Algebra Examples

x^{2} + y^{2} = 25

$x^{2} + y^{2} = 25$

Step 1

Since

x = r \cos (θ)

$x = r \cos (θ)$ , replace

x

$x$ with

r \cos (θ)

$r \cos (θ)$ .

{(r \cos (θ))}^{2} + y^{2} = 25

${(r \cos (θ))}^{2} + y^{2} = 25$

Step 2

Since

y = r \sin (θ)

$y = r \sin (θ)$ , replace

y

$y$ with

r \sin (θ)

$r \sin (θ)$ .

{(r \cos (θ))}^{2} + {(r \sin (θ))}^{2} = 25

${(r \cos (θ))}^{2} + {(r \sin (θ))}^{2} = 25$

Step 3

Solve for

r

$r$ .

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Step 3.1

Subtract

25

$25$ from both sides of the equation.

{(r \cos (θ))}^{2} + {(r \sin (θ))}^{2} - 25 = 0

${(r \cos (θ))}^{2} + {(r \sin (θ))}^{2} - 25 = 0$

Step 3.2

Simplify the left side of the equation.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Apply the product rule to

r \cos (θ)

$r \cos (θ)$ .

r^{2} \cos^{2} (θ) + {(r \sin (θ))}^{2} - 25 = 0

$r^{2} \cos^{2} (θ) + {(r \sin (θ))}^{2} - 25 = 0$

Step 3.2.1.2

Apply the product rule to

r \sin (θ)

$r \sin (θ)$ .

r^{2} \cos^{2} (θ) + r^{2} \sin^{2} (θ) - 25 = 0

$r^{2} \cos^{2} (θ) + r^{2} \sin^{2} (θ) - 25 = 0$

r^{2} \cos^{2} (θ) + r^{2} \sin^{2} (θ) - 25 = 0

$r^{2} \cos^{2} (θ) + r^{2} \sin^{2} (θ) - 25 = 0$

Step 3.2.2

Simplify with factoring out.

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Step 3.2.2.1

Factor

r^{2}

$r^{2}$ out of

r^{2} \cos^{2} (θ)

$r^{2} \cos^{2} (θ)$ .

r^{2} (\cos^{2} (θ)) + r^{2} \sin^{2} (θ) - 25 = 0

$r^{2} (\cos^{2} (θ)) + r^{2} \sin^{2} (θ) - 25 = 0$

Step 3.2.2.2

Factor

r^{2}

$r^{2}$ out of

r^{2} \sin^{2} (θ)

$r^{2} \sin^{2} (θ)$ .

r^{2} (\cos^{2} (θ)) + r^{2} (\sin^{2} (θ)) - 25 = 0

$r^{2} (\cos^{2} (θ)) + r^{2} (\sin^{2} (θ)) - 25 = 0$

Step 3.2.2.3

Factor

r^{2}

$r^{2}$ out of

r^{2} (\cos^{2} (θ)) + r^{2} (\sin^{2} (θ))

$r^{2} (\cos^{2} (θ)) + r^{2} (\sin^{2} (θ))$ .

r^{2} (\cos^{2} (θ) + \sin^{2} (θ)) - 25 = 0

$r^{2} (\cos^{2} (θ) + \sin^{2} (θ)) - 25 = 0$

r^{2} (\cos^{2} (θ) + \sin^{2} (θ)) - 25 = 0

$r^{2} (\cos^{2} (θ) + \sin^{2} (θ)) - 25 = 0$

Step 3.2.3

Rearrange terms.

r^{2} (\sin^{2} (θ) + \cos^{2} (θ)) - 25 = 0

$r^{2} (\sin^{2} (θ) + \cos^{2} (θ)) - 25 = 0$

Step 3.2.4

Apply pythagorean identity.

r^{2} \cdot 1 - 25 = 0

$r^{2} \cdot 1 - 25 = 0$

Step 3.2.5

Multiply

r^{2}

$r^{2}$ by

1

$1$ .

r^{2} - 25 = 0

$r^{2} - 25 = 0$

r^{2} - 25 = 0

$r^{2} - 25 = 0$

Step 3.3

Factor

r^{2} - 25

$r^{2} - 25$ .

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Step 3.3.1

Rewrite

25

$25$ as

5^{2}

$5^{2}$ .

r^{2} - 5^{2} = 0

$r^{2} - 5^{2} = 0$

Step 3.3.2

Since both terms are perfect squares, factor using the difference of squares formula,

a^{2} - b^{2} = (a + b) (a - b)

$a^{2} - b^{2} = (a + b) (a - b)$ where

a = r

$a = r$ and

b = 5

$b = 5$ .

(r + 5) (r - 5) = 0

$(r + 5) (r - 5) = 0$

(r + 5) (r - 5) = 0

$(r + 5) (r - 5) = 0$

Step 3.4

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

r + 5 = 0

$r + 5 = 0$

r - 5 = 0

$r - 5 = 0$

Step 3.5

Set

r + 5

$r + 5$ equal to

0

$0$ and solve for

r

$r$ .

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Step 3.5.1

Set

r + 5

$r + 5$ equal to

0

$0$ .

r + 5 = 0

$r + 5 = 0$

Step 3.5.2

Subtract

5

$5$ from both sides of the equation.

r = - 5

$r = - 5$

r = - 5

$r = - 5$

Step 3.6

Set

r - 5

$r - 5$ equal to

0

$0$ and solve for

r

$r$ .

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Step 3.6.1

Set

r - 5

$r - 5$ equal to

0

$0$ .

r - 5 = 0

$r - 5 = 0$

Step 3.6.2

Add

5

$5$ to both sides of the equation.

r = 5

$r = 5$

r = 5

$r = 5$

Step 3.7

The final solution is all the values that make

(r + 5) (r - 5) = 0

$(r + 5) (r - 5) = 0$ true.

r = - 5, 5

$r = - 5, 5$

r = - 5, 5

$r = - 5, 5$