Algebra Examples

{(x - 5)}^{2} + {(y - 3)}^{2} = 29

${(x - 5)}^{2} + {(y - 3)}^{2} = 29$

$- x + y = - 9$

Step 1

Add

$x$ to both sides of the equation.

$y = - 9 + x$

${(x - 5)}^{2} + {(y - 3)}^{2} = 29$

Step 2

Replace all occurrences of

$y$ with

$- 9 + x$ in each equation.

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Step 2.1

Replace all occurrences of

$y$ in

${(x - 5)}^{2} + {(y - 3)}^{2} = 29$ with

$- 9 + x$ .

${(x - 5)}^{2} + {((- 9 + x) - 3)}^{2} = 29$

$y = - 9 + x$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify

${(x - 5)}^{2} + {((- 9 + x) - 3)}^{2}$ .

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Step 2.2.1.1

Simplify each term.

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Step 2.2.1.1.1

Rewrite

${(x - 5)}^{2}$ as

$(x - 5) (x - 5)$ .

$(x - 5) (x - 5) + {((- 9 + x) - 3)}^{2} = 29$

$y = - 9 + x$

Step 2.2.1.1.2

Expand

$(x - 5) (x - 5)$ using the FOIL Method.

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Step 2.2.1.1.2.1

Apply the distributive property.

$x (x - 5) - 5 (x - 5) + {((- 9 + x) - 3)}^{2} = 29$

$y = - 9 + x$

Step 2.2.1.1.2.2

Apply the distributive property.

$x \cdot x + x \cdot - 5 - 5 (x - 5) + {((- 9 + x) - 3)}^{2} = 29$

$y = - 9 + x$

Step 2.2.1.1.2.3

Apply the distributive property.

$x \cdot x + x \cdot - 5 - 5 x - 5 \cdot - 5 + {((- 9 + x) - 3)}^{2} = 29$

$y = - 9 + x$

$x \cdot x + x \cdot - 5 - 5 x - 5 \cdot - 5 + {((- 9 + x) - 3)}^{2} = 29$

$y = - 9 + x$

Step 2.2.1.1.3

Simplify and combine like terms.

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Step 2.2.1.1.3.1

Simplify each term.

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Step 2.2.1.1.3.1.1

Multiply

$x$ by

$x$ .

$x^{2} + x \cdot - 5 - 5 x - 5 \cdot - 5 + {((- 9 + x) - 3)}^{2} = 29$

$y = - 9 + x$

Step 2.2.1.1.3.1.2

Move

$- 5$ to the left of

$x$ .

$x^{2} - 5 \cdot x - 5 x - 5 \cdot - 5 + {((- 9 + x) - 3)}^{2} = 29$

$y = - 9 + x$

Step 2.2.1.1.3.1.3

Multiply

$- 5$ by

$- 5$ .

$x^{2} - 5 x - 5 x + 25 + {((- 9 + x) - 3)}^{2} = 29$

$y = - 9 + x$

$x^{2} - 5 x - 5 x + 25 + {((- 9 + x) - 3)}^{2} = 29$

$y = - 9 + x$

Step 2.2.1.1.3.2

Subtract

$5 x$ from

$- 5 x$ .

$x^{2} - 10 x + 25 + {((- 9 + x) - 3)}^{2} = 29$

$y = - 9 + x$

$x^{2} - 10 x + 25 + {((- 9 + x) - 3)}^{2} = 29$

$y = - 9 + x$

Step 2.2.1.1.4

Subtract

$3$ from

$- 9$ .

$x^{2} - 10 x + 25 + {(x - 12)}^{2} = 29$

$y = - 9 + x$

Step 2.2.1.1.5

Rewrite

${(x - 12)}^{2}$ as

$(x - 12) (x - 12)$ .

$x^{2} - 10 x + 25 + (x - 12) (x - 12) = 29$

$y = - 9 + x$

Step 2.2.1.1.6

Expand

$(x - 12) (x - 12)$ using the FOIL Method.

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Step 2.2.1.1.6.1

Apply the distributive property.

$x^{2} - 10 x + 25 + x (x - 12) - 12 (x - 12) = 29$

$y = - 9 + x$

Step 2.2.1.1.6.2

Apply the distributive property.

$x^{2} - 10 x + 25 + x \cdot x + x \cdot - 12 - 12 (x - 12) = 29$

$y = - 9 + x$

Step 2.2.1.1.6.3

Apply the distributive property.

$x^{2} - 10 x + 25 + x \cdot x + x \cdot - 12 - 12 x - 12 \cdot - 12 = 29$

$y = - 9 + x$

$x^{2} - 10 x + 25 + x \cdot x + x \cdot - 12 - 12 x - 12 \cdot - 12 = 29$

$y = - 9 + x$

Step 2.2.1.1.7

Simplify and combine like terms.

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Step 2.2.1.1.7.1

Simplify each term.

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Step 2.2.1.1.7.1.1

Multiply

$x$ by

$x$ .

$x^{2} - 10 x + 25 + x^{2} + x \cdot - 12 - 12 x - 12 \cdot - 12 = 29$

$y = - 9 + x$

Step 2.2.1.1.7.1.2

Move

$- 12$ to the left of

$x$ .

$x^{2} - 10 x + 25 + x^{2} - 12 \cdot x - 12 x - 12 \cdot - 12 = 29$

$y = - 9 + x$

Step 2.2.1.1.7.1.3

Multiply

$- 12$ by

$- 12$ .

$x^{2} - 10 x + 25 + x^{2} - 12 x - 12 x + 144 = 29$

$y = - 9 + x$

$x^{2} - 10 x + 25 + x^{2} - 12 x - 12 x + 144 = 29$

$y = - 9 + x$

Step 2.2.1.1.7.2

Subtract

$12 x$ from

$- 12 x$ .

$x^{2} - 10 x + 25 + x^{2} - 24 x + 144 = 29$

$y = - 9 + x$

$x^{2} - 10 x + 25 + x^{2} - 24 x + 144 = 29$

$y = - 9 + x$

$x^{2} - 10 x + 25 + x^{2} - 24 x + 144 = 29$

$y = - 9 + x$

Step 2.2.1.2

Simplify by adding terms.

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Step 2.2.1.2.1

Add

$x^{2}$ and

$x^{2}$ .

$2 x^{2} - 10 x + 25 - 24 x + 144 = 29$

$y = - 9 + x$

Step 2.2.1.2.2

Subtract

$24 x$ from

$- 10 x$ .

$2 x^{2} - 34 x + 25 + 144 = 29$

$y = - 9 + x$

Step 2.2.1.2.3

Add

$25$ and

$144$ .

$2 x^{2} - 34 x + 169 = 29$

$y = - 9 + x$

$2 x^{2} - 34 x + 169 = 29$

$y = - 9 + x$

$2 x^{2} - 34 x + 169 = 29$

$y = - 9 + x$

$2 x^{2} - 34 x + 169 = 29$

$y = - 9 + x$

$2 x^{2} - 34 x + 169 = 29$

$y = - 9 + x$

Step 3

Solve for

$x$ in

$2 x^{2} - 34 x + 169 = 29$ .

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Step 3.1

Subtract

$29$ from both sides of the equation.

$2 x^{2} - 34 x + 169 - 29 = 0$

$y = - 9 + x$

Step 3.2

Subtract

$29$ from

$169$ .

$2 x^{2} - 34 x + 140 = 0$

$y = - 9 + x$

Step 3.3

Factor the left side of the equation.

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Step 3.3.1

Factor

$2$ out of

$2 x^{2} - 34 x + 140$ .

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Step 3.3.1.1

Factor

$2$ out of

$2 x^{2}$ .

$2 (x^{2}) - 34 x + 140 = 0$

$y = - 9 + x$

Step 3.3.1.2

Factor

$2$ out of

$- 34 x$ .

$2 (x^{2}) + 2 (- 17 x) + 140 = 0$

$y = - 9 + x$

Step 3.3.1.3

Factor

$2$ out of

$140$ .

$2 x^{2} + 2 (- 17 x) + 2 \cdot 70 = 0$

$y = - 9 + x$

Step 3.3.1.4

Factor

$2$ out of

$2 x^{2} + 2 (- 17 x)$ .

$2 (x^{2} - 17 x) + 2 \cdot 70 = 0$

$y = - 9 + x$

Step 3.3.1.5

Factor

$2$ out of

$2 (x^{2} - 17 x) + 2 \cdot 70$ .

$2 (x^{2} - 17 x + 70) = 0$

$y = - 9 + x$

$2 (x^{2} - 17 x + 70) = 0$

$y = - 9 + x$

Step 3.3.2

Factor.

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Step 3.3.2.1

Factor

$x^{2} - 17 x + 70$ using the AC method.

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Step 3.3.2.1.1

Consider the form

$x^{2} + b x + c$ . Find a pair of integers whose product is

$c$ and whose sum is

$b$ . In this case, whose product is

$70$ and whose sum is

$- 17$ .

$- 10, - 7$

$y = - 9 + x$

Step 3.3.2.1.2

Write the factored form using these integers.

$2 ((x - 10) (x - 7)) = 0$

$y = - 9 + x$

$2 ((x - 10) (x - 7)) = 0$

$y = - 9 + x$

Step 3.3.2.2

Remove unnecessary parentheses.

$2 (x - 10) (x - 7) = 0$

$y = - 9 + x$

$2 (x - 10) (x - 7) = 0$

$y = - 9 + x$

$2 (x - 10) (x - 7) = 0$

$y = - 9 + x$

Step 3.4

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

$0$ , the entire expression will be equal to

$0$ .

$x - 10 = 0$

$x - 7 = 0$

$y = - 9 + x$

Step 3.5

Set

$x - 10$ equal to

$0$ and solve for

$x$ .

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Step 3.5.1

Set

$x - 10$ equal to

$0$ .

$x - 10 = 0$

$y = - 9 + x$

Step 3.5.2

Add

$10$ to both sides of the equation.

$x = 10$

$y = - 9 + x$

$x = 10$

$y = - 9 + x$

Step 3.6

Set

$x - 7$ equal to

$0$ and solve for

$x$ .

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Step 3.6.1

Set

$x - 7$ equal to

$0$ .

$x - 7 = 0$

$y = - 9 + x$

Step 3.6.2

Add

$7$ to both sides of the equation.

$x = 7$

$y = - 9 + x$

$x = 7$

$y = - 9 + x$

Step 3.7

The final solution is all the values that make

$2 (x - 10) (x - 7) = 0$ true.

$x = 10, 7$

$y = - 9 + x$

$x = 10, 7$

$y = - 9 + x$

Step 4

Replace all occurrences of

$x$ with

$10$ in each equation.

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Step 4.1

Replace all occurrences of

$x$ in

$y = - 9 + x$ with

$10$ .

$y = - 9 + 10$

$x = 10$

Step 4.2

Simplify

$y = - 9 + 10$ .

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Step 4.2.1

Simplify the left side.

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Step 4.2.1.1

Remove parentheses.

$y = - 9 + 10$

$x = 10$

$y = - 9 + 10$

$x = 10$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

Add

$- 9$ and

$10$ .

$y = 1$

$x = 10$

$y = 1$

$x = 10$

$y = 1$

$x = 10$

$y = 1$

$x = 10$

Step 5

Replace all occurrences of

$x$ with

$7$ in each equation.

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Step 5.1

Replace all occurrences of

$x$ in

$y = - 9 + x$ with

$7$ .

$y = - 9 + 7$

$x = 7$

Step 5.2

Simplify

$y = - 9 + 7$ .

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Step 5.2.1

Simplify the left side.

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Step 5.2.1.1

Remove parentheses.

$y = - 9 + 7$

$x = 7$

$y = - 9 + 7$

$x = 7$

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

Add

$- 9$ and

$7$ .

$y = - 2$

$x = 7$

$y = - 2$

$x = 7$

$y = - 2$

$x = 7$

$y = - 2$

$x = 7$

Step 6

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(10, 1)$

$(7, - 2)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(10, 1), (7, - 2)$

Equation Form:

$x = 10, y = 1$

$x = 7, y = - 2$

Step 8