Algebra Examples

$x^{2} - y = 1$ $2 x^{2} + y^{2} = 17$

Step 1

Solve for $y$ in $x^{2} - y = 1$ .

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Step 1.1

Subtract $x^{2}$ from both sides of the equation.

$- y = 1 - x^{2}$

$2 x^{2} + y^{2} = 17$

Step 1.2

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

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Step 1.2.1

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

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

$2 x^{2} + y^{2} = 17$

Step 1.2.2

Simplify the left side.

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Step 1.2.2.1

Dividing two negative values results in a positive value.

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

$2 x^{2} + y^{2} = 17$

Step 1.2.2.2

Divide $y$ by $1$ .

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

$2 x^{2} + y^{2} = 17$

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

$2 x^{2} + y^{2} = 17$

Step 1.2.3

Simplify the right side.

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Step 1.2.3.1

Simplify each term.

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Step 1.2.3.1.1

Divide $1$ by $- 1$ .

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

$2 x^{2} + y^{2} = 17$

Step 1.2.3.1.2

Dividing two negative values results in a positive value.

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

$2 x^{2} + y^{2} = 17$

Step 1.2.3.1.3

Divide $x^{2}$ by $1$ .

$y = - 1 + x^{2}$

$2 x^{2} + y^{2} = 17$

$y = - 1 + x^{2}$

$2 x^{2} + y^{2} = 17$

$y = - 1 + x^{2}$

$2 x^{2} + y^{2} = 17$

$y = - 1 + x^{2}$

$2 x^{2} + y^{2} = 17$

$y = - 1 + x^{2}$

$2 x^{2} + y^{2} = 17$

Step 2

Replace all occurrences of $y$ with $- 1 + x^{2}$ in each equation.

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Step 2.1

Replace all occurrences of $y$ in $2 x^{2} + y^{2} = 17$ with $- 1 + x^{2}$ .

$2 x^{2} + {(- 1 + x^{2})}^{2} = 17$

$y = - 1 + x^{2}$

Step 2.2

Simplify the left side.

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Step 2.2.1

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

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Step 2.2.1.1

Simplify each term.

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Step 2.2.1.1.1

Rewrite ${(- 1 + x^{2})}^{2}$ as $(- 1 + x^{2}) (- 1 + x^{2})$ .

$2 x^{2} + (- 1 + x^{2}) (- 1 + x^{2}) = 17$

$y = - 1 + x^{2}$

Step 2.2.1.1.2

Expand $(- 1 + x^{2}) (- 1 + x^{2})$ using the FOIL Method.

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Step 2.2.1.1.2.1

Apply the distributive property.

$2 x^{2} - 1 (- 1 + x^{2}) + x^{2} (- 1 + x^{2}) = 17$

$y = - 1 + x^{2}$

Step 2.2.1.1.2.2

Apply the distributive property.

$2 x^{2} - 1 \cdot - 1 - 1 x^{2} + x^{2} (- 1 + x^{2}) = 17$

$y = - 1 + x^{2}$

Step 2.2.1.1.2.3

Apply the distributive property.

$2 x^{2} - 1 \cdot - 1 - 1 x^{2} + x^{2} \cdot - 1 + x^{2} x^{2} = 17$

$y = - 1 + x^{2}$

$2 x^{2} - 1 \cdot - 1 - 1 x^{2} + x^{2} \cdot - 1 + x^{2} x^{2} = 17$

$y = - 1 + x^{2}$

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 $- 1$ by $- 1$ .

$2 x^{2} + 1 - 1 x^{2} + x^{2} \cdot - 1 + x^{2} x^{2} = 17$

$y = - 1 + x^{2}$

Step 2.2.1.1.3.1.2

Rewrite $- 1 x^{2}$ as $- x^{2}$ .

$2 x^{2} + 1 - x^{2} + x^{2} \cdot - 1 + x^{2} x^{2} = 17$

$y = - 1 + x^{2}$

Step 2.2.1.1.3.1.3

Move $- 1$ to the left of $x^{2}$ .

$2 x^{2} + 1 - x^{2} - 1 \cdot x^{2} + x^{2} x^{2} = 17$

$y = - 1 + x^{2}$

Step 2.2.1.1.3.1.4

Rewrite $- 1 x^{2}$ as $- x^{2}$ .

$2 x^{2} + 1 - x^{2} - x^{2} + x^{2} x^{2} = 17$

$y = - 1 + x^{2}$

Step 2.2.1.1.3.1.5

Multiply $x^{2}$ by $x^{2}$ by adding the exponents.

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Step 2.2.1.1.3.1.5.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$2 x^{2} + 1 - x^{2} - x^{2} + x^{2 + 2} = 17$

$y = - 1 + x^{2}$

Step 2.2.1.1.3.1.5.2

Add $2$ and $2$ .

$2 x^{2} + 1 - x^{2} - x^{2} + x^{4} = 17$

$y = - 1 + x^{2}$

$2 x^{2} + 1 - x^{2} - x^{2} + x^{4} = 17$

$y = - 1 + x^{2}$

$2 x^{2} + 1 - x^{2} - x^{2} + x^{4} = 17$

$y = - 1 + x^{2}$

Step 2.2.1.1.3.2

Subtract $x^{2}$ from $- x^{2}$ .

$2 x^{2} + 1 - 2 x^{2} + x^{4} = 17$

$y = - 1 + x^{2}$

$2 x^{2} + 1 - 2 x^{2} + x^{4} = 17$

$y = - 1 + x^{2}$

$2 x^{2} + 1 - 2 x^{2} + x^{4} = 17$

$y = - 1 + x^{2}$

Step 2.2.1.2

Combine the opposite terms in $2 x^{2} + 1 - 2 x^{2} + x^{4}$ .

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Step 2.2.1.2.1

Subtract $2 x^{2}$ from $2 x^{2}$ .

$0 + 1 + x^{4} = 17$

$y = - 1 + x^{2}$

Step 2.2.1.2.2

Add $0$ and $1$ .

$1 + x^{4} = 17$

$y = - 1 + x^{2}$

$1 + x^{4} = 17$

$y = - 1 + x^{2}$

$1 + x^{4} = 17$

$y = - 1 + x^{2}$

$1 + x^{4} = 17$

$y = - 1 + x^{2}$

$1 + x^{4} = 17$

$y = - 1 + x^{2}$

Step 3

Solve for $x$ in $1 + x^{4} = 17$ .

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Step 3.1

Move all terms not containing $x$ to the right side of the equation.

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Step 3.1.1

Subtract $1$ from both sides of the equation.

$x^{4} = 17 - 1$

$y = - 1 + x^{2}$

Step 3.1.2

Subtract $1$ from $17$ .

$x^{4} = 16$

$y = - 1 + x^{2}$

$x^{4} = 16$

$y = - 1 + x^{2}$

Step 3.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt[4]{16}$

$y = - 1 + x^{2}$

Step 3.3

Simplify $\pm \sqrt[4]{16}$ .

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Step 3.3.1

Rewrite $16$ as $2^{4}$ .

$x = \pm \sqrt[4]{2^{4}}$

$y = - 1 + x^{2}$

Step 3.3.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 2$

$y = - 1 + x^{2}$

$x = \pm 2$

$y = - 1 + x^{2}$

Step 3.4

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

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Step 3.4.1

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

$x = 2$

$y = - 1 + x^{2}$

Step 3.4.2

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

$x = - 2$

$y = - 1 + x^{2}$

Step 3.4.3

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

$x = 2, - 2$

$y = - 1 + x^{2}$

$x = 2, - 2$

$y = - 1 + x^{2}$

$x = 2, - 2$

$y = - 1 + x^{2}$

Step 4

Replace all occurrences of $x$ with $2$ in each equation.

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Step 4.1

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

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

$x = 2$

Step 4.2

Simplify the right side.

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Step 4.2.1

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

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Step 4.2.1.1

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

$y = - 1 + 4$

$x = 2$

Step 4.2.1.2

Add $- 1$ and $4$ .

$y = 3$

$x = 2$

$y = 3$

$x = 2$

$y = 3$

$x = 2$

$y = 3$

$x = 2$

Step 5

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

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Step 5.1

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

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

$x = - 2$

Step 5.2

Simplify the right side.

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Step 5.2.1

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

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Step 5.2.1.1

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

$y = - 1 + 4$

$x = - 2$

Step 5.2.1.2

Add $- 1$ and $4$ .

$y = 3$

$x = - 2$

$y = 3$

$x = - 2$

$y = 3$

$x = - 2$

$y = 3$

$x = - 2$

Step 6

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

$(2, 3)$

$(- 2, 3)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(2, 3), (- 2, 3)$

Equation Form:

$x = 2, y = 3$

$x = - 2, y = 3$

Step 8