Algebra Examples

$y = - {(x - 2)}^{2} + 5$ $y = - x + 1$

Step 1

Eliminate the equal sides of each equation and combine.

$- {(x - 2)}^{2} + 5 = - x + 1$

Step 2

Solve $- {(x - 2)}^{2} + 5 = - x + 1$ for $x$ .

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Step 2.1

Move all terms containing $x$ to the left side of the equation.

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Step 2.1.1

Add $x$ to both sides of the equation.

$- {(x - 2)}^{2} + 5 + x = 1$

Step 2.1.2

Simplify each term.

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Step 2.1.2.1

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

$- ((x - 2) (x - 2)) + 5 + x = 1$

Step 2.1.2.2

Expand $(x - 2) (x - 2)$ using the FOIL Method.

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Step 2.1.2.2.1

Apply the distributive property.

$- (x (x - 2) - 2 (x - 2)) + 5 + x = 1$

Step 2.1.2.2.2

Apply the distributive property.

$- (x \cdot x + x \cdot - 2 - 2 (x - 2)) + 5 + x = 1$

Step 2.1.2.2.3

Apply the distributive property.

$- (x \cdot x + x \cdot - 2 - 2 x - 2 \cdot - 2) + 5 + x = 1$

Step 2.1.2.3

Simplify and combine like terms.

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Step 2.1.2.3.1

Simplify each term.

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Step 2.1.2.3.1.1

Multiply $x$ by $x$ .

$- (x^{2} + x \cdot - 2 - 2 x - 2 \cdot - 2) + 5 + x = 1$

Step 2.1.2.3.1.2

Move $- 2$ to the left of $x$ .

$- (x^{2} - 2 \cdot x - 2 x - 2 \cdot - 2) + 5 + x = 1$

Step 2.1.2.3.1.3

Multiply $- 2$ by $- 2$ .

$- (x^{2} - 2 x - 2 x + 4) + 5 + x = 1$

Step 2.1.2.3.2

Subtract $2 x$ from $- 2 x$ .

$- (x^{2} - 4 x + 4) + 5 + x = 1$

Step 2.1.2.4

Apply the distributive property.

$- x^{2} - (- 4 x) - 1 \cdot 4 + 5 + x = 1$

Step 2.1.2.5

Simplify.

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Step 2.1.2.5.1

Multiply $- 4$ by $- 1$ .

$- x^{2} + 4 x - 1 \cdot 4 + 5 + x = 1$

Step 2.1.2.5.2

Multiply $- 1$ by $4$ .

$- x^{2} + 4 x - 4 + 5 + x = 1$

Step 2.1.3

Add $4 x$ and $x$ .

$- x^{2} + 5 x - 4 + 5 = 1$

Step 2.1.4

Add $- 4$ and $5$ .

$- x^{2} + 5 x + 1 = 1$

Step 2.2

Subtract $1$ from both sides of the equation.

$- x^{2} + 5 x + 1 - 1 = 0$

Step 2.3

Combine the opposite terms in $- x^{2} + 5 x + 1 - 1$ .

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Step 2.3.1

Subtract $1$ from $1$ .

$- x^{2} + 5 x + 0 = 0$

Step 2.3.2

Add $- x^{2} + 5 x$ and $0$ .

$- x^{2} + 5 x = 0$

Step 2.4

Factor $- x$ out of $- x^{2} + 5 x$ .

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Step 2.4.1

Factor $- x$ out of $- x^{2}$ .

$- x \cdot x + 5 x = 0$

Step 2.4.2

Factor $- x$ out of $5 x$ .

$- x \cdot x - x \cdot - 5 = 0$

Step 2.4.3

Factor $- x$ out of $- x (x) - x (- 5)$ .

$- x (x - 5) = 0$

Step 2.5

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

$x = 0$

$x - 5 = 0 + y = - x + 1$

Step 2.6

Set $x$ equal to $0$ .

$x = 0$

Step 2.7

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 2.7.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 2.7.2

Add $5$ to both sides of the equation.

$x = 5$

Step 2.8

The final solution is all the values that make $- x (x - 5) = 0$ true.

$x = 0, 5$

Step 3

Evaluate $y$ when $x = 0$ .

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Step 3.1

Substitute $0$ for $x$ .

$y = - (0) + 1$

Step 3.2

Simplify $- (0) + 1$ .

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Step 3.2.1

Multiply $- 1$ by $0$ .

$y = 0 + 1$

Step 3.2.2

Add $0$ and $1$ .

$y = 1$

Step 4

Evaluate $y$ when $x = 5$ .

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Step 4.1

Substitute $5$ for $x$ .

$y = - (5) + 1$

Step 4.2

Simplify $- (5) + 1$ .

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Step 4.2.1

Multiply $- 1$ by $5$ .

$y = - 5 + 1$

Step 4.2.2

Add $- 5$ and $1$ .

$y = - 4$

Step 5

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

$(0, 1)$

$(5, - 4)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(0, 1), (5, - 4)$

Equation Form:

$x = 0, y = 1$

$x = 5, y = - 4$

Step 7