ and

10.24

$10.24$ .

x^{2} - 0.6 x + y^{2} - 6.4 y + 10.33 = 10 - 1

$x^{2} - 0.6 x + y^{2} - 6.4 y + 10.33 = 10 - 1$

Step 1.2.2

Move

- 0.6 x

$- 0.6 x$ .

x^{2} + y^{2} - 0.6 x - 6.4 y + 10.33 = 10 - 1

$x^{2} + y^{2} - 0.6 x - 6.4 y + 10.33 = 10 - 1$

x^{2} + y^{2} - 0.6 x - 6.4 y + 10.33 = 10 - 1

$x^{2} + y^{2} - 0.6 x - 6.4 y + 10.33 = 10 - 1$

x^{2} + y^{2} - 0.6 x - 6.4 y + 10.33 = 10 - 1

$x^{2} + y^{2} - 0.6 x - 6.4 y + 10.33 = 10 - 1$

Step 2

Subtract

1

$1$ from

10

$10$ .

x^{2} + y^{2} - 0.6 x - 6.4 y + 10.33 = 9

$x^{2} + y^{2} - 0.6 x - 6.4 y + 10.33 = 9$

Step 3

Subtract

9

$9$ from both sides of the equation.

x^{2} + y^{2} - 0.6 x - 6.4 y + 10.33 - 9 = 0

$x^{2} + y^{2} - 0.6 x - 6.4 y + 10.33 - 9 = 0$

Step 4

Subtract

9

$9$ from

10.33

$10.33$ .

x^{2} + y^{2} - 0.6 x - 6.4 y + 1.33 = 0

$x^{2} + y^{2} - 0.6 x - 6.4 y + 1.33 = 0$

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