Algebra Examples

${(x - 1)}^{2} - 7 > {(x - 2)}^{2}$

Step 1

Add $7$ to both sides of the inequality.

${(x - 1)}^{2} > {(x - 2)}^{2} + 7$

Step 2

Subtract ${(x - 2)}^{2}$ from both sides of the inequality.

${(x - 1)}^{2} - {(x - 2)}^{2} > 7$

Step 3

Convert the inequality to an equation.

${(x - 1)}^{2} - {(x - 2)}^{2} = 7$

Step 4

Subtract $7$ from both sides of the equation.

${(x - 1)}^{2} - {(x - 2)}^{2} - 7 = 0$

Step 5

Simplify ${(x - 1)}^{2} - {(x - 2)}^{2} - 7$ .

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Step 5.1

Simplify each term.

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Step 5.1.1

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

$(x - 1) (x - 1) - {(x - 2)}^{2} - 7 = 0$

Step 5.1.2

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

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Step 5.1.2.1

Apply the distributive property.

$x (x - 1) - 1 (x - 1) - {(x - 2)}^{2} - 7 = 0$

Step 5.1.2.2

Apply the distributive property.

$x \cdot x + x \cdot - 1 - 1 (x - 1) - {(x - 2)}^{2} - 7 = 0$

Step 5.1.2.3

Apply the distributive property.

$x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1 - {(x - 2)}^{2} - 7 = 0$

Step 5.1.3

Simplify and combine like terms.

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Step 5.1.3.1

Simplify each term.

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Step 5.1.3.1.1

Multiply $x$ by $x$ .

$x^{2} + x \cdot - 1 - 1 x - 1 \cdot - 1 - {(x - 2)}^{2} - 7 = 0$

Step 5.1.3.1.2

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

$x^{2} - 1 \cdot x - 1 x - 1 \cdot - 1 - {(x - 2)}^{2} - 7 = 0$

Step 5.1.3.1.3

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

$x^{2} - x - 1 x - 1 \cdot - 1 - {(x - 2)}^{2} - 7 = 0$

Step 5.1.3.1.4

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

$x^{2} - x - x - 1 \cdot - 1 - {(x - 2)}^{2} - 7 = 0$

Step 5.1.3.1.5

Multiply $- 1$ by $- 1$ .

$x^{2} - x - x + 1 - {(x - 2)}^{2} - 7 = 0$

Step 5.1.3.2

Subtract $x$ from $- x$ .

$x^{2} - 2 x + 1 - {(x - 2)}^{2} - 7 = 0$

Step 5.1.4

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

$x^{2} - 2 x + 1 - ((x - 2) (x - 2)) - 7 = 0$

Step 5.1.5

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

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Step 5.1.5.1

Apply the distributive property.

$x^{2} - 2 x + 1 - (x (x - 2) - 2 (x - 2)) - 7 = 0$

Step 5.1.5.2

Apply the distributive property.

$x^{2} - 2 x + 1 - (x \cdot x + x \cdot - 2 - 2 (x - 2)) - 7 = 0$

Step 5.1.5.3

Apply the distributive property.

$x^{2} - 2 x + 1 - (x \cdot x + x \cdot - 2 - 2 x - 2 \cdot - 2) - 7 = 0$

Step 5.1.6

Simplify and combine like terms.

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Step 5.1.6.1

Simplify each term.

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Step 5.1.6.1.1

Multiply $x$ by $x$ .

$x^{2} - 2 x + 1 - (x^{2} + x \cdot - 2 - 2 x - 2 \cdot - 2) - 7 = 0$

Step 5.1.6.1.2

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

$x^{2} - 2 x + 1 - (x^{2} - 2 \cdot x - 2 x - 2 \cdot - 2) - 7 = 0$

Step 5.1.6.1.3

Multiply $- 2$ by $- 2$ .

$x^{2} - 2 x + 1 - (x^{2} - 2 x - 2 x + 4) - 7 = 0$

Step 5.1.6.2

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

$x^{2} - 2 x + 1 - (x^{2} - 4 x + 4) - 7 = 0$

Step 5.1.7

Apply the distributive property.

$x^{2} - 2 x + 1 - x^{2} - (- 4 x) - 1 \cdot 4 - 7 = 0$

Step 5.1.8

Simplify.

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Step 5.1.8.1

Multiply $- 4$ by $- 1$ .

$x^{2} - 2 x + 1 - x^{2} + 4 x - 1 \cdot 4 - 7 = 0$

Step 5.1.8.2

Multiply $- 1$ by $4$ .

$x^{2} - 2 x + 1 - x^{2} + 4 x - 4 - 7 = 0$

Step 5.2

Combine the opposite terms in $x^{2} - 2 x + 1 - x^{2} + 4 x - 4 - 7$ .

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Step 5.2.1

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

$- 2 x + 1 + 0 + 4 x - 4 - 7 = 0$

Step 5.2.2

Add $- 2 x + 1$ and $0$ .

$- 2 x + 1 + 4 x - 4 - 7 = 0$

Step 5.3

Add $- 2 x$ and $4 x$ .

$2 x + 1 - 4 - 7 = 0$

Step 5.4

Subtract $4$ from $1$ .

$2 x - 3 - 7 = 0$

Step 5.5

Subtract $7$ from $- 3$ .

$2 x - 10 = 0$

Step 6

Factor $2$ out of $2 x - 10$ .

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Step 6.1

Factor $2$ out of $2 x$ .

$2 (x) - 10 = 0$

Step 6.2

Factor $2$ out of $- 10$ .

$2 x + 2 \cdot - 5 = 0$

Step 6.3

Factor $2$ out of $2 x + 2 \cdot - 5$ .

$2 (x - 5) = 0$

Step 7

Divide each term in $2 (x - 5) = 0$ by $2$ and simplify.

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Step 7.1

Divide each term in $2 (x - 5) = 0$ by $2$ .

$\frac{2 (x - 5)}{2} = \frac{0}{2}$

Step 7.2

Simplify the left side.

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Step 7.2.1

Cancel the common factor of $2$ .

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Step 7.2.1.1

Cancel the common factor.

$\frac{2 (x - 5)}{2} = \frac{0}{2}$

Step 7.2.1.2

Divide $x - 5$ by $1$ .

$x - 5 = \frac{0}{2}$

Step 7.3

Simplify the right side.

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Step 7.3.1

Divide $0$ by $2$ .

$x - 5 = 0$

Step 8

Add $5$ to both sides of the equation.

$x = 5$

Step 9

The solution consists of all of the true intervals.

$x > 5$

Step 10

The result can be shown in multiple forms.

Inequality Form:

$x > 5$

Interval Notation:

$(5, \infty)$

Step 11