Algebra Examples

${(x - 3)}^{2} + y^{2} = 40$ $x - y = - 1$

Step 1

Add $y$ to both sides of the equation.

$x = - 1 + y$

${(x - 3)}^{2} + y^{2} = 40$

Step 2

Replace all occurrences of $x$ with $- 1 + y$ in each equation.

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Step 2.1

Replace all occurrences of $x$ in ${(x - 3)}^{2} + y^{2} = 40$ with $- 1 + y$ .

${((- 1 + y) - 3)}^{2} + y^{2} = 40$

$x = - 1 + y$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify ${((- 1 + y) - 3)}^{2} + y^{2}$ .

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Step 2.2.1.1

Simplify each term.

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Step 2.2.1.1.1

Subtract $3$ from $- 1$ .

${(y - 4)}^{2} + y^{2} = 40$

$x = - 1 + y$

Step 2.2.1.1.2

Rewrite ${(y - 4)}^{2}$ as $(y - 4) (y - 4)$ .

$(y - 4) (y - 4) + y^{2} = 40$

$x = - 1 + y$

Step 2.2.1.1.3

Expand $(y - 4) (y - 4)$ using the FOIL Method.

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Step 2.2.1.1.3.1

Apply the distributive property.

$y (y - 4) - 4 (y - 4) + y^{2} = 40$

$x = - 1 + y$

Step 2.2.1.1.3.2

Apply the distributive property.

$y \cdot y + y \cdot - 4 - 4 (y - 4) + y^{2} = 40$

$x = - 1 + y$

Step 2.2.1.1.3.3

Apply the distributive property.

$y \cdot y + y \cdot - 4 - 4 y - 4 \cdot - 4 + y^{2} = 40$

$x = - 1 + y$

$y \cdot y + y \cdot - 4 - 4 y - 4 \cdot - 4 + y^{2} = 40$

$x = - 1 + y$

Step 2.2.1.1.4

Simplify and combine like terms.

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Step 2.2.1.1.4.1

Simplify each term.

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Step 2.2.1.1.4.1.1

Multiply $y$ by $y$ .

$y^{2} + y \cdot - 4 - 4 y - 4 \cdot - 4 + y^{2} = 40$

$x = - 1 + y$

Step 2.2.1.1.4.1.2

Move $- 4$ to the left of $y$ .

$y^{2} - 4 \cdot y - 4 y - 4 \cdot - 4 + y^{2} = 40$

$x = - 1 + y$

Step 2.2.1.1.4.1.3

Multiply $- 4$ by $- 4$ .

$y^{2} - 4 y - 4 y + 16 + y^{2} = 40$

$x = - 1 + y$

$y^{2} - 4 y - 4 y + 16 + y^{2} = 40$

$x = - 1 + y$

Step 2.2.1.1.4.2

Subtract $4 y$ from $- 4 y$ .

$y^{2} - 8 y + 16 + y^{2} = 40$

$x = - 1 + y$

$y^{2} - 8 y + 16 + y^{2} = 40$

$x = - 1 + y$

$y^{2} - 8 y + 16 + y^{2} = 40$

$x = - 1 + y$

Step 2.2.1.2

Add $y^{2}$ and $y^{2}$ .

$2 y^{2} - 8 y + 16 = 40$

$x = - 1 + y$

$2 y^{2} - 8 y + 16 = 40$

$x = - 1 + y$

$2 y^{2} - 8 y + 16 = 40$

$x = - 1 + y$

$2 y^{2} - 8 y + 16 = 40$

$x = - 1 + y$

Step 3

Solve for $y$ in $2 y^{2} - 8 y + 16 = 40$ .

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Step 3.1

Subtract $40$ from both sides of the equation.

$2 y^{2} - 8 y + 16 - 40 = 0$

$x = - 1 + y$

Step 3.2

Subtract $40$ from $16$ .

$2 y^{2} - 8 y - 24 = 0$

$x = - 1 + y$

Step 3.3

Factor the left side of the equation.

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Step 3.3.1

Factor $2$ out of $2 y^{2} - 8 y - 24$ .

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Step 3.3.1.1

Factor $2$ out of $2 y^{2}$ .

$2 (y^{2}) - 8 y - 24 = 0$

$x = - 1 + y$

Step 3.3.1.2

Factor $2$ out of $- 8 y$ .

$2 (y^{2}) + 2 (- 4 y) - 24 = 0$

$x = - 1 + y$

Step 3.3.1.3

Factor $2$ out of $- 24$ .

$2 y^{2} + 2 (- 4 y) + 2 \cdot - 12 = 0$

$x = - 1 + y$

Step 3.3.1.4

Factor $2$ out of $2 y^{2} + 2 (- 4 y)$ .

$2 (y^{2} - 4 y) + 2 \cdot - 12 = 0$

$x = - 1 + y$

Step 3.3.1.5

Factor $2$ out of $2 (y^{2} - 4 y) + 2 \cdot - 12$ .

$2 (y^{2} - 4 y - 12) = 0$

$x = - 1 + y$

$2 (y^{2} - 4 y - 12) = 0$

$x = - 1 + y$

Step 3.3.2

Factor.

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Step 3.3.2.1

Factor $y^{2} - 4 y - 12$ 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 $- 12$ and whose sum is $- 4$ .

$- 6, 2$

$x = - 1 + y$

Step 3.3.2.1.2

Write the factored form using these integers.

$2 ((y - 6) (y + 2)) = 0$

$x = - 1 + y$

$2 ((y - 6) (y + 2)) = 0$

$x = - 1 + y$

Step 3.3.2.2

Remove unnecessary parentheses.

$2 (y - 6) (y + 2) = 0$

$x = - 1 + y$

$2 (y - 6) (y + 2) = 0$

$x = - 1 + y$

$2 (y - 6) (y + 2) = 0$

$x = - 1 + y$

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 $0$ .

$y - 6 = 0$

$y + 2 = 0$

$x = - 1 + y$

Step 3.5

Set $y - 6$ equal to $0$ and solve for $y$ .

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Step 3.5.1

Set $y - 6$ equal to $0$ .

$y - 6 = 0$

$x = - 1 + y$

Step 3.5.2

Add $6$ to both sides of the equation.

$y = 6$

$x = - 1 + y$

$y = 6$

$x = - 1 + y$

Step 3.6

Set $y + 2$ equal to $0$ and solve for $y$ .

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Step 3.6.1

Set $y + 2$ equal to $0$ .

$y + 2 = 0$

$x = - 1 + y$

Step 3.6.2

Subtract $2$ from both sides of the equation.

$y = - 2$

$x = - 1 + y$

$y = - 2$

$x = - 1 + y$

Step 3.7

The final solution is all the values that make $2 (y - 6) (y + 2) = 0$ true.

$y = 6, - 2$

$x = - 1 + y$

$y = 6, - 2$

$x = - 1 + y$

Step 4

Replace all occurrences of $y$ with $6$ in each equation.

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Step 4.1

Replace all occurrences of $y$ in $x = - 1 + y$ with $6$ .

$x = - 1 + 6$

$y = 6$

Step 4.2

Simplify $x = - 1 + 6$ .

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Step 4.2.1

Simplify the left side.

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Step 4.2.1.1

Remove parentheses.

$x = - 1 + 6$

$y = 6$

$x = - 1 + 6$

$y = 6$

Step 4.2.2

Simplify the right side.

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Step 4.2.2.1

Add $- 1$ and $6$ .

$x = 5$

$y = 6$

$x = 5$

$y = 6$

$x = 5$

$y = 6$

$x = 5$

$y = 6$

Step 5

Replace all occurrences of $y$ with $- 2$ in each equation.

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Step 5.1

Replace all occurrences of $y$ in $x = - 1 + y$ with $- 2$ .

$x = - 1 - 2$

$y = - 2$

Step 5.2

Simplify $x = - 1 - 2$ .

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Step 5.2.1

Simplify the left side.

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Step 5.2.1.1

Remove parentheses.

$x = - 1 - 2$

$y = - 2$

$x = - 1 - 2$

$y = - 2$

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

Subtract $2$ from $- 1$ .

$x = - 3$

$y = - 2$

$x = - 3$

$y = - 2$

$x = - 3$

$y = - 2$

$x = - 3$

$y = - 2$

Step 6

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

$(5, 6)$

$(- 3, - 2)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(5, 6), (- 3, - 2)$

Equation Form:

$x = 5, y = 6$

$x = - 3, y = - 2$

Step 8