Algebra Examples

$x^{3} + 2 x^{2} - y - 1 = 0$ , $2 - y + x - x^{2} = 0$

Step 1

Solve for $y$ in $x^{3} + 2 x^{2} - y - 1 = 0$ .

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Step 1.1

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

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Step 1.1.1

Subtract $x^{3}$ from both sides of the equation.

$2 x^{2} - y - 1 = - x^{3}$

$2 - y + x - x^{2} = 0$

Step 1.1.2

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

$- y - 1 = - x^{3} - 2 x^{2}$

$2 - y + x - x^{2} = 0$

Step 1.1.3

Add $1$ to both sides of the equation.

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

$2 - y + x - x^{2} = 0$

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

$2 - y + x - x^{2} = 0$

Step 1.2

Divide each term in $- y = - x^{3} - 2 x^{2} + 1$ by $- 1$ and simplify.

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Step 1.2.1

Divide each term in $- y = - x^{3} - 2 x^{2} + 1$ by $- 1$ .

$\frac{- y}{- 1} = \frac{- x^{3}}{- 1} + \frac{- 2 x^{2}}{- 1} + \frac{1}{- 1}$

$2 - y + x - x^{2} = 0$

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.

$\frac{y}{1} = \frac{- x^{3}}{- 1} + \frac{- 2 x^{2}}{- 1} + \frac{1}{- 1}$

$2 - y + x - x^{2} = 0$

Step 1.2.2.2

Divide $y$ by $1$ .

$y = \frac{- x^{3}}{- 1} + \frac{- 2 x^{2}}{- 1} + \frac{1}{- 1}$

$2 - y + x - x^{2} = 0$

$y = \frac{- x^{3}}{- 1} + \frac{- 2 x^{2}}{- 1} + \frac{1}{- 1}$

$2 - y + x - x^{2} = 0$

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

Dividing two negative values results in a positive value.

$y = \frac{x^{3}}{1} + \frac{- 2 x^{2}}{- 1} + \frac{1}{- 1}$

$2 - y + x - x^{2} = 0$

Step 1.2.3.1.2

Divide $x^{3}$ by $1$ .

$y = x^{3} + \frac{- 2 x^{2}}{- 1} + \frac{1}{- 1}$

$2 - y + x - x^{2} = 0$

Step 1.2.3.1.3

Move the negative one from the denominator of $\frac{- 2 x^{2}}{- 1}$ .

$y = x^{3} - 1 \cdot (- 2 x^{2}) + \frac{1}{- 1}$

$2 - y + x - x^{2} = 0$

Step 1.2.3.1.4

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

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

$2 - y + x - x^{2} = 0$

Step 1.2.3.1.5

Multiply $- 2$ by $- 1$ .

$y = x^{3} + 2 x^{2} + \frac{1}{- 1}$

$2 - y + x - x^{2} = 0$

Step 1.2.3.1.6

Divide $1$ by $- 1$ .

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

$2 - y + x - x^{2} = 0$

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

$2 - y + x - x^{2} = 0$

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

$2 - y + x - x^{2} = 0$

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

$2 - y + x - x^{2} = 0$

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

$2 - y + x - x^{2} = 0$

Step 2

Replace all occurrences of $y$ with $x^{3} + 2 x^{2} - 1$ in each equation.

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Step 2.1

Replace all occurrences of $y$ in $2 - y + x - x^{2} = 0$ with $x^{3} + 2 x^{2} - 1$ .

$2 - (x^{3} + 2 x^{2} - 1) + x - x^{2} = 0$

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

Step 2.2

Simplify the left side.

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Step 2.2.1

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

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Step 2.2.1.1

Simplify each term.

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Step 2.2.1.1.1

Apply the distributive property.

$2 - x^{3} - (2 x^{2}) + 1 + x - x^{2} = 0$

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

Step 2.2.1.1.2

Simplify.

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Step 2.2.1.1.2.1

Multiply $2$ by $- 1$ .

$2 - x^{3} - 2 x^{2} + 1 + x - x^{2} = 0$

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

Step 2.2.1.1.2.2

Multiply $- 1$ by $- 1$ .

$2 - x^{3} - 2 x^{2} + 1 + x - x^{2} = 0$

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

$2 - x^{3} - 2 x^{2} + 1 + x - x^{2} = 0$

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

$2 - x^{3} - 2 x^{2} + 1 + x - x^{2} = 0$

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

Step 2.2.1.2

Simplify by adding terms.

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Step 2.2.1.2.1

Add $2$ and $1$ .

$- x^{3} - 2 x^{2} + 3 + x - x^{2} = 0$

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

Step 2.2.1.2.2

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

$- x^{3} - 3 x^{2} + 3 + x = 0$

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

$- x^{3} - 3 x^{2} + 3 + x = 0$

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

$- x^{3} - 3 x^{2} + 3 + x = 0$

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

$- x^{3} - 3 x^{2} + 3 + x = 0$

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

$- x^{3} - 3 x^{2} + 3 + x = 0$

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

Step 3

Solve for $x$ in $- x^{3} - 3 x^{2} + 3 + x = 0$ .

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Step 3.1

Factor the left side of the equation.

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Step 3.1.1

Reorder terms.

$- x^{3} - 3 x^{2} + x + 3 = 0$

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

Step 3.1.2

Factor out the greatest common factor from each group.

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Step 3.1.2.1

Group the first two terms and the last two terms.

$(- x^{3} - 3 x^{2}) + x + 3 = 0$

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

Step 3.1.2.2

Factor out the greatest common factor (GCF) from each group.

$x^{2} (- x - 3) - 1 (- x - 3) = 0$

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

$x^{2} (- x - 3) - 1 (- x - 3) = 0$

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

Step 3.1.3

Factor the polynomial by factoring out the greatest common factor, $- x - 3$ .

$(- x - 3) (x^{2} - 1) = 0$

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

Step 3.1.4

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

$(- x - 3) (x^{2} - 1^{2}) = 0$

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

Step 3.1.5

Factor.

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Step 3.1.5.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$(- x - 3) ((x + 1) (x - 1)) = 0$

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

Step 3.1.5.2

Remove unnecessary parentheses.

$(- x - 3) (x + 1) (x - 1) = 0$

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

$(- x - 3) (x + 1) (x - 1) = 0$

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

$(- x - 3) (x + 1) (x - 1) = 0$

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

Step 3.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$- x - 3 = 0$

$x + 1 = 0$

$x - 1 = 0$

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

Step 3.3

Set $- x - 3$ equal to $0$ and solve for $x$ .

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Step 3.3.1

Set $- x - 3$ equal to $0$ .

$- x - 3 = 0$

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

Step 3.3.2

Solve $- x - 3 = 0$ for $x$ .

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Step 3.3.2.1

Add $3$ to both sides of the equation.

$- x = 3$

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

Step 3.3.2.2

Divide each term in $- x = 3$ by $- 1$ and simplify.

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Step 3.3.2.2.1

Divide each term in $- x = 3$ by $- 1$ .

$\frac{- x}{- 1} = \frac{3}{- 1}$

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

Step 3.3.2.2.2

Simplify the left side.

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Step 3.3.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{3}{- 1}$

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

Step 3.3.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{3}{- 1}$

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

$x = \frac{3}{- 1}$

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

Step 3.3.2.2.3

Simplify the right side.

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Step 3.3.2.2.3.1

Divide $3$ by $- 1$ .

$x = - 3$

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

$x = - 3$

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

$x = - 3$

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

$x = - 3$

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

$x = - 3$

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

Step 3.4

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 3.4.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

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

Step 3.4.2

Subtract $1$ from both sides of the equation.

$x = - 1$

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

$x = - 1$

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

Step 3.5

Set $x - 1$ equal to $0$ and solve for $x$ .

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Step 3.5.1

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

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

Step 3.5.2

Add $1$ to both sides of the equation.

$x = 1$

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

$x = 1$

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

Step 3.6

The final solution is all the values that make $(- x - 3) (x + 1) (x - 1) = 0$ true.

$x = - 3, - 1, 1$

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

$x = - 3, - 1, 1$

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

Step 4

Replace all occurrences of $x$ with $- 3$ in each equation.

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Step 4.1

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

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

$x = - 3$

Step 4.2

Simplify the right side.

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Step 4.2.1

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

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Step 4.2.1.1

Simplify each term.

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Step 4.2.1.1.1

Raise $- 3$ to the power of $3$ .

$y = - 27 + 2 {(- 3)}^{2} - 1$

$x = - 3$

Step 4.2.1.1.2

Raise $- 3$ to the power of $2$ .

$y = - 27 + 2 \cdot 9 - 1$

$x = - 3$

Step 4.2.1.1.3

Multiply $2$ by $9$ .

$y = - 27 + 18 - 1$

$x = - 3$

$y = - 27 + 18 - 1$

$x = - 3$

Step 4.2.1.2

Simplify by adding and subtracting.

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Step 4.2.1.2.1

Add $- 27$ and $18$ .

$y = - 9 - 1$

$x = - 3$

Step 4.2.1.2.2

Subtract $1$ from $- 9$ .

$y = - 10$

$x = - 3$

$y = - 10$

$x = - 3$

$y = - 10$

$x = - 3$

$y = - 10$

$x = - 3$

$y = - 10$

$x = - 3$

Step 5

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

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Step 5.1

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

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

$x = - 1$

Step 5.2

Simplify the right side.

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Step 5.2.1

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

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Step 5.2.1.1

Simplify each term.

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Step 5.2.1.1.1

Raise $- 1$ to the power of $3$ .

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

$x = - 1$

Step 5.2.1.1.2

Raise $- 1$ to the power of $2$ .

$y = - 1 + 2 \cdot 1 - 1$

$x = - 1$

Step 5.2.1.1.3

Multiply $2$ by $1$ .

$y = - 1 + 2 - 1$

$x = - 1$

$y = - 1 + 2 - 1$

$x = - 1$

Step 5.2.1.2

Simplify by adding and subtracting.

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Step 5.2.1.2.1

Add $- 1$ and $2$ .

$y = 1 - 1$

$x = - 1$

Step 5.2.1.2.2

Subtract $1$ from $1$ .

$y = 0$

$x = - 1$

$y = 0$

$x = - 1$

$y = 0$

$x = - 1$

$y = 0$

$x = - 1$

$y = 0$

$x = - 1$

Step 6

Replace all occurrences of $x$ with $- 3$ in each equation.

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Step 6.1

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

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

$x = - 3$

Step 6.2

Simplify the right side.

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Step 6.2.1

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

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Step 6.2.1.1

Simplify each term.

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Step 6.2.1.1.1

Raise $- 3$ to the power of $3$ .

$y = - 27 + 2 {(- 3)}^{2} - 1$

$x = - 3$

Step 6.2.1.1.2

Raise $- 3$ to the power of $2$ .

$y = - 27 + 2 \cdot 9 - 1$

$x = - 3$

Step 6.2.1.1.3

Multiply $2$ by $9$ .

$y = - 27 + 18 - 1$

$x = - 3$

$y = - 27 + 18 - 1$

$x = - 3$

Step 6.2.1.2

Simplify by adding and subtracting.

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Step 6.2.1.2.1

Add $- 27$ and $18$ .

$y = - 9 - 1$

$x = - 3$

Step 6.2.1.2.2

Subtract $1$ from $- 9$ .

$y = - 10$

$x = - 3$

$y = - 10$

$x = - 3$

$y = - 10$

$x = - 3$

$y = - 10$

$x = - 3$

$y = - 10$

$x = - 3$

Step 7

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

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Step 7.1

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

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

$x = - 1$

Step 7.2

Simplify the right side.

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Step 7.2.1

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

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Step 7.2.1.1

Simplify each term.

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Step 7.2.1.1.1

Raise $- 1$ to the power of $3$ .

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

$x = - 1$

Step 7.2.1.1.2

Raise $- 1$ to the power of $2$ .

$y = - 1 + 2 \cdot 1 - 1$

$x = - 1$

Step 7.2.1.1.3

Multiply $2$ by $1$ .

$y = - 1 + 2 - 1$

$x = - 1$

$y = - 1 + 2 - 1$

$x = - 1$

Step 7.2.1.2

Simplify by adding and subtracting.

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Step 7.2.1.2.1

Add $- 1$ and $2$ .

$y = 1 - 1$

$x = - 1$

Step 7.2.1.2.2

Subtract $1$ from $1$ .

$y = 0$

$x = - 1$

$y = 0$

$x = - 1$

$y = 0$

$x = - 1$

$y = 0$

$x = - 1$

$y = 0$

$x = - 1$

Step 8

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

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Step 8.1

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

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

$x = 1$

Step 8.2

Simplify the right side.

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Step 8.2.1

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

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Step 8.2.1.1

Simplify each term.

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Step 8.2.1.1.1

One to any power is one.

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

$x = 1$

Step 8.2.1.1.2

One to any power is one.

$y = 1 + 2 \cdot 1 - 1$

$x = 1$

Step 8.2.1.1.3

Multiply $2$ by $1$ .

$y = 1 + 2 - 1$

$x = 1$

$y = 1 + 2 - 1$

$x = 1$

Step 8.2.1.2

Simplify by adding and subtracting.

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Step 8.2.1.2.1

Add $1$ and $2$ .

$y = 3 - 1$

$x = 1$

Step 8.2.1.2.2

Subtract $1$ from $3$ .

$y = 2$

$x = 1$

$y = 2$

$x = 1$

$y = 2$

$x = 1$

$y = 2$

$x = 1$

$y = 2$

$x = 1$

Step 9

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

$(- 3, - 10)$

$(- 1, 0)$

$(1, 2)$

Step 10

The result can be shown in multiple forms.

Point Form:

$(- 3, - 10), (- 1, 0), (1, 2)$

Equation Form:

$x = - 3, y = - 10$

$x = - 1, y = 0$

$x = 1, y = 2$

Step 11