Algebra Examples

$x (49 x + 13) \leq {(7 x + 1)}^{2}$

Step 1

Solve $x \geq - 1$ .

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Step 1.1

Simplify $x (49 x + 13)$ .

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Step 1.1.1

Rewrite.

$0 + 0 + x (49 x + 13) \leq {(7 x + 1)}^{2}$

Step 1.1.2

Simplify by multiplying through.

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Step 1.1.2.1

Apply the distributive property.

$x (49 x) + x \cdot 13 \leq {(7 x + 1)}^{2}$

Step 1.1.2.2

Reorder.

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Step 1.1.2.2.1

Rewrite using the commutative property of multiplication.

$49 x \cdot x + x \cdot 13 \leq {(7 x + 1)}^{2}$

Step 1.1.2.2.2

Move $13$ to the left of $x$ .

$49 x \cdot x + 13 \cdot x \leq {(7 x + 1)}^{2}$

Step 1.1.3

Multiply $x$ by $x$ by adding the exponents.

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Step 1.1.3.1

Move $x$ .

$49 (x \cdot x) + 13 \cdot x \leq {(7 x + 1)}^{2}$

Step 1.1.3.2

Multiply $x$ by $x$ .

$49 x^{2} + 13 \cdot x \leq {(7 x + 1)}^{2}$

$49 x^{2} + 13 x \leq {(7 x + 1)}^{2}$

Step 1.2

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

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Step 1.2.1

Rewrite ${(7 x + 1)}^{2}$ as $(7 x + 1) (7 x + 1)$ .

$49 x^{2} + 13 x \leq (7 x + 1) (7 x + 1)$

Step 1.2.2

Expand $(7 x + 1) (7 x + 1)$ using the FOIL Method.

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Step 1.2.2.1

Apply the distributive property.

$49 x^{2} + 13 x \leq 7 x (7 x + 1) + 1 (7 x + 1)$

Step 1.2.2.2

Apply the distributive property.

$49 x^{2} + 13 x \leq 7 x (7 x) + 7 x \cdot 1 + 1 (7 x + 1)$

Step 1.2.2.3

Apply the distributive property.

$49 x^{2} + 13 x \leq 7 x (7 x) + 7 x \cdot 1 + 1 (7 x) + 1 \cdot 1$

Step 1.2.3

Simplify and combine like terms.

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Step 1.2.3.1

Simplify each term.

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Step 1.2.3.1.1

Rewrite using the commutative property of multiplication.

$49 x^{2} + 13 x \leq 7 \cdot 7 x \cdot x + 7 x \cdot 1 + 1 (7 x) + 1 \cdot 1$

Step 1.2.3.1.2

Multiply $x$ by $x$ by adding the exponents.

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Step 1.2.3.1.2.1

Move $x$ .

$49 x^{2} + 13 x \leq 7 \cdot 7 (x \cdot x) + 7 x \cdot 1 + 1 (7 x) + 1 \cdot 1$

Step 1.2.3.1.2.2

Multiply $x$ by $x$ .

$49 x^{2} + 13 x \leq 7 \cdot 7 x^{2} + 7 x \cdot 1 + 1 (7 x) + 1 \cdot 1$

Step 1.2.3.1.3

Multiply $7$ by $7$ .

$49 x^{2} + 13 x \leq 49 x^{2} + 7 x \cdot 1 + 1 (7 x) + 1 \cdot 1$

Step 1.2.3.1.4

Multiply $7$ by $1$ .

$49 x^{2} + 13 x \leq 49 x^{2} + 7 x + 1 (7 x) + 1 \cdot 1$

Step 1.2.3.1.5

Multiply $7 x$ by $1$ .

$49 x^{2} + 13 x \leq 49 x^{2} + 7 x + 7 x + 1 \cdot 1$

Step 1.2.3.1.6

Multiply $1$ by $1$ .

$49 x^{2} + 13 x \leq 49 x^{2} + 7 x + 7 x + 1$

Step 1.2.3.2

Add $7 x$ and $7 x$ .

$49 x^{2} + 13 x \leq 49 x^{2} + 14 x + 1$

Step 1.3

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

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Step 1.3.1

Subtract $49 x^{2}$ from both sides of the inequality.

$49 x^{2} + 13 x - 49 x^{2} \leq 14 x + 1$

Step 1.3.2

Subtract $14 x$ from both sides of the inequality.

$49 x^{2} + 13 x - 49 x^{2} - 14 x \leq 1$

Step 1.3.3

Combine the opposite terms in $49 x^{2} + 13 x - 49 x^{2} - 14 x$ .

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Step 1.3.3.1

Subtract $49 x^{2}$ from $49 x^{2}$ .

$13 x + 0 - 14 x \leq 1$

Step 1.3.3.2

Add $13 x$ and $0$ .

$13 x - 14 x \leq 1$

Step 1.3.4

Subtract $14 x$ from $13 x$ .

$- x \leq 1$

Step 1.4

Divide each term in $- x \leq 1$ by $- 1$ and simplify.

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Step 1.4.1

Divide each term in $- x \leq 1$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{1}{- 1}$

Step 1.4.2

Simplify the left side.

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Step 1.4.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{1}{- 1}$

Step 1.4.2.2

Divide $x$ by $1$ .

$x \geq \frac{1}{- 1}$

Step 1.4.3

Simplify the right side.

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Step 1.4.3.1

Divide $1$ by $- 1$ .

$x \geq - 1$

Step 2

Use the inequality $x \geq - 1$ to build the set notation.

${x | x \geq - 1}$

Step 3