Algebra Examples

$y - 3 = {(x - 1)}^{2}$ $2 x + y = 5$

Step 1

Subtract $2 x$ from both sides of the equation.

$y = 5 - 2 x$

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

Step 2

Replace all occurrences of $y$ with $5 - 2 x$ in each equation.

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Step 2.1

Replace all occurrences of $y$ in $y - 3 = {(x - 1)}^{2}$ with $5 - 2 x$ .

$(5 - 2 x) - 3 = {(x - 1)}^{2}$

$y = 5 - 2 x$

Step 2.2

Simplify $(5 - 2 x) - 3 = {(x - 1)}^{2}$ .

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Step 2.2.1

Simplify the left side.

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Step 2.2.1.1

Subtract $3$ from $5$ .

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

$y = 5 - 2 x$

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

$y = 5 - 2 x$

Step 2.2.2

Simplify the right side.

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Step 2.2.2.1

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

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Step 2.2.2.1.1

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

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

$y = 5 - 2 x$

Step 2.2.2.1.2

Expand $(x - 1) (x - 1)$ using the FOIL Method.

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Step 2.2.2.1.2.1

Apply the distributive property.

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

$y = 5 - 2 x$

Step 2.2.2.1.2.2

Apply the distributive property.

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

$y = 5 - 2 x$

Step 2.2.2.1.2.3

Apply the distributive property.

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

$y = 5 - 2 x$

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

$y = 5 - 2 x$

Step 2.2.2.1.3

Simplify and combine like terms.

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Step 2.2.2.1.3.1

Simplify each term.

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Step 2.2.2.1.3.1.1

Multiply $x$ by $x$ .

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

$y = 5 - 2 x$

Step 2.2.2.1.3.1.2

Move $- 1$ to the left of $x$ .

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

$y = 5 - 2 x$

Step 2.2.2.1.3.1.3

Rewrite $- 1 x$ as $- x$ .

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

$y = 5 - 2 x$

Step 2.2.2.1.3.1.4

Rewrite $- 1 x$ as $- x$ .

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

$y = 5 - 2 x$

Step 2.2.2.1.3.1.5

Multiply $- 1$ by $- 1$ .

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

$y = 5 - 2 x$

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

$y = 5 - 2 x$

Step 2.2.2.1.3.2

Subtract $x$ from $- x$ .

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

$y = 5 - 2 x$

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

$y = 5 - 2 x$

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

$y = 5 - 2 x$

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

$y = 5 - 2 x$

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

$y = 5 - 2 x$

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

$y = 5 - 2 x$

Step 3

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

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Step 3.1

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

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

$y = 5 - 2 x$

Step 3.2

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

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Step 3.2.1

Add $2 x$ to both sides of the equation.

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

$y = 5 - 2 x$

Step 3.2.2

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

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Step 3.2.2.1

Add $- 2 x$ and $2 x$ .

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

$y = 5 - 2 x$

Step 3.2.2.2

Add $x^{2}$ and $0$ .

$x^{2} + 1 = 2$

$y = 5 - 2 x$

$x^{2} + 1 = 2$

$y = 5 - 2 x$

$x^{2} + 1 = 2$

$y = 5 - 2 x$

Step 3.3

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

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Step 3.3.1

Subtract $1$ from both sides of the equation.

$x^{2} = 2 - 1$

$y = 5 - 2 x$

Step 3.3.2

Subtract $1$ from $2$ .

$x^{2} = 1$

$y = 5 - 2 x$

$x^{2} = 1$

$y = 5 - 2 x$

Step 3.4

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

$x = \pm \sqrt{1}$

$y = 5 - 2 x$

Step 3.5

Any root of $1$ is $1$ .

$x = \pm 1$

$y = 5 - 2 x$

Step 3.6

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

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Step 3.6.1

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

$x = 1$

$y = 5 - 2 x$

Step 3.6.2

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

$x = - 1$

$y = 5 - 2 x$

Step 3.6.3

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

$x = 1, - 1$

$y = 5 - 2 x$

$x = 1, - 1$

$y = 5 - 2 x$

$x = 1, - 1$

$y = 5 - 2 x$

Step 4

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

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Step 4.1

Replace all occurrences of $x$ in $y = 5 - 2 x$ with $1$ .

$y = 5 - 2 \cdot 1$

$x = 1$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $5 - 2 \cdot 1$ .

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Step 4.2.1.1

Multiply $- 2$ by $1$ .

$y = 5 - 2$

$x = 1$

Step 4.2.1.2

Subtract $2$ from $5$ .

$y = 3$

$x = 1$

$y = 3$

$x = 1$

$y = 3$

$x = 1$

$y = 3$

$x = 1$

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 $y = 5 - 2 x$ with $- 1$ .

$y = 5 - 2 \cdot - 1$

$x = - 1$

Step 5.2

Simplify the right side.

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Step 5.2.1

Simplify $5 - 2 \cdot - 1$ .

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Step 5.2.1.1

Multiply $- 2$ by $- 1$ .

$y = 5 + 2$

$x = - 1$

Step 5.2.1.2

Add $5$ and $2$ .

$y = 7$

$x = - 1$

$y = 7$

$x = - 1$

$y = 7$

$x = - 1$

$y = 7$

$x = - 1$

Step 6

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

$(1, 3)$

$(- 1, 7)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(1, 3), (- 1, 7)$

Equation Form:

$x = 1, y = 3$

$x = - 1, y = 7$

Step 8